Solutions

9.27

a.

The posterior density is

$$\pi(\lambda\vert X) \propto \frac{1}{\lambda^n}e^{-\sum x_i/\lambda} \frac{1}{\lambda^{a+1}} e^{-1/(b\lambda)}$$

$$= \frac{1}{\lambda^{n+a+1}} e^{-[1/b + \sum x_i]/\lambda}$$

$$= (n+a, [1/b + \sum x_i]^{-1})$$

The inverse gamma density is unimodal, so the HPD region is an interval $[c_1,c_2]$ with $c_1, c_2$ chosen to have equal density values and $P(Y > 1/c_1)+P(Y<1/c_2) = \alpha$, with $Y \sim$   Gamma$(n+a, [1/b + \sum x_i]^{-1})$.

.b

The distribution of $S^2$ is Gamma$((n-1)/2, 2 \sigma^2 /(n-1))$. The resulting posterior density is therefore

$$\pi(\sigma^2\vert s^2) \propto \frac{(s^2)^{(n-1)/2 - 1}}{(\sigma^2/(n-1))^{(n-1)/2}} e^{-(n-1)s^2/\sigma^2} \frac{1}{(\sigma^2)^{a+1}} e^{-1/(b \sigma^2)}$$

$$\propto \frac{1}{(\sigma^2)^{(n-1)/2 + a + 1}} e^{-[1/b + (n-1)s^2]/\sigma^2}$$

$$= ((n-1)/2 + a, [1/b + (n-1)s^2]^{-1})$$

As in the previous part, the HPD region is an interval that can be determined by solving a corresponding set of equations.

c.

The limiting posterior distribution is IG$((n-1)/2,[(n-1)s^2]^{-1})$. The limiting HPD region is an interval $[c_1,c_2]$ with $c_1 = (n-1)s^2/\chi^2_{n-1,\alpha_1}$ and $c_2 = (n-1)s^2/\chi^2_{n-1,1-\alpha_2}$ where $\alpha_1+\alpha_2=\alpha$ and $c_1, c_2$ have equal posterior density values.

9.33

a.

Since $0 \in C_{a}(x)$ for all $a,x$,

$$P_{\mu=0}(0 \in C_{a}(X)) = 1$$

For $\mu < 0$,

$$P_{\mu}(\mu \in C_{a}(X)) = P_{\mu}(\min\{0,X-a\} \le \mu)$$

$$= P_{\mu}(X-a\le\mu) = P_{\mu}(X-\mu\le a) = 1-\alpha$$

if $a = z_{\alpha}$. For $\mu > 0$,

$$P_{\mu}(\mu \in C_{a}(X)) = P_{\mu}(\max\{0,X-a\} \ge \mu)$$

$$= P_{\mu}(X+a\ge\mu) = P_{\mu}(X-\mu\ge-a) = 1-\alpha$$

if $a = z_{\alpha}$.

b.

For $\pi(\mu) \equiv 1$, $f(\mu\vert x) \sim N(x, 1)$.

$$P(\min\{0,x-a\}\le\mu\le\max\{0,x-a\}\vert X=x) = P(x-a \le \mu \le x+a\vert X=x)$$

$$= 1-2\alpha$$

if $a = z_{\alpha}$ and $-z_{\alpha}\le x \le z_{\alpha}$. For $a = z_{\alpha}$ and $x > z_{\alpha}$,

$$P(\min\{0,x-a\}\le\mu\le\max\{0,x-a\}\vert X=x) = P(-x \le Z \le a)$$

$$\rightarrow P(Z \le z) = 1-\alpha$$

as $x \rightarrow \infty$.

10.1

The mean is $\mu = \theta/3$, so the method of moments estimator is $W_n = 3 \overline{X}_n$. By the law of large numbers $\overline{X}_n \overset{P}{\rightarrow}\mu = \theta/3$, so $W_n = 3 \overline{X}_n \overset{P}{\rightarrow}\theta$.