Chapter 16: Introduction to Bayesian Methods of Inference

16.1

Refer to Table 16.1.

a. β (10,30)

b. n = 25

c. β (10,30) , n = 25

d. Yes

e. Posterior for the β (1,3) prior.

16.2

a.-d. Refer to Section 16.2

16.3

a.-e. Applet exercise, so answers vary.

16.4

a.-d. Applex exercise, so answers vary.

16.5

It should take more trials with a beta(10, 30) prior.

16.6

⎛n⎞

Here, L( y | p ) = p( y | p ) = ⎜⎜ ⎟⎟ p y (1 − p ) n− y , where y = 0, 1, …, n and 0 < p < 1. So,

⎝ y⎠

⎛n⎞

Γ( α + β) α −1

f ( y , p ) = ⎜⎜ ⎟⎟ p y (1 − p ) n − y ×

p (1 − p )β−1

Γ(α)Γ(β)

⎝ y⎠

so that

1

⎛ n ⎞ Γ(α + β) y + α −1

Γ(α + β) Γ( y + α )Γ( n − y + β)

.

m( y ) = ∫ ⎜⎜ ⎟⎟

p

(1 − p ) n − y +β −1 dp =

y Γ(α )Γ(β)

Γ( α )Γ(β)

Γ( n + α + β)

0⎝ ⎠

The posterior density of p is then

Γ( n + α + β)

g * ( p | y) =

p y + α −1 (1 − p ) n − y +β−1 , 0 < p < 1.

Γ( y + α)Γ( n − y + β)

This is the identical beta density as in Example 16.1 (recall that the sum of n i.i.d.

Bernoulli random variables is binomial with n trials and success probability p).

16.7

a. The Bayes estimator is the mean of the posterior distribution, so with a beta posterior

with α = y + 1 and β = n – y + 3 in the prior, the posterior mean is

1

Y +1

Y

=

+

pˆ B =

.

n+4 n+4 n+4

E (Y ) + 1 np + 1

V (Y )

np(1 − p )

b. E ( pˆ B ) =

=

=

≠ p , V ( pˆ ) =

2

n+4

n+4

( n + 4)

( n + 4) 2

16.8

a. From Ex. 16.6, the Bayes estimator for p is pˆ B = E ( p | Y ) =

Y +1

.

n+2

b. This is the uniform distribution in the interval (0, 1).

c. We know that pˆ = Y / n is an unbiased estimator for p. However, for the Bayes

estimator,

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E ( pˆ B ) =

E (Y ) + 1 np + 1

V (Y )

np(1 − p )

=

and V ( pˆ B ) =

.

=

2

n+2

n+2

( n + 2)

( n + 2) 2

2

np(1 − p ) ⎛ np + 1

np(1 − p ) + (1 − 2 p ) 2

⎞

.

+⎜

− p⎟ =

( n + 2) 2 ⎝ n + 2

( n + 2) 2

⎠

d. For the unbiased estimator pˆ , MSE( pˆ ) = V( pˆ ) = p(1 – p)/n. So, holding n fixed, we

must determine the values of p such that

np(1 − p ) + (1 − 2 p ) 2 p(1 − p )

.

<

n

( n + 2) 2

The range of values of p where this is satisfied is solved in Ex. 8.17(c).

Thus, MSE ( pˆ B ) = V ( pˆ B ) + [ B( pˆ B )]2 =

16.9

a. Here, L( y | p ) = p( y | p ) = (1 − p ) y −1 p , where y = 1, 2, … and 0 < p < 1. So,

Γ( α + β) α −1

f ( y , p ) = (1 − p ) y −1 p ×

p (1 − p )β −1

Γ(α)Γ(β)

so that

1

Γ(α + β) Γ( α + 1)Γ( y + β − 1)

Γ(α + β) α

.

m( y ) = ∫

p (1 − p )β+ y − 2 dp =

Γ

(

α

)

Γ

(

β

)

Γ

(

y

+

α

+

β

)

Γ

(

α

)

Γ

(

β

)

0

The posterior density of p is then

Γ(α + β + y )

g * ( p | y) =

pα (1 − p) β + y − 2 , 0 < p < 1.

Γ(α + 1)Γ( β + y − 1)

This is a beta density with shape parameters α* = α + 1 and β* = β + y – 1.

b. The Bayes estimators are

α +1

(1) pˆ B = E ( p | Y ) =

,

α +β +Y

( 2) [ p(1 − p )] B

= E( p | Y ) − E( p 2 | Y ) =

=

(α + 2)(α + 1)

α +1

−

α + β + Y (α + β + Y + 1)(α + β + Y )

(α + 1)(β + Y − 1)

,

(α + β + Y + 1)(α + β + Y )

where the second expectation was solved using the result from Ex. 4.200. (Alternately,

1

the answer could be found by solving E[ p(1 − p ) | Y ] = ∫ p(1 − p ) g * ( p | Y )dp .

0

16.10 a. The joint density of the random sample and θ is given by the product of the marginal

densities multiplied by the gamma prior:

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f ( y 1 , … , y n , θ) =

=

[∏

n

i =1

θ exp( −θy i )

]Γ(α1)β

α

θ α −1 exp(−θ / β)

⎞

⎛

n

θ n + α −1

θ n + α −1

β

⎟

⎜− θ

exp

y

/

exp

−

θ

−

θ

β

=

∑

n

α

α

i =1 i

⎟

⎜

Γ(α )β

Γ(α )β

⎜

β∑i =1 y i + 1 ⎟⎠

⎝

(

)

∞

⎛

⎞

β

1

⎜

⎟dθ , but this integral resembles

n + α −1

θ

exp − θ

b. m( y1 ,…, y n ) =

α ∫

n

⎜

Γ( α)β 0

β∑i =1 y i + 1 ⎟⎠

⎝

β

.

that of a gamma density with shape parameter n + α and scale parameter

n

β∑i =1 y i + 1

⎞

⎛

1

β

⎟

⎜

Thus, the solution is m( y1 ,…, y n ) =

(

n

)

Γ

+

β

⎜ β n y +1⎟

Γ(α )β α

⎠

⎝ ∑i =1 i

n+α

.

c. The solution follows from parts (a) and (b) above.

d. Using the result in Ex. 4.111,

⎡

⎤

β

1

⎥

(

)

μˆ B = E (μ | Y ) = E (1 / θ | Y ) = * *

=⎢

n

+

α

−

1

β (α − 1) ⎢ β∑n Yi + 1

⎥

i =1

⎣

⎦

β∑i =1Yi + 1

n

=

e. The prior mean for 1/θ is E (1 / θ) =

β(n + α − 1)

=

∑

n

Y

−1

1

n + α − 1 β(n + α − 1)

i =1 i

+

1

(again by Ex. 4.111). Thus, μˆ B can be

β( α − 1)

written as

n

1 ⎛ α −1 ⎞

⎛

⎞

μˆ B = Y ⎜

⎟+

⎜

⎟,

⎝ n + α − 1 ⎠ β(α − 1) ⎝ n + α − 1 ⎠

which is a weighted average of the MLE and the prior mean.

f. We know that Y is unbiased; thus E(Y ) = μ = 1/θ. Therefore,

n

1 ⎛ α −1 ⎞ 1 ⎛

n

1 ⎛ α −1 ⎞

⎞

⎛

⎞

E (μˆ B ) = E (Y )⎜

⎟+

⎜

⎟= ⎜

⎟+

⎜

⎟.

⎝ n + α − 1 ⎠ β(α − 1) ⎝ n + α − 1 ⎠ θ ⎝ n + α − 1 ⎠ β(α − 1) ⎝ n + α − 1 ⎠

Therefore, μˆ B is biased. However, it is asymptotically unbiased since

E (μˆ B ) − 1 / θ → 0 .

Also,

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2

2

1 ⎛

1

n

n

n

⎞

⎞

⎛

→0.

V (μˆ B ) = V (Y )⎜

⎟ = 2

⎟ = 2 ⎜

θ (n + α − 1)2

θ n ⎝ n + α −1⎠

⎝ n + α −1⎠

p

So, μˆ B ⎯

⎯→

1 / θ and thus it is consistent.

16.11 a. The joint density of U and λ is

( nλ ) u exp( − nλ )

1

×

λα −1 exp(−λ / β)

α

u!

Γ(α )β

u

n

=

λu + α−1 exp(− nλ − λ / β)

α

u!Γ(α)β

f ( u, λ ) = p( u | λ ) g ( λ ) =

⎡

nu

u + α −1

exp

=

λ

⎢− λ

u!Γ(α)β α

⎣

⎛ β ⎞⎤

⎜⎜

⎟⎟⎥

⎝ nβ + 1 ⎠ ⎦

⎛ β ⎞⎤

⎜⎜

⎟⎟⎥ dλ , but this integral resembles that of a

⎝ nβ + 1 ⎠ ⎦

β

. Thus, the

gamma density with shape parameter u + α and scale parameter

nβ + 1

b. m(u ) =

∞

⎡

nu

λu + α −1 exp⎢− λ

α ∫

u!Γ( α )β 0

⎣

⎛ β ⎞

nu

⎟⎟

Γ(u + α )⎜⎜

solution is m(u ) =

α

u! Γ(α )β

⎝ nβ + 1 ⎠

u+α

.

c. The result follows from parts (a) and (b) above.

⎛ β ⎞

⎟⎟ .

d. λˆ B = E (λ | U ) = α *β * = (U + α)⎜⎜

⎝ nβ + 1 ⎠

e. The prior mean for λ is E(λ) = αβ. From the above,

⎛ β ⎞

⎛ nβ ⎞

⎛ 1 ⎞

n

⎟⎟ = Y ⎜⎜

⎟⎟ + αβ⎜⎜

⎟⎟ ,

λˆ B = ∑i =1 Yi + α ⎜⎜

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

which is a weighted average of the MLE and the prior mean.

(

)

f. We know that Y is unbiased; thus E(Y ) = λ Therefore,

⎛ 1 ⎞

⎛ nβ ⎞

⎛ 1 ⎞

⎛ nβ ⎞

⎟⎟ .

⎟⎟ + αβ⎜⎜

⎟⎟ = λ⎜⎜

⎟⎟ + αβ⎜⎜

E (λˆ B ) = E (Y )⎜⎜

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

So, λˆ is biased but it is asymptotically unbiased since

B

E (λˆ B ) – λ → 0.

Also,

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2

2

⎛ nβ ⎞

λ ⎛ nβ ⎞

nβ

⎟⎟ = λ

⎟⎟ = ⎜⎜

V (λˆ B ) = V (Y )⎜⎜

→ 0.

n ⎝ nβ + 1 ⎠

(nβ + 1)2

⎝ nβ + 1 ⎠

p

So, λˆ B ⎯

⎯→

λ and thus it is consistent.

16.12 First, it is given that W = vU = v ∑i =1 (Yi − μ 0 ) 2 is chi–square with n degrees of freedom.

n

Then, the density function for U (conditioned on v) is given by

1

1

(uv )n / 2−1 e −uv / 2 =

f U (u | v ) = v fW ( uv ) = v

u n / 2−1 v n / 2 e − uv / 2 .

n/2

n/2

Γ( n / 2)2

Γ( n / 2)2

a. The joint density of U and v is then

1

1

f ( u, v ) = f U ( u | v ) g ( v ) =

u n / 2−1 v n / 2 exp(−uv / 2) ×

v α −1 exp(− v / β)

n/2

Γ( n / 2)2

Γ( α)β α

1

u n / 2−1 v n / 2+ α−1 exp( −uv / 2 − v / β)

=

Γ( n / 2)Γ(α )2 n / 2 β α

=

⎡

1

u n / 2−1 v n / 2+ α−1 exp⎢− v

n/2 α

Γ( n / 2)Γ(α )2 β

⎣

⎛ 2β ⎞⎤

⎜⎜

⎟⎟⎥ .

⎝ uβ + 2 ⎠ ⎦

∞

⎡

⎛ 2β ⎞⎤

1

n / 2 −1

⎟⎟⎥ dv , but this integral

u

v n / 2+ α−1 exp⎢− v ⎜⎜

n/2 α

∫

u

2

β

+

Γ( n / 2)Γ( α )2 β

⎝

⎠⎦

0

⎣

resembles that of a gamma density with shape parameter n/2 + α and scale parameter

b. m( u ) =

⎛ 2β ⎞

u n / 2 −1

2β

⎟⎟

Γ( n / 2 + α)⎜⎜

. Thus, the solution is m(u ) =

n/2 α

uβ + 2

Γ( n / 2)Γ( α)2 β

⎝ uβ + 2 ⎠

n / 2+ α

.

c. The result follows from parts (a) and (b) above.

d. Using the result in Ex. 4.111(e),

σˆ 2B = E (σ 2 | U ) = E (1 / v | U ) =

⎛ Uβ + 2 ⎞

1

1

Uβ + 2

⎟⎟ =

⎜⎜

.

=

*

β ( α − 1) n / 2 + α − 1 ⎝ 2β ⎠ β(n + 2α − 2 )

*

1

. From the above,

β( α − 1)

Uβ + 2

U⎛

n

1 ⎛ 2(α − 1) ⎞

⎞

σˆ 2B =

= ⎜

⎟+

⎜

⎟.

β(n + 2α − 2 ) n ⎝ n + 2α − 2 ⎠ β(α − 1) ⎝ n + 2α − 2 ⎠

e. The prior mean for σ 2 = 1 / v =

16.13 a. (.099, .710)

b. Both probabilities are .025.

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Instructor’s Solutions Manual

c. P(.099 < p < .710) = .95.

d.-g. Answers vary.

h. The credible intervals should decrease in width with larger sample sizes.

16.14 a.-b. Answers vary.

16.15 With y = 4, n = 25, and a beta(1, 3) prior, the posterior distribution for p is beta(5, 24).

Using R, the lower and upper endpoints of the 95% credible interval are given by:

> qbeta(.025,5,24)

[1] 0.06064291

> qbeta(.975,5,24)

[1] 0.3266527

16.16 With y = 4, n = 25, and a beta(1, 1) prior, the posterior distribution for p is beta(5, 22).

Using R, the lower and upper endpoints of the 95% credible interval are given by:

> qbeta(.025,5,22)

[1] 0.06554811

> qbeta(.975,5,22)

[1] 0.3486788

This is a wider interval than what was obtained in Ex. 16.15.

16.17 With y = 6 and a beta(10, 5) prior, the posterior distribution for p is beta(11, 10). Using

R, the lower and upper endpoints of the 80% credible interval for p are given by:

> qbeta(.10,11,10)

[1] 0.3847514

> qbeta(.90,11,10)

[1] 0.6618291

16.18 With n = 15,

∑

n

i =1

y i = 30.27, and a gamma(2.3, 0.4) prior, the posterior distribution for

θ is gamma(17.3, .030516). Using R, the lower and upper endpoints of the 80% credible

interval for θ are given by

> qgamma(.10,shape=17.3,scale=.0305167)

[1] 0.3731982

> qgamma(.90,shape=17.3,scale=.0305167)

[1] 0.6957321

The 80% credible interval for θ is (.3732, .6957). To create a 80% credible interval for

1/θ, the end points of the previous interval can be inverted:

.3732 < θ < .6957

1/(.3732) > 1/θ > 1/(.6957)

Since 1/(.6957) = 1.4374 and 1/(.3732) = 2.6795, the 80% credible interval for 1/θ is

(1.4374, 2.6795).

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16.19 With n = 25,

∑

n

i =1

y i = 174, and a gamma(2, 3) prior, the posterior distribution for λ is

gamma(176, .0394739). Using R, the lower and upper endpoints of the 95% credible

interval for λ are given by

> qgamma(.025,shape=176,scale=.0394739)

[1] 5.958895

> qgamma(.975,shape=176,scale=.0394739)

[1] 8.010663

16.20 With n = 8, u = .8579, and a gamma(5, 2) prior, the posterior distribution for v is

gamma(9, 1.0764842). Using R, the lower and upper endpoints of the 90% credible

interval for v are given by

> qgamma(.05,shape=9,scale=1.0764842)

[1] 5.054338

> qgamma(.95,shape=9,scale=1.0764842)

[1] 15.53867

The 90% credible interval for v is (5.054, 15.539). Similar to Ex. 16.18, the 90% credible

interval for σ2 = 1/v is found by inverting the endpoints of the credible interval for v,

given by (.0644, .1979).

16.21 From Ex. 6.15, the posterior distribution of p is beta(5, 24). Now, we can find

P * ( p ∈ Ω 0 ) = P * ( p < .3) by (in R):

> pbeta(.3,5,24)

[1] 0.9525731

Therefore, P * ( p ∈ Ω a ) = P * ( p ≥ .3) = 1 – .9525731 = .0474269. Since the probability

associated with H0 is much larger, our decision is to not reject H0.

16.22 From Ex. 6.16, the posterior distribution of p is beta(5, 22). We can find

P * ( p ∈ Ω 0 ) = P * ( p < .3) by (in R):

> pbeta(.3,5,22)

[1] 0.9266975

Therefore, P * ( p ∈ Ω a ) = P * ( p ≥ .3) = 1 – .9266975 = .0733025. Since the probability

associated with H0 is much larger, our decision is to not reject H0.

16.23 From Ex. 6.17, the posterior distribution of p is beta(11, 10). Thus,

P * ( p ∈ Ω 0 ) = P * ( p < .4) is given by (in R):

> pbeta(.4,11,10)

[1] 0.1275212

Therefore, P * ( p ∈ Ω a ) = P * ( p ≥ .4) = 1 – .1275212 = .8724788. Since the probability

associated with Ha is much larger, our decision is to reject H0.

16.24 From Ex. 16.18, the posterior distribution for θ is gamma(17.3, .0305). To test

H0: θ > .5 vs. Ha: θ ≤ .5,

*

*

we calculate P (θ ∈ Ω 0 ) = P (θ > .5) as:

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Instructor’s Solutions Manual

> 1 - pgamma(.5,shape=17.3,scale=.0305)

[1] 0.5561767

Therefore, P * (θ ∈ Ω a ) = P * (θ ≥ .5) = 1 – .5561767 = .4438233. The probability

associated with H0 is larger (but only marginally so), so our decision is to not reject H0.

16.25 From Ex. 16.19, the posterior distribution for λ is gamma(176, .0395). Thus,

P * (λ ∈ Ω 0 ) = P * (λ > 6) is found by

> 1 - pgamma(6,shape=176,scale=.0395)

[1] 0.9700498

Therefore, P * (λ ∈ Ω a ) = P * ( λ ≤ 6) = 1 – .9700498 = .0299502. Since the probability

associated with H0 is much larger, our decision is to not reject H0.

16.26 From Ex. 16.20, the posterior distribution for v is gamma(9, 1.0765). To test:

H0: v < 10 vs. Ha: v ≥ 10,

*

*

we calculate P ( v ∈ Ω 0 ) = P ( v < 10) as

> pgamma(10,9, 1.0765)

[1] 0.7464786

Therefore, P * (λ ∈ Ω a ) = P * ( v ≥ 10) = 1 – .7464786 = .2535214. Since the probability

associated with H0 is larger, our decision is to not reject H0.

16.1

Refer to Table 16.1.

a. β (10,30)

b. n = 25

c. β (10,30) , n = 25

d. Yes

e. Posterior for the β (1,3) prior.

16.2

a.-d. Refer to Section 16.2

16.3

a.-e. Applet exercise, so answers vary.

16.4

a.-d. Applex exercise, so answers vary.

16.5

It should take more trials with a beta(10, 30) prior.

16.6

⎛n⎞

Here, L( y | p ) = p( y | p ) = ⎜⎜ ⎟⎟ p y (1 − p ) n− y , where y = 0, 1, …, n and 0 < p < 1. So,

⎝ y⎠

⎛n⎞

Γ( α + β) α −1

f ( y , p ) = ⎜⎜ ⎟⎟ p y (1 − p ) n − y ×

p (1 − p )β−1

Γ(α)Γ(β)

⎝ y⎠

so that

1

⎛ n ⎞ Γ(α + β) y + α −1

Γ(α + β) Γ( y + α )Γ( n − y + β)

.

m( y ) = ∫ ⎜⎜ ⎟⎟

p

(1 − p ) n − y +β −1 dp =

y Γ(α )Γ(β)

Γ( α )Γ(β)

Γ( n + α + β)

0⎝ ⎠

The posterior density of p is then

Γ( n + α + β)

g * ( p | y) =

p y + α −1 (1 − p ) n − y +β−1 , 0 < p < 1.

Γ( y + α)Γ( n − y + β)

This is the identical beta density as in Example 16.1 (recall that the sum of n i.i.d.

Bernoulli random variables is binomial with n trials and success probability p).

16.7

a. The Bayes estimator is the mean of the posterior distribution, so with a beta posterior

with α = y + 1 and β = n – y + 3 in the prior, the posterior mean is

1

Y +1

Y

=

+

pˆ B =

.

n+4 n+4 n+4

E (Y ) + 1 np + 1

V (Y )

np(1 − p )

b. E ( pˆ B ) =

=

=

≠ p , V ( pˆ ) =

2

n+4

n+4

( n + 4)

( n + 4) 2

16.8

a. From Ex. 16.6, the Bayes estimator for p is pˆ B = E ( p | Y ) =

Y +1

.

n+2

b. This is the uniform distribution in the interval (0, 1).

c. We know that pˆ = Y / n is an unbiased estimator for p. However, for the Bayes

estimator,

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Instructor’s Solutions Manual

E ( pˆ B ) =

E (Y ) + 1 np + 1

V (Y )

np(1 − p )

=

and V ( pˆ B ) =

.

=

2

n+2

n+2

( n + 2)

( n + 2) 2

2

np(1 − p ) ⎛ np + 1

np(1 − p ) + (1 − 2 p ) 2

⎞

.

+⎜

− p⎟ =

( n + 2) 2 ⎝ n + 2

( n + 2) 2

⎠

d. For the unbiased estimator pˆ , MSE( pˆ ) = V( pˆ ) = p(1 – p)/n. So, holding n fixed, we

must determine the values of p such that

np(1 − p ) + (1 − 2 p ) 2 p(1 − p )

.

<

n

( n + 2) 2

The range of values of p where this is satisfied is solved in Ex. 8.17(c).

Thus, MSE ( pˆ B ) = V ( pˆ B ) + [ B( pˆ B )]2 =

16.9

a. Here, L( y | p ) = p( y | p ) = (1 − p ) y −1 p , where y = 1, 2, … and 0 < p < 1. So,

Γ( α + β) α −1

f ( y , p ) = (1 − p ) y −1 p ×

p (1 − p )β −1

Γ(α)Γ(β)

so that

1

Γ(α + β) Γ( α + 1)Γ( y + β − 1)

Γ(α + β) α

.

m( y ) = ∫

p (1 − p )β+ y − 2 dp =

Γ

(

α

)

Γ

(

β

)

Γ

(

y

+

α

+

β

)

Γ

(

α

)

Γ

(

β

)

0

The posterior density of p is then

Γ(α + β + y )

g * ( p | y) =

pα (1 − p) β + y − 2 , 0 < p < 1.

Γ(α + 1)Γ( β + y − 1)

This is a beta density with shape parameters α* = α + 1 and β* = β + y – 1.

b. The Bayes estimators are

α +1

(1) pˆ B = E ( p | Y ) =

,

α +β +Y

( 2) [ p(1 − p )] B

= E( p | Y ) − E( p 2 | Y ) =

=

(α + 2)(α + 1)

α +1

−

α + β + Y (α + β + Y + 1)(α + β + Y )

(α + 1)(β + Y − 1)

,

(α + β + Y + 1)(α + β + Y )

where the second expectation was solved using the result from Ex. 4.200. (Alternately,

1

the answer could be found by solving E[ p(1 − p ) | Y ] = ∫ p(1 − p ) g * ( p | Y )dp .

0

16.10 a. The joint density of the random sample and θ is given by the product of the marginal

densities multiplied by the gamma prior:

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Instructor’s Solutions Manual

f ( y 1 , … , y n , θ) =

=

[∏

n

i =1

θ exp( −θy i )

]Γ(α1)β

α

θ α −1 exp(−θ / β)

⎞

⎛

n

θ n + α −1

θ n + α −1

β

⎟

⎜− θ

exp

y

/

exp

−

θ

−

θ

β

=

∑

n

α

α

i =1 i

⎟

⎜

Γ(α )β

Γ(α )β

⎜

β∑i =1 y i + 1 ⎟⎠

⎝

(

)

∞

⎛

⎞

β

1

⎜

⎟dθ , but this integral resembles

n + α −1

θ

exp − θ

b. m( y1 ,…, y n ) =

α ∫

n

⎜

Γ( α)β 0

β∑i =1 y i + 1 ⎟⎠

⎝

β

.

that of a gamma density with shape parameter n + α and scale parameter

n

β∑i =1 y i + 1

⎞

⎛

1

β

⎟

⎜

Thus, the solution is m( y1 ,…, y n ) =

(

n

)

Γ

+

β

⎜ β n y +1⎟

Γ(α )β α

⎠

⎝ ∑i =1 i

n+α

.

c. The solution follows from parts (a) and (b) above.

d. Using the result in Ex. 4.111,

⎡

⎤

β

1

⎥

(

)

μˆ B = E (μ | Y ) = E (1 / θ | Y ) = * *

=⎢

n

+

α

−

1

β (α − 1) ⎢ β∑n Yi + 1

⎥

i =1

⎣

⎦

β∑i =1Yi + 1

n

=

e. The prior mean for 1/θ is E (1 / θ) =

β(n + α − 1)

=

∑

n

Y

−1

1

n + α − 1 β(n + α − 1)

i =1 i

+

1

(again by Ex. 4.111). Thus, μˆ B can be

β( α − 1)

written as

n

1 ⎛ α −1 ⎞

⎛

⎞

μˆ B = Y ⎜

⎟+

⎜

⎟,

⎝ n + α − 1 ⎠ β(α − 1) ⎝ n + α − 1 ⎠

which is a weighted average of the MLE and the prior mean.

f. We know that Y is unbiased; thus E(Y ) = μ = 1/θ. Therefore,

n

1 ⎛ α −1 ⎞ 1 ⎛

n

1 ⎛ α −1 ⎞

⎞

⎛

⎞

E (μˆ B ) = E (Y )⎜

⎟+

⎜

⎟= ⎜

⎟+

⎜

⎟.

⎝ n + α − 1 ⎠ β(α − 1) ⎝ n + α − 1 ⎠ θ ⎝ n + α − 1 ⎠ β(α − 1) ⎝ n + α − 1 ⎠

Therefore, μˆ B is biased. However, it is asymptotically unbiased since

E (μˆ B ) − 1 / θ → 0 .

Also,

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329

Instructor’s Solutions Manual

2

2

1 ⎛

1

n

n

n

⎞

⎞

⎛

→0.

V (μˆ B ) = V (Y )⎜

⎟ = 2

⎟ = 2 ⎜

θ (n + α − 1)2

θ n ⎝ n + α −1⎠

⎝ n + α −1⎠

p

So, μˆ B ⎯

⎯→

1 / θ and thus it is consistent.

16.11 a. The joint density of U and λ is

( nλ ) u exp( − nλ )

1

×

λα −1 exp(−λ / β)

α

u!

Γ(α )β

u

n

=

λu + α−1 exp(− nλ − λ / β)

α

u!Γ(α)β

f ( u, λ ) = p( u | λ ) g ( λ ) =

⎡

nu

u + α −1

exp

=

λ

⎢− λ

u!Γ(α)β α

⎣

⎛ β ⎞⎤

⎜⎜

⎟⎟⎥

⎝ nβ + 1 ⎠ ⎦

⎛ β ⎞⎤

⎜⎜

⎟⎟⎥ dλ , but this integral resembles that of a

⎝ nβ + 1 ⎠ ⎦

β

. Thus, the

gamma density with shape parameter u + α and scale parameter

nβ + 1

b. m(u ) =

∞

⎡

nu

λu + α −1 exp⎢− λ

α ∫

u!Γ( α )β 0

⎣

⎛ β ⎞

nu

⎟⎟

Γ(u + α )⎜⎜

solution is m(u ) =

α

u! Γ(α )β

⎝ nβ + 1 ⎠

u+α

.

c. The result follows from parts (a) and (b) above.

⎛ β ⎞

⎟⎟ .

d. λˆ B = E (λ | U ) = α *β * = (U + α)⎜⎜

⎝ nβ + 1 ⎠

e. The prior mean for λ is E(λ) = αβ. From the above,

⎛ β ⎞

⎛ nβ ⎞

⎛ 1 ⎞

n

⎟⎟ = Y ⎜⎜

⎟⎟ + αβ⎜⎜

⎟⎟ ,

λˆ B = ∑i =1 Yi + α ⎜⎜

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

which is a weighted average of the MLE and the prior mean.

(

)

f. We know that Y is unbiased; thus E(Y ) = λ Therefore,

⎛ 1 ⎞

⎛ nβ ⎞

⎛ 1 ⎞

⎛ nβ ⎞

⎟⎟ .

⎟⎟ + αβ⎜⎜

⎟⎟ = λ⎜⎜

⎟⎟ + αβ⎜⎜

E (λˆ B ) = E (Y )⎜⎜

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

⎝ nβ + 1 ⎠

So, λˆ is biased but it is asymptotically unbiased since

B

E (λˆ B ) – λ → 0.

Also,

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Instructor’s Solutions Manual

2

2

⎛ nβ ⎞

λ ⎛ nβ ⎞

nβ

⎟⎟ = λ

⎟⎟ = ⎜⎜

V (λˆ B ) = V (Y )⎜⎜

→ 0.

n ⎝ nβ + 1 ⎠

(nβ + 1)2

⎝ nβ + 1 ⎠

p

So, λˆ B ⎯

⎯→

λ and thus it is consistent.

16.12 First, it is given that W = vU = v ∑i =1 (Yi − μ 0 ) 2 is chi–square with n degrees of freedom.

n

Then, the density function for U (conditioned on v) is given by

1

1

(uv )n / 2−1 e −uv / 2 =

f U (u | v ) = v fW ( uv ) = v

u n / 2−1 v n / 2 e − uv / 2 .

n/2

n/2

Γ( n / 2)2

Γ( n / 2)2

a. The joint density of U and v is then

1

1

f ( u, v ) = f U ( u | v ) g ( v ) =

u n / 2−1 v n / 2 exp(−uv / 2) ×

v α −1 exp(− v / β)

n/2

Γ( n / 2)2

Γ( α)β α

1

u n / 2−1 v n / 2+ α−1 exp( −uv / 2 − v / β)

=

Γ( n / 2)Γ(α )2 n / 2 β α

=

⎡

1

u n / 2−1 v n / 2+ α−1 exp⎢− v

n/2 α

Γ( n / 2)Γ(α )2 β

⎣

⎛ 2β ⎞⎤

⎜⎜

⎟⎟⎥ .

⎝ uβ + 2 ⎠ ⎦

∞

⎡

⎛ 2β ⎞⎤

1

n / 2 −1

⎟⎟⎥ dv , but this integral

u

v n / 2+ α−1 exp⎢− v ⎜⎜

n/2 α

∫

u

2

β

+

Γ( n / 2)Γ( α )2 β

⎝

⎠⎦

0

⎣

resembles that of a gamma density with shape parameter n/2 + α and scale parameter

b. m( u ) =

⎛ 2β ⎞

u n / 2 −1

2β

⎟⎟

Γ( n / 2 + α)⎜⎜

. Thus, the solution is m(u ) =

n/2 α

uβ + 2

Γ( n / 2)Γ( α)2 β

⎝ uβ + 2 ⎠

n / 2+ α

.

c. The result follows from parts (a) and (b) above.

d. Using the result in Ex. 4.111(e),

σˆ 2B = E (σ 2 | U ) = E (1 / v | U ) =

⎛ Uβ + 2 ⎞

1

1

Uβ + 2

⎟⎟ =

⎜⎜

.

=

*

β ( α − 1) n / 2 + α − 1 ⎝ 2β ⎠ β(n + 2α − 2 )

*

1

. From the above,

β( α − 1)

Uβ + 2

U⎛

n

1 ⎛ 2(α − 1) ⎞

⎞

σˆ 2B =

= ⎜

⎟+

⎜

⎟.

β(n + 2α − 2 ) n ⎝ n + 2α − 2 ⎠ β(α − 1) ⎝ n + 2α − 2 ⎠

e. The prior mean for σ 2 = 1 / v =

16.13 a. (.099, .710)

b. Both probabilities are .025.

Chapter 16: Introduction to Bayesian Methods of Inference

331

Instructor’s Solutions Manual

c. P(.099 < p < .710) = .95.

d.-g. Answers vary.

h. The credible intervals should decrease in width with larger sample sizes.

16.14 a.-b. Answers vary.

16.15 With y = 4, n = 25, and a beta(1, 3) prior, the posterior distribution for p is beta(5, 24).

Using R, the lower and upper endpoints of the 95% credible interval are given by:

> qbeta(.025,5,24)

[1] 0.06064291

> qbeta(.975,5,24)

[1] 0.3266527

16.16 With y = 4, n = 25, and a beta(1, 1) prior, the posterior distribution for p is beta(5, 22).

Using R, the lower and upper endpoints of the 95% credible interval are given by:

> qbeta(.025,5,22)

[1] 0.06554811

> qbeta(.975,5,22)

[1] 0.3486788

This is a wider interval than what was obtained in Ex. 16.15.

16.17 With y = 6 and a beta(10, 5) prior, the posterior distribution for p is beta(11, 10). Using

R, the lower and upper endpoints of the 80% credible interval for p are given by:

> qbeta(.10,11,10)

[1] 0.3847514

> qbeta(.90,11,10)

[1] 0.6618291

16.18 With n = 15,

∑

n

i =1

y i = 30.27, and a gamma(2.3, 0.4) prior, the posterior distribution for

θ is gamma(17.3, .030516). Using R, the lower and upper endpoints of the 80% credible

interval for θ are given by

> qgamma(.10,shape=17.3,scale=.0305167)

[1] 0.3731982

> qgamma(.90,shape=17.3,scale=.0305167)

[1] 0.6957321

The 80% credible interval for θ is (.3732, .6957). To create a 80% credible interval for

1/θ, the end points of the previous interval can be inverted:

.3732 < θ < .6957

1/(.3732) > 1/θ > 1/(.6957)

Since 1/(.6957) = 1.4374 and 1/(.3732) = 2.6795, the 80% credible interval for 1/θ is

(1.4374, 2.6795).

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16.19 With n = 25,

∑

n

i =1

y i = 174, and a gamma(2, 3) prior, the posterior distribution for λ is

gamma(176, .0394739). Using R, the lower and upper endpoints of the 95% credible

interval for λ are given by

> qgamma(.025,shape=176,scale=.0394739)

[1] 5.958895

> qgamma(.975,shape=176,scale=.0394739)

[1] 8.010663

16.20 With n = 8, u = .8579, and a gamma(5, 2) prior, the posterior distribution for v is

gamma(9, 1.0764842). Using R, the lower and upper endpoints of the 90% credible

interval for v are given by

> qgamma(.05,shape=9,scale=1.0764842)

[1] 5.054338

> qgamma(.95,shape=9,scale=1.0764842)

[1] 15.53867

The 90% credible interval for v is (5.054, 15.539). Similar to Ex. 16.18, the 90% credible

interval for σ2 = 1/v is found by inverting the endpoints of the credible interval for v,

given by (.0644, .1979).

16.21 From Ex. 6.15, the posterior distribution of p is beta(5, 24). Now, we can find

P * ( p ∈ Ω 0 ) = P * ( p < .3) by (in R):

> pbeta(.3,5,24)

[1] 0.9525731

Therefore, P * ( p ∈ Ω a ) = P * ( p ≥ .3) = 1 – .9525731 = .0474269. Since the probability

associated with H0 is much larger, our decision is to not reject H0.

16.22 From Ex. 6.16, the posterior distribution of p is beta(5, 22). We can find

P * ( p ∈ Ω 0 ) = P * ( p < .3) by (in R):

> pbeta(.3,5,22)

[1] 0.9266975

Therefore, P * ( p ∈ Ω a ) = P * ( p ≥ .3) = 1 – .9266975 = .0733025. Since the probability

associated with H0 is much larger, our decision is to not reject H0.

16.23 From Ex. 6.17, the posterior distribution of p is beta(11, 10). Thus,

P * ( p ∈ Ω 0 ) = P * ( p < .4) is given by (in R):

> pbeta(.4,11,10)

[1] 0.1275212

Therefore, P * ( p ∈ Ω a ) = P * ( p ≥ .4) = 1 – .1275212 = .8724788. Since the probability

associated with Ha is much larger, our decision is to reject H0.

16.24 From Ex. 16.18, the posterior distribution for θ is gamma(17.3, .0305). To test

H0: θ > .5 vs. Ha: θ ≤ .5,

*

*

we calculate P (θ ∈ Ω 0 ) = P (θ > .5) as:

Chapter 16: Introduction to Bayesian Methods of Inference

333

Instructor’s Solutions Manual

> 1 - pgamma(.5,shape=17.3,scale=.0305)

[1] 0.5561767

Therefore, P * (θ ∈ Ω a ) = P * (θ ≥ .5) = 1 – .5561767 = .4438233. The probability

associated with H0 is larger (but only marginally so), so our decision is to not reject H0.

16.25 From Ex. 16.19, the posterior distribution for λ is gamma(176, .0395). Thus,

P * (λ ∈ Ω 0 ) = P * (λ > 6) is found by

> 1 - pgamma(6,shape=176,scale=.0395)

[1] 0.9700498

Therefore, P * (λ ∈ Ω a ) = P * ( λ ≤ 6) = 1 – .9700498 = .0299502. Since the probability

associated with H0 is much larger, our decision is to not reject H0.

16.26 From Ex. 16.20, the posterior distribution for v is gamma(9, 1.0765). To test:

H0: v < 10 vs. Ha: v ≥ 10,

*

*

we calculate P ( v ∈ Ω 0 ) = P ( v < 10) as

> pgamma(10,9, 1.0765)

[1] 0.7464786

Therefore, P * (λ ∈ Ω a ) = P * ( v ≥ 10) = 1 – .7464786 = .2535214. Since the probability

associated with H0 is larger, our decision is to not reject H0.

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