Solutions to Homework 3

1. 5.2.1

(solution)

L(θ) =8∏

i=1

θki(1− θ)1−ki = θP8

i=1 ki(1− θ)8−P8

i=1 ki = θ5(1− θ)3

dL(θ)dθ

= θ53(1− θ)2(−1) + 5θ4(1− θ)3 = θ4(1− θ)2(−8θ + 5)dL(θ)

dθ= 0 implies θ̂ = 5

8

2. 5.2.3

(solution)

L(θ) =4∏

i=1

λe−λyi = λ4e−λP4

i=1 yi = λ4e−32.8λ

dL(θ)dθ

= λ4(−32.8)e−32.8λ + 4λ3e−32.8λ = λ3e−32.8λ(4− 32.8λ)dL(θ)

dθ= 0 implies λ̂ = 4

32.8= 0.122

3. 5.2.8

(solution) From Example 5.2.1, p̂ = nPni=1 ki

. For these data, n=1011,

and∑n

i=1 ki = 1(678)+2(227)+3(56)+4(28)+5(8)+6(14) or∑n

i=1 ki =1536. Then p̂ = 1011

1536= 0.658

Number of Occupants Frequency Expected frequency1 678 665.22 227 227.53 56 77.84 28 26.65 8 9.16 14 3.4

1

4. 5.2.9

(solution)(a) L(θ) = (1

θ)n, if 0 ≤ y1, y2, y3, . . . , yn ≤ θ, and 0 otherwise. Thus

θ̂ = ymax, which for these data is 14.2(b) L(θ) = ( 1

θ2−θ1)n, if θ1 ≤ y1, y2, y3, . . . , yn ≤ θ2, and 0 otherwise.

Thus θ̂1 = ymin, and θ̂2 = ymax For these data, θ̂1 = 1.8, θ̂2 = 14.2

5. 5.2.14

(solution) Let θ = σ2, so L(θ) =∏n

i=11√2πθ

e−12

(yi−µ)2

θ = 2π−n2 θ−

n2 e−

12

1θ

Pni=1(yi−µ)2

ln L(θ) = −n2ln2π−n

2lnθ−1

21θ

∑ni=1(yi−µ)2 d ln L(θ)

dθ= − n

2θ+1

21θ2

∑ni=1(yi−

µ)2 = 12

−nθ+Pn

i=1(yi−µ)2

θ2 Setting d ln L(θ)dθ

= 0 gives θ̂ = σ̂2 = 1n

∑ni=1(yi −

µ)2.

6. 5.2.17

(solution) For Y Poisson, E(Y ) = λ. Then λ̂ = y = 136. The maximum

likelihood estimator is the same.

7. 5.2.19

(solution) E(Y ) = θ1 so θ̂1 = y. E(Y 2) =∫ θ1+θ2

θ1−θ2y(2) 1

2θ2 dy = 12θ2 [

y3

3]θ1+θ2θ1−θ2

=

θ21 + 1

3θ22 Substitute ˆθ1 = y into the equation θ2

1 + 13θ22 = 1

n

∑ni=1 y

∑2i

to obtain θ̂2 =√

3( 1n

∑ni=1 y2

i − y2)

8. 5.2.21

(solution) E(X) = 0θ0(1 − θ)1−0 + 1θ1(1 − θ)1−1 = θ. Then θ̂ = y,which for the given data is 2

5.

9. 5.2.24

(solution) From Chapter 4(Theorem 4.4.2), E(X) = 1p. Setting 1

p= x,

gives p̂ = 1x. For the given data, p̂ = 0.479. The expected frequency

are:

2

Number of Clusters/Song Frequency Expected frequency1 132 119.82 52 62.43 34 32.54 9 16.95 7 8.86 5 4.67 5 2.48 6 2.6

10. 5.3.2

(solution) The confidence interval is (y− zα2

σ√n, y + zα

2

σ√n) = (70.833−

1.96 8.0√6, 70.833 + 1.96 8.0√

6) = (64.432, 77.234). Since 80 does not fall

within the confidence interval, that men and women metabolize methylmer-cury at the same rate is not believable.

11. 5.3.3

(solution) The length of the confidence interval is 2zα2

σ√n

= 2(1.96(14.3))√n

=56.056√

n. For 56.056√

n≤ 3.06, n ≥ (56.056

3.06)2 = 335.58, so take n = 336.

12. 5.3.5

(solution) The probability that the given interval will contain µ isP (−0.96 < Z < 1.06) = 0.6869. The probability of four or five suchintervals is binomial with n = 5 and p = 0.6869, so the probability is5(0.6869)4(0.3131) + (0.6869)5 = 0.501.

13. 5.3.11

(solution) Budweiser would use the sample proportion 0.54 alone as theestimate. Schlitz would construct the 95% confidence interval (0.36,0.56) to claim that values¡0.50 are believable.

14. 5.3.13

(solution) 2.58√

p(1−p)n

≤ 2.58√

14n≤ 0.01, so take n ≥ (2.58)2

4(0.01)2=

16, 641.

3

15. 5.3.22

(solution) For candidate A, the believable values for the probabilityof winning fall in the range (0.52-0.05, 0.52+0.05)=(0.47, 0.57). Forcandidate B, the believable values for the probability of winning fall inthe range (0.48-0.05, 0.48+0.05)=(0.43, 0.53). Since 0.05 falls in bothintervals, there is a sense in which the candidates can be consideredtied.

16. 5.3.26

(solution)

(a) Take n to be the smallest integer≥ z20.075

4(0.03)2= 1.442

4(0.03)2= 576.

(b)Take n to be the smallest integer≥ z20.075p(1−p)

(0.03)2= 1.442(0.10)(0.90)

(0.03)2=

207.36, so n = 208.

17. 5.4.5

(solution) E(X) = E( 1n

∑ni=1 Xi) = 1

n

∑ni=1 E(Xi) = 1

n

∑ni=1 λ = λ In

general, the sample mean is an unbiased estimator of the mean µ.

18. 5.4.11

(solution) Since E(W ) = θ, E(W 2) = V ar(W )+[E(W )]2 = V ar(W )+θ2. Thus, W 2 cannot be an unbiased estimator for θ2 if V ar(W ) 6= 0.

19. 5.4.12

(solution) For each i, E[(Yi−µ)2] = σ2 by definition of σ2. E[ 1n

∑ni=1(Yi−

µ)2] = 1n

∑ni=1 E[(Yi − µ)2] = 1

nnσ2σ2

20. 5.4.14

(solution) fYmin(y) = nfY (y)(1−FY (y))n−1 = n1

θe−

yθ [1−(1−e−

yθ )]n−1 =

n1θe−

nyθ .Then fnYmin

(y) = 1nfYmin

( yn) = 1

nn1

θe−n y

n1θ = 1

θe−

yθ . E(nYmin) =

θ, so nYmin is unbiased. Also, 1n

∑ni=1 Yi is unbiased since each Yi is

unbiased since each Yiis unbiased(See 5.4.5).

21. 5.4.15

(solution) Since limn→∞E(W2) = limn→∞(V ar(W )+[E(W )]2) = limn→∞(σ2

n+

µ2) = µ2, W2

is an asymptotically unbiased estimator for µ2.

4

22. 5.4.16

(solution) θ̂n = 1n

∑ni=1(Yi − Y )2, so E(θ̂n) = n−1

nσ2. This estimator is

asymptotically unbiased since limn→∞E(θ̂n) = limn→∞n−1

nσ2 = σ2.

23. 5.4.17

(solution)(a) E(p̂1) = E(X1) = p, since X1 is binomial with n = 1. E(p̂2) =E(X

n) = 1

nnp = p, since X is binomial.

(b) V ar(p̂1) = p(1−p); V ar(X) = np(1−p). Then V ar(p̂2) = np(1−p)n2 =

p(1−p)n

, which is smaller than V ar(p̂1) by a factor of n.

24. 5.4.19

(solution)(a) See 5.4.14(b) V ar(Y1 = θ2), since Y1 is a gamma variable with parameters 1 ans1θ. V ar(Y ) = Y1

n= θ2

nFrom 5.4.14, V ar(nYmin) = V ar(Y ) = θ2

(c) V ar(θ̂3)

V ar(θ̂1)= θ2

θ2 = 1 and V ar(θ̂3)

V ar(θ̂2)= θ2

( θ2

n)

= n

25. 5.4.20

(solution) V ar(λ̂1) = V ar(X1) = λ. V ar(λ̂2) = V ar(X) = λn. V ar(λ̂2)

V ar(λ̂1)=

( λn

)

λ= 1

n.

26. 5.4.21

(solution) From Example 5.4.7, V ar(θ̂1) = V ar((n+1n

)Ymax) = θ2

n(n+2).

Generalizing Question 5.4.18, we obtain V ar(θ̂2) = V ar((n+1)Ymin) =

nθ2

(n+2)and V ar(θ̂2)

V ar(θ̂1)=

nθ2

(n+2)

θ2

n(n+2)

= n2

27. 5.5.2

(solution) ln fY (Y ; θ) = − ln θ− Yθ. ∂ ln fY (Y ;θ)

∂θ= −1

θ+ Y

θ2 .∂2 ln fY (Y ;θ)

∂θ2 =1θ2 − 2Y

θ3 . E[∂2 ln fY (Y ;θ)∂θ2 ] = 1

θ2 − 2θθ3 = − 1

θ2 , so the Cramer-Rao bound isθ2

n. Also, V ar(θ̂) = V ar(Y ) = V ar(Y )

n= θ2

n, so θ̂ is a best estimator.

28. 5.5.4

5

(solution) ln fY (Y ; µ) = − ln√

2πσ − 12

(Y−µ)2

σ2 . ∂ ln fY (Y ;µ)∂µ

= (Y−µ)σ2 .

∂2 ln fY (Y ;µ)∂µ2 = − 1

σ2 . E[∂2 ln fY (Y ;µ)∂µ2 ] = − 1

σ2 , so the Cramer-Rao bound

is σ2

n. Also, V ar(µ̂) = V ar(Y ) = V ar(Y )

n= σ2

n, so µ̂ is an efficient

estimator.

29. 5.5.5

(solution) ln fY (Y ; θ) = − ln θ. ∂ ln fY (Y ;θ)∂θ

= −1θ. E[(∂ ln fY (Y ;θ)

∂θ)2] = 1

θ2 ,

so the Cramer-Rao bound is θ2

n. From 5.4.21, V ar(θ̂) = θ2

n(n+2), which is

smaller than the the Cramer-Rao bound. This occurs because Theorem5.5.1 is not necessarily valid if the range of the pdf depends on theparameter.

Also, V ar(θ̂) = V ar(Y ) = V ar(Y )n

= θ2

n, so θ̂ is a best estimator.

6