Probability And Statistics Book Solutions 3h2y1u

Z

∞

y

=

Z

1

Z

y

f (x, y) dx dy = −∞

12 = 5

Z

−∞

Z

1

y

y(1 + y) 0

0

0

12 x dx dy = 10

0

Z

12 xy(1 + y) dx dy 5 1

y 3 (1 + y) dy =

0

27 . 50

Here we used that f (x, y) = 0 for (x, y) outside the unit square.

9.12 a Since the integral over R2 of the t probability density function is equal to 1, we find from Z 1Z 2 (3u2 + 8uv) du dv = 10 0

0

that K =

1 . 10

9.12 b

Z Z

P(2X ≤ Y ) = Z

(x,y):2x
Z

2

= x=0

y=2x

1 (3x2 + 8xy) dx dy 10

1 (3x2 + 8xy) dx dy 10

= 0.45. 9.13 Ra For r ≥ 0, let DRR r be the disc with origin (0, 0) and radius r. Since 1 = R ∞ ∞ −∞

−∞

f (x, y) dx dy =

D1

c dx dy = c · area of D1 , we find that c = 1/π.

9.13 b Clearly FR (r) = 0 for r < 0, and FR (r) = 1 for r > 1. For r between 0 and 1, ZZ 1 πr 2 dx dy = = r2 . FR (r) = P((X, Y ) ∈ Dr ) = π π Dr

482

Full solutions from MIPS: DO NOT DISTRIBUTE

9.13 c For x between −1 and 1, Z √1−x2

fX (x) = −

√

1−x2

1 2p dy = 1 − x2 . π π

For x outside the interval −1, 1] we have that fX (x) = 0. 9.14 a Let f be the t probability density function of the pair (X, Y ), and F 2 their t distribution function. Since f (x, y) = ∂x∂ ∂y F (x, y), we first determine F . Setting G = (−∞, a] × (−∞, b], for −1 ≤ a, b ≤ 1, we have that (a + 1)(b + 1) . 4

F (a, b) = P((X, Y ) ∈ G) =

Furthermore, if a ≤ −1 or b ≤ −1, we have that F (a, b) = 0, since G ∩ = ∅. In a similar way, we find for a > 1 and −1 ≤ b ≤ 1 that F (a, b) = (b + 1)/2, while for −1 ≤ a ≤ 1 and b > 1 we have that F (a, b) = (a + 1)/2. Finally, if a, b > 1, then F (a, b) = 1. Taking derivatives, it then follows that f (x, y) = 1/4 for a and b between −1 and 1, and f (x, y) = 0 for all other values of x and y. 9.14 b Note that for x between −1 and 1, the marginal probability density function fX of X is given by Z 1 1 fX (x) = f (x, y) dy = , 2 −1 and that fX (x) = 0 for all other values of x. Similarly, fY (y) = 1/2 for y between −1 and 1, and fY (y) = 0 otherwise. But then we find that f (x, y) = fX (x)fY (y), for all possible xs and ys, and we find that X and Y are independent, U (−1, 1) distributed random variables. 9.15 a Setting have that

(a, b) as the set of points (x, y), for which x ≤ a and y ≤ b, we area (∆ ∩ (a, b)) F (a, b) = . area of ∆

ˆ ˆ ˆ ˆ ˆ

If If If If If

a < 0 or if b < 0 (or both), then area (∆ ∩ (a, b)) = ∅, so F (a, b) = 0, (a, b) ∈ ∆, then area (∆ ∩ (a, b)) = a(b − 21 a), so F (a, b) = a(2b − a), 0 ≤ b ≤ 1, and a > b, then area (∆ ∩ (a, b)) = 21 b2 , so F (a, b) = b2 , 0 ≤ a ≤ 1, and b > 1, then area (∆ ∩ (a, b)) = a − 12 a2 , so F (a, b) = 2a − a2 , both a > 1 and b > 1, then area (∆ ∩ (a, b)) = 12 , so F (a, b) = 1. 2

9.15 b Since f (x, y) = ∂x∂ ∂y F (x, y), we find for (x, y) ∈ ∆ that f (x, y) = 2. Furthermore, f (x, y) = 0 for (x, y) outside the triangle ∆. 9.15 c For x between 0 and 1, Z

Z

∞

fX (x) =

1

f (x, y) dy = −∞

2 dy = 2(1 − x). x

For y between 0 and 1, Z

Z

∞

fY (y) =

y

f (x, y) dy = −∞

2 dx = 2y. 0

29.1 Full solutions

483

9.16 Following the solution of Exercise 9.14 one finds that U and V are independent U (0, 1) distributed random variables. Using the results from Section 8.4 we find that FU (x) = 1 − (1 − x)2 for x between 0 and 1, and that FV (y) = y 2 for y between 0 and 1. Differentiating yields the desired result. 9.17 For 0 ≤ s ≤ t ≤ a, it follows from the fact that U1 and U2 are independent uniformly distributed random variables over [0, a], that P(U1 ≤ t, U2 ≤ t) = t2 /a2 , and that

P(s < U1 ≤ t, s < U2 ≤ t) = (t − s)2 /a2 .

But then the answer is an immediate consequence of the hint. The statement can also be obtained as follows. Note that P(V ≤ s, Z ≤ t) = P(U2 ≤ s, s < U1 ≤ t) + P(U1 ≤ s, s < U2 ≤ t) +P(U1 ≤ s, U2 ≤ s) . Using independence and that fact that Ui has distribution function FU1 (u) = u/a, we find for 0 ≤ s ≤ t ≤ a: P(V ≤ s, Z ≤ t) =

s(t − s) s(t − s) t2 − (t − s)2 s2 + + = . a2 a2 a2 a2

9.18 a By definition E [Xi ] =

N X

kpXi (k) =

k=1

1 1 N +1 1 (1 + 2 + · · · + N ) = · N (N + 1) = . N N 2 2

9.18 b Using the identity, we find

E Xi2 =

N X

N (N + 1)(2N + 1) 1 X 2 k = . N 6

k2 P(Xi = k) =

k=1

k=1

But then we have that

Var(Xi ) = E Xi2 − E [Xi ]

2

=

N2 − 1 . 12

√ √ 9.19 a Clearly we must √ have that a = 50 = 5 2, but then it follows that 2ab = 80, implying that b = 4 2. Since 32 + c = 50, we find that c = 18. 9.19 b Note that Z

∞ −∞

√ √ 2y−4 2x)2

e−(5

Z

∞

dy =

e− 2 (

y−µ σ

) dy

−∞

√ Z = σ 2π √ = σ 2π,

since

1

∞ −∞

1 y−µ 1 √ e− 2 ( σ ) dy σ 2π

1 y−µ 1 √ e− 2 σ for − ∞ < y < ∞ σ 2π is the probability density function of an N (µ, σ 2 ) distributed random variable (and therefore integrates to 1).

484


9.19 c

Z

Z

∞

fX (x) =

f (x, y) dy = −∞

30 = π =

Z

∞

∞

30 −50x2 −50y2 +80xy e dy π

−∞ √ √ −(5 2y−4 2x)2 −18x2

e

e

−∞

√ 30 2π −18x2 = e π 10

1 −1 √ e 2 1 2π 6

for −∞ < x < ∞. So we see that X has a N (0,

1 ) 36

dy

x 1/6

2

,

distribution.

9.20 a Since the needle hits the sheet of paper at an random position, the midpoint (X, Y ) falls completely randomly between some lines. Consequently, the distance Z between (X, Y ) and the line “under” (X, Y ) has a U (0, 1) distribution. Also the orientation of the needle is completely random, so the angle between the needle and the positive x-axis can be anything between 0 and 180 degrees. But then H has a U (0, π) distribution. 9.20 b From Figure 29.1 we see that—in case Z ≤ 1/2—the needle hits the line under it when Z ≤ 12 sin H. In case Z > 1/2, we have a similar picture, but then the needle hits the line above it when 1 − Z ≤ 12 sin H.

(X, Y .)....

... ..... ..... ..... ..... ....... ......... ............. . .............................. .. ............. .... ........ ..... ..... .......................................1 . . ..... . . . . 2

Z

sin H

Fig. 29.1. Solution of Exercise 9.20, case Z ≤ 1/2

9.20 c Since Z and H are independent, uniformly distributed random variables, the probability we are looking for is equal to the area of the two regions in [0, π) × [0, 1] for which either 1 1 Z ≤ sin H or 1 − Z ≤ sin H, 2 2 divided (!) by the total area π. Note that these two regions both have the same are, and that this area is equal to Z

π 0

1 1 sin H dH = · 2 = 1. 2 2

29.1 Full solutions

485

So the probability we are after is equal to 1 1 2 + = . π π π 10.1 a The t probability distribution of X and Y is given by a b

1

2

3

P(Y = b)

1 2

0.217 0.153 0.057 0.106 0.244 0.223

0.427 0.573

P(X = a)

0.323 0.397 0.280

1

This means that E [X] = 1 · 0.323 + 2 · 0.397 + 3 · 0.280 = 1.957 E [Y ] = 1 · 0.427 + 2 · 0.573 = 1.573 E [XY ] = 1 · 1 · 0.217 + · · · + 2 · 3 · 0.223 = 3.220. It follows that Cov(X, Y ) = E [XY ] − E [X] E [Y ] = 3.220 − 1.957 · 0.245 = 0.142. 10.1 b From the t probability distribution of X and Y we compute

E X 2 = 1 · 0.323 + 4 · 0.397 + 9 · 0.280 = 4.431

Var(X) = E X 2 − (E [X])2 = 0.601

E Y

2

= 1 · 0.427 + 4 · 0.573 = 2.719

Var(Y ) = E Y 2 − (E [Y ])2 = 0.245. Using the result from a we find ρ (X, Y ) = p

Cov(X, Y ) 0.142 = 0.369. = √ 0.601 · 0.245 Var(X) Var(Y )

10.2 a From Exercise 9.2: x y

so that E [XY ] = 10.2 b E [Y ] = 0 − 0 = 0.

1 2

1 6

0

1

2

pY (y)

−1 1

1/6 1/6 1/6 0 1/2 0

1/2 1/2

pX (x)

1/6 2/3 1/6

1

· 1 · (−1) +

· (−1) +

1 2

1 6

· 2 · (−1) +

1 2

· 1 · 1 = 0.

· 1 = 0, so that Cov(X, Y ) = E [XY ] − E [X] · E [Y ] =

10.2 c From the marginal distributions we find

486

Full solutions from MIPS: DO NOT DISTRIBUTE 1 2 1 +1· +2· =1 6 3 6 1 2 1 E X2 = 0 · + 1 · + 4 · = 1 6 3 6 4 1 Var(X) = E X 2 − (E [X])2 = − 12 = 3 3 1 1 E Y 2 = (−1)2 · + 12 · = 1 2 2 Var(Y ) = E Y 2 − (E [Y ])2 = 1 − 02 = 1. E [X] = 0 ·

Since Cov(X, Y ) = 0 we have Var(X + Y ) = Var(X) + Var(Y ) =

1 3

+ 1 = 43 .

10.2 d Since Cov(X, −Y ) = −Cov(X, Y ) = 0 we have Var(X − Y ) = Var(X) + Var(−Y ) = Var(X) + (−1)2 Var(Y ) = 13 + 1 = 43 .

10.3 We have E [U ] = 1, E [V ] = 12 , E U 2 = 32 , E V 2 = 12 , Var(U ) = 12 , Var(V ) = 1 , and E [U V ] = 12 . Hence Cov(U, V ) = 0 and ρ (U, V ) = 0. 4 10.4 Both X and Y have marginal probabilities ( 14 , 41 , 14 , 14 ), so that 1 1 1 1 + 2 · + 3 · + 4 · = 2.5 4 4 4 4 1 1 1 1 E [Y ] = 1 · + 2 · + 3 · + 4 · = 2.5 4 4 4 4 16 1 E [XY ] = 1 · 1 · + ··· + 4 · 4 · = 6.25. 136 136 E [X] = 1 ·

Hence Cov(X, Y ) = E [XY ] − E [X] E [Y ] = 6.25 − 2.5 · 2.5 = 0. 8 10 6 − 72 = 72 , and similarly P(X = 0, Y = 2) = 10.5 a First find P(X = 1, Y = 0) = 13 − 72 4 1 1 5 and P(Y = 2) = 6 . Then P(X = 1) = 4 and P(X = 2) = 12 , and finally 72 5 P(X = 2, Y = 2) = 71 :

a b 0 1 2

0

1

2

8/72 6/72 10/72 12/72 9/72 15/72 4/72 3/72 5/72 1/3

1/4

1/3 1/2 1/6

5/12

1

10.5 b With a we find: 1 1 5 13 +1· +2· = 3 4 12 12 1 1 1 5 E [Y ] = 0 · + 1 · + 2 · = 3 2 6 6 9 5 65 E [XY ] = 1 · 1 · + ··· + 2 · 2 · = . 72 72 72 E [X] = 0 ·

Hence Cov(X, Y ) = E [XY ] − E [X] E [Y ] =

65 72

−

13 12

·

5 6

= 0.

10.5 c Yes, for all combinations (a, b) it holds that P(X = a, Y = b) = P(X = a) P(Y = b).

29.1 Full solutions

487

10.6 a When c = 0, the t distribution becomes a b

−1

−1 0 1

0

1

2/45 9/45 4/45 7/45 5/45 3/45 6/45 1/45 8/45

P(X = a)

1/3

1/3

P(Y = b) 1/3 1/3 1/3

1/3

1

We find E [X] = (−1) · 13 + 0 · 13 + 1 · 31 = 0, and similarly E [Y ] = 0. By leaving out where either X = 0 or Y = 0, we find E [XY ] = (−1) · (−1) ·

2 4 6 8 + (−1) · 1 · + 1 · (−1) · +1·1· = 0, 45 45 45 45

which implies that Cov(X, Y ) = E [XY ] − E [X] E [Y ] = 0. 10.6 b Note that the variables X and Y in part b are equal to the ones from part a, shifted by c. If we write U and V for the variables from a, then X = U + c and Y = V + c. According to the rule on the covariance under change of units, we then immediately find Cov(X, Y ) = Cov(U + c, V + c) = Cov(U, V ) = 0. Alternatively, one could also compute the covariance from Cov(X, Y ) = E [XY ] − E [X] E [Y ]. We find E [X] = (c − 1) · 31 + c · 13 + (c + 1) · 13 = c, and similarly E [Y ] = c. Since 2 9 4 + (c − 1) · c · + (c + 1) · (c + 1) · 45 45 45 7 5 3 +c · (c − 1) · +c·c· + c · (c + 1) · 45 45 45 6 1 8 +(c + 1) · (c − 1) · + (c + 1) · c · + (c + 1) · (c + 1) · = c2 , 45 45 45

E [XY ] = (c − 1) · (c − 1) ·

we find Cov(X, Y ) = E [XY ] − E [X] E [Y ] = c2 − c · c = 0. 10.6 c No, X and Y are not independent. For instance, P(X = c, Y = c + 1) = 1/45, which differs from P(X = c) P(Y = c + 1) = 1/9. 10.7 a E [XY ] = 81 , E [X] = E [Y ] = 12 , so that Cov(X, Y ) = E [XY ] − E [X] E [Y ] = 1 − ( 12 )2 = − 18 . 8 10.7 b Since X has a Ber ( 12 ) distribution, Var(X) = 14 . Similarly, Var(Y ) = 41 , so that Cov(X, Y ) −1/8 1 ρ (X, Y ) = p = =− . 1/4 2 Var(X) Var(Y ) 10.7 c For any ε between − 14 and 14 , Cov(X, Y ) = E [XY ] − ( 12 )2 = ( 14 − ε) − ( 12 )2 = −ε, so that Cov(X, Y ) −ε ρ (X, Y ) = p = = −4ε. 1/4 Var(X) Var(Y ) Hence, ρ (X, Y ) is equal to −1, 0, or 1, for ε equal to 1/4, 0, or −1/4.

10.8 a E X 2 = Var(X) + (E [X])2 = 8.

10.8 b E −2X 2 + Y = −2E X 2 + E [Y ] = −2 · 8 + 3 = −13.

488


10.9 a If the aggregated blood sample tests negative, we do not have to perform additional tests, so that Xi takes on the value 1. If the aggregated blood sample tests positive, we have to perform 40 additional tests for the blood sample of each person in the group, so that Xi takes on the value 41. We first find that P(Xi = 1) = P(no infections in group of 40) = (1 − 0.001)40 = 0.96, and therefore P(Xi = 41) = 1 − P(Xi = 1) = 0.04. 10.9 b First compute E [Xi ] = 1·0.96+41·0.04 = 2.6. The expected total number of tests is E [X1 + X2 + · · · + X25 ] = E [X1 ]+E [X2 ]+· · ·+E [X25 ] = 25·2.6 = 65. With the original procedure of blood testing, the total number of tests is 25·40 = 1000. On average the alternative procedure would only require 65 tests. Only with very small probability one would end up with doing more than 1000 tests, so the alternative procedure is better. 10.10 a We find Z

Z

∞

E [X] =

2 2 9 4 7 3 9x3 + 7x2 dx = x + x 225 225 4 3

3

xfX (x) dx = Z

−∞

E [Y ] =

0

Z

∞

2

1 3 4 1 (3y 3 + 12y 2 ) dy = y + 4y 3 25 25 4

yfY (y) dy = −∞

1

3

= 0

2

109 , 50

157 , 100

= 1

so that E [X + Y ] = E [X] + E [Y ] = 15/4. 10.10 b We find

E X2 =

E Y2 =

Z Z Z

∞

x2 fX (x) dx =

−∞ ∞

y 2 fY (y) dy =

Z Z

−∞ 3Z 2

3 0 2

1

2 2 9 5 7 4 9x4 + 7x3 dx = x + x 225 225 5 4

1 3 5 1 (3y 4 + 12y 3 ) dy = y + 3y 4 25 25 5 Z

3

Z

2

xyf (x, y) dy dx =

E [XY ] = 0

4 = 75

Z

4 7 75 3

=

1 3

Z

2

3

x

y dy

0

Z

2

1 3

x3 dx +

0

2

so that E (X + Y )

2 15 75 4

=E X

2

1

0

2 dx + 75 Z

3

Z

= 0

= 1

1287 , 250

318 , 125

2 2x3 y 2 + x2 y 3 dy dx 75 3

Z

2

x2

0

y 3 dy

dx

1

x2 dx =

0

2

3

171 , 50

+ E Y 2 + 2E [XY ] = 3633/250.

10.10 c We find

1287 − 250

318 − 125

Var(X) = E X 2 − (E [X])2 = Var(Y ) = E Y 2 − (E [Y ])2 =

109 50

157 100

Var(X + Y ) = E (X + Y )2 − (E [X + Y ])2 =

2

=

2

=

989 , 2500

791 , 10 000

3633 − 250

15 4

2

=

939 . 2000

Hence, Var(X) + Var(Y ) = 0.4747, which differs from Var(X + Y ) = 0.4695.

9 10.11 Cov(T, S) = Cov X + 32, 95 X + 32 = ( 59 )2 Cov(X, Y ) = 9.72 and ρ (T, S) = 5 9 9 ρ 5 X + 32, 5 X + 32 = ρ (X, Y ) = 0.8.

29.1 Full solutions

489

10.12 Since H has a U (25, 35) distribution, it follows immediately that E [H] = 30. Furthermore 3 1 Z 12.5 1 r E R2 = r2 dr = 2.5 = 102.0833. 5 15 7.5 7.5

Hence πE [H] E R2 = π · 30 · 102.0833 = 9621.128. 10.13 a You can apply Exercise 8.15 directly for the case n = 2 to conclude that E [X] = 13 and E [Y ] = 23 . In order to compute the variances we need the probability densities of X and Y . Using the rules about the distribution of the minimum and maximum of independent random variables in Section 8.4, we find fX (x) = 2(1 − x) fY (y) = 2y

for 0 ≤ x ≤ 1 for 0 ≤ y ≤ 1.

R1

This means that E X 2 = 0 2x2 (1 − x) dx = 61 , so that Var(X) = R1 1 Similarly, E Y 2 = 0 2y 3 dy = 12 , so that Var(Y ) = 21 − ( 23 )2 = 18 . 10.13 b Since U and V are independent, each with variance Var(X + Y ) = Var(U + V ) = Var(U ) + Var(V ) = 16 .

1 , 12

1 6

− ( 13 )2 =

1 . 18

it follows that

10.13 c To compute Cov(X, Y ) we need E [XY ]. This can be solved from 1 = Var(X + Y ) = E (X + Y )2 − (E [X + Y ])2 6 = E X 2 + 2E [XY ] + E Y 2 − (E [X] + E [Y ])2 .

1 3 Substitute E [X] = 13 and E X 2 = Var(X) + (E [X])2 = 18 + 19 = 18 , and similarly 1 E [Y ] = 23 and E Y 2 = Var(Y ) + (E [Y ])2 = 18 + 49 = 12 . This leads to E [XY ] = 41 .

10.14 a By using the alternative expression for the covariance and linearity of expectations, we find Cov(X + s, Y + u) = E [(X + s)(Y + u)] − E [X + s] E [Y + u] = E [XY + sY + uX + su] − (E [X] + s)(E [Y ] + u) = (E [XY ] + sE [Y ] + uE [X] + su) − (E [X] E [Y ] + sE [Y ] + uE [X] + su) = E [XY ] − E [X] E [Y ] = Cov(X, Y ) . 10.14 b By using the alternative expression for the covariance and the rule on expectations under change of units, we find Cov(rX, tY ) = E [(rX)(tY )] − E [rX] E [tY ] = E [rtXY ] − (rE [X])(tE [Y ]) = rtE [XY ] − rtE [X] E [Y ] = rt (E [XY ] − E [X] E [Y ]) = rtCov(X, Y ) . 10.14 c First applying part a and then part b yields Cov(rX + s, tY + u) = Cov(rX, tY ) = rtCov(X, Y ) .

490


10.15 a Left plot: looks like 500 realizations from a pair (X, Y ) whose 2-dimensional distribution has contourlines that are circles. This means X and Y are uncorrelated. Middle plot: looks like 500 realizations from a pair (X, Y ) whose 2-dimensional distribution has contourlines that are ellipsoids with the line y = x as main axis. This means X and Y are positively correlated; Right plot: looks like 500 realizations from a pair (X, Y ) whose 2-dimensional distribution has contourlines that are ellipsoids with the line y = x as main axis. This means X and Y are negatively correlated. 10.15 b In the right picture the points are concentrated more closely than in the other pictures. Hence |ρ(X, Y )| will be the largest for the right picture. 10.16 a Cov(X, X + Y ) = Cov(X, X) + Cov(X, Y ) = Var(X) + Cov(X, Y ) 10.16 b Anything can happen, depending on the sign of Cov(X, Y ) and its magnitude compared to Var(X). 10.16 c If X and Y are uncorrelated, Cov(X, Y ) = Var(X) ≥ 0, so apart from the special case where X is constant, X and X + Y are positively correlated. 10.17 a Since all expectations are zero, it is sufficient to show

E (X + Y + Z)2 = E X 2 + E Y 2 + E Y 2 + 2E [XY ] + 2E [XZ] + 2E [Y Z] . This follows immediately from the hint with n = 3, and using linearity of expectations. ˜ = X − E [X], and similarly Y˜ and Z. ˜ Then part a applies to X, ˜ 10.17 b Write X ˜ Y˜ , and Z:

˜ + Y˜ + Z˜ = Var X ˜ + Var Y˜ + Var Z˜ Var X

˜ Y˜ + 2Cov X, ˜ Z˜ + 2Cov Y˜ , Z˜ . + 2Cov X, According to the rules on pages 104 and and 151 about the variance covariance under ˜ = Var(X), Cov X, ˜ Y˜ = Cov(X, Y ), and a change of units, it follows that Var X similarly for all other variances and covariances. 10.17 c Similar to part a first consider the case with all random variables having expectation zero. As in part a, the hint in a together with linearity of expectations ˜ i = Xi − E [Xi ]. yields the desired equality. Then argue as in part b by introducing X 10.17 d Use part c and note that there are n Var(Xi ) and n(n − 1) Cov(Xi , Xj ) with i 6= j. 10.18 First note that X1 + X2 + · · · + XN is the sum of all numbers, which is a nonrandom constant. Therefore, Var(X1 + X2 + · · · + XN ) = 0. In Section 9.3 we argued that, although we draw without replacement, each Xi has the same distribution. By the same reasoning, we find that each pair (Xi , Xj ), with i 6= j, has the same t distribution, so that Cov(Xi , Xj ) = Cov(X1 , X2 ) for all pairs with i 6= j. Direct application of Exercise 10.17 with σ 2 = (N − 1)(N + 1) and γ = Cov(X1 , X2 ) gives 0 = Var(X1 + X2 + · · · + XN ) = N ·

(N − 1)(N + 1) + N (N − 1)Cov(X1 , X2 ) . 12

Solving this identity gives Cov(X1 , X2 ) = −(N + 1)/12.

29.1 Full solutions

491

10.19 By definition and linearity of expectations: Cov(X, Y ) = E [(X − E [X])(Y − E [Y ])] = E [XY − XE [Y ] − E [X] Y + E [X] E [Y ]] = E [XY ] − E [XE [Y ]] − E [E [X] Y ] + E [X] E [Y ] = E [XY ] − E [X] E [Y ] − E [X] E [Y ] + E [X] E [Y ] = E [XY ] − E [X] E [Y ] . 10.20 To compute the correlation, we need:

Cov U, U 2 = E U 3 − E [U ] E U 2 , Var(U ) = E U 2 − (E [U ])2 ,

Var U 2 = E U 4 − E U 2 Hence we determine

h

Z

i

E Uk =

a

uk

0

2

.

1 ak du = , a k+1

which yields

Cov U, U 2 = Var(U ) =

Var U 2 = Hence

a3 a a2 − = 4 2 3

a2 a 2 − = 3 2 a4 − 5

a2 3

1 11 − 4 23

2

=

1 1 − 3 4

1 1 − 5 9

1 3 a , 12

a3 =

1 2 a , 12

a2 =

a4 =

4 4 a . 45

1√ 1√ 135 = 15 = 0.968. 12 4 Note that the answer does not depend on a.

ρ U, U 2 =

11.1 a In the addition rule, k is between 2 and 12. We must always have 1 ≤ k−` ≤ 6 and 1 ≤ ` ≤ 6, or equivalently, k − 6 ≤ ` ≤ k − 1 and 1 ≤ ` ≤ 6. For k = 2, . . . , 6, this means pZ (k) = P(X + Y = k) =

k−1 X

pX (k − `)pY (`) =

`=1

k−1 X `=1

1 1 k−1 · = 6 6 36

and for k = 7, . . . , 12, pZ (k) = P(X + Y = k) =

6 X

pX (k − `)pY (`) =

`=k−6

6 X 1 `=k−6

6

·

1 13 − k = . 6 36

11.1 b In the addition rule, k is between 2 and 2N . We must always have 1 ≤ k − ` ≤ N and 1 ≤ ` ≤ N , or equivalently, k − N ≤ ` ≤ k − 1 and 1 ≤ ` ≤ N . For k = 2, . . . , N , this means pZ (k) = P(X + Y = k) =

k−1 X `=1

pX (k − `)pY (`) =

k−1 X `=1

k−1 1 1 · = N N N2

492


and for k = N + 1, . . . , 2N , N X

pZ (k) = P(X + Y = k) =

pX (k − `)pY (`) =

`=k−N

N X `=k−N

1 1 2N − k + 1 · = . N N N2

11.2 a By using the rule on addition of two independent discrete random variables, we have P(X + Y = k) = pZ (k) =

∞ X

pX (k − `)pY (`).

`=0

Because pX (a) = 0 for a ≤ −1, all with ` ≥ k + 1 vanish, so that P(X + Y = k) = also using

Pk `=0

k `

k X `=0

k 1k−` −1 1` −1 e−2 X k e · e = (k − `)! `! k! `

!

=

`=0

2k −2 e , k!

k

= 2 in the last equality.

11.2 b Similar to part a, by using the rule on addition of two independent discrete random variables and leaving out for which pX (a) = 0, we have P(X + Y = k) =

k X λk−` `=0

(k − `)!

!

(λ + µ)k −(λ+µ) X k λk−` µ` µ` −µ · e = e . `! k! ` (λ + µ)k k

e

−λ

`=0

Next, write λk−` µ` = (λ + µ)k

µ λ+µ

`

λ λ+µ

k−`

=

µ λ+µ

`

1−

µ λ+µ

k−`

= p` (1 − p)k−`

with p = µ/(λ + µ). This means that !

(λ + µ)k −(λ+µ) X k ` (λ + µ)k −(λ+µ) P(X + Y = k) = e p (1 − p)k−` = e , k! ` k! k

using that

Pk `=0

k `

`=0

`

k−`

p (1 − p)

11.3 pZ (k) =

k X `=0

=

k X `=0

= 1. !

n k−`

!

n k−`

!

m 1 k−` 1 ` 1 n−(k−`) 3 m−` (2) (4) (2) (4) ` !

m 1 n 1 ` 3 m−` (2) (4) (4) . `

This cannot be simplified and is not equal to a binomial probability of the type n+m k r (1 − r)n+m−k for some 0 < r < 1. k 11.4 a From the fact that X has an N (2, 5) distribution, it follows that E [X] = 2 and Var(X) = 5. Similarly, E [Y ] = 5 and Var(Y ) = 9. Hence by linearity of expectations, E [Z] = E [3X − 2Y + 1] = 3E [X] − 2E [Y ] + 1 = 3 · 2 − 2 · 5 + 1 = −3. By the rules for the variance and covariance, Var(Z) = 9Var(X) + 4Var(Y ) − 12Cov(X, Y ) = 9 · 5 + 4 · 9 − 12 · 0 = 81, using that Cov(X, Y ) = 0, due to independence of X and Y .

29.1 Full solutions

493

11.4 b The random variables 3X and −2Y + 1 are independent and, according to the rule for the normal distribution under a change of units (page 112), it follows that they both have a normal distribution. Next, the sum rule for independent normal random variables then yields that Z = (3X) + (−2Y + 1) also has a normal distribution. Its parameters are the expectation and variance of Z. From a it follows that Z has an N (−3, 81) distribution. 11.4 c From b we know that Z has an N (−3, 81) distribution, so that (Z + 3)/9 has a standard normal distribution. Therefore

P(Z ≤ 6) = P

6+3 Z +3 ≤ 9 9

= Φ(1),

where Φ is the standard normal distribution function. From Table ?? we find that Φ(1) = 1 − 0.1587 = 0.8413. 11.5 According to the addition rule, the probability density of Z = X + Y is given by Z 1 Z ∞ fX (z − y)fY (y) dy, fX (z − y)fY (y) dy = fZ (z) = 0

−∞

where we use that fY (y) = 0 for y ∈ / [0, 1]. When 0 ≤ y ≤ 1, the following holds. For z < 0, also z − y < 0 so that fX (z − y) = 0, and for z > 2, z − y > 1 so that fX (z − y) = 0. For 0 ≤ z < 1: Z

Z

1

z

fX (z − y)fY (y) dy =

fZ (z) =

1 dy = z, 0

0

whereas for 1 ≤ z ≤ 2: Z

Z

1

1

fX (z − y)fY (y) dy =

fZ (z) =

1 dy = 2 − z. z−1

0

11.6 According to the addition rule Z

Z

z

fZ (z) = 0

=

z

fX (z − y)fY (y) dy = 1 −z/2 e 16

0

Z

z

(z − y)y dy = 0

1 1 (z − y)e−(z−y)/2 ye−y/2 dy. 4 4

1 −z/2 1 2 1 3 e zy − y 16 2 3

z

= 0

z 3 −z/2 e . 96

11.7 Each Xi has the same distribution as the sum of k independent Exp (λ) distributed random variables. Since the Xi are independent, X1 + X2 + · · · + Xn has the same distribution as the sum of nk independent Exp (λ) distributed random variables, which is a Gam (nk, λ) distribution. 11.8 a Y = rX + s has probability density function fY (y) =

y − s 1 1 fX = r r rπ 1 +

y−s 2 r

=

r . π (r2 + (y − s)2 )

We see that Y = rX + s has a Cau (s, r) distribution. 11.8 b If Y = (X − β)/α then Y has a Cau (0, 1) distribution.

494


11.9 a According to the product rule on page 172, Z

z

z

fZ (z) =

fY 1

Z

3 = 2 z

3 = 2

x

z 1

fX (x)

Z

1 dx = x

z

1

1 3 1 dx = 2 − x−2 x3 z 2

1 1 − 4 z2 z

z 1

3 1 dx x4 x

1

z 2 x

3 1 = 2 z2

1 1− 2 z

.

11.9 b According to the product rule, Z

z

fZ (z) =

fY 1

αβ = β+1 z

z

1 fX (x) dx = x x

Z

z

z 1

αβ xβ−α−1 dx = β+1 z

1

αβ = β−α

Z

1

1

−

z β+1

z α+1

β

α

x β−α

=

z β+1 xα+1 x β−α z

1 dx x

1 αβ β−α 1 − z α − β z β+1

1

.

11.10 a In the quotient rule for Z = X/Y for 0 < z < ∞ fixed, we must have 1 ≤ zx < ∞ and 1 ≤ x < ∞. Hence for 0 < z < 1, Z

Z

∞

∞

fX (zx)fY (y)x dx =

fZ (z) = 1/z

4 = 3 z

Z

∞

−5

x 1/z

4 dx = 3 z

1/z

−x−4 4

2 2 x dx (zx)3 x3

∞

1 0 − (−z 4 ) = z. z3

= 1/z

For z ≥ 1, we find Z

Z

∞

fZ (z) =

fX (zx)fY (y)x dx = 1

=

4 z3

Z

∞

x−5 dx =

1

4 z3

∞

2

(zx)3 1 −4 ∞

−x 4

= 1

2 x dx x3

1 . z3

11.10 b In the quotient rule for Z = X/Y for 0 < z < ∞ fixed, we must have 1 ≤ zx < ∞ and 1 ≤ x < ∞. Hence for 0 < z < 1, Z

Z

∞

fZ (z) = 1/z

=

∞

fX (zx)fY (y)x dx = αβ z α+1

Z

1/z ∞

x−α−β−1 dx =

1/z

α β x dx (zx)α+1 xβ+1

αβ z α+1

−x−α−β α+β

∞

αβ 1 αβ β−1 α+β = 0 − (−z ) = z . α + β z α+1 α+β

1/z

For z ≥ 1, we find Z

Z

∞

fZ (z) =

∞

fX (zx)fY (y)x dx = 1/z

αβ = α+1 z

Z 1

1 ∞

x−α−β−1 dx =

α β dx (zx)α+1 xβ+1

αβ z α+1

−x−α−β α+β

∞

= 1

1 αβ . α + β z α+1

29.1 Full solutions

495

11.11 a Put T3 = X1 + X2 + X3 . Then T3 can be interpreted as the time of the third success in a series of independent experiments with probability p of success. Then T3 = k means that the first (k − 1) experiments contained exactly 2 successes, and the k-th experiment was also a success. Since the number of successes in (k − 1) experiments has a Bin (k − 1, p) distribution, it follows that !

k−1 2 1 p (1 − p)(k−1−2) · p = (k − 1)(k − 2)p3 (1 − p)k−3 . 2 2

P(T3 = k) =

11.11 b This follows from a smart calculation: ∞ X

⇔

k=3 ∞ X k=3

pZ (k) = 1 1 (k − 1)(k − 2)p3 (1 − p)k−3 = 1 2

1 2X p (k − 1)(k − 2)p(1 − p)k−3 = 1 ( now put k − 2 = m) 2 ∞

⇔

k=3 ∞ X

⇔

1 2 p m(m + 1)p(1 − p)m−1 = 1 2 m=1

⇔

1 2X p (m2 + m)P(X1 = m) = 1 2 m=1 ∞

1 2 2 p (E X1 + E [X1 ]) = 1 2 2 ⇔ p (E X12 + E [X1 ]) = 2. ⇔

11.11 c The first part follows directly from b:

E X12 =

2 2 1 2−p − E [X1 ] = 2 − = . p2 p p p2

For the second part:

Var(X1 ) = E X12 − (E [X1 ])2 = 11.12 Γ(1) =

R∞ 0

2−p − p2

2

1 p

=

1−p . p2

e−t dt = 1 and Z

∞

Γ(x + 1) = 0

tx e−t dt = −tx e−t

t=∞ t=0

Z

∞

+

xtx−1 e−t dt.

0

Since x > 0, the first term on the right hand side is zero and the second term is equal to xΓ(x). 11.13 a

496

Full solutions from MIPS: DO NOT DISTRIBUTE Z

a

P(Zn ≤ a) =

0

= −

λn z n−1 e−λz dz (n − 1)!

λn−1 z n−1 e−λz (n − 1)!

(λa)n−1 −λa e + =− (n − 1)! =−

z=a

Z

+ Z

a

λn−1 z n−2 e−λz dz (n − 2)!

0 z=0 a n−1 n−2 −λz

λ

0

z e (n − 2)!

dz

(λa)n−1 −λa e + P(Zn−1 ≤ a) . (n − 1)!

11.13 b Use part a recursively: P(Zn ≤ a) = − =−

(λa)n−1 −λa e + P(Zn−1 ≤ a) (n − 1)! (λa)n−1 −λa (λa)n−2 −λa e − e + P(Zn−2 ≤ a) (n − 1)! (n − 2)!

.. . =−

n−1 X i=1

(λa)i −λa e + P(Z1 ≤ a) . i!

11.13 c Since P(Z1 ≤ a) = 1 − e−λa , we find P(Zn ≤ a) = P(Z1 ≤ a) −

n−1 X i=1

= 1 − e−λa −

n−1 X i=1

(λa)i −λa e i!

X (λa)i −λa (λa)i −λa e =1− e . i! i! i=0 n−1

12.1 e This is certainly open to discussion. Bankruptcies: no (they come in clusters, don’t they?). Eggs: no (I suppose after one egg it takes the chicken some time to produce another). Examples 3 and 4 are the best candidates. Example 5 could be modeled by the Poisson process if the crossing is not a dangerous one; otherwise authorities might take measures and destroy the homogeneity. 12.2 Let X be the number of customers on a day. Given is that P(X = 0) = 10−5 . Since X is Poisson distributed, P(X = 0) = e−λ . So e−λ = 10−5 , which implies −λ = −5 ln(10), and hence λ = 11.5. Then also E [X] = λ = 11.5. 12.3 When N has a Pois (4) distribution, P(N = 4) = 44 e−4 /4! = 0.195. 12.4 When X has a Pois (2) distribution, P(X ≤ 1) = P(X = 0) + P(X = 1) = e−2 + 2e−2 = 3e−2 = 0.406. 12.5 a Model the errors in the bytes of the hard disk as a Poisson process, with intensity λ per byte. Given is that λ · 220 = 1, or λ = 2−20 . The expected number of errors in 512 = 29 bytes is λ29 = 2−11 = 0.00049. 12.5 b Let Y be the number of errors on the hard disk. Then Y has a Poisson distribution with parameter µ = 39054015 × 0.00049 = 19069.34. Then P(Y ≥ 1) = 1 − P(Y = 0) = 1 − e−19069.34 = 1.00000 · · · . For all practical purposes this happens with probability 1.

29.1 Full solutions

497

12.6 The expected numbers of flaws in 1 meter is 100/40 = 2.5, and hence the number of flaws X has a Pois (2.5) distribution. The answer is P(X = 2) = 1 (2.5)2 e−2.5 = 0.256. 2! 12.7 a It is reasonable to estimate λ with (nr. of cars)/(total time in sec.) = 0.192. 12.7 b 19/120 = 0.1583, and if λ = 0.192 then P(N (10) = 0) = e−0.192·10 = 0.147. 12.7 c P(N (10) = 10) with λ from a seems a reasonable approximation of this probability. It equals e−1.92 · (0.192 · 10)10 /10! = 2.71 · 10−5 . 12.8 a We have E [X(X − 1)] =

∞ X

k(k − 1)e−λ λk /k! =

k=0

=λ

∞ X 2

∞ X

e−λ λk /(k − 2)!

k=2

e−λ λk−2 /(k − 2)! = λ2

∞ X

e−λ λj /j! = λ2 .

j=0

k=2

12.8 b Since Var(X) = E [X(X − 1)] + E [X] − (E [X])2 , we have Var(X) = λ2 + λ − λ2 = λ. 12.9 In a Poisson process with intensity 1, the number of points in a interval of length t has a Poisson distribution with parameter t. So the number Y1 of points in [0, µ1 ] is Pois (µ1 ) distributed, and the number Y2 in [µ1 + µ1 + µ2 ] is Pois (µ2 ) distributed. But the sum Y1 + Y2 of these is equal to the number of points in [0, µ1 + µ2 ] , and so is Pois (µ1 + µ2 ) distributed. k

12.10 We have to consider the numbers pk = P(X = k) = µk! e−µ . To compare them, divide: pk+1 /pk = µ/(k + 1), and note that pk < ( or > )pk+1 is equivalent to pk+1 /pk > ( or < )1. So it follows immediately that — for µ < 1 the probabilities P(X = k) are decreasing, and µ — for µ > 1 the probabilities are increasing as long as k+1 > 1, and decreasing from the moment where this fraction has become less than 1. Lastly if µ = 1, then p0 = p1 > p2 > p3 > . . . . 12.11 Following the hint, we obtain: P(N ([0, s] = k, N ([0, 2s]) = n) = P(N ([0, s]) = k, N ((s, 2s]) = n − k) = P(N ([0, s]) = k) · P(N ((s, 2s]) = n − k) = (λs)k e−λs /(k!) · (λs)n−k e−λs /((n − k)!) = (λs)n e−λ2s /(k!(n − k)!). So P(N ([0, s]) = k, N ([0, 2s]) = n) P(N ([0, 2s]) = n) = n!/(k!(n − k)!) · (λs)n /(2λs)n

P(N ([0, s]) = k | N ([0, 2s]) = n) =

= n!/(k!(n − k)!) · (1/2)n . This holds for k = 0, . . . , n, so we find the Bin (n, 21 ) distribution. 12.12 a The event {X2 ≤ t} is a dist union of the events {X1 ≤ s, X2 ≤ t} and {X1 > s, X2 ≤ t}.

498


12.12 b If X2 ≤ t and Na = 2, then the first two points of the Poisson process must lie in [0, t], and no points lie in (t, a], and conversely. Therefore P(X2 ≤ t, Na = 2) = P(N ([0, t]) = 2, N ((t, a]) = 0) 2 2 = λ2!t e−λt · e−λ(a−t) 1 2 2 −λa . = 2λ t e Similarly, P(X1 > s, X2 ≤ t, Na = 2) = P(N ([0, s)) = 0, N ([s, t]) = 2, N ((t, a]) = 0) 2 2 = e−λs · λ (t−s) e−λ(t−s) · e−λ(a−t) 2! 2 −λa 1 2 = 2 λ (t − s) e . The result now follows by using part a. 12.12 c This follows immediately from part b since P(X1 ≤ s, X2 ≤ t, Na = 2) , P(X1 ≤ s, X2 ≤ t | Na = 2) = P(Na = 2) where the denominator equals

1 2 2 λ a 2

e−λa .

12.13 a Given that Nt = n + m, the number of daisies Xt has a binomial distribution with parameters n + m and 1/4. So P(Xt = n, Yt = m|Nt = n + m) = 1 n 3 m P(Xt = n|Nt = n + m) = n+m . n 4 4 12.13 b Using part a we find that P(Xt = t =m) = P(Xt = n, Yt = m|N Y t = n + m) P(Nt = n + m) = n,

n+m n

1 4

n

3 4

m

(λt)n+m e−λt (n+m)!

=

1 1 n! m!

n

1 4

3 4

m

(λt)n+m e−λt .

12.13 c UsingPb we find that P(Xt = n, Yt = m) = P(Xt = n) = ∞ m=0 P m −(3λ/4)t 1 1 n 1 1 ( ) (λt)n e−(λ/4)t ∞ = n! ((λ/4)t)n e−(λ/4)t , m=0 m! (3λ/4) e n! 4 since the sum over m just adds all the probabilities P(Z = m) where Z has a Pois (3λ/4) distribution. 12.14 a This follows since 1 − 1/n → 1 and 1/n → 0. 12.14 b This is an easy computation: E [Xn ] = (1 − n1 ) · 0 + ( n1 ) · 7n = 7 for all n. √ 13.1 For U (−1, 1), µ = 0 and σ = 1/ 3, and we obtain the following table: k

1

2 3 4

P(|Y − µ| < kσ) 0.577 1 1 1 √ For U (−a, a), µ = 0 and σ = a/ 3, and we obtain the following table: k

1

2 3 4

P(|Y − µ| < kσ) 0.577 1 1 1 For N (0, 1), µ = 0 and σ = 1, and we obtain the following table: k

1

2

3

4

P(|Y − µ| < kσ) 0.6826 0.9544 0.9974 1

29.1 Full solutions

499

For N (µ, σ 2 ), P(|Y − µ| < kσ) = P(|Y − µ|/σ < k), and we obtain the same result as for N (0, 1). p For Par (3), µ = 3/2, σ = 3/4, and Z

3/2+k

√

3/2

P(|Y − µ| < kσ) =

√ 3x−4 dx = 1 − (3/2 + k 3/2)−3 ,

1

and we obtain the following table: k

1

2

3

4

P(|Y − µ| < kσ) 0.925 0.970 0.985 0.992 √ For Geo (3), µ = 2, σ = 2, and P(|Y − µ| < kσ) = P |Y − 2| < k 2 = √ √ √ P 2 − k 2 < Y < 2 + k 2 = P Y < 2 + k 2 , and we obtain the following table: k

1

2

3

4

P(|Y − µ| < kσ) 0.875 0.9375 0.9844 0.9922 13.2 a From the formulas for the U (a, b) distribution, substituting a = −1/2 and b = 1/2, we derive that E [Xi ] = 0 and Var(Xi ) = 1/12. 13.2 b We write S = X1 + X2 + · · · + X100 , for which we find E [S] = E [X1 ] + · · · + 1 E [X100 ] = 0 and, by independence, Var(S) = Var(X1 ) + · · · + Var(X100 ) = 100 · 12 = 100/12. We find from Chebyshev’s inequality: P(|S| > 10) = P(|S − 0| > 10) ≤

Var(S) 1 = . 102 12

13.3 We can apply the law of large numbers (Yi ), with Yi = |Xi | R 0.5 to the sequence for i = 1, . . . . Since E [Yi ] = E [|Xi |] = 2 0 P x dx = 2 · 21 · (0.5)2 = 0.25, and since the Yi have finite variance, it follows that n1 n i=1 |Xi | → 0.25 as n → ∞. 13.4 a Because Xi has a Ber (p) distribution, E [Xi ] = p and Var(Xi ) = p (1 − p), ¯ n = p and Var X ¯ n = Var(Xi ) /n = p (1 − p)/n. By Chebyshev’s and so E X inequality: ¯ n − p| ≥ 0.2 ≤ p (1 − p)/n = 25p(1 − p) . P |X (0.2)2 n The right-hand side should be at most 0.1 (note that we switched to the complement). If p = 1/2 we therefore require 25/(4n) ≤ 0.1, or n ≥ 25/(4 · 0.1) = 62.5, i.e., n ≥ 63. Now, suppose p 6= 1/2, using n = 63 and p(1 − p) ≤ 1/4 we conclude that 25p(1 − p)/n ≤ 25 · (1/4)/63 = 0.0992 < 0.1, so (because of the inequality) the computed value satisfies for other values of p as well. 13.4 b For arbitrary a > 0 we conclude from Chebyshev’s inequality:

¯ n − p| ≥ a ≤ P |X

p(1 − p) p (1 − p)/n 1 = ≤ , a2 na2 4na2

where we used p (1 − p) ≤ 1/4 again. The question now becomes: when a = 0.1, for what n is 1/(4na2 ) ≤ 0.1? We find: n ≥ 1/(4 · 0.1 · (0.1)2 ) = 250, so n = 250 is large enough.

500


13.4 c From part a we know that an error of size 0.2 or occur with a probability of at most 25/4n, regardless of the values of p. So, we need 25/(4n) ≤ 0.05, i.e., n ≥ 25/(4 · 0.05) = 125. ¯ n ≤ 0.5 for the case that p = 0.6. Then E X ¯ n = 0.6 13.4 d We compute P X ¯ n = 0.6 · 0.4/n. Chebyshev’s inequality cannot be used directly, we need and Var X ¯ n ≤ 0.5 is contained in the event “the an intermediate step: the probability that X prediction is off by at least 0.1, in either direction.” So ¯ n ≤ 0.5 ≤ P |X ¯ n − 0.6| ≥ 0.1 ≤ 0.6 · 0.4/n = 24 P X (0.1)2 n

For n ≥ 240 this probability is 0.1 or smaller. ¯ n − c| ≤ 0.5 ≥ 0.9, we must have P |U ¯n | ≤ 0.5 ≥ 0.9. Now 13.5 To get P |M 3/n ¯n | ≥ 0.5 ≤ Var(Ui )/n P |U = 0.25 = 12 . If we want this below 0.1, we should take n (0.5)2 n ≥ 120. 13.6 The probability distribution for an individual game has P(X = −1) = 18 , and 37 1 P(X = 1) = 19 . From this we compute E [X] = (−1) · 18 + 1 · 19 = 37 . If the game is 37 37 37 played 365 · 1000 times, and we suppose that the roulette machine is “fair”, the total 1 gain will, according to the law of large numbers, be close to 365 · 1000 · 37 = 9865 (Euro). (Using Chebyshev’s inequality one can, for example, find a lower bound for the probability that the total gain will be between 9865 − 250 and 9865 + 250.) 13.7 Following the hint we define Yi = 1Pwhen Xi ∈ (−∞, a], and 0 otherwise. In this way we have written Fn (a) = n1 n i=1 Yi . The expectation of Yi equals E [Yi ] = P(Xi ∈ (−∞, a]) = F (a). Moreover, Var(Yi ) = F (a)(1 − F (a)) is finite. Since Fn (a) = Y¯n , the law of large numbers tells us that lim P(|Fn (a) − F (a)| > ε) = 0.

n→∞

R a+h

13.8 a We have p = E [Y ] = a−h xe−x dx ≈ 2h · 2e−2 . (Of course the integral can be computed exactly, but this approximation is excellent: for h = 0.25 we get p = 0.13534, while the exact result is p = 0.13533.) Now Var(Y ) = p(1 − p) = 4he−2 (1 − 4he−2 ), and

Var Y¯n /2h =

1 4he−2 (1 − 4he−2 ). 4h2 n

13.8 b What is required is that

P Y¯n /2h − f (a) ≥ 0.2f (a) ≤ 0.2. Chebyshev’s inequality yields that

Var Y¯n /2h , P Y¯n /2h − f (a) ≥ 0.2f (a) ≤ 0.04f (a)2

so we want Var Y¯n /2h ≤ 0.008 · (2e−2 )2 . Using part a this leads to 1 0.008 · e−2 ≤ n 1 − e−2

=⇒

n≥

e2 − 1 = 798.6 0.008

=⇒

n = 799.

29.1 Full solutions

501

13.9 a The statement looks like the law of large numbers, and indeed, if we look more closely, we see that Tn is the average of an i.i.d. sequence: define Yi = Xi2 , then Tn = Y¯n . The law of large numbers now states: if Y¯n is the average of n independent random variables with expectation µ and variance σ 2 , then for any ε > 0: limn→∞ P |Y¯n − µ| > ε = 0. So, if a = µ and the variance σ 2 is finite, then it is true.

13.9 b We compute expectation and variance of Yi : E [Yi ] = E Xi2 =

2

4

R1

1 4 x −1 2

R1

1 2 x −1 2 2

dx =

1/3. And: E Yi = E Xi = dx = 1/5, so Var(Yi ) = 1/5 − (1/3) = 4/45. The variance is finite, so indeed, the law of large numbers applies, and the statement is true if a = E Xi2 = 1/3. 13.10 a For 0 ≤ ε ≤ 1: P(|Mn − 1| > ε) = P(1 − ε < Mn < 1 + ε) = P(1 − ε < Mn ) = (1 − ε)n . 13.10 b For any 0 < ε: limn→∞ (1 − ε)n = 0. The conclusion is that Mn converges to 1, as n goes to infinity. This cannot be obtained by a straightforward application of Chebyshev’s inequality or the law of large numbers. 13.11 a We have P(|X − 1| > 8) = P(X = 10) = 1/10. On the other hand we get from Chebyshev’s inequality P(|X − 1| > 8) ≤ (t − 1)/64 = 9/64. 13.11 b For a = 5 we have EXACT: P(|X − 1| > 5) = P(X = 10) = 1/10, CHEB: P(|X − 1| > 5) ≤ 9/25, so the Chebyshev gap equals 9/25-1/10=0.26. For a = 10 we have EXACT: P(|X − 1| > 10) = 0, CHEB: P(|X − 1| > 10) ≤ 9/100, so the Chebyshev gap equals 9/100=0.09. 13.11 c Choosing a = t as in the previous question, we have P(|X − 1| > t) = 0, and P(|X − 1| > t) ≤ (t − 1)/t2 < 1/t. So to answer the question we can simply choose t = 100, t = 1000 and t = 10 000. 13.11 d As we have seen that the Chebyshev gap can be made arbitrarily small (taking arbitrarily large t in part c) we can not find a closer bound for this family of probability distributions. ¯ n = 1 Pn µi , and Var X ¯ n = 12 Pn Var(Xi ), this inequal13.12 a Since E X i=1 i=1 n n ity follows directly from Chebyshev’s. 13.12 b This follows directly from n X 1 ¯n = 1 Var X Var(Xi ) ≤ M → 0 n2 i=1 n

as

n → ∞.

14.1 Since µ = 2 and σ = 2, we find that

¯ 144 > 1 P(X1 + X2 + · · · + X144 > 144) = P X ¯ 144 − µ √ √ X 1−µ =P 144 > 144 σ σ

1−2 2 ≈ 1 − Φ(−6) = 1. = P Z144 > 12

502


14.2 Since E [Xi ] = 1/4 and Var(Xi ) = 3/80, using the central limit theorem we find √

P(X1 + X2 + · · · + X625 < 170) = P

¯ 625 − 1/4 √ X 170/625 − 1/4 p 625 p < 625 3/80 3/80

!

≈ Φ(2.8402) = 1 − 0.0023 = 0.9977.

¯ n − p| < 0.2 = 1−P X ¯ n − p ≥ 0.2 −P X ¯ n − p ≤ −0.2 . 14.3 First note that P |X Because µ = p and σ 2 = p(1 − p), we find, using the central limit theorem:

¯ n − p ≥ 0.2 = P P X

√

¯n − p √ 0.2 X np ≥ np p(1 − p) p(1 − p)

= P Zn ≥

√

0.2 np p(1 − p)

!

!

≈P Z≥

√

0.2 np p(1 − p)

!

,

where Z has an N (0, 1) distribution. Similarly,

¯ n − p ≤ −0.2 ≈ P Z ≥ P X

√

0.2 np p(1 − p)

!

,

so we are looking for the smallest positive integer n such that 1 − 2P Z ≥

√

0.2 np p(1 − p)

!

≥ 0.9,

i.e., the smallest positive integer n such that P Z≥

√

0.2 np p(1 − p)

!

≤ 0.05.

From Table ?? it follows that √

np

0.2 ≥ 1.645. p(1 − p)

Since p(1 − p) ≤ 1/4 for all p between 0 and 1, we see that n should be at least 17. 14.4 Using the central limit theorem, with µ = 2, and σ 2 = 4,

√ 2−2 T¯30 − 2 ≤ 30 2 2 1 ≈ Φ(0) = . 2

P(T1 + T2 + · · · + T30 ≤ 60) = P

√

30

14.5 In Section 4.3 we have seen that X has the same probability distribution as X1 + X2 + · · · + Xn , where X1 , X2 , . . . , Xn are independent Ber (p) distributed random variables. Recall that E [Xi ] = p, and Var(Xi ) = p(1 − p). But then we have for any real number a that X − np ≤a P p np(1 − p)

!

=P

X1 + X2 + · · · + Xn − np p ≤a np(1 − p)

!

= P(Zn ≤ a) ;

29.1 Full solutions

503

see also (14.1). It follows from the central limit theorem that X − np P p ≤a np(1 − p) i.e., the random variable √X−np

np(1−p)

!

≈ Φ(a),

has a distribution that is approximately standard

normal. 14.6 a Using the result of Exercise 14.5 yields that X − 25 25 − 25 P(X ≤ 25) = P p ≤ p 75/4 75/4 X − 25 ≤0 =P p 75/4

!

!

≈ P(Z ≤ 0) 1 = , 2 where Z has a standard normal distribution. In the same way, X − 25 26 − 25 P(X < 26) = P p ≤ p 75/4 75/4 X − 25 2 =P p ≤ √ 5 3 75/4

!

!

≈ P(Z ≤ 0.2309) = 0.591. 14.6 b P(X ≤ 2) ≈ P(Z ≤ −5.31) ≈ 0 (the table ends with z = 3.49). 1/4

1/4

¯ n , one easily shows that E [Y ] = n µ, and Var(Y ) = 14.7 Setting Y = n σ X σ √ 1/ n. Since ¯n − µ 1X ≥ a = P(|Y − E [Y ]| ≥ a) , P n 4 σ it follows from Chebyshev’s inequality that

¯n − µ 1X ≥a P n 4 σ

≤

1 √ . a2 n

As a consequence, if n goes to infinity, we see that most of the probability mass 1 ¯ of the random variable n 4 Xnσ−µ is concentrated in the interval between −a and a, for every a > 0. Since we can choose a arbitrarily small, this explains the spike in Figure 14.1.

14.8 a 1 = Var(Xi ) = E Xi2 − (E [Xi ])2 = E Xi2 − 0, so E Xi2 = 1.

14.8 b (integration by parts: E Xi4 = R

1

2

R

1

2

x4 √12π e− 2 x dx =

2 3 xe− 2x dx = 3E Xi = 23) 2 4 Var Xi = E Xi − (E Xi )2 = 3 − 1 = 2.

−1 √ 2π

1

2

[x3 e− 2 x ] −

R

1

2

3x2 e− 2 x dx =

504


14.8 c P(Y100 > 110) ≈ P Z >

√1 2

≈ 0.242.

14.9 a The probability that for a chain of at least 50 meters more than 1002 links are needed is the same as the probability that a chain of 1002 chains is shorter than 50 meters. Assuming that the random variables X1 , X2 , . . . , X1002 are independent, and using the central limit theorem, we have that

P(X1 + X2 + · · · + X1002 < 5000) ≈ P Z <

5000 √ −5 √ 1002 · 1002 0.04

= 0.0571,

where Z has an N (0, 1) distribution. So about 6% of the customers will receive a free chain. 14.9 b We now have that P(X1 + X2 + · · · + X1002 < 5000) ≈ P(Z < 0.0032) , which is slightly larger than 1/2. So about half of the customers will receive a free chain. Clearly something has to be done: a seemingly minor change of expected value has major consequences! 14.10 a Note that

√ √ √ ¯ √ ¯ n − c| ≤ 0.5 = P − 3 n Mn − c < 3 n P |M 6 σ 6 √ √ − 3√ 3√ ≈P n < n 6 6 √ 3√ = 1 − 2P Z ≥ n , 6

√ √ where Z is N (0, 1). Now choose n such that 1 − 2P Z ≥ 63 n = 0.9, i.e., let n √ √ √ √ be such that P Z ≥ 63 n = 0.05. Then 63 n = 2.75 yielding that n = 90.75. Since n is an integer, we find n = 91. ¯ n − c)/σ 14.10 b In case the Ui are normally distributed, the random variable (M has an N (0, 1) distribution, and the calculations in the answer of a are ‘exact’ (cf. Section 5.5). In all other cases this random variable has a distribution which is by approximation equal to an N (0, 1) distribution (see also the discussion in Section 14.2 on the size of n).

15.1 a For bin (32.5, 33.5], the height equals the number of xj in Bi divided by n|Bi |: 3/(5732 · 1) = 0.00052. Similar computations for the other bins give bin (32.5, 33.5] (33.5, 34.5] (34.5, 35.5] (35.5, 36.5] (36.5, 37.5] (37.5, 38.5] (38.5, 39.5] (39.5, 40.5]

height 0.00052 0.00331 0.01413 0.03297 0.07135 0.13137 0.18526 0.18876

bin (40.5, 41.5] (41.5, 42.5] (42.5, 43.5] (43.5, 44.5] (44.5, 45.5] (45.5, 46.5] (46.5, 47.5] (47.5, 48.5]

height 0.16312 0.11270 0.05461 0.02931 0.00872 0.00314 0.00052 0.00017

29.1 Full solutions

505

15.1 b Symmetric, since the histogram has one mode with more or less the same decay in both directions. 0.20

0.15

0.10

0.05

0 32

34

36

38

40

42

44

46

48

50

15.2 a For bin (50, 55], the height equals the number of xj in Bi divided by n|Bi |: 1/(23 · 5) = 0.0087. The heights of the other bins can be computed similarly. bin

height

(50,55] (55,60] (60,65] (65,70] (70,75] (75,80] (80,85]

0.0087 0.0174 0.0087 0.0870 0.0348 0.0348 0.0087

15.2 b No, 31 degrees is extremely low compared to the temperature experienced on earlier takeoffs. 15.3 a For bin (0, 250], the height equals the number of xj in Bi divided by n|Bi |: 141/(190 · 250) = 0.00297. The heights of the other bins can be computed similarly. Bin

Height

(0,250] (250,500] (500,750] (750,1000] (1000,1250] (1250,1500] (1500,1750] (1750,2000] (2250,2500] (2250,2500]

0.00297 0.00067 0.00015 0.00008 0.00002 0.00004 0.00004 0 0 0.00002

506


15.3 b Skewed, since the histogram has one mode and only decays to the right. 0.003

0.002

0.001

0 0

500 1000 1500 2000 2500

15.4 a For bin (0, 500], the height equals the number of xj in Bi divided by n|Bi |: 86/(135 · 500) = 0.0012741. The heights of the other bins can be computed similarly.

Bin [0,500] (500,1000] (1000,1500] (1500,2000] (2000,2500] (2500,3000] (3000,3500] (3500,4000] (4000,4500] (4500,5000] (5000,5500] (5500,6000] (6000,6500]

Height 0.0012741 0.0003556 0.0001778 0.0000741 0.0000148 0.0000148 0.0000296 0 0.0000148 0 0.0000148 0.0000148 0.0000148

15.4 b Since all elements (not rounded) are strictly positive, the value of Fn at zero is zero. At 500 the value of Fn is the number of xi ≤ 500 divided by n: 86/135 = 0.6370. The values at the other endpoints can be computed similarly: t

Fn (t)

t

Fn (t)

0 500 1000 1500 2000 2500 3000

0 0.6370 0.8148 0.9037 0.9407 0.9481 0.9556

3500 4000 4500 5000 5500 6000 6500

0.9704 0.9704 0.9778 0.9778 0.9852 0.9926 1

15.4 c The area under the histogram on bin (1000, 1500] is 500 · 0.000178 = 0.0889, and Fn (1500) − Fn (1000) = 0.9037 − 0.8148 = 0.0889.

29.1 Full solutions

507

15.5 Since the increase of Fn on bin (0, 1] is equal to the area under the histogram on the same bin, the height on this bin must be 0.2250/(1 − 0) = 0.2250. Similarly, the height on bin (1, 3] is (0.445-0.225)/(3-1)=0.1100, and so on: Bin

Height

(0, 1] (1, 3] (3, 5] (5, 8] (8, 11] (11, 14] (14, 18]

0.2250 0.1100 0.0850 0.0400 0.0230 0.0350 0.0225

15.6 Because (2 − 0) · 0.245 + (4 − 2) · 0.130 + (7 − 4) · 0.050 + (11 − 7) · 0.020 + (15 − 11) · 0.005 = 1, there are no data points outside the listed bins. Hence number of xi ≤ 7 n number of xi in bins (0, 2], (2, 4] and (4, 7] = n n · (2 − 0) · 0.245 + n · (4 − 2) · 0.130 + n · (7 − 4) · 0.050 = n = 0.490 + 0.260 + 0.150 = 0.9.

Fn (7) =

15.7 Note that the increase of Fn on each bin is equal to the area under the histogram on the same bin. Since all bins have the same width, it follows from the results of Exercise 15.2 that the increase of Fn is the largest (5 · 0.0.870 = 0.435) on (65, 70]. 15.8 K does not satisfy (K1), since it is negative between 0 and 1 and does not integrate to one. 15.9 a The scatterplot indicates that larger durations correspond to longer waiting times. 15.9 b By judgement by the eye, the authors predicted a waiting time of about 80. 15.9 c If you project the points on the vertical axes you get the dataset of waiting times. Since the two groups of points in the scatterplot are separated in North East direction (both vertical as well as horizontal), you will see two modes. 15.10 Each steep part of Fn corresponds to a mode of the dataset. One counts four of such parts. 15.11 The height of the histogram on a bin (a, b] is number of xi in (a, b] (number of xi ≤ b) – (number of xi ≤ a) = n(b − a) n(b − a) Fn (b) − Fn (a) . = b−a

508


15.12 a By inserting the expression for fn,h (t), we get n t − xi 1 X t · fn,h (t) dt = t· K dt nh i=1 h −∞ −∞

Z

Z

∞

=

∞

n Z 1X ∞ t t − xi K dt. n i=1 −∞ h h

For each i fixed we find with change of integration variables u = (t − xi )/h, Z

∞ −∞

t K h

t − xi h

Z

∞

dt =

(xi + hu)K (u) du −∞

Z

Z

∞

= xi

∞

K (u) du + h −∞

uK (u) du = xi , −∞

R∞

using that K integrates to one and that −∞ uK (u) du = 0, because K is symmetric. Hence Z ∞ n Z n t − xi 1X 1X ∞ t K xi . dt = t · fn,h (t) dt = n i=1 −∞ h h n i=1 −∞ 15.12 b By means of a similar reasoning Z

∞

t2 · fn,h (t) dt =

Z

−∞

∞

t2 ·

−∞

=

n 1 X t − xi K dt nh i=1 h

n Z t − xi 1 X ∞ t2 K dt. n i=1 −∞ h h

For each i: Z

∞ −∞

Z

t2 K h ∞

=

t − xi h

dt

(xi + hu)2 K (u) du =

−∞

=

x2i

=

x2i

Z

Z

∞

Z

+h

2

Z

uK (u) du + h −∞

∞

(x2i + 2xi hu + h2 u2 )K (u) du

−∞ ∞

K (u) du + 2xi h −∞

∞

2

Z

∞

u2 K (u) du

−∞

2

u K (u) du, −∞

again using that K integrates to one and that K is symmetric. 16.1 a The dataset has 135 elements, so the sample median is the 68th element in order of magnitude. From the table in Exercise 15.4 we see that this is 290. 16.1 b The dataset has n = 135 elements. The lower quartile is the 25th empirical percentile. We have k = b0.25 · (135 + 1)c = 34, so that α = 0, and qn (0.25) = x(34) = 81, according to the table in Exercise 15.4. Similarly, the upper quartile is qn (0.75) = x(102) = 843, and the IQR is 843 − 81 = 762. 16.1 c We have k = b0.37 · (135 + 1)c = b50.32c = 50, so that α = 0.32, and qn (0.37) = x(50) + 0.32 · (x(51) − x(50) ) = 143 + 0.32 · (148 − 143) = 144.6.

29.1 Full solutions

509

16.2 To compute the lower quartile, we use that n = 272, so that k = b0.25 · 273c = b68.25c = 68 and α = 0.25, and hence from Table 15.2: qn (0.25) = x(68) + 0.25 · (x(69) − x(68) ) = 129 + 0.25 · (130 − 129) = 129.25. Similarly, qn (0.5) = 240 and qn (0.75) = 267.75. Hence, |qn (0.75) − qn (0.5)| = 27.75 and |qn (0.25) − qn (0.5)| = 110.75. The kernel density estimate (see Figure 15.3) has a ‘peak’ just right of 240, so there is relatively a lot of probability mass just right of the sample median, namely about 25% between 240 and 110.75. 16.3 a Because n = 24, the sample median is the average of the 12th and 13th element. Since these are both equal to 70, the sample median is also 70. The lower quartile is the pth empirical quantile for p = 1/4. We get k = bp(n + 1)c = b0.25 · (24 + 1)c = b6.25c = 6, so that qn (0.25) = x(6) + 0.25 · (x(7) − x(6) ) = 66 + 0.25 · (67 − 66) = 66.25. Similarly, the upper quartile is the pth empirical quantile for p = 3/4: qn (0.75) = x(18) + 0.75 · (x(19) − x(18) ) = 75 + 0.75 · (75 − 75) = 75. 16.3 b In part a we found the sample median and the two quartiles. From this we compute the IQR: qn (0.75) − qn (0.25) = 75 − 66.25 = 8.75. This means that qn (0.25) − 1.5 · IQR = 66.25 − 1.5 · 8.75 = 53.125, qn (0.75) + 1.5 · IQR = 75 + 1.5 · 8.75 = 88.125. Hence, the last element below 88.125 is 88, and the first element above 53.125 is 57. Therefore, the upper whisker runs until 88 and the lower whisker until 57, with two elements 53 and 31 below. This leads to the following boxplot: 81 75 70 66.25 57 ◦

◦

31

16.3 c The values 53 and 31 are outliers. Value 31 is far away from the bulk of the data and appears to be an extreme outlier. 16.4 a Check that y¯n = 700/99 and x ¯n = 492/11, so that 5 5 (¯ xn − 32) = 9 9

492 − 32 11

=

700 . 99

16.4 b Since n = 11, for both datasets the sample median is 6th element in order of magnitude. Check that Med(y1 , . . . , yn ) = 50/9 and Med(x1 , . . . , xn ) = 42, so that 5 5 50 (Med(x1 , . . . , xn ) − 32) = (42 − 32) = , 9 9 9 which is equal to Med(y1 , . . . , yn ).

510


16.4 c For any real number a, we have n n 1X 1X y¯n = yi = (axi + b) = a n i=1 n i=1

n 1X xi n i=1

!

+ b = a¯ xn + b.

For a ≥ 0, the order of magnitude of yi = axi + b is the same as that of the xi ’s, and therefore Med(y1 , . . . , yn ) = aMed(x1 , . . . , xn ) + b. When a < 0, the order of yi = axi + b is reverse to that of the xi ’s, but the position of the middle number in order of magnitude (or average of the two middle numbers) remains the same, so that the above rule still holds. 16.5 a Check that the sample variance of the yi ’s is (sC )2 = 132550/992 , so that 2

5 9

(sF )2 =

2

5 9

132550 482 = , 11 992

which is equal to (sC )2 , and hence sC = 59 sF . 16.5 b For the xi ’s we have Med(x1 , . . . , xn ) = 42. This leads to the following table 43 43 41 41 41 42 43 58 58 41 41 xi |xi − 42| 1 1 1 1 1 0 1 16 16 1 1 |xi − 42| ordered 0 1 1 1 1 1 1 1 1 16 16 so that MADF = Med|xi − 42| = 1. Similarly, we have Med(y1 , . . . , yn ) = yi |yi −

50 | 9

|yi −

50 | 9

ordered

so that MADC = Med|yi −

55 9

55 9

5

5

5

50 9

55 9

130 9

130 9

5

5

5 9

5 9

5 9

5 9

5 9

0

5 9

80 9

80 9

5 9

5 9

0

5 9

5 9

5 9

5 9

5 9

5 9

5 9

5 9

80 9

80 9

50 | 9

50 , 9

and

= 59 . Therefore, MADC = 59 MADF .

16.5 c Since, y¯n = a¯ xn + b, for any real number a we have for the sample variance of the yi ’s s2Y =

n n 1 X 1 X (yi − y¯n )2 = ((axi + b) − (a¯ xn + b))2 n − 1 i=1 n − 1 i=1

n 1 X =a (xi − x ¯n )2 = a2 s2X n − 1 i=1

,

2

so that sY = |a| sX . Because MedY = aMedX + b, it follows that MADY = Med|yi − MedY | = Med|(axi + b) − (aMedX + b)| = |a|Med|xi − MedX | = |a|MADX . 16.6 a Yes, we find x ¯ = (1 + 5 + 9)/3 = 15/3 = 5, y¯ = (2 + 4 + 6 + 8)/4 = 20/4 = 5, so that (¯ x + y¯)/2 = 5. The average for the combined dataset is also equal to 5: (15 + 20)/7 = 5.

29.1 Full solutions

511

16.6 b The mean of x1 , x2 , . . . , xn , y1 , y2 , . . . , ym equals x 1 + · · · + x n + y1 + · · · + ym n¯ xn + m¯ ym n m = = x ¯n + y¯m . n+m n+m n+m n+m In general, this is not equal to (¯ xn + y¯m )/2. For instance, replace 1 in the first dataset by 4. Then x ¯n = 6 and y¯m = 5, so that (¯ xn + y¯m )/2 = 5 21 . However, the average of the combined dataset is 38/7 = 5 27 . 16.6 c Yes, m = n implies n/(n + m) = m/(n + m) = 1/2. From the expressions found in part b we see that the sample mean of the combined dataset equals (¯ xn + y¯m )/2. 16.7 a We have Medx = 5 and Medy = (4 + 6)/2 = 5, whereas for the combined dataset 1 2 4 5 6 8 9 the sample median is the fourth number in order of magnitude: 5. 16.7 b This will not be true in general: take 1, 2, 3 and 5, 7 with sample medians 2 and 6. The combined dataset has sample median 3, whereas (2 + 6)/2 = 4. 16.7 c This will not be true in general: take 1, 9 and 2, 4 with sample medians 5 and 3. The combined dataset has sample median 3, whereas (5 + 3)/2 = 4. 16.8 The ordered combined dataset is 1, 2, 4, 5, 6, 8, 9, so that the sample median equals 5. The absolute deviations from 5 are: 4, 3, 1, 0, 1, 3, 4, and if we put them in order: 0, 1, 1, 3, 3, 4, 4. The MAD is the sample median of the absolute deviations, which is 3. 16.9 a One can easily check that x ¯n = 10/3. The average of the yi ’s: 1 3

−

1 1 +1+ 6 15

=

1 54 3 · = , 3 60 10

so that y¯7 = 1/¯ x7 . 16.9 b No, for instance take 1, 2, and 3. Then x ¯n = 2, and y¯n = (1 + 11/18.

1 2

+ 13 )/3 =

16.9 c First note that the transformation y = 1/x puts the yi ’s in reverse order compared to the xi ’s. If n is odd, then the middle element in order of magnitude remains the middle element under the transformation y = 1/x. Therefore, if n is odd, the answer is yes. If n is even, the answer is no, for example take 1 and 2. Then MedX = 23 and MedY = Med(1, 32 ) = (1 + 32 )/2 = 54 . 16.10 a The sum of the elements is 16.8 + y which goes to infinity as y → ∞, and therefore the sample mean also goes to infinity, because we only divide the sum by the sample size n. When y → ∞, the ordered data will be 3.0 4.2 4.6 5.0 y Therefore, no matter how large y gets, the sample median, which is the middle number, remains 4.6. 16.10 b In order to let the middle number go to infinity, we must replace at least three numbers. For instance, replace 3.2, 4.2, and 5.0 by some real number y that goes to infinity. In that case, the ordered data will be 3.0 4.6 y y y of which the middle number y goes to infinity.

512


16.10 c Without loss of generality suppose that the dataset is already ordered. By the same reasoning as in part a, one can argue that we only have to replace one element in order to let the sample mean go to infinity. When n = 2k + 1 is odd, the middle number is the (k + 1)st number. By the same reasoning as in part b one can argue that in order to let that go to infinity, we must replace the middle number, as well as k other elements by y, which means we have to replace k + 1 = (n − 1)/2 = b(n + 1)/2c elements. When n = 2k is even, the middle number is the average of the kth and (k + 1)st number. In order to let that go to infinity, it suffices to replace k elements, that include either the kth or the (k + 1)st number, by y, This means we have to replace k = n/2 = b(n + 1)/2c elements. 16.11 a. From Exercise 16.10 we already know that the sample mean goes to infinity. This implies that (4.6 − x ¯n )2 also goes to infinity and therefore also the sample variance, as well as the sample standard deviation. From Exercise 16.10 we know that the sample median remains 4.6. Hence, the ordered absolute deviations are 0 0.4 0.4 1.6 |y − 4.6| Therefore, no matter how large y gets, the MAD, which is the middle number of the ordered absolute deviations, remains 0.4. b. In order to let the middle number the ordered absolute deviations go to infinity, we must at least replace three numbers. For instance, replace 3.2 and 4.2 by y and 5.0 by −y, where y is some real number that goes to infinity. In that case, the ordered data will be −y 3.0 4.6 y y of which the middle number is 4.6. The ordered absolute deviations are 0 0.4 |y − 4.6| |y − 4.6| |4.6 + y| The MAD is the middle number, which goes to infinity. c. Without loss of generality suppose that the dataset is already ordered. By the same reasoning as in part a we only need to replace one element in order to let the sample standard deviation go to infinity. When n = 2k + 1 is odd, the middle number is the (k + 1)st number. By the same reasoning as in part b one can argue that in order to let the MAD go to infinity, we must replace the middle number by y, as well as k other numbers. This means we have to replace k + 1 = (n − 1)/2 = b(n + 1)/2c elements, for instance one by −y and the rest by y. In that case, the sample median remains finite and the majority of the absolute deviations from the median tends to infinity. When n = 2k is even, the middle number is the average of the kth and (k + 1)st number. In order to let the MAD go to infinity, it suffices to replace k elements, that include either the kth or the (k + 1)st number, by y. This means we have to replace k = n/2 = b(n + 1)/2c elements, for instance one by −y and the rest by y. In that case, the sample median remains finite and half of the absolute deviations from the median tends to infinity. 16.12 The sample mean is 1 1 N (N + 1) N +1 (1 + 2 + · · · + N ) = · = N N 2 2

29.1 Full solutions

513

When N = 2k + 1 is odd, the sample median is the (k + 1)st number, which is k + 1 = (N − 1)/2 + 1 = (N + 1)/2. When N = 2k is even, the sample median is the average of the kth and (k + 1)st number: (N/2 + (N/2 + 1)) N +1 = . 2 2 16.13 First note that the sample mean is zero, and n = 2N + 1. Therefore the sample variance is 1 (−N )2 + · · · + (−1)2 + 02 + 12 + · · · N 2 n−1 (N + 1)(2N + 1) 1 1 N (N + 1)(2N + 1) = · 2 · 12 + · · · N 2 = · = , 2N N 6 6

s2n =

q

1 so that sn = (N + 1)(2N + 1). Since n = 2N + 1 is always odd, the sample 6 median is the middle number, which is zero. Therefore we have the following

−N · · · −1 0 1 ··· N xi |xi | N ··· 1 0 1 ··· N |xi | ordered 0 1 1 ··· ··· N N Since n = 2N + 1, which is odd, MAD = Med|xi | is the (N + 1)st number in the bottom row: (N + 1)/2. 16.14 When n = 2i + 1 is odd, then in the formula qn ( 21 ) = x(k) + α[x(k+1) − x(k) ] for the 50th empirical percentile, we find k = b 21 (n + 1)c = bi + 1c = i + 1 and α = 0, so that qn ( 21 ) = x(i+1) , which is the middle number in order of magnitude. When n = 2i is even, we find k = b 12 (n + 1)c = bi + 12 c = i and α = 21 , so that qn ( 21 ) = x(i) +

1 2

x(i+1) − x(i) =

x(i+1) +x(i) , 2

which is the average of the two middle numbers in order of magnitude. 16.15 First write n n n n n 1X 2 1X 2 1X 1X 2 1X (xi − x ¯n )2 = xi − 2¯ xn xi + x ¯2n = xi − 2¯ xn xi + x ¯n . n i=1 n i=1 n i=1 n i=1 n i=1

Next, by inserting n 1X xi = x ¯n n i=1

we find

and

n 1X 2 1 x ¯n = · n · x ¯2n = x ¯2n , n i=1 n

n n n 1X 2 1X 2 1X (xi − x ¯n )2 = xi − 2¯ x2n + x ¯2n = xi − x ¯2n . n i=1 n i=1 n i=1

16.16 According to Exercise 15.12 Z

∞

tfn,h (t) dt = x ¯n −∞

and

514

Full solutions from MIPS: DO NOT DISTRIBUTE Z

∞

t2 fn,h (t) dt =

−∞

Z ∞ n 1X 2 xi + h2 u2 K(u) du. n i=1 −∞

Therefore by the rule for the variance, Var(X) = E X 2 − (E [X])2 , the variance corresponding to fn,h is Z

∞ −∞

Z

t2 fn,h (t) dt −

2

∞

tfn,h (t) dt

Z

∞

=

−∞

=

t2 fn,h (t) dt − (¯ xn )2

−∞ n X

1 n

x2i − (¯ xn )2 + h2

Z

∞

u2 K(u) du

−∞

i=1

Z ∞ n 1X = (xi − x ¯ n ) 2 + h2 u2 K(u) du, n i=1 −∞

according to Exercise 16.15. 16.17 For p = i/(n + 1) we find that k = bp(n + 1)c = i so that

and

α = p(n + 1) − k = 0,

qn (p) = x(k) + α x(k+1) − x(k) = x(i) . 17.1 Before one starts looking at the figures, it is useful to recall a couple features of the two distributions involved: 1. most of the probability mass of the N (0, 1) distribution lies between ±3, and between ±9 for the N (0, 9) distribution. 2. the height at zero of the exponential density is λ and the median of the Exp (λ) is ln(2)/λ (see Exercise 5.11). We then have the following: 1. N (3, 1), since the mode is about 3. 2. N (0, 1), since Fn is 0.5 at about zero and all elements are between about ±2. 3. N (0, 1), since the mode is about zero and all elements are between about −4 and 3. 4. N (3, 1), since the mode is about 3. 5. Exp (1/3), since the sample median is about 2. 6. Exp (1), since the sample median is less than 1. 7. N (0, 1), since the mode is about zero and all elements are between about ±4. 8. N (0, 9), since the mode is about zero, and all elements are between about −6 and 9. 9. Exp (1), since the height of the histogram at zero is about 0.7. Moreover, almost all elements are less than 6, whereas the probability of exceeding 6 for the Exp (1/3) distribution is 0.135. 10. N (3, 1), since Fn is 0.5 at about 3. 11. N (0, 9), since the mode is about zero and all elements are between about ±12. 12. Exp (1/3), since the height of the histogram at zero is about 0.24. Moreover, there are several elements beyond 10, which has probability 0.000045 for the Exp (1) distribution. 13. N (0, 9), since the mode is about zero and all elements are between about −5 and 8.

29.1 Full solutions

515

14. Exp (1/3), since the height of the kernel density estimate at zero is about 0.35. 15. Exp (1), since the sample median is about 1. 17.2 We continue as in Exercise 17.1: 1. Exp (1/3), since the sample median is about 2. 2. N (0, 9), since the sample median is about zero and about −6 and 8. 3. Exp (1/3), since the sample median is about 2. 4. N (0, 1), since the sample median is about zero and about −2 and 3. 5. N (3, 1), since the sample median is about 3. 6. Exp (1), since the sample median is less than one. 7. N (0, 9), since the sample median is about zero and about −9 and 7. 8. N (0, 9), since the sample median is about zero and about −7 and 9. 9. N (3, 1), since the sample median is about 3. 10. Exp (1), since the sample median is less than one. 11. N (3, 1), since the sample median is about 3. 12. Exp (1), since the sample median is less than one. 13. N (0, 1), since the sample median is about zero and about ±3. 14. N (0, 1), since the sample median is about zero and about −3 and 4. 15. Exp (1/3), since the sample median is about 2.

all elements are between all elements are between



17.3 a The model distribution corresponds to the number of women in a queue. A queue has 10 positions. The occurrence of a woman in any position is independent of the occurrence of a woman in other positions. At each position a woman occurs with probability p. Counting the occurrence of a woman as a “success,” the number of women in a queue corresponds to the number of successes in 10 independent experiments with probability p of success and is therefore modeled by a Bin (10, p) distribution. 17.3 b We have 100 queues and the number of women xi in the ith queue is a realization of a Bin (10, p) random variable. Hence, according to Table 17.2, the average number of women x ¯100 resembles the expectation 10p of the Bin (10, p) distribution. We find x ¯100 = 435/100 = 4.35, so an estimate for p is 4.35/10 = 0.435. 17.4 a Recall that the parameter µ is the expectation of the Pois (µ) distribution. Hence, according to Table 17.2 the sample mean seems a reasonable estimate. Since, the dataset contains 229 zero’s, 211 ones, etc., x ¯n =

211 · 1 + 93 · 2 + 35 · 3 + 7 · 4 + 1 · 7 537 = = 0.9323. 576 576

17.4 b From part a the parameter µ = 0.9323. Then P(X = 0) = e−0.9323 = 0.393, and the other Poisson probabilities can be computed similarly. The following table compares the relative frequencies to the Poisson probabilities. hits

0

1

2

3

4

5

6

7

rel.freq. 0.397 0.367 0.162 0.06 0.012 0.0000 0.0000 0.0017 prob. 0.393 0.367 0.171 0.053 0.012 0.0023 0.0004 0.0000

516


17.5 a One possibility is to use the fact that the geometric distribution has expectation 1/p, so that the sample mean will be close to 1/p. From the average number of cycles needed, 331/93, we then estimate p by 93/331 = 0.2809. Another possibility is to use that fact that p is equal to the probability of getting pregnant in the first cycle. We can estimate this by the relative frequency of women that got pregnant in the first cycle: 29/93 = 0.3118. 17.5 b From the average number of cycles needed, 1285/474, we could estimate p by 474/1285 = 0.3689. Another possibility is to estimate p by the relative frequency of women that got pregnant in the first cycle: 198/474 = 0.4177. 17.5 c p

p + p(1 − p) + p(1 − p)2

0.2809 0.3118

0.6281 0.6741

nonsmokers 0.3689 0.4177

0.7486 0.8026

smokers

17.6 a The parameters µ and σ are the expectation and standard deviation of the N (µ, σ 2 ) distribution. Hence, according to Table 17.2 we can estimate µ by the sample mean: n 2283772 1X xi = µ= = 39.84, n i=1 5732 and σ by the sample standard deviation sn . To compute this, use Exercise 16.15 to get s2n

n 1 X n = (xi − x ¯ n )2 = n − 1 i=1 n−1

=

n n−1

"

Hence, take sn =

√

n 1X 2 xi n i=1

!2

"

#

n 1X (xi − x ¯n )2 n i=1

− (¯ xn )2 =

#

5732 9124064 − (39.8)2 = 4.35. 5731 5732

4.35 = 2.09 as an estimate for σ.

17.6 b Since we model the chest circumference by means of a random variable X with a N (µ, σ 2 ) distribution, the required probability is equal to P(38.5 < X < 42.5) = P(X < 42.5) − P(X < 38.5) 42.5 − µ 38.5 − µ =P Z< −P Z < σ σ

=Φ

42.5 − µ σ

−Φ

38.5 − µ σ

where Z = (X − µ)/σ has a standard normal distribution. We can estimate this by plugging in the estimates 39.84 and 2.09 for µ and σ. This gives Φ (1.27) − Φ (−0.64) = 0.8980 − 0.2611 = 0.6369. Another possibility is to estimate P(38.5 < X < 42.5) by the relative frequency of chest circumferences between 38.5 and 42.5. Using the information in Exercise 15.1 we find 3725/5732 = 0.6499.

29.1 Full solutions

517

17.7 a If we model the series of disasters by a Poisson process, then as a property of the Poisson process, the interdisaster times should follow an exponential distribution (see Section 12.3). This is indeed confirmed by the histogram and empirical distribution of the observed interdisaster times; they resemble the probability density and distribution function of an exponential distribution. 17.7 b The average length of a time interval is 40 549/190 = 213.4 days. Following Table 17.2 this should resemble the expectation of the Exp (λ) distribution, which is 1/λ. Hence, as an estimate for λ we could take 190/40 549 = 0.00469. 17.8 a The distribution function of Y is given by

α

FY (a) = P(Y ≤ a) = P X ≤ y 1/α = 1 − eλ

y

.

This is the distribution function of the Exp (λα ) distribution, which has expectation 1/λα . Therefore, E [X α ] = E [Y ] = 1/λα . 17.8 b Take α-powers of the data: yi = xα i . Then, according to part a, the yi ’s are a realization of a sample from an Exp (λα ) distribution, with expectation 1/λα . Hence, 1/λα can be estimated by the average of these numbers: n 1 1X α xi . ≈ λα n i=1

Next, solve for λ to get an estimate for λ: λ=

n 1X α xi n i=1

!−1/α

.

When we plug in α = 2.102 and apply this formula to the dataset, we get λ = (10654.85)−1/2.102 = 0.0121. 17.9 a A (perfect) cylindrical cone with diameter d (at the base) and height h has volume πd2 h/12, or about 0.26d2 h. The effective wood of a tree is the trunk without the branches. Since the trunk is similar to a cylindrical cone, one can expect a linear relation between the effective wood and d2 h. 17.9 b We find

P

yi /xi 9.369 = = 0.3022 n 31 P ( yi )/n 26.486/31 y¯/¯ x= P = = 0.3028 ( xi )/n 87.456/31 P xi yi 95.498 least squares = P 2 = = 0.3035. 314.644 xi z¯n =

17.10 a G(y) = P(Y ≤ y) = P(|X − m| ≤ y) = P(m − y ≤ X ≤ m + y) = P(X ≤ m + y) − P(X ≤ m − y) = F (m + y) − F (m − y).

518


17.10 b If f is symmetric around its median, then f (m − y) = f (m + y) for all y. This implies that the area under f on (−∞, m−y] is equal to the area under f on [m+y, ∞), i.e. F (m−y) = 1−F (m+y). By a this means G(y) = 2F (m+y)−1. To derive Ginv ( 12 ) put 12 = G(y). Then, since m = F inv ( 12 ) 1 3 = 2F (m + y) − 1 ⇔ = F (m + y) 2 4 ⇔ y = F inv ( 34 ) − m = F inv ( 43 ) − F inv ( 12 ). 17.10 c First determine an expression for F inv (x), by putting x = F (u). Then F inv (x) = µ + σΦinv (x). With b it follows that the MAD of a N (µ, σ 2 ) is equal to F inv ( 43 ) − F inv ( 12 ) = σΦinv ( 34 ) − σΦinv ( 12 ) = σΦinv ( 34 ), using that Φinv ( 12 ) = 0. For the N (5, 4) distribution the MAD is 2Φinv ( 34 ) = 2 · 0.6745 = 1.3490. 17.11 a Using Exercise 17.10 a it follows that G(y) = F (m + y) − F (m − y)

= 1 − e−λ(m−y) − 1 − e−λ(m+y)

= e−λm eλy − e−λy =

1 λy e − e−λy , 2

using that m is the median of the Exp (λ) distribution, and satisfies 1 − e−λm = 21 . 17.11 b Combining a with Exercise 17.10 b the MAD of the Exp (λ) distribution is a solution of G(y) = 12 , so that 12 eλy − e−λy = 12 . Multiplying this equation with eλy , yields that the MAD must satisfy e2λy − eλy − 1 = 0. 17.11 c Put x = eλy and solve x2 − x − 1 = 0 for x. This gives √ 1± 5 x= . 2 Since the MAD must be positive, it can be found from the relation √ 1+ 5 . eλy = 2 If follows that the MAD of the Exp (λ) distribution is equal to √ √ ln(1 + 5) − ln 2 ln(1 + 5) = − F inv ( 12 ). λ λ 18.1 If we view a bootstrap dataset as a vector (x∗1 , . . . , x∗n ), we have five possibilities 1, 2, 3, 4, and 6 for each of the six positions. Therefore there are 56 different vectors possible. They are not equally likely. For instance, (1, 1, 1, 1, 1, 1) has probability ( 13 )6 , whereas (2, 2, 2, 2, 2, 2) has probability ( 16 )6 to occur.

29.1 Full solutions

519

18.2 a Because the bootstrap sample can only be 1 if and only if all elements in the bootstrap random sample are 1, we get:

¯ n∗ = 1 = P(X1∗ = 1, . . . , X4∗ = 1) P X = P(X1∗ = 1) · · · P(X4∗ = 1) =

4

1 4

= 0.0039.

18.2 b Because the maximum is less than 4, is equivalent to all numbers being less than 4, we get P(max Xi∗ = 6) = 1 − P(max Xi∗ ≤ 4) = 1 − P(X1∗ ≤ 4, . . . , X4∗ ≤ 4) = 1 − P(X1∗ ≤ 4) · · · P(X4∗ ≤ 4) = 1 −

4

3 4

= 0.6836

18.3 a Note that generating from the empirical distribution function is the same as choosing one of the elements of the original dataset with equal probability. Hence, an element in the bootstrap dataset equals 0.35 with probability 0.1. The number of ways to have exactly three out of ten elements equal to 0.35 is 10 , and each has 3 probability (0.1)3 (0.9)7 . Therefore, the probability that the bootstrap dataset has exactly three elements equal to 0.35 is equal to 10 (0.1)3 (0.9)7 = 0.0574. 3 18.3 b Having at most two elements less than or equal to 0.38 means that 0, 1, or 2 elements are less than or equal to 0.38. Five elements of the original dataset are smaller than or equal to 0.38, so that an element in the bootstrap dataset is less than or equal to 0.38 with probability 0.5. Hence, the probability that the bootstrap dataset has at most two elements less than or equal to 0.38 is equal to 10 (0.5)10 + 10 (0.5) + 10 (0.5)10 = 0.0547. 1 2 18.3 c Five elements of the dataset are smaller than or equal to 0.38 and two are greater than 0.42. Therefore, obtaining a bootstrap dataset with two elements less than or equal to 0.38, and the other elements greater than 0.42 has probabil. So the answer is ity (0.5)2 (0.2)8 . The number of such bootstrap datasets is 10 2 10 (0.5)2 (0.2)8 = 0.000029. 2 ∗ 18.4 a There are 9 elements strictly less than 0.46, so that P(M10 < 0.46) = 9 10 ( 10 ) = 0.3487.

18.4 b There are n − 1 elements strictly less than mn , so that P(Mn∗ < mn ) = ( n−1 )n = (1 − 1/n)n . n 18.5 Since each Xi∗ is either 0, 3, or 6, it is not so difficult to see that X1∗ +X2∗ +X3∗ can only take the values 0, 3, 6, 9, 12, 15, and 18. The sample mean of the ¯ 3∗ − x dataset is x ¯n = 3. Therefore X ¯n = (X1∗ + X2∗ + X3∗ )/3 − 3 can only take ¯ 3∗ − x the values −3, −2, −1, 0, 1, 2, and 3. The value X ¯n = −3 corresponds to (X1∗ , X2∗ , X3∗ ) = (0, 0, 0). Because there are 33 = 27 possibilities for (X1∗ , X2∗ , X3∗ ), ¯ 3∗ − x ¯ 3∗ − x the probability P X ¯n = −3 = 1/27. Similarly, the value X ¯n = −2 ∗ ∗ ∗ corresponds to (X , X , X ) being equal to (3, 0, 0), (0, 3, 0), or (0, 0, 3), so that 1 2 3 ¯ 3∗ − x P X ¯n = −2 = 3/27. The other probabilities can be computed in the same way, which leads to a ¯ n∗ − x P X ¯n = a

−3 −2 −1

0

1

2

3

3 27

7 27

6 27

3 27

1 27

1 27

6 27

520


18.6 a We have Fλ (m) =

1 2

⇔ 1 − e−λm =

1 2

⇔ −λm = ln( 12 ) = − ln(2) ⇔ m =

ln 2 . λ

18.6 b Since we know that the dataset is a realization of a sample from an Exp (λ) distribution, we are dealing with a parametric bootstrap. Therefore we must generate the bootstrap datasets from the distribution function of the Exp (λ) distribution with the parameter λ estimated by 1/¯ xn = 0.2. Because the bootstrap simulation is for Med(X1 , X2 , . . . , Xn )−mλ , in each iteration we must compute Med(x∗1 , x∗2 , . . . , x∗n )− m∗ , where m∗ denotes the median of the estimated exponential distribution: m∗ = ln(2)/(1/¯ xn ) = 5 ln(2). This leads to the following parametric bootstrap simulation procedure: 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from Fˆ (x) = 1 − e−0.2x . 2. Compute the centered sample median for the bootstrap dataset: Med(x∗1 , x∗2 , . . . , x∗n ) − 5 ln(2), where Med(x∗1 , x∗2 , . . . , x∗n ) is the sample median of x∗1 , x∗2 , . . . , x∗n . Repeat steps 1 and 2 many times. 18.7 For the parametric bootstrap, we must estimate the parameter θ by θˆ = ˆ distribution. This (n + 1)mn /n, and generate bootstrap samples from the U (0, θ) ˆ = (n + 1)mn /(2n). Hence, for each bootstrap distribution has expectation µθˆ = θ/2 sample x∗1 , x∗2 , . . . , x∗n compute x ¯∗n − µθˆ = x ¯∗n − (n + 1)mn /(2n). Note that this is different from the empirical bootstrap simulation, where one would estimate µ by x ¯n and compute x ¯∗n − x ¯n . 18.8 a Since we know nothing about the distribution of the interfailure times, we estimate F by the empirical distribution function Fn of the software data and we estimate the expectation µ of F by the expectation µ∗ = x ¯n = 656.8815 of Fn . ¯ n∗ − 656.8815. The The bootstrapped centered sample mean is the random variable X corresponding empirical bootstrap simulation is described as follows: 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from Fn , i.e., draw with replacement 135 numbers from the software data. 2. Compute the centered sample mean for the bootstrap dataset: x ¯∗n − 656.8815 where x ¯∗n is the sample mean of x∗1 , x∗2 , . . . , x∗n . Repeat steps 1 and 2 one thousand times. 18.8 b Because the interfailure times are now assumed to have an Exp (λ) distribuˆ = 1/¯ tion, we must estimate λ by λ xn = 0.0015 and estimate F by the distribution function of the Exp (0.0015) distribution. Estimate the expectation µ = 1/λ of the ˆ = x Exp (λ) distribution by µ∗ = 1/λ ¯n = 656.8815. Also now, the bootstrapped ¯ n∗ − 656.8815. The corresponding centered sample mean is the random variable X parametric bootstrap simulation is described as follows: 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from the Exp (0.0015) distribution.

29.1 Full solutions

521

2. Compute the centered sample mean for the bootstrap dataset: x ¯∗n − 656.8815, where x ¯∗n is the sample mean of x∗1 , x∗2 , . . . , x∗n . Repeat steps 1 and 2 one thousand times. We see that in this simulation the boot¯ n∗ − x strapped centered sample mean is the same in both cases: X ¯n , but the corresponding simulation procedures differ in step 1. ˆ = ln 2/mn = 0.0024 and estimate F by the distribution 18.8 c Estimate λ by λ function of the Exp (0.0024) distribution. Estimate the expectation µ = 1/λ of the ˆ = 418.3816. The corresponding parametric bootExp (λ) distribution by µ∗ = 1/λ strap simulation is described as follows: 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from the Exp (0.0024) distribution. 2. Compute the centered sample mean for the bootstrap dataset: x ¯∗n − 418.3816, where x ¯∗n is the sample mean of x∗1 , x∗2 , . . . , x∗n . Repeat steps 1 and 2 one thousand times. We see that in this parametric bootstrap simulation the bootstrapped centered sample mean is different from the one in the ¯ n∗ − (ln 2)/mn instead of X ¯ n∗ − x empirical bootstrap simulation: X ¯n . 18.9 Estimate µ by x ¯n = 39.85, σ by sn = 2.09, and estimate F by the distribution function of the N (39.85, 4.37) distribution. 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from the N (39.85, 4.37) distribution. 2. Compute the centered sample mean for the bootstrap dataset: x ¯∗n − 39.85 where x ¯n is the sample mean of x∗1 , x∗2 , . . . , x∗n .

¯ n − µ| > 1 by the relative Repeat steps 1 and 2 one thousand times. Estimate P |X frequency of bootstrapped centered sample means that are greater than 1 in absolute value: number of x ¯∗n with |¯ x∗n − 39.85| greater than 1 . 1000 18.10 a Perform the empirical bootstrap simulation as in part a of Exercise 18.8. 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from Fn , i.e., draw with replacement 135 numbers from the software data. 2. Compute the centered sample mean for the bootstrap dataset: x ¯∗n − 656.8815 where x ¯∗n is the sample mean of x∗1 , x∗2 , . . . , x∗n . Repeat steps 1 and 2 one thousand times. ¯ n − µ| > 10 by the relative frequency of bootstrapped centered samEstimate P |X ple means that are greater than 1 in absolute value: number of x ¯∗n with |¯ x∗n − 656.8815| greater than 10 . 1000

522


18.10 b Perform the parametric bootstrap simulation as in part b of Exercise 18.8 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from the Exp (0.0015) distribution. 2. Compute the centered sample mean for the bootstrap dataset: x ¯∗n − 656.8815, where x ¯∗n is the sample mean of x∗1 , x∗2 , . . . , x∗n . Repeat steps 1 and 2 one thousand times. ¯ n − µ| > 10 as in part a. Estimate P |X 18.11 Estimate µ by x ¯n = 39.85, σ by sn = 2.09, and estimate F by the distribution function of the N (39.85, 4.37) distribution. 1. Generate a bootstrap dataset x∗1 , x∗2 , . . . , x∗n from the N (39.85, 4.37) distribution. 2. Compute the sample mean x ¯∗n , sample standard deviation s∗n and empirical dis∗ ∗ tribution function Fn of x1 , x∗2 , . . . , x∗n . Use these to compute the bootstrapped KS distance t∗ks = sup |Fn∗ (a) − Fx¯∗n ,s∗n (a)|

R

a∈

Repeat steps 1 and 2 a large number of times. To investigate to which degree the value 0.0987 agrees with the assumed normality of the dataset, one may compute the percentage of t∗ks values that are greater then 0.0987. The closer this percentage is to one, the better the normal distribution fits the data. 18.12 a Note that P(Tn ≤ 0) = P(Mn ≥ θ) = 0, because all Xi are less than θ with probability one. To compute P(Tn∗ ≤ 0), note that, since the bootstrap random sample is from the original dataset, we always have that Mn∗ ≤ mn . Hence P(Tn∗ ≤ 0) = P(Mn∗ ≥ mn ) = P(Mn∗ = mn ) = 1 − P(Mn∗ < mn ) . Furthermore, from Exercise 18.4 we know that P(Mn∗ < mn ) = 1 −

1 n . n

18.12 b Note that Gn (0) = P(Tn ≤ 0) = P(Mn ≤ 0) = P(X1 ≤ 0) · · · P(Xn ≤ 0) , which is zero because the Xi have a U (0, θ) distribution. With part a it follows that

sup |G∗n (t) − Gn (t)| ≥ |G∗n (0) − Gn (0)| = P(Tn∗ ≤ 0) = 1 − 1 −

R

t∈

1 n

n

.

18.12 c From the inequality e−x ≥ 1 − x, it follows that

1− 1−

1 n

n

≥ 1 − e−1/n

n

= 1 − e−1 .

18.13 a The density of a U (0, θ) distribution is given by fθ (x) = 1/θ, for 0 ≤ x ≤ θ. Hence for 0 ≤ a ≤ θ, Z a 1 a Fθ (a) = dx = . θ θ 0

29.1 Full solutions

523

18.13 b We have P(Tn ≤ t) = P(1 − Mn /θ ≤ t) = P(Mn ≥ θ(1 − t)) = 1 − P(Mn ≤ θ(1 − t)) . Using the rule on page 115 about the distribution function of the maximum, it follows that for 0 ≤ t ≤ 1, Gn (t) = 1 − P(Mn ≤ θ(1 − t)) = 1 − Fθ (θ(1 − t))n = 1 − (1 − t)n . 18.13 c By the same argument as before

ˆ − t) . G∗n (t) = P(Tn∗ ≤ t) = 1 − P Mn∗ ≤ θ(1 ˆ distriSince Mn∗ is the maximum of a random sample X1∗ , X2∗ , . . . , Xn∗ from a U (0, θ) bution, again the rule on page 115 about the distribution function of the maximum yields that for 0 ≤ t ≤ 1,

ˆ − t) = 1 − F ˆ(θ(1 ˆ − t))n = 1 − (1 − t)n . G∗n (t) = 1 − P Mn∗ ≤ θ(1 θ 19.1 a We must show that E [T ] = θ2 . From the formulas for the expectation and variance of uniform random we deduce that E [Xi ] = 0 and Var(Xi ) = variables (2θ)2 /12 = θ2 /3. Hence E Xi2 = Var(Xi )+(E [Xi ])2 = θ2 /3. Therefore, by linearity of expectations 3 2 (E X1 + E X22 + · · · + E Xn2 ) n 3 θ2 θ2 3 θ2 = + ··· + = ·n· = θ2 . n 3 3 n 3

E [T ] =

Since E [T ] = θ2 , the random variable T is an unbiased estimator for θ2 . √ 19.1 b The function g(x) = − x is a strictly convex function,h because g 00 (x) = p √ i −3/2 (x )/4 > 0. Therefore, by Jensen’s inequality, − E [T ] < −E T . Since, from h√ i √ part a we know that E [T ] = θ2 , this means that E T < θ. In other words, T is a biased estimator for θ, with negative bias. 19.2 a We must check whether E [S] = µ. By linearity of expectations, we have

E [S] = E

1 1 1 1 1 1 X1 + X2 + X3 = E [X1 ] + E [X2 ] + E [X3 ] 2 3 6 2 3 6

1 1 1 = µ+ µ+ µ= 2 3 6

1 1 1 + + 2 3 6

µ = µ.

So indeed, S is an unbiased estimator for µ. 19.2 b We must check under what conditions E [T ] = µ for all µ. By linearity of expectations, we have E [T ] = E [a1 X1 + a2 X2 + · · · + an Xn ] = a1 E [X1 ] + a2 E [X2 ] + · · · + an E [Xn ] = a1 µ + a2 µ + · · · + an µ = (a1 + a2 + · · · + an ) µ. This is equal to µ if and only if a1 + a2 + · · · + an = 1. Therefore, T is an unbiased estimator for µ if and only if a1 + a2 + · · · + an = 1.

524


19.3 We must check under what conditions E [T ] = µ for all µ. By linearity of expectations, we have E [T ] = E [a(X1 + X2 + · · · + Xn ) + b] = a (E [X1 ] + E [X2 ] + · · · + E [Xn ]) + b = a (µ + µ + · · · + µ) + b = anµ + b. For T to be an unbiased estimator for µ, one must have anµ + b = µ for all µ, with a and b being constants not depending on µ. This can only be the case, when b = 0 and a = 1/n. 19.4 a The function g(x) = 1/x is convex, so that according to Jensen’s inequality:

E [T ] = E

1 ¯n X

>

1 ¯n . E X

¯ n = E [X1 ] = 1/p, so that E [T ] > 1/(1/p) = p. We conclude that Furthermore, E X T is a biased estimator for p with positive bias. 19.4 b For each i label ‘Xi ≤ 3’ as a success, then the total number Y of all Xi ≤ 3 has a Bin (n, θ) distribution where θ = p + (1 − p)p + (1 − p)2 p represents the probability that a woman becomes pregnant within three or fewer cycles. This implies that E [Y ] = nθ and therefore E [S] = E [Y /n] = E [Y ] /n = θ. This means that S is an unbiased estimator for θ. 19.5 We must find out for which c, E [T ] = µ, where µ = 1/λ is the expectation of the Exp (λ) distribution. Because Mn has an Exp (nλ) distribution with expectation 1/(nλ), we have 1 E [T ] = E [cMn ] = cE [Mn ] = c · . nλ Hence, with the choice c = n, we get E [T ] = 1/λ = µ, which means that T is an unbiased estimator for µ. 19.6 a We must show that E [T ] = 1/λ. By linearity of expectations E [T ] =

n ¯ n − E [Mn ] = n E X n−1 n−1

=

n 1 1 n − = n−1 λ nλ n−1

1−

1 n

δ+

·

1 λ

− δ+

1 nλ

1 1 = . λ λ

¯ n and Mn of which the expectation is δ. From 19.6 b Find a linear combination of X ¯ the for E Xn and E [Mn ] we see that we can eliminate λ by subtracting expressions ¯ n from nE [Mn ]. Therefore, first consider nMn − X ¯ n , which has expectation E X

¯ n = nE [Mn ] − E X ¯n = n δ + 1 E nMn − X nλ

This means that

− δ+

1 λ

= (n − 1)δ.

¯n nMn − X n−1 ¯ n /(n − 1) = δ, so that T is an unbiased has expectation δ: E [T ] = E nMn − X estimator for δ. T =

29.1 Full solutions

525

19.6 c Plug in x ¯n = 8563.5, mn = 2398 and n = 20 in the estimator of part b: t=

20 · 2398 − 8563.5 = 2073.5. 19

19.7 a Note that by linearity of expectations E [T1 ] =

4 E [N1 ] − 2. n

Because N1 has a Bin (n, p1 ) distribution, with p1 = (θ+2)/4, it follows that E [N1 ] = np1 = n(θ + 2)/4, so that E [T1 ] = θ. The argument for T2 is similar. 19.7 b Plug in n = 3839, n1 = 1997, and n2 = 32 in the estimators T1 and T2 : 4 · 1997 − 2 = 0.0808, 3839 4 t2 = · 32 = 0.0333. 3839 t1 =

19.8 From the model assumptions it follows that E [Yi ] = βxi for each i. Using linearity of expectations, this implies that

E [Yn ] 1 E [Y1 ] 1 βx1 βxn + ··· + = + ··· + n x1 xn n x1 xn E [Y1 ] + · · · + E [Yn ] βx1 + · · · + βxn E [B2 ] = = = β, x1 + · · · + xn x1 + · · · + xn x1 E [Y1 ] + · · · + xn E [Yn ] βx21 + · · · + βx2n E [B3 ] = = = β. 2 2 x1 + · · · + xn x21 + · · · + x2n

E [B1 ] =

= β,

19.9 Write T = e−Z/n , where Z = X1 + X2 + · · · + Xn has a Pois (nµ) distribution, with probabilities P(Z = k) = e−nµ (nµ)k /k!. Therefore h

i

E [T ] = E e−Z/n =

∞ X

e−k/n · P(Z = k)

k=0

=e

−nµ

∞ X k=0

Use that

P∞ k=0

e

−k/n

∞ nµe−1/n X (nµ)k = e−nµ · k! k!

k

k=0

xk /k! = ex , with x = nµe−1/n , and conclude that E [T ] = e−nµ · enµe

−1/n

= e−nµ(1−e

−1/n

)

.

¯ n = MSE(Xn ) = σ 2 /n, which is decreasing in n, so the 20.1 We have Var X ¯ n is. In particular Var X ¯ n /Var X ¯ 2n = larger the sample, the more efficient X 2 2 ¯ 2n is twice as efficient as X ¯n . (σ /n)/(σ /2n) = 2 shows that X 20.2 a Compute the mean squared errors of S and T : MSE(S) = Var(S) + [bias(S)]2 = 40 + 0 = 40; MSE(T ) = Var(T ) + [bias(T )]2 = 4 + 9 = 13. We prefer T , because it has a smaller MSE. 20.2 b Compute the mean squared errors of S and T : MSE(S) = 40, as in a; MSE(T ) = Var(T ) + [bias(T )]2 = 4 + a2 . So, if a < 6: prefer T . If a ≥ 6: prefer S. The preferences are based on the MSE criterion.

526


20.3 Var(T1 ) = 1/(nλ2 ), Var(T2 ) = 1/λ2 ; hence we prefer T1 , because of its smaller variance. 20.4 a Since L has the same distribution as N + 1 − M , we find E [T3 ] = E [3L−1] = E [3N +3−3M −1] = 3N +2− E [2T2 + 2] = 3N −2N = N. Here we use that E [2T2 ] = 2N , since T2 is unbiased. 20.4 b We have Var(T3 ) = Var(3L − 1) = 9Var(L) = 9Var(M ) =

1 (N + 1)(N − 2). 2

20.4 c We compute: Var(T3 ) /Var(T2 ) = 9Var(M ) /Var

3 M 2

− 1 = 9Var(M ) / 94 Var(M ) = 4.

So using the maximum is 4 times as efficient as using the minimum! 20.5 From the variance of the sum rule: Var((U + V )/2) = 41 (Var(U ) + Var(V ) + 2Cov(U, V )). Using that Var(V )=Var(U ), we get from this that that the relative efficiency of U with respect to W is equal to Var(W ) Var((U + V )/2) Cov(U, V ) 1 1 1 p = = 1 + 1 + 2p = + ρ (U, V ) . Var(U ) Var(U ) 4 2 2 Var(V ) Var(U )

Since the correlation coefficient is always between -1 and 1, it follows that the relative efficiency of U with respect to W is always between 0 and 1. Hence it ia always better (in the MSE sense) to use W . 20.6 a By linearity of expectations, E [U1 ] = E [T1 ] + 13 (π − E [T1 ] − E [T2 ] − E [T3 ]), which equals α1 + 13 (π − α1 − α2 − α3 ) = α1 . To compute the variances, rewrite U1 as U1 = 23 T1 − 31 T2 − 13 T3 + 13 π. Then, by independence, Var(U1 ) = 49 Var(T1 ) + 19 Var(T2 ) + 91 Var(T3 ) = 23 σ 2 . 20.6 b We have Var(T1 ) /Var(U1 ) = 3/2, so U1 is 50% more efficient than T1 . 20.6 c There are at least two ways to obtain an efficient estimator for α1 = α2 . The first is via the insight that T1 and T2 both estimate α1 , so (T1 + T2 )/2 is an efficient estimator for α1 (c.f. Exercise 20.1). Then we can improve the efficiency in the same way as in part a. This yields the estimator V1 = 12 (T1 + T2 ) + 13 (π − T1 − T2 − T3 ) = 16 T1 + 61 T2 − 13 T3 + 31 π. The second way is to find the linear estimator V1 = uT1 + vT2 + wT3 + t with the smallest MSE, optimising over u, v, w and t. This will result in the same estimator as obtained with the first way. 20.7 We compare the MSE’s, which by unbiasedness are equal to the variances. Both N1 and N2 are binomally distributed, so Var(N1 ) = np1 (1 − p1 ) = n(θ + 2)(1 − (θ + 2)/4)/4, and Var(N2 ) = np2 (1 − p2 ) = nθ(1 − θ)/4. It follows that Var(T1 ) =

1 16 1 16 Var(N1 ) = (4 − θ2 ); Var(T2 ) = 2 Var(N2 ) = θ(4 − θ). n2 n n n

Since (4 − θ2 ) > θ(4 − θ) for all θ, we prefer T2 .

29.1 Full solutions

527

20.8 a This follows directly from linearity of expectations:

¯ n +(1 − r)Y¯m = rE X ¯ n + (1 − r)E Y¯m = rµ + (1 − r)µ = µ. E [T ] = E rX ¯ n and Y¯m are independent, we find MSE(T )=Var(T ) = 20.8 b Using that X ¯ n + (1 − r)2 Var Y¯m = r2 · σ 2 /n + (1 − r)2 · σ 2 /m. r2 Var X To find the minimum of this parabola we differentiate with respect to r and equate the result to 0: 2r/n − 2(1 − r)/m = 0. This gives the minimum value: 2rm − 2n(1 − r) = 0 or r = n/(n + m). 20.9 a Since E [T1 ] = p, T1 is unbiased. The estimator T2 takes only the values 0 and 1, the latter with probability pn . So E [T2 ] = pn , and T2 is biased for all n > 1. 20.9 b Since T1 is unbiased, MSE(T1 ) = Var(T1 ) = np(1 − p)/n2 = n1 p(1 − p). Now n for T2 : this random variable has a Ber (p ) distribution, hence MSE(T2 ) = E (T2 − θ)2 = E (T2 − p)2 = p2 · P(T2 = 0) + (1 − p)2 · P(T2 = 1) = p2 (1 − pn ) + (1 − p)2 pn = pn − 2pn+1 + p2 . 20.9 c For n = 2: MSE(T2 ) = 4p MSE(T1 ), so for p < T1 , but otherwise T1 is more efficient.

1 4

T2 is more efficient than

20.10 a Recall that the variance of an Exp (λ) distribution equals 1/λ2 , hence the mean squared error of T equals MSE(T ) = Var(T ) + E [T ] − λ−1

2

= c2 · n · λ−2 + c · n · λ−1 − λ−1 −2

=λ

c2 · n + (c · n − 1)

=λ

c2 · (n2 + n) − 2cn + 1 .

−2

2

2

20.10 b This is a parabola in c, taking its smallest value when 2c(n2 + n) = 2n, which happens for c = 1/(n + 1). So the estimator with the smallest MSE is U=

1 · (X1 + X2 + · · · + Xn ) . n+1

Substituting c = 1/n, and c = 1/(n + 1) in the formula for the MSE, we find that ¯n) = MSE(X

1 , n

and

MSE(U ) =

1 . n+1

So U performs better in of MSE, but not much. 20.11 We have MSE(T1 ) = Var(T1 ), which equals Var(T1 ) = Var

n X

xi Yi

i=1

n .X

!

x2i

=

i=1

n X

.

x2i Var(Yi )

i=1

n X

!2

x2i

=σ

2

.

i=1

i=1

Also, MSE(T2 ) = Var(T2 ), which equals 1 X Yi n i=1 xi n

Var(T2 ) = Var

!

1 X Yi Var n2 i=1 xi n

=

Finally, MSE(T3 ) = Var(T3 ), which equals

=

n X

n σ2 X 1/x2i . 2 n i=1

!

x2i

.

528


Var(T3 ) = Var

n X

Yi

n .X

i=1

i=1

!

xi

=

n X

Var(Yi )

i=1

.

n X

!2

xi

= nσ

2

.

i=1

n X

!2

xi

.

i=1

To show that Var(T3 ) < Var(T2 ): introduce a random variable X by P(X = xi ) = pi for i = 1 . . . n, and apply Jensen with the function g(x) = 1/x2 , which is strictly convex on (0, ∞) (note that all xi in the cherry tree example are positive). 20.12 a Following the hint: P(Mn ≤ k) = P(X1 ≤ k, X2 ≤ k, . . . , Xn ≤ k) k−n+1 k k−1 · ,···· · = N N −1 N −n+1 (N − n)! k! = . (k − n)! N! 20.12 b To have Mn = n we should have drawn all numbers 1, 2 . . . , n, and conversely, so n!(N − n)! P(Mn = n) = P(Mn ≤ n) = . N! 20.12 c This also follows directly from part a: P(Mn = k) = P(Mn ≤ k) − P(Mn ≤ k − 1) (N − n)! (k − 1)! (N − n)! k! = − (k − n)! N! (k − 1 − n)! N! (k − 1)![k − (k − n)] (N − n)! = (k − 1 − n)! N! (k − 1)! (N − n)! = n· . (k − n)! N! 21.1 Setting Xi = j if red appears in the ith experiment for the first time on the jth throw, we have that X1 , X2 , and X3 are independent Geo (p) distributed random variables, where p is the probability that red appears when throwing the selected die. The likelihood function is L(p) = P(X1 = 3, X2 = 5, X3 = 4) = (1 − p)2 p · (1 − p)4 p · (1 − p)3 p = p3 (1 − p)9 , 3

9

so for D1 one has that L(p) = L( 56 ) = 56 1 − 65 , whereas for D2 one has that 6 1 1 3 1 9 5 L(p) = L( 6 ) = 6 1 − 6 = 5 · L( 6 ). It is very likely that we picked D2 . 21.2 As in the solution of Exercise 21.1, the likelihood is given by L(p) = p3 (1−p)9 , where p is between 0 and 1. So the loglikelihood is given by `(p) = 3 ln p + 9 ln(1 − p), and differentiating the loglikelihood with respect to p gives: d 3 9 `(p) = − . dp p 1−p

d `(p) = 0 if and only if p = 1/4, and since `(p) (and also L(p)!) atWe find that dp tains its maximum for this value of p, we find that the maximum likelihood estimate of p is pˆ = 1/4.

29.1 Full solutions

529

21.3 a The likelihood L(µ) is given by L(µ) = (X = 0)229 P(X = 1)211 P(X = 2)93 P(X = 3)35 P(X = 4)7 P(X = 7) ˜ 537 e−576µ . = Cµ But then the loglikelihood `(µ) satisfies ˜ + 537 ln µ − 576µ. `(µ) = ln C But then we have, that `0 (µ) = 537/µ − 576, from which it follows that `0 (µ) = 0 if and only if µ = 537/576 = 0.93229. 21.3 b Now the likelihood L(µ) is given by L(µ) = (X = 0, 1)440 P(X = 2)93 P(X = 3)35 P(X = 4)7 P(X = 7) ˜ + µ)440 µ316 e−576µ . = C(1 But then the loglikelihood `(µ) satisfies ˜ + 440 ln(1 + µ) + 316 ln µ − 576µ. `(µ) = ln C But then we have, that `0 (µ) =

440 + 316/µ − 576, 1+µ

from which it follows that `0 (µ) = 0 if and only if 576µ2 − 180µ + 260 = 0. We find that µ = 0.9351086. 21.4 a The likelihood L(µ) is given by L(µ) = P(X1 = x1 , . . . , Xn = xn ) = P(X1 = x1 ) · · · P(Xn = xn ) =

µx1 −µ µxn −µ e−nµ · e ··· ·e = µx1 +x2 +···+xn . x1 ! xn ! x1 ! · · · xn !

21.4 b We find that the loglikelihood `(µ) is given by `(µ) =

n X

!

xi

ln(µ) − ln (x1 ! · · · xn !) − nµ.

i=1

Hence

P

xi d` = − n, dµ µ and we find—after checking that we indeed have a maximum!—that x ¯n is the maximum likelihood estimate for µ. 21.4 c In b we have seen that x ¯n is the maximum likelihood estimate for µ. Due to the invariance principle from Section 21.4 we thus find that e−¯xn is the maximum likelihood estimate for e−µ . 21.5 a By definition, the likelihood L(µ) is given by L(µ) = fµ (x1 ) · · · fµ (xn ) 2 2 1 1 1 1 = √ e− 2 (x1 −µ) · · · √ e− 2 (xn −µ) 2π 2π 1

= (2π)−n/2 · e− 2

Pn

2 i=1 (xi −µ)

.

530


But then the loglikelihood `(µ) satisfies `(µ) = −

n n 1X ln(2π) − (xi − µ)2 . 2 2 i=1

Differentiating `(µ) with respect to µ yields `0 (µ) =

n X

(xi − µ) = −nµ +

i=1

n X

xi .

i=1

So `0 (µ) = 0 if and only if µ = x ¯n . Since `(µ) attains a maximum at this value of µ (check this!), we find that x ¯n is the maximum likelihood estimate for µ. 21.5 b Again by the definition of the likelihood, we find that L(σ) is given by L(σ) = fσ (x1 ) · · · fσ (xn ) 2 2 1 2 1 2 1 1 = √ e− 2 x1 /σ · · · √ e− 2 xn /σ σ 2π σ 2π = σ −n (2π)−n/2 e

Pn 2 − 12 i=1 xi 2σ .

But then the loglikelihood `(σ) satisfies `(σ) = −n ln σ −

n n 1 X 2 ln(2π) − xi . 2 2σ 2 i=1

Differentiating `(σ) with respect to σ yields `0 (σ) = −

n n 1 X 2 + 3 xi . σ σ i=1

r

1 Xn 2 xi . Since `(σ) attains a maximum at i=1 n r 1 Xn 2 this value of σ (check this!), we find that xi is the maximum likelihood i=1 n estimate for σ. 0

So ` (σ) = 0 if and only if σ =

21.6 a In this case the likelihood function L(δ) = 0 for δ > x(1) , and P

L(δ) = enδ−

xi

for δ ≤ x(1) .

21.6 b It follows from the graph of L(δ) in a that x(1) is the maximum likelihood estimate for δ. 21.7 By definition, the likelihood L(θ) is given by L(θ) = fθ (x1 ) · · · fθ (xn ) x1 1 2 2 xn 1 2 2 = 2 e− 2 x1 /θ · · · 2 e− 2 xn /θ θ θ ! = θ−2n

n Y

xi

i=1

But then the loglikelihood `(θ) is equal to

e

Pn 2 − 12 i=1 xi 2θ

.

29.1 Full solutions n Y

`(θ) = −2n ln θ + ln

!

xi

−

i=1

531

n 1 X 2 xi . 2 2θ i=1

Differentiating `(θ) with respect to θ yields n −2n 1 X 2 + 3 xi , θ θ i=1

`0 (θ) =

and we find that `0 ()θ) = 0 if and only if v u n u 1 X θ=t x2i .

2n

i=1

Since `(θ) attains a maximum at this value of this!), we find that the q θ (check Pn 1 2 maximum likelihood estimate for θ is given by i=1 xi . 2n 21.8 a The likelihood L(θ) is given by

L(θ) = C · =

1 (2 + θ) 4

1997

·

1 θ 4

32

·

906

1 (1 − θ) 4

·

904

1 (1 − θ) 4

C · (2 + θ)1997 · θ32 · (1 − θ)1810 , 43839

where C is the number of ways we can assign 1997 starchy-greens, 32 sugary-whites, 906 starchy-whites, and 904 sugary-greens to 3839 plants. Hence the loglikelihood `(θ) is given by `(θ) = ln(C) − 3839 ln(4) + 1997 ln(2 + θ) + 32 ln(θ) + 1810 ln(1 − θ). 21.8 b A short calculation shows that d`(θ) =0 dθ

3810θ2 − 1655θ − 64 = 0,

⇔

so the maximum likelihood estimate of θ is (after checking that L(θ) indeed attains a maximum for this value of θ): √ −1655 + 3714385 = 0.0357. 7620 21.8 c In this general case the likelihood L(θ) is given by

L(θ) = C · =

1 (2 + θ) 4

n1

·

1 θ 4

n2

·

n3

1 (1 − θ) 4

·

n4

1 (1 − θ) 4

·

C · (2 + θ)n1 · θn2 · (1 − θ)n3 +n4 , 4n

where C is the number of ways we can assign n1 starchy-greens, n2 sugary-whites, n3 starchy-whites, and n4 sugary-greens to n plants. Hence the loglikelihood `(θ) is given by `(θ) = ln(C) − n ln(4) + n1 ln(2 + θ) + n2 ln(θ) + (n3 + n4 ) ln(1 − θ).

532


A short calculation shows that d`(θ) =0 dθ

nθ2 − (n1 − n2 − 2n3 − 2n4 )θ − 2n2 = 0,

⇔

so the maximum likelihood estimate of θ is (after checking that L(θ) indeed attains a maximum for this value of θ): n1 − n2 − 2n3 − 2n4 +

p

(n1 − n2 − 2n3 − 2n4 )2 + 8nn2 . 2n

21.9 The probability density of this distribution is given by fα,β (x) = 0 if x is not between α and β, and 1 β−α

fα,β (x) =

for

α ≤ x ≤ β.

Since the xi must be in the interval between α and β, the likelihood (which is a function of two variables!) is given by

L(α, β) =

1 β−α

n

for

α ≤ x(1)

and

β ≥ x(n) ,

and L(α, β) = 0 for all other values of α and β. So outside the ‘rectangle’ (−∞, x(1) ]× [x(n) , ∞) the likelihood is zero, and clearly on this ‘rectangle’ it attains its maximum in (x(1) , x(n) ). The maximum likelihood estimates of α and β are therefore α ˆ = x(1) and βˆ = x(n) . 21.10 The likelihood is L(α) =

α

α

···

xα+1 xα+1 1 2

α xα+1 n

= αn

Y n

−(α+1)

xi

,

i=1

so the loglikelihood is `(α) = n ln α − (α + 1) ln

Y n

xi .

i=1

Differentiating `(α) to α yields as maximum likelihood α ˆ = n/ ln

Qn i=1

xi .

21.11 a Since the dataset is a realization of a random sample from a Geo (1/N ) distribution, the likelihood is L(N ) = P(X1 = x1 , X2 = x2 , . . . , Xn = xn ), where each Xi has a Geo (1/N ) distribution. So

L(N ) =

1−

=

1 N

1 1− N

x1 −1

1 1 1− N N Pn

−n+

i=1

xi

x2 −1

1 N

1 1 ··· 1 − N N

xn −1

n

.

But then the loglikelihood is equal to

`(N ) = −n ln N +

−n+

n X i=1

xi ln 1 −

1 . N

1 N

29.1 Full solutions

533

Differentiating to N yields d −n `(N ) = + dN N d dN

−n+

n X

xi

i=1

1 , N (N − 1)

Now `(N ) = 0 if and only if N = x ¯n . Because `(N ) attains its maximum at ˆ =x x ¯n , we find that the maximum likelihood estimate of N is N ¯n . 21.11 b Since P(Y = k) = 1/N for k = 1, 2, . . . , N , the likelihood is given by

L(N ) =

1 N

n

for N ≥ y(n) ,

and L(N ) = 0 for N < y(n) . So L(N ) attains its maximum at y(n) ; the maximum ˆ = y(n) . likelihood estimate of N is N 21.12 Since L(N ) = P(Z = k), it follows from Exercise 4.13c that

L(N ) =

m k

N −m r−k N r

.

In order to see that L(N ) increases for N < mr , and decreases for N > mr , consider k k the ratio L(N )/L(N − 1). After some elementary calculations one finds that (N − m)(n − r) L(N ) = . L(N − 1) N (N − m − r + k) L(N ) −r) So L(N ) is increasing if L(N = N(N(N−m)(N > 1, and another elementary calcu−1) −m−r+k) mr lation shows that this is when N < k .

21.13 Let N (t1 , t2 ) be the number of customers arriving in the show between time t1 and time t2 . Then it follows from the assumption that customers arrive at the shop according to a Poisson process with rate λ, that P(N (t1 , t2 ) = k) =

(λ(t2 − t1 ))k −λ(t2 −t1 ) e , k!

for

k = 0, 1, 2, . . . .

The likelihood L(λ) is given by L(λ) = P(N (12.00, 12.15) = 2, N (12.15, 12.45) = 0, N (12.45, 13.00) = 1) +P(N (12.00, 12.15) = 1, N (12.15, 12.45) = 1, N (12.45, 13.00) = 0) . Since N (12.00, 12.15), N (12.15, 12.45), and N (12.45, 13.00) are independent random variables, we find that ( 14 λ)2 − 1 λ ( 21 λ)0 − 1 λ ( 14 λ)1 − 1 λ e 4 · e 2 · e 4 2! 0! 1! ( 1 λ)1 − 1 λ ( 12 λ)1 − 1 λ ( 14 λ)0 − 1 λ + 4 e 4 · e 2 · e 4 1! 1! 0! 1 3 1 2 −λ = λ + λ e . 128 8

L(λ) =

Now L0 (λ) = 0 if and only if λ3 + 13λ2 − 32λ = 0. Since λ > 0, we find that √ −13+ 297 λ= = 2.1168439. 2

534


21.14 For i = 1, 2, . . . , n, ri is the realization of a continuous random variable Ri . Since the shots at the disc do not influence each other, the Ri are independent, and all have the same distribution function Fθ (x) = x2 /θ2 if x is between 0 and θ. But then fθ (x) = 2x/θ2 for x between 0 and θ (and f (x) = 0 otherwise). Since the disc is hit each of the n shots, the likelihood is: Q

L(θ) = fθ (r1 )fθ (r2 ) · · · fθ (rn ) =

2n n 2r1 2r2 2rn i=1 ri · · · = , θ2 θ2 θ2 θ2n

for θ ≥ r(n) , and L(θ) = 0 otherwise. But then we at once see that L(θ) attains its maximum at θ = r(n) , i.e., the maximum likelihood estimate for θ is θˆ = r(n) . 21.15 At temperature t, the probability of failure of an O-ring is given by p(t), so the probability of a failure of k O-rings at this temperature, for k = 0, 1, . . . , 6, is given by ! 6 (p(t))k (1 − p(t))6−k . k Setting

!16

C=

6 0

6 1

!5

6 2

!2

,

we find that the likelihood L(a, b) is given by L(a, b) = C · (p(53))2 (1 − p(53))4 · · · (p(53))0 (1 − p(53))6 =C·

23 Y

(p(ti ))ni (1 − p(ti ))6−ni ,

i=1

where ti is the temperature and ni is the number of failing O-rings during the ith launch, for i = 1, 2 . . . , 23. But then the loglikelihood `(a, b) is given by `(a, b) = ln C +

23 X

ni ln p(ti ) +

i=1

23 X

(6 − ni ) ln(1 − p(ti )).

i=1

21.16 Since sn is the realization of a Bin (n, p) distributed random variable Sn , we find that the likelihood L(p) is given by !

L(p) =

n psn (1 − p)n−sn , sn

from which we see that the loglikelihood `(p) satisfies n `(p) = ln sn

!

+ sn ln p + (n − sn ) ln(1 − p).

But then differentiating `(p) with respect to p yields that `0 (p) =

sn n − sn − , p 1−p

and we find that `0 (p) = 0 if and only if p = sn /n. Since `(p) attains a maximum at this value of p (check this!), we have obtained that sn /n is the maximum likelihood

29.1 Full solutions

535

estimate (and Sn /n the maximum likelihood estimator) for p. Due to the invariance principle from Section 21.4 we find that 2n/sn is the maximum likelihood estimate for π, and that 2n T = Sn is the maximum likelihood estimator for π. P

22.1 a Since xi yi = 12.4, (c.f. (22.1) and (22.2)), that n βˆ =

P

P

P

xi = 9,

P

yi = 4.8,

P

x2i = 35, and n = 3, we find

P

xi yi − ( xi )( yi ) 3 · 12.4 − 9 · 4.8 1 P P = =− , n x2i − ( xi )2 3 · 35 − 92 4

and α ˆ = y¯n − βˆx ¯n = 2.35. ˆ i , for i = 1, . . . , n, we find r1 = 2 − 2.35 + 0.25 = −0.1, 22.1 b Since ri = yi − α ˆ − βx r2 = 1.8 − 2.35 + 0.75 = 0.2, r3 = 1 − 2.35 + 1.25 = −0.1, and r1 + r2 + r3 = −0.1 + 0.2 − 0.1 = 0. 22.1 c See Figure 29.2. 3 .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .∗ .......... ............. .......... . . . .......... ... . . . .......... . ... . .......... . . . .......... ... . . . . .......... .. .......... ..... .......... ........

·

2

1

·

·

(¯ x3 , y¯3 )

0 −1

0

1

2

3

4

5

6

Fig. 29.2. Solution of Exercise 22.1 c. P

22.2 As in the previous exercise, we have that xi yi = 12.4, P 2 xi = 35. However, now n = 4, and we find that n βˆ =

P

P

P

xi = 9,

P

yi = 4.8,

P

xi yi − ( xi )( yi ) 4 · 12.4 − 9 · 4.8 P P = 0.10847. = n x2i − ( xi )2 4 · 35 − 92

Since we now have that x ¯n = 9/4, and y¯n = 1.2, we find that α ˆ = y¯n − βˆx ¯n = 0.9559. 22.3 a Ignoring the subscripts to the sums, we have X

xi = 10,

X

x2i = 21.84,

X

yi = 20,

and

X

xi yi = 41.61

From (22.1) we find that the least squares estimate βˆ of β is given by

536

Full solutions from MIPS: DO NOT DISTRIBUTE 5 · 41.61 − 10 · 20 βˆ = = 0.875, 5 · 21.84 − 102

while from (22.2) it follows that the least squares estimate α ˆ of α is given by α ˆ=

20 10 − 0.875 · = 2.25 5 5

22.3 b See Figure 29.3. 5

··

. .... .... ... ... . . . ... ... ... .... .... . . . . ... .......... ...∗ ... ........... ... .... . . . . .... .... .... .... ... .... ... . . .... . .. .... . . . ... ... . . .. . . . .. . . . ... . . 5 ... ... ... ... .... . . . . .... .... .... .... . . . .... .... .... ... . . . ... .... ... .... .... . . ..

··

4

·

3

2

(¯ x , y¯5 )

1

0 −1

0

1

2

3

4

Fig. 29.3. Solution of Exercise 22.3 b.

22.4 Since the least squares estimate βˆ of β is given by (22.1), we find that 100 · 5189 − 231.7 · 321 βˆ = = 2.385. 100 · 2400.8 − 231.72 From (22.2) we find that α ˆ = 3.21 − 2.317cdotβˆ = −2.316. 22.5 With the assumption that α = 0, the method of least squares tells us now to minimize S(β) =

n X

(yi − βxi )2 .

i=1

Now

n n n X X X dS(β) = −2 (yi − βxi )xi = −2 xi yi − β x2i dβ i=1 i=1 i=1

!

,

29.1 Full solutions so

dS(β) =0 dβ

537

Pn xi yi β = Pi=1 n 2 .

⇔

i=1

xi

Because S(β) has a minimum for this last value of β, we see that the least squares estimator βˆ of β is given by Pn xi Yi βˆ = Pi=1 n 2 . i=1 xi 22.6 In Exercise 22.5 we have seen that n X

βˆ =

xi Yi

i=1 n X

, x2i

i=1

so for the timber dataset we find that βˆ = 34.13. We use the estimated regression line y = 34.13x to predict the Janka hardness. For density x = 65 we find as a prediction for the Janka hardness y = 2218.45. 22.7 The sum of squares function S(α, β) is now given by S(α, β) =

n X

yi − eα+βxi

2

.

i=1

In order to find the values of α and β for which S(α, β) attains a maximum, one could differentiate S(α, β) to α and β. However, this does not yield workable expressions such as those in (22.1) and (22.2). In order to find α and β for a given bivariate dataset (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ), an iterative method is best suited. 22.8 In the model with intercept, α and β are chosen in such a way, that S(α, β) = 2 minimal. In the model with no intercept we chose β in such a way, that i=1 ri isP n 2 S(0, β) = i=1 ri is minimal. Clearly, in this last model the residual sum of squares Pn 2 i=1 ri is greater than, or equal to the residual sum of squares in the model with intercept. Pn

22.9 With the assumption that β = 0, the method of least squares tells us now to minimize S(α) =

n X

(yi − α)2 .

i=1

Now

X dS(α) = −2 (yi − α) = −2(n¯ yn − nα) = −2n(¯ yn − α). dα i=1 n

= 0 if and only if α = y¯n . Since S(α) has a minimum for this last value of So dS(α) dα α, we see that Y¯n is the least squares estimator α ˆ of α. P

P

P

P

22.10 a One has that n = 3, xi = 3, xi yi = 2, yi = 4, x2i = 5, and 3·2−3·4 ¯3 = 1 and y¯3 = 4/3, we find that α ˆ = 7/3. Since so βˆ = 3·5−32 = −1. Since x ˆ = |r1 | + |r2 | + |r3 | = 4/3. r1 = −1/3, r2 = 2/3, and r3 = −1/3, we find that A(α, ˆ β) ˆ = r12 + r22 + r32 = 1/3. Note that S(α, ˆ β)

538


22.10 b A(α, −1) = |2 − α + 0| + |2 − α + 1| + |0 − α + 2|. For 2 ≤ α < 3 we find that A(α, −1) = α − 1. For α ≥ 3 we find that A(α, −1) = 3α − 7, while for α < 2 ˆ when α is between we find that A(α, −1) = 7 − 3α. We find that A(α, −1) < (α, ˆ β) 17/9 and 7/3. Note that A(α, −1) is minimal (and equal to 1) for α − 2. 22.10 c Since A(2, −1) = 1, we must have that 1 < α < 3 (since otherwise A(α, β) ≥ |2 − α + 0| > 1). Clearly we must have that β < 0 (since otherwise A(α, β) ≥ |0 − α − 2β| > 1). Considering the various cases for α yields that A(α, β) attains its minimum at α = 2 and β = −1. 22.11 a In the present set-up one has to minimize S(β, γ) =

n X

2

yi − (βxi + γx2i )

.

i=1

Differentiating S(β, γ) to β and γ yields X ∂S = −2 xi yi − βxi − γx2i ∂β i=1 n

and

X 2 ∂S = −2 xi yi − βxi − γxi2 . ∂γ i=1 n

We find that ∂S =0 ∂β

⇔

∂S =0 ∂γ

⇔

β

n X

x2i + γ

i=1

and β

n X i=1

n X

x3i =

i=1

x3i + γ

n X i=1

n X

xi yi

i=1

x4i =

n X

x2i yi .

i=1

22.11 b Using Cramer’s rule from linear algebra now yields that P P 3 P x2i yi P xi4 P P P P xi yi xi ( xi Yi )( x4i ) − ( x3i )( x2i Yi ) P 2 P 4 P 3 2 β = P 2 P 3 = ( xi )( xi ) − ( xi ) P x3i P x4i xi xi

and

P 2 P P x3i P x2i yi P P P P xi xi yi ( x2i )( x2i Yi ) − ( x3i )( xi Yi ) P 2 P 4 P 3 2 . γ = P 2 P 3 = ( xi )( xi ) − ( xi ) P x3i P x4i xi xi

Since S(β, γ) is a ‘vase’, the above stationary point (β, γ) is a global minimum for S(β, γ). This finishes the exercise. 22.12 a Since the denominator of βˆ is a number, not a random variable, one has that P P P h i E [n( xi Yi ) − ( xi )( Yi )] P 2 P E βˆ = . x xi − ( xi )2 Furthermore, the numerator of this last fraction can be written as

29.1 Full solutions h X

E n which is equal to n

X

i

h X

xi Yi − E (

X

(xi E [Yi ]) − (

X

xi )(

xi )

X

539

i

Yi ) ,

E [Yi ] .

22.12 b Substituting E [Yi ] = α + βxi in the last expression, we find that h i n E βˆ =

P

P

P

(xi (α + βxi )) − ( xi ) [ (α + βxi )] P P . x x2i − ( xi )2 h i

22.12 c The numerator of the previous expression for E βˆ can be simplified to nα

P

xi + nβ

which is equal to

P

P

x2i − nα xi − β( P P n x2i − ( xi )2 P

P

xi )(

P

xi )

,

P

β(n x2i − ( xi )2 ) . P P n x2i − ( xi )2 h i

22.12 d From c it now follows that E βˆ = β, i.e., βˆ is an unbiased estimator for β. 23.1 This is the case: normal data with variance known. So we should use the formula from Section 23.2 (the case variance known):

σ σ ¯n + 1.96 √ x ¯n − 1.96 √ , x n n

,

where x ¯n = 743, σ = 5 and n = 16. Because zα/2 = z0.025 = 1.96, the 95% confidence interval is:

5 5 743 − 1.96 · √ , 743 + 1.96 · √ 16 16

= (740.55, 745.45).

23.2 This is the case: normal data with variance unknown. So we should use the formula from Section 23.2 (the case variance unknown):

sn sn x ¯n − tn−1,α/2 √ , x ¯n + tn−1,α/2 √ n n

,

where x ¯n = 3.54, sn = 0.13 and n = 34. Because tn−1,α/2 = t33,0.01 ≈ t30,0.01 = 2.457, the 98% confidence interval is:

0.13 0.13 3.54 − 2.457 · √ , 3.54 + 2.457 · √ 34 34

= (3.485, 3.595).

One can redo the same calculation using t33,0.01 = 2.445 (obtained from a software package), and find (3.4855, 3.5945). 23.3 This is the case: normal data with variance unknown. So we should use the formula from Section 23.2 (the case variance unknown):

sn sn x ¯n − tn−1,α/2 √ , x ¯n + tn−1,α/2 √ n n

,

where x ¯n = 93.5, sn = 0.75 and n = 10. Because tn−1,α/2 = t9,0.025 = 2.262, the 95% confidence interval is:

0.75 0.75 93.5 − 2.262 · √ , 93.5 + 2.262 · √ 10 10

= (92.96, 94.036).

540


23.4 This is the case: normal data with variance unknown. So we should use the formula from Section 23.2 (the case variance unknown):

sn sn ¯n + tn−1,α/2 √ x ¯n − tn−1,α/2 √ , x n n

,

where x ¯n = 195.3, sn = 16.7 and n = 18. Because tn−1,α/2 = t17,0.025 = 2.110, the 95% confidence interval is:

16.7 16.7 195.3 − 2.110 · √ , 195.3 + 2.110 · √ 18 18

= (186.99, 203.61).

23.5 a The standard confidence interval for the mean of a normal sample with unknown variance applies, with n = 23, x ¯ = 0.82 and s = 1.78, so:

s s x ¯ − t22,0.025 · √ , x ¯ + t22,0.025 · √ 23 23

.

The critical values come from the t (22) distribution: t22,0.025 = 2.074. The actual interval becomes:

1.78 1.78 0.82 − 2.074 · √ , 0.82 + 2.074 · √ 23 23

= (0.050, 1.590).

23.5 b Generate one thousand samples of size 23, by drawing with replacement from the 23 numbers 1.06, For q each sample 1 22

P

x∗1 , x∗2 , . . . , x∗23

1.04,

2.62,

...,

∗

x ¯∗23

compute: t =

2.01. √ − 0.82/(s∗23 / 23), where s∗23 =

(x∗i − x ¯∗23 )2 .

23.5 c We need to estimate the critical value c∗l such that P(T ∗ ≤ c∗l ) ≈ 0.025. We take c∗l = −2.101, the 25th of the ordered values, an estimate for the 25/1000 = 0.025 quantile. Similarly, c∗l is estimated by the 976th, which is 2.088. The bootstrap confidence interval uses the c∗ values instead of the t-distribution values ±tn−1,α/2 , but beware: c∗l is from the left tail and appears on the right-hand side of the interval and c∗u on the left-hand side:

x ¯n −

sn c∗u √

n

,x ¯n −

sn c∗l √

n

.

Substituting c∗l = −2.101 and c∗u = 2.088, the confidence interval becomes:

1.78 1.78 0.82 − 2.088 · √ , 0.82 + 2.101 · √ 23 23

= (0.045, 1.600).

23.6 a Because events described by inequalities do not change when we multiply the inqualities by a positive constant or add or subtract a constant, the ñ < θ < U ñ = P(3Ln + 7 < 3µ + 7 < 3Un + 7) = following equalities hold: P L P(3Ln < 3µ < 3Un ) = P(Ln < µ < Un ), and this equals 0.95, as is given. ñ, U ñ ), that 23.6 b The confidence interval for θ is obtained as the realization of (L ˜ is: (ln , u ñ ) = (3ln + 7, 3un + 7). This is obtained by transforming the confidence interval for µ (using the transformation that is applied to µ to get θ).

29.1 Full solutions

541

23.6 c We start with P(Ln < µ < Un ) = 0.95 and try to get 1 − µ in the middle: P(Ln < µ < Un ) = P(−Ln > −µ > −Un ) = P(1 − Ln > 1 − µ > 1 − Un ) = P(1 − Un < 1 − µ < 1 − Ln ), where we see that the minus sign causes an inter˜ n = 1 − Un and U ñ = 1 − Ln . The confidence interval: (1 − 5, 1 − (−2)) = change: L (−4, 3). 23.6 d If we knew that Ln and Un were always positive, then we could conclude: P(Ln < µ < Un ) = P L2n < µ2 < Un2 and we could just square the numbers in the confidence interval for µ to get the one for θ. Without the positivity assumption, the sharpest conclusion you can draw from Ln < µ < Un is that µ2 is smaller than the maximum of L2n and Un2 . So, 0.95 = P(Ln < µ < Un ) ≤ P 0 ≤ µ2 < max{L2n , Un2 } and the confidence interval [0, max{ln2 , u2n }) = [0, 25) has a confidence of at least 95%. This kind of problem may occur when the transformation is not one-to-one (both −1 and 1 are mapped to 1 by squaring). 23.7 We know that (ln , un ) = (2, 3), where ln and un are the realizations of Ln and Un , that have the property P(Ln < µ < Un ) = 0.95. This is equivalent with

P e−Un < e−µ < e−Ln

= 0.95,

so that e−3 , e−2 = (0.050, 0.135) is a 95% confidence interval for P(X = 0) = e−µ . 23.8 Define random variables Xi = weight of ith bottle together with filling amount Wi = weight of ith bottle alone Yi = weight of ith filling amount so that Xi = Wi +Yi . It is given that Wi has a N (250, 152 ) distribution and according to Exercise 23.1, Yi has a N (µy , 52 ). Since they are independent Xi = Wi + Yi has a normal distribution with expectation µ = 250 + µy and variance 152 + 52 = 250: Xi ∼ N (250 + µy , 250). Our data consist of the weights of 16 filled bottles of wine, x1 , . . . , x16 . On the basis of these we can construct a confidence interval for µ. Since we are in the case: normal data with known variance, this is

where x ¯n = 998, σ = confidence interval is:

√

σ σ x ¯n − zα/2 √ , x ¯n + zα/2 √ n n

,

250 and n = 16. Because zα/2 = z0.025 = 1.96, the 95%

√ √ 250 250 998 − 1.96 · √ , 998 + 1.96 · √ = (990.25, 1005.75). 16 16

Since, this is a 95% confidence interval for µ = 250 + µy , the 95% confidence interval for µy , is given by (990.25 − 250, 1005.75 − 50) = (740.25, 755.75). 23.9 a Since we do not assume a particular parametric model, we are dealing with an empirical bootstrap simulation. Generate bootstrap samples x∗1 , x∗2 , . . . , x∗n size n = 2608, from the empirical distribution function of the dataset, which is equivalent to generate from the discrete distribution with probability mass function:

542

Full solutions from MIPS: DO NOT DISTRIBUTE a p(a)

0 57/2608

1 2 3 4 203/2608 383/2608 525/2608 532/2608

a 5 6 7 p(a) 408/2608 273/2608 139/2608

8 45/2608

9 27/2608

a p(a)

13 1/2608

14 1/2608

10 10/2608

11 4/2608

12 0

Then determine:

x ¯∗n − 3.8715 √ . s∗n / 2608 Repeat this one thousand times. Next estimate the values c∗l and c∗u such that t∗ =

P(T ∗ ≤ c∗l ) = 0.025

and

P(T ∗ ≥ c∗u ) = 0.025.

The bootstrap confidence interval is then given by:

x ¯n −

sn c∗u √

n

,x ¯n −

sn c∗l √

n

.

23.9 b For a 95% confidence interval we need the empirical 0.025-quantile and the 0.975-th quantile, which we estimate by the 25th and 976th order statistic: cl ≈ 1.862 and cu ≈ −1.888. This results in

1.9225 1.9225 3.8715 − 1.862 √ , 3.8715 − (−1.888) √ 2608 2608

= (3.801, 3.943).

23.9 c The (estimated) critical values that we would obtain from the table are −2.228 and 2.234, instead of −1.888 and 1.862 we used in part b. Hence, the resulting interval would be larger. 23.10 a This interval has been obtained from

sn sn ¯n + tn−1,α/2 √ x ¯n − tn−1,α/2 √ , x n n

,

so that the sample mean is in the middle and must be equal to (1.6 + 7.8)/2 = 4.7. 23.10 b From the formula we see that half the width of the 95% confidence interval is 7.8 − 1.6 sn sn = 3.1 = t15,0.025 √ = 2.131 √ . 2 n n Similarly, half the width of the 99% confidence interval is sn sn 2.947 sn 2.947 t15,0.005 √ = 2.947 √ = · 2.131 √ = · 3.1 = 4.287. 2.131 2.131 n n n Hence the 99% confidence interval is (4.7 − 4.287, 4.7 + 4.287) = (0.413, 8.987). 23.11 a For the 98% confidence interval the same formula is used as for the 95% interval, replacing the critical values by larger ones. This is the case, no matter whether the critical values are from the normal or t-distribution, or from a bootstrap experiment. Therefore, the 98% interval contains the 95%, and so must also contain the number 0.

29.1 Full solutions

543

23.11 b From a new bootstrap experiment we would obtain new and, most probably, different values c∗u and c∗l . It therefore could be, if the number 0 is close to the edge of the first bootstrap confidence interval, that it is just outside the new interval. 23.11 c The new dataset will resemble the old one in many ways, but things like the sample mean would most likely differ from the old one, and so there is no guarantee that the number 0 will again be in the confidence interval. 23.12 a This follows immediately from the change of units rule for normal random variables on page 112 and the fact that if the Zi ’s are independent, so are the µ + σZi ’s. 23.12 b We have n n n X 1X 1X ¯ = 1 ¯ X Xi = (µ + σZi ) = µ + σ · Zi = µ + σ Z. n i=1 n i=1 n i=1

From this we also find that ¯ = (µ + σZi ) − (µ + σ Z) ¯ = σ(Zi − Z) ¯ Xi − X It follows that 2 SX =

n n X 1 X 2 ¯ 2= 1 (Xi − X) σ 2 (Zi − Z¯i )2 = σ 2 SZ , n − 1 i=1 n − 1 i=1

so that SX = σSZ . ¯ − µ = σ Z, ¯ 23.12 c The equality follows immediately from part b, by inserting X and SX = σSZ . The right hand side is the studentized mean for a random sample from a N (0, 1) distribution, and therefore its distribution does not depend on µ and σ. The left hand side is the studentized mean for a random sample from a N (µ, σ 2 ) distribution. Since the two are equal, also their distributions are, which means that the distribution of the studentized mean for a random sample from a N (µ, σ 2 ) distribution (the left hand side) also does not depend on µ and σ. 24.1 From Section 24.1, using z0.05 = 1.645 we find the equation:

This reduces to

70 −p 100

2

−

(1.645)2 p(1 − p) < 0. 100

1.0271 p2 − 1.4271 p + 0.49 < 0.

The zeroes of this parabola are p1,2 =

−(−1.4271) ±

p

(−1.4271)2 − 4 · 1.0271 · 0.49 = 0.6947 ± 0.0746 2 · 1.0271

and the 90% confidence interval is (0.6202, 0.7693). 24.2 a Since n = 6 is too small to expect a good approximation from the central limit theorem, we cannot apply the Wilson method.

544


24.2 b Solve

or

140 −p 250

2

−

(1.96)2 p(1 − p) = 0 250

1.0154 p2 − 1.1354 p + 0.3136 = 0.

That is p1,2 =

−(−1.1354) ±

p

(−1.1354)2 − 4 · 1.0154 · 0.3136 = 0.5591 ± 0.0611 2 · 1.0154

and the 90% confidence interval is (0.4980, 0.6202).

√ 24.3 The width of the confidence interval is 2 · 2.576 · σ/ n, where σ = 5. So, we require (see Section 24.4):

n≥

2 · 2.576 · 5 1

2

= (25.76)2 = 663.6,

that is, at least a sample size of 664.

√ 24.4 a The width of the confidence interval will be about 2tn−1,0.05 s/ n. For s we substitute our current estimate of σ, 0.75, and we use tn−1,0.05 = z0.05 = 1.645, for the moment assuming that n will be large. This results in

n≥

2 · 1.645 · 0.75 0.1

2

= 608.9,

so we use n = 609 (which is indeed large, so it is appropriate to use the critical value from the normal distribution). 24.4 b In our computation, we used s = 0.75. From the new dataset of size 609 we are going to compute s609 and use that in the computation. If s609 > 0.75 then the confidence interval will be too wide. 24.5 a From P −zα/2 we deduce

P

X − np < p zα/2 n/4

!

=1−α

zα/2 zα/2 X X − √

< + √ n n 2 n 2 n

= 1 − α.

So, the approximate 95% confidence interval is

x z0.05 x z0.05 − √ , + √ . n 2 n n 2 n √ √ √ The width is 2z0.05 /(2 n) = z0.05 / n and so n should satisfy 1.96/ n ≤ 0.01 or 2 n ≥ (196) = 38416. 24.5 b The confidence interval is 19477 1.96 √ ± = 0.5070 ± 0.005, 38416 2 · 38416 so of the intended width.

29.1 Full solutions

545

24.6 a The environmentalists are interested in a lower confidence bound, because they would like to make a statement like “We are 97.5% confidence that the concentration exceeds 1.68 ppm [and √ that is much too high.]” We have normal data, with σ unknown so we use s16 = 1.12 = 1.058 as an estimate and use the critical value corresponding to 2.5% from the t (15) √ distribution: t15,0.025 = 2.131. The lower confidence bound is 2.24 − 2.131 · 1.058/ 16 = 2.24 − 0.56 = 1.68, the interval: (1.68, ∞). 24.6 b For similar reasons, the plant management constructs an upper confidence bound (“We are 97.5% confident pollution does not exceed 2.80 [and this is acceptable.]”). The computation is the same except for a minus sign: 2.24 + 2.131 · √ 1.058/ 16 = 2.24 + 0.56 = 2.80, so the interval is [0, 2.80). Note that the computed upper and lower bounds are in fact the endpoints of the 95% two-sided confidence interval. 24.7 a From the normal approximation we know

¯n − µ X P −z0.025 < √ √ < z0.025 µ/ n or

P i.e.,

P

¯n − µ X √ √ µ/ n

¯n − µ X

2

≈ 0.95

!

2

<

2 z0.025

µ < (1.96)2 √ n

≈ 0.95,

≈ 0.95.

Just as with the derivation of the Wilson method (Section 24.1) we now conclude that the 95% confidence interval contains those µ for which µ (¯ xn − µ)2 ≤ (1.96)2 . n 24.7 b We need to solve (¯ xn − µ)2 − (1.96)2 µ/n = 0, where x ¯n = 3.8715 and n = 2608, resulting in

µ2 − 2 · 3.8715 + or

(1.96)2 2608

µ + (3.8715)2 = 0,

µ2 − 7.7446µ + 14.9889 = 0.

From the roots we find the confidence interval (3.7967, 3.9478). 24.7 c The confidence interval (3.7967, 3.9478)is almost the same as the one in Exercise 23.9 b. This is not surprising: n is very large, so the normal approximation should be very good. 24.8 a We solve

or

15 −p 23

2

−

(1.96)2 p(1 − p) = 0 23

1.1670 p2 − 1.4714 p + 0.4253 = 0,

from which we find the confidence interval (0.4489, 0.8119).

546


24.8 b Replacing the 1.96 in the equation above by z0.05 = 1.645, we find (0.4808, 0.7915). The appropriate one-sided confidence interval will provide us with a lower bound on the outer lane winning probability p, so it is (0.4808, 1]. 24.9 a From Section 8.4 we know: P(M ≤ a) = [FX (a)]12 , so P(M/θ ≤ t) = P(M ≤ θt) = [FX (θt)]12 . Since Xi has a U (0, θ) distribution, FX (θt) = t, for 0 ≤ t ≤ 1. Substituting this shows the result. 24.9 b For cl we need to solve (cl )12 = α/2, or cl = (α/2)1/12 = (0.05)1/12 = 0.7791. For cu we need to solve (cu )12 = 1−α/2, or cu = (1−α/2)1/12 = (0.95)1/12 = 0.9958. 24.9 c From b we know that P(cl < M/θ < cu ) = P(0.7790 < M/θ < 0.9958) = 0.90. Rewriting this equation, we get: P(0.7790 θ < M < 0.9958 θ) = 0.90 and P(M/0.9958 < θ < M/0.7790) = 0.90. This means that (m/0.9958, m/0.7790) = (3.013, 3.851) is a 90% confidence interval for θ. 24.9 d From b we derive the general formula:

P (α/2)1/n <

M < (1 − α/2)1/n θ

= 1 − α.

The left hand inequality can be rewritten as θ < M/(α/2)1/n and the right hand one as M/(1 − α/2)1/n < θ. So, the statement above can be rewritten as:

P

M M <θ< (1 − α/2)1/n (α/2)1/n

= 1 − α,

so that the general formula for the confidence interval becomes:

m m , (1 − α/2)1/n (α/2)1/n

.

24.10 a From Section 11.2 we know that Sn , being the sum of n independent Exp (λ) random variables, has a Gam (n, λ) distribution. From Exercise 8.4 we know: λXi has an Exp (1) distribution. Combining these facts, it follows that λSn = λX1 +· · ·+λXn has a Gam (n, 1) distribution. 24.10 b From the quantiles we see 0.9 = P(q0.05 ≤ λS20 ≤ q0.95 ) = P(13.25 ≤ λS20 ≤ 27.88) = P(13.25/S20 ≤ λ ≤ 27.88/S20 ) . Noting that the realization of S20 is x1 + · · · + x20 = 20 x ¯20 , we conclude that

13.25 27.88 , 20 x ¯20 20 x ¯20

=

0.6625 1.394 , x ¯20 x ¯20

is a 95% confidence interval for λ. 25.1 The alternative hypothesis should reflect the belief that arrival delays of trains exhibit more variation during rush hours than during quiet hours. Therefore take H1 : σ1 > σ2 . 25.2 The alternative hypothesis should reflect the belief that the number of babies born in Cleveland, Ohio, in the month of September in 1977 is higher than 1472. Therefore, take H1 : µ > 1472.

29.1 Full solutions

547

25.3 When the regression line runs through the origin, then α = 0. One possible testing problem is to test H0 : α = 0 against H1 : α 6= 0. If H0 : α = 0 is rejected in favor of H1 : α 6= 0, the parameter α should be left in the model. 25.4 a Denote the observed numbers of cycles for the smokers by X1 , X2 , . . . , Xn1 and similarly Y1 , Y2 , . . . , Yn2 for the nonsmokers. A test statistic should compare estimators for p1 and p2 . Since the geometric distributions have expectations 1/p1 ¯ n1 for p1 with the estimator 1/Y¯n2 for and 1/p2 , we could compare the estimator 1/X ¯ n1 with Y¯n2 . For instance, take test statistic T = X ¯ n1 − Y¯n2 . p2 , or simply compare X Values of T close to zero are in favor of H0 , and values far away from zero are in ¯ n1 /Y¯n2 . favor of H1 . Another possibility is T = X 25.4 b In this case, the maximum likelihood estimators pˆ1 and pˆ2 give better indications about p1 and p2 . They can be compared in the same way as the estimators in a. 25.4 c The probability of getting pregnant during a cycle is p1 for the smoking women and p2 for the nonsmokers. The alternative hypothesis should express the belief that smoking women are less likely to get pregnant than nonsmoking women. Therefore take H1 : p1 < p2 . 25.5 a When the maximum is greater than 5, at least one Xi is greater than 5 so that θ must be greater than 5, and we know for sure that the null hypothesis is false. Therefore the set of relevant values of T1 = max{X1 , X2 , . . . , Xn } is the interval [0, 5]. Similar to Exercise 8.15, one can argue that E [T1 ] = nθ/(n + 1). Hence values of T1 close to 5n/(n + 1) are in favor of H0 . Values of T1 in the neighborhood of 0 indicate that θ < 5, and values of T1 very close to 5 indicate that θ > 5. Both these regions are in favor of H1 .

values of T1 in favor of H1 0

values of T1 in favor of H0 5n n+1


¯ n and 5 is greater than 5, then 2X ¯ n must be 25.5 b When the distance between 2X greater than 10, which means that at least one Xi is greater than 5. In that case we know for sure that the null hypothesis is false. Therefore the set of relevant values ¯ n − 5| is the interval [0, 5]. Since X ¯ n will be close to θ/2, values of T2 of T2 = |2X close to zero are in favor of H0 . Values of T2 far away from zero either correspond ¯ n far below 5, which indicates θ < 5, or correspond to 2X ¯ n far above 5, which to 2X indicates θ > 5. Hence values of T2 far away from zero are in favor of H1 .



25.6 a The p-value P(T ≥ 2.34) = 0.23 is larger than 0.05, so do not reject. 25.6 b The p-value P(T ≥ 2.34) = 1 − P(T ≤ 2.34) = 0.77 is larger than 0.05, so do not reject. 25.6 c The p-value P(T ≥ 0.03) = 0.968 is larger than 0.05, so do not reject.

548


25.6 d The p-value P(T ≥ 1.07) = 1 − P(T ≤ 1.07) = 0.019 is less than 0.05, so reject. 25.6 e The p-value P(T ≥ 1.07) ≥ P(T ≥ 2.34) = 0.99, which is larger than 0.05, so do not reject. 25.6 f The p-value P(T ≥ 2.34) ≤ P(T ≥ 1.07) = 0.0.019, which is smaller than 0.05, so reject. 25.6 g The p-value P(T ≥ 2.34) ≤ P(T ≥ 1.07) = 0.200. Therefore, the p-value is smaller than 0.200, but that does not give enough information to decide about the null hypothesis. 25.7 a Since the parameter µ is the expectation of T , values of T much larger than 1472 suggest that µ > 1472. Because we test H0 : µ = 1472 against H1 : µ > 1472, the more values of T are to the right, the stronger evidence they provide in favor of H1 . 25.7 b According to part a, values to the right of t = 1718 bare stronger evidence in favor of H1 . Therefore, the p-value is P(T ≥ 1718), where T has a Poisson distribution with µ = 1472. Because the distribution of T can be approximated by a normal distribution with mean 1472 and variance 1472, we can approximate this probability as follows

P(T ≥ 1718) = P

T − 1472 1718 − 1472 √ √ ≥ 1472 1472

≈ P(Z ≥ 6.412)

where Z has an N (0, 1) distribution. From Table ?? we see that the latter probability is almost zero (to be precise, 7.28 · 10−11 , which was obtained using a statistical software package). 25.8 The values of Fn and Φ lie between 0 and 1, so that the maximal distance between the two graphs also lies between 0 and 1. In fact, since Fn has jumps of size 1/n, the minimal value of T must be half the size of a jump: 1/(2n). This would correspond to the situation, where at each observation, the graph of Φ precisely runs through the middle of the two heights of Fn . When the graph of Fn lies far to the right (or to the left) of that of Φ, the maximum distance between the two graphs can be arbitrary close to 1. When the dataset is a realization from a distribution different from the standard normal, the corresponding distribution function F will differ from Φ. Since Fn ≈ F (recall Table 17.2), the graph of Fn will show large differences with that of Φ resulting in a relatively large value of T . On the other hand, when the dataset is indeed a realization from the standard normal, then Fn ≈ Φ, resulting in a relative small value of T . We conclude that only large values of T close to 1 are evidence against the null hypothesis. 25.9 Only values of Tks close to 1 are evidence against the null hypothesis. Therefore the p-value is P(Tks ≥ 0.176). On the basis of the bootstrap results, this probability is approximated by the relative frequency of Tks -values greater than or equal to 0.176, which is zero. 25.10 a The alternative hypothesis should express the belief that the gross calorific exceeds 23.75 MJ/kg. Therefore take H1 : µ > 23.75.

29.1 Full solutions

549

¯ n ≥ 23.788 under the null hypothesis. 25.10 b The p-value is the probability P X ¯ n has a We can compute this probability by using that under the null hypothesis X N (23.75, (0.1)2 /23) distribution:

¯ 23.75 23.788 − 23.75 ¯ n ≥ 23.788 = P Xn − √ √ P X ≥ 0.1/ 23 0.1/ 23

= P(Z ≥ 1.82) ,

where Z has an N (0, 1) distribution. From Table ?? we find P(Z ≥ 1.82) = 0.0344. 25.11 A type I error occurs when µ = 0 and |t| ≥ 2. When µ = 0, then T has an N (0, 1) distribution. Hence, by symmetry of the N (0, 1) distribution and Table ??, we find that the probability of committing a type I error is P(|T | ≥ 2) = P(T ≤ −2) + P(T ≥ 2) = 2 · P(T ≥ 2) = 2 · 0.0228 = 0.0456. 26.1 A type I error is to falsely reject the null hypothesis, i.e., to falsely conclude “suspect is guilty”. This happened in 9 out of 140 cases. Hence, the probability of a type I error is 9/140 = 0.064. A type II error is to falsely accept the null hypothesis, i.e., to falsely conclude “suspect is innocent”. This happened in 15 out of 140 cases. Hence, the probability of a type II error is 15/140 = 0.107. 26.2 According to Exercise 25.11, we do not reject if |T | < 2. Therefore the probability of a type II error is P(|T | < 2) = P(−2 < T < 2) = P(T < 2) − P(T < −2) , where T has a N (1, 1) distribution. Using that T − 1 has a N (0, 1) distribution, we find that P(T < 2) − P(T < −2) = P(T − 1 < 1) − P(T − 1 < −3) = 0.8413 − 0.0013 = 0.84. 26.3 a A type I error is to falsely reject the null hypothesis H0 : θ = 2. We reject when X ≤ 0.1 or X ≥ 1.9. Because under the null hypothesis, X has a U (0, 2) distribution, the probability of committing a type I error is P(X ≤ 0.1 | θ = 2) + P(X ≥ 1.9 | θ = 2) = 0.05 + 0.05 = 0.1. 26.3 b In this case a type II error is to falsely accept the null hypothesis when θ = 2.5 We accept when 0.1 < X < 1.9. Because under θ = 2.5, X has a U (0, 2.5) distribution, the probability of committing a type I error is P(0.1 < X < 1.9 | θ = 2.5) =

1.9 − 0.1 = 0.72. 2.5

26.4 a Since T has a Bin (144, p) distribution, values of T close to 144/8 = 18 are in favor of the null hypothesis. Values of T far above 18 indicate that p > 1/8, whereas values far below 18 indicate that p < 1/8. This means we reject only for values of T far above 18. Hence we are only dealing with a right critical value. 26.4 b Denote the right critical value by c. Then we must solve P(T ≥ c) = 0.01. Using the normal approximation for the binomial distribution,

550

Full solutions from MIPS: DO NOT DISTRIBUTE c − np P(T ≥ c) ≈ P Z ≥ p np(1 − p)

!

0

1

c − 18 A , = P @Z ≥ q 18 · 78

where Z has a N (0, 1) distribution. We find c by solving c − 18

q

18 ·

7 8

= z0.01 = 2.326,

which gives c = 27.2. Because T can only take integer values, the right critical value is taken to be 28, and the critical region is {28, 29, · · · , 144}. Since the observed number t = 29 falls in the critical region, we reject H0 : p = 1/8 in favor of H1 : p > 1/8. 26.5 a The p-value is P(X ≥ 15) under the null hypothesis H0 : p = 1/2. Using Table 26.3 we find P(X ≥ 15) = 1 − P(X ≤ 14) = 1 − 0.8950 = 0.1050. 26.5 b Only values close to 23 are in favor of H1 : p > 1/2, so the critical region is of the form K = {c, c + 1, . . . , 23}. The critical value c is the smallest value, such that P(X ≥ c) ≤ 0.05 under H0 : p = 1/2, or equivalently, 1 − P(X ≤ c − 1) ≤ 0.05, which means P(X ≤ c − 1) ≥ 0.95. From Table 26.3 we conclude that c − 1 = 15, so that K = {16, 17, . . . , 23}. 26.5 c A type I error occurs if p = 1/2 and X ≥ 16. The probability that this happens is P(X ≥ 16 | p = 1/2) = 1 − P(X ≤ 15 | p = 1/2) = 1 − 0.9534 = 0.0466, where we have used Table 26.3 once more. 26.5 d In this case, a type II error occurs if p = 0.6 and X ≤ 15. To approximate P(X ≤ 15 | p = 0.6), we use the same reasoning as in Section 14.2, but now with n = 23 and p = 0.6. Write X as the sum of independent Bernoulli random variables: X = R1 + · · · + Rn , and apply the central limit theorem with µ = p = 0.6 and σ 2 = p(1 − p) = 0.24. Then P(X ≤ 15) = P(R1 + · · · + Rn ≤ 15) R1 + · · · + Rn − nµ 15 − nµ √ √ =P ≤ σ n σ n 15 − 13.8 √ = P Z23 ≥ √ ≈ Φ(0.51) = 0.6950. 0.24 23 26.6 a Because T has a Pois (µ) distribution (see Exercise 25.7), we always have E [T ] = µ. Therefore values of T around 1472 are in favor of the null hypothesis, values of T far to the left of 1472 are in favor of µ < 1472, and values of T far to the right 1472 are in favor of µ > 1472. Therefore, only values far to the right of 1472 are in favor of H1 : µ > 1472, so that we only have a right critical value. 26.6 b Since, according to part a, we only have a right critical value c, we must solve P(T ≥ c) = 0.05. Using the normal approximation

P(T ≥ c) = P

T −µ c−µ ≥ √ √ µ µ

c − 1472 ≈P Z≥ √ 1472

where Z has a N (0, 1) distribution. We find c by solving c − 1472 √ = z0.05 = 1.645, 1472

,

29.1 Full solutions

551

which gives c = 1535.1. Because T can only take integer values we take right critical value 1536, and critical region {1536, 1537, ∞}. The observed number 1718 falls into the critical region, so that we reject H0 : µ = 1472 in favor of H1 : µ > 1472. 26.7 We must solve P(X1 + X2 ≤ c) = 0.05 under the null hypothesis. Under the null hypothesis, X1 and X2 are independent random variables with a U (0, 1) distribution. According to Exercise 11.5, the random variable T = X1 + X2 has density f (t) = t, for 0 ≤ t ≤ 1, and f (t) = 2 − t, for 1 ≤ t ≤ 2. Since this density integrates to 0.5 for 0 ≤ t ≤ 1, we can find c by solving Z

c

t dt = P(T ≤ c) = 0.05, 0

or equivalently 21 c2 = 0.05, which gives left critical value c = 0.316. The corresponding critical region for T = X1 + X2 is [0, 0.316]. ¯ n takes values in (0, ∞). Recall that the Exp (λ) distri26.8 a Test statistic T = X ¯ n will bution has expectation 1/λ, and that according to the law of large numbers X ¯ n close to 1 are in favor of H0 : λ = 1, and only be close to 1/λ. Hence, values of X ¯ n close to zero are in favor H1 : λ > 1. Large values of X ¯ n also provide values of X evidence against H0 : λ = 1, but even stronger evidence against H1 : λ > 1. We ¯ n has critical region K = (0, cl ]. This is an example in which conclude that T = X the alternative hypothesis and the test statistic deviate from the null hypothesis in opposite directions. ¯ ¯ n close to zero correspond Test statistic T 0 = e−Xn takes values in (0, 1). Values of X 0 ¯ to values of T close to 1, and large values of Xn correspond to values of T 0 close to 0. Hence, only values of T 0 close to 1 are in favor H1 : λ > 1. We conclude that T 0 has critical region K 0 = [cu , 1). Here the alternative hypothesis and the test statistic deviate from the null hypothesis in the same direction. ¯ n close to 1 are in favor of H0 : λ = 1. Values of X ¯ n close 26.8 b Again, values of X ¯ n suggest λ < 1. Hence, both small to zero suggest λ > 1, whereas large values of X ¯ n are in favor of H1 : λ 6= 1. We conclude that T = X ¯ n has and large values of X critical region K = (0, cl ] ∪ [cu , ∞). ¯ n correspond to values of T 0 close to 1 and 0. Hence, Small and large values of X values of T 0 both close to 0 and close 1 are in favor of H1 : λ 6= 1. We conclude that T 0 has critical region K 0 = (0, c0l ] ∪ [c0u , 1). Both test statistics deviate from the null hypothesis in the same directions as the alternative hypothesis. ¯ n )2 takes values in [0, ∞). Since µ is the expectation 26.9 a Test statistic T = (X ¯ n is close to µ. of the N (µ, 1) distribution, according to the law of large numbers, X ¯ n close to zero are in favor of H0 : µ = 0. Large negative values Hence, values of X ¯ n suggest µ < 0, and large positive values of X ¯ n suggest µ > 0. Therefore, both of X ¯ n are in favor of H1 : µ 6= 0. These large negative and large positive values of X values correspond to large positive values of T , so T has critical region K = [cu , ∞). This is an example in which the test statistic deviates from the null hypothesis in one direction, whereas the alternative hypothesis deviates in two directions. Test statistic T 0 takes values in (−∞, 0) ∪ (0, ∞). Large negative values and large ¯ n correspond to values of T 0 close to zero. Therefore, T 0 has positive values of X 0 critical region K = [c0l , 0) ∪ (0, c0u ]. This is an example in which the test statistic deviates from the null hypothesis for small values, whereas the alternative hypothesis deviates for large values.

552


¯ n are in favor of µ > 0, which correspond to 26.9 b Only large positive values of X large values of T . Hence, T has critical region K = [cu , ∞). This is an example where the test statistic has the same type of critical region with a one-sided or two-sided alternative. Of course, the critical value cu in part b is different from the one in part a. ¯ n correspond to small positive values of T 0 . Hence, T 0 has Large positive values of X critical region K 0 = (0, c0u ]. This is another example where the test statistic deviates from the null hypothesis for small values, whereas the alternative hypothesis deviates for large values. 27.1 a The value of the t-test statistic is t=

x ¯n − 10 11 − 10 √ = √ = 2. sn / n 2/ 4

The right critical value is tn−1,α/2 = t15,0.025 = 2.131. The observed value t = 2 is smaller than this, so we do not reject H0 : µ = 10 in favor of H1 : µ 6= 10. 27.1 b The right critical value is now tn−1,α = t15,0.05 = 1.753. The observed value t = 2 is larger than this, so we reject H0 : µ = 10 in favor of H1 : µ > 10. 27.2 a The belief that the pouring temperature is at the right target setting is put to the test. The alternative hypothesis should represent the belief that the pouring temperature differs from the target setting. Hence, test H0 : µ = 2550 against H1 : µ 6= 2550. 27.2 b The value of the t-test statistic is t=

x ¯n − 2550 2558.7 − 2550 √ = 1.21. √ = √ sn / n 517.34/ 10

Because H1 : µ 6= 2550, both small and large values of T are in favor of H1 . Therefore, the right critical value is tn−1,α/2 = t9,0.025 = 3.169. The observed value t = 1.21 is smaller than this, so we do not reject H0 : µ = 2550 in favor of H1 : µ 6= 2550. 27.3 a The alternative hypothesis should represent the belief that the load at failure exceeds 10 MPa. Therefore, take H1 : µ > 10. 27.3 b The value of the t-test statistic is t=

13.71 − 10 x ¯n − 10 √ √ = = 4.902. sn / n 3.55/ 22

Because H1 : µ > 10, only large values of T are in favor of H1 . Therefore, the right critical value is tn−1,α = t21,0.05 = 1.721. The observed value t = 4.902 is larger than this, so we reject H0 : µ = 10 in favor of H1 : µ > 10. 27.4 The value of the t-test statistic is t=

x ¯n − 31 31.012 − 31 √ = √ = 0.435. sn / n 0.1294/ 22

Because H1 : µ > 31, only large values of T are in favor of H1 . Therefore, the right critical value is tn−1,α = t21,0.01 = 2.518. The observed value t = 0.435 is smaller than this, so we do not reject H0 : µ = 31 in favor of H1 : µ > 31.

29.1 Full solutions

553

27.5 a The interest is whether the inbreeding coefficient exceeds 0. Let µ represent this coefficient for the species of wasps. The value 0 is the a priori specified value of the parameter, so test null hypothesis H0 : µ = 0. The alternative hypothesis should express the belief that the inbreeding coefficient exceeds 0. Hence, we take alternative hypothesis H1 : µ > 0. The value of the test statistic is t=

0.044 √ = 0.70. 0.884/ 197

27.5 b Because n = 197 is large, we approximate the distribution of T under the null hypothesis by an N (0, 1) distribution. The value t = 0.70 lies to the right of zero, so the p-value is the right tail probability P(T ≥ 0.70). By means of the normal approximation we find from Table ?? that the right tail probability P(T ≥ 0.70) ≈ 1 − Φ(0.70) = 0.2420. This means that the value of the test statistic is not very far in the (right) tail of the distribution and is therefore not to be considered exceptionally large. We do not reject the null hypothesis. 27.6 The belief that the intercept is zero is put to the test. The alternative hypothesis should represent the belief that the intercept differs from zero. Therefore, test H0 : α = 0 against H1 : α 6= 0. The value of the t-test statistic is ta =

α ˆ 5.388 = = 2.875. sa 1.874

Because H1 : α 6= 0, both small and large values of Ta are in favor of H1 . Therefore, the right critical value is tn−1,α/2 = t5,0.05 = 2.015. The observed value t = 2.875 is larger than this, so we reject H0 : α = 0 in favor of H1 : α 6= 0. 27.7 a The data are modeled by a simple linear regression model: Yi = α + βxi , where Yi is the gas consumption and xi is the average outside temperature in the ith week. Higher gas consumption as a consequence of smaller temperatures corresponds to β < 0. It is natural to consider the value 0 as the a priori specified value of the parameter (it corresponds to no change of gas consumption). Therefore, we take null hypothesis H0 : β = 0. The alternative hypothesis should express the belief that the gas consumption increases as a consequence of smaller temperatures. Hence, we take alternative hypothesis H1 : β < 0. The value of the test statistic is tb =

βˆ −0.3932 = = −20.06. sb 0.0196

The test statistic Tb has a t-distribution with n − 2 = 24 degrees of freedom. The value −20.06 is smaller than the left critical value t24,0.05 = −1.711, so we reject. 27.7 b For the data after insulation, the value of the test statistic is tb =

−0.2779 = −11.03, 0.0252

and Tb has a t (28) distribution. The value −11.03 is smaller than the left critical value t28,0.05 = −1.701, so we reject.

554


28.1 a H0 : µ1 = µ2 and H1 : µ1 6= µ2 ; The value of the test statistic is tp = −2.130. The critical values are ±t142,0.025 . These are not in Table ??, but 1.96 < t142,0.025 < 2.009, so that we reject the null hypothesis. 28.1 b The value of the test statistic is the same (see Exercise 28.4). The normal approximation yields that the critical values are ±1.96. The observed value td = −2.130 is smaller than the left critical value −1.96, so that we reject the null hypothesis. 28.1 c The observed value td = −2.130 is smaller than the left critical value −2.004, so that we reject the null hypothesis. The salaries differ significantly. 28.2 First consider testing H0 : µ1 = µ2 against H1 : µ1 6= µ2 . In view of the observed sample variances, there is no reason to assume equal variances, so compute s2d =

s2y s2x 7.77 25.33 + = + = 0.1071, n m 775 261

and

x ¯n − y¯m 39.08 − 37.59 √ = = 4.553. sd 0.1071 With sample sizes n1 = 775 and n2 = 261 we can use the normal approximation. The p-value is P(Td ≥ 4.553) ≈ 0, so that we reject H0 : µ1 = µ2 in favor of H1 : µ1 6= µ2 . The other testing problems are handled in the same way. For testing H0 : µ1 = µ3 against H1 : µ1 6= µ3 , we find td =

s2d =

7.77 4.95 + = 0.0178 775 633

and

td =

39.08 − 39.60 √ = −3.898 0.0178

with p-value P(Td ≤ −3.898) ≈ 0, so that we reject H0 : µ1 = µ3 in favor of H1 : µ1 6= µ3 . For testing H0 : µ2 = µ3 against H1 : µ2 = 6 µ3 , we find s2d =

25.33 4.95 + = 0.1049 261 633

and

td =

37.59 − 39.60 √ = −6.206 0.1049

with p-value P(Td ≤ −6.206) ≈ 0, so that we reject H0 : µ2 = µ3 in favor of H1 : µ2 6= µ3 . 28.3 a The value of the test statistic is tp =

22.43 − 11.01 = 2.492. 4.58

Under the assumption of normal data with equal variances, we must compare this with the right critical value t43,0.025 . This is not Table ??, but 2.009 < t43,0.025 < 2.021. Hence, tp > t43,0.025 , so that we reject the null hypothesis. 28.3 b The value tp = 2.492 is greater than the right critical value 1.959, so that we reject on the basis of the bootstrap simulation. 28.3 c The value of the test statistic is td =

22.43 − 11.01 = 2.463. 4.64

Without the assumption of equal variances, and using the normal approximation, we must compare this with the right critical value z0.025 = 1.96. Since tp > 1.96, we reject the null hypothesis.

29.1 Full solutions

555

28.3 d Because we test at level 0.05, we reject if the right tail probability corresponding to td = 2.463 is smaller than 0.025. Since this is the case, we reject on the basis of the bootstrap simulation. 28.4 When n = m, then Sp2 = =

2 (n − 1)SX + (n − 1)SY2 n+n−2

1 1 + n n

2 + SY2 ) 2 (n − 1)(SX S2 S2 = X + Y = Sd2 . 2(n − 1) n n n

2 2 2 2 28.5 a When aSX + bSY2 is unbiased for σ 2 , we should have E 2aSX + bS Y2 = σ 2 . 2 2 2 2 Using that SX and SY are both unbiased for σ , i.e., E SX = σ and E SY = σ , we get 2 2 + bSY2 = aE SX + bE SY2 = (a + b)σ 2 . E aSX

2 Hence, E aSX + bSY2 = σ 2 for all σ > 0 if and only if a + b = 1. 2 28.5 b By independence of SX and SY2 write 2 Var aSX + (1 − a)SY2

2 = a2 Var SX + (1 − a)2 Var SY2

=

(1 − a)2 a2 + n−1 m−1

2σ 4 .

To find the value of a that minimizes this, differentiate with respect to a and put the derivative equal to zero. This leads to 2(1 − a) 2a − = 0. n−1 m−1 Solving for a yields a = (n − 1)/(n + m − 2). Note that the second derivative of 2 Var aSX + (1 − a)SY2 is positive so that this is indeed a minimum. 28.6 a By independence of X1 , X2 , . . . , Xn and Y1 , Y2 , . . . , Ym we have 2 2 ¯ n − Y¯m = Var X ¯ n + Var Y¯m = σX + σY . Var X n m

28.6 b

E Sp2 = =

2 (n − 1)E SX + (m − 1)E SY2 n+m−2 2 (n − 1)σX + (m − 1)σY2 n+m−2

1 1 + n m

1 1 + n m

.

2 ¯ n − Y¯m = σX In principle this may differ from Var X /n + σY2 /m.

28.6 c First note that

2 2 2 E aSX + bSY2 = aE SX + bE SY2 = aσX + bσY2 . 2 For unbiasedness this must equal σX /n + σY2 /m for all σX > 0 and σY > 0, with a, b not depending on σX and σY . This is only possible for a = 1/n and b = 1/m.

556


2 28.6 d From part a together with σX = σY2 = σ 2 , we have

E Sp2 =

2 (n − 1)σX + (m − 1)σY2 n+m−2

(n − 1) + (m − 1) =σ · n+m−2 2

1 1 + n m

1 1 + n m

=σ

2

1 1 + n m

.

28.6 e No, not in general, see part b. If n = m, then according to (the computations in) Exercise 28.4, Sp2 = Sd2 . Since, according to part c, Sd2 is always an unbiased ¯ n − Y¯m it follows that Sp2 is also an unbiased estimator for estimator for Var X ¯ ¯ Var Xn − Ym . One may also check this as follows:

E Sp2 = =

2 (n − 1)σX + (m − 1)σY2 n+m−2 2 (n − 1)σX + (n − 1)σY2 n+n−2

2 = (σX + σY2 ) ·

n−1 n+n−2

1 1 + n m

1 1 + n n

1 1 + n n

2 = (σX + σY2 )

1 σ2 σ2 = X + Y. n n n

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