Solutions

7.22

We have

$$\overline{X}\vert\theta \sim N(\theta,\sigma^{2}/n)$$

$$\theta \sim N(\mu,\tau^{2})$$

a.

The joint density of $\overline{X}, \theta$ is

$$f(\overline{x},\theta) = f(\overline{x}\vert\theta) f(\theta) \pr... ...igma^{2}}(\overline{x}-\theta)^{2} -\frac{1}{2\tau^{2}}(\theta-\mu)^{2}\right\}$$

This is a joint normal distribution. The means and variances are

$$E[\theta] = \mu E[\overline{X}] = E[\theta] = \mu$$

$$(\theta) = \tau^{2} (\overline{X}) = (\theta) + \frac{\sigma^{2}}{n} = \tau^{2} + \frac{\sigma^{2}}{n}$$

The covariance and correlation are

$$(\theta,\overline{X}) = E[\overline{X}\theta]-\mu^{2} = E[\theta^{2}]-\mu^{2} = \mu^{2}+\tau^{2}-\mu^{2} = \tau^{2}$$

$$\rho = \frac{\tau^{2}}{\sqrt{(\tau^{2}+\sigma^{2}/n)\sigma^{2}/n}}$$

b.

This means that the marginal distribution of $\overline{X}$ is $N(\mu, \tau^{2}+\sigma^{2}/n)$.

c.

The posterior distribuiton of $\theta$ is

$$f(\theta\vert\overline{x}) \propto \exp\left\{-\frac{n}{2\sigma^{2}}(\overline{X}-\theta)^{2} -\frac{1}{2\tau^{2}}(\theta-\mu)^{2}\right\}$$

$$\propto \exp\left\{ \left(\frac{n}{\sigma^{2}}\overline{x}+\frac{... ...ac{1}{2} \left(\frac{n}{\sigma^{2}}+\frac{1}{\tau^{2}}\right)\theta^{2}\right\}$$

This is a normal distribution with mean and variance

$$(\theta\vert\overline{X}) = \left(\frac{n}{\sigma^{2}}+\frac{1}{\tau^{2}}\right)^{-1} = \frac{\tau^{2}\sigma^{2}/n}{\tau^{2}+\sigma^{2}/n}$$

$$E[\theta\vert\overline{X}] = (\theta\vert\overline{X}) \left(\frac{n}{\sigma^{2}}\overline{X}+... ...u^{2}+\sigma^{2}/n}\overline{X} + \frac{\sigma^{2}/n}{\tau^{2}+\sigma^{2}/n}\mu$$

7.23

We have $S^{2}\vert\sigma^{2} \sim$   Gamma$((n-1)/2, 2 \sigma^{2}/(n-1))$ and

$$f(\sigma^{2}) = \frac{1}{\Gamma(\alpha)\beta^{\alpha}} \frac{1}{(\sigma^{2})^{\alpha+1}}e^{-1/(\beta\sigma^{2})}$$

The posterior distribution $\sigma^{2}\vert S^{2}$ is therefore

$$f(\sigma^{2}\vert s^{2}) \propto \frac{1}{(\sigma^{2})^{(n-1)/2}}e^{-s^{2}(n-1)/(2\sigma^{2})} \frac{1}{(\sigma^{2})^{\alpha+1}}e^{-1/(\beta\sigma^{2})}$$

$$= (\alpha+(n-1)/2, (1/\beta+(n-1)s^{2}/2)^{-1})$$

If $Y \sim$   IG$(a,b)$, then $V = 1/Y \sim$   Gamma$(a,b)$. So

$$E[Y] = E[1/V] = \int_{0}^{\infty}\frac{1}{v}\frac{1}{\Gamma(a)b^{a}}v^{a-1}e^{-v/b}dv$$

$$= \frac{1}{b\Gamma(a)}\int_{0}^{\infty}z^{a-2}e^{-z}dz$$

$$= \frac{\Gamma(a-1)}{b\Gamma(a)} = \frac{1}{b(a-1)}$$

So the posterior mean of $\sigma^{2}$ is

$$E[\sigma^{2}\vert S^{2}] = \frac{1/\beta + (n-1)S^{2}/2}{\alpha+(n-3)/2}$$

7.33

From Example 7.3.5 the MSE of $\widehat{p}_B$ is

$$E[(\widehat{p}_B-p)^2] = \frac{n p (1-p)}{(\alpha + \beta + n)^2} + \left(\frac{np + \alpha}{\alpha+\beta+n} - p\right)^2$$

$$= \frac{n p (1-p)}{(\sqrt{n/4} + \sqrt{n/4} + n)^2} + \left(\frac{np + \sqrt{n/4}}{\sqrt{n/4}+\sqrt{n/4}+n} - p\right)^2$$

$$= \frac{np(1-p) + (np + \sqrt{n/4} - p(\sqrt{n} + n))^2}{(\sqrt{n} + n)^2}$$

$$= \frac{np(1-p)+(\sqrt{n/4}-p\sqrt{n})^2}{(\sqrt{n} + n)^2}$$

$$= \frac{n}{(\sqrt{n} + n)^2}\left(p(1-p)+(1/2-p)^2\right)$$

$$= \frac{n}{(\sqrt{n} + n)^2}\left(p -p^2+1/4 + p^2 - p)^2\right)$$

$$= \frac{n/4}{(\sqrt{n} + n)^2}$$

which is constant in $p$.

7.38

a.

The population density is

$$\theta x^{\theta-1} = \theta x^{-1}e^{\theta\log x}$$

So $T(X) = \frac{1}{n}\log X_{i}$ is efficient for $\tau(\theta)=E_{\theta}[\log X_{1}]$.

$$\tau(\theta) = \int_{0}^{1}\log x \theta x^{\theta-1}dx$$

$$= - \int_{0}^{\infty} y \theta e^{-\theta y} dy = 1/\theta$$

b.

The population density is

$$\frac{\log \theta}{\theta -1} \theta^{x} = \frac{\log \theta}{\theta-1}e^{x\log\theta}$$

So

$$\sum \log f(x_{i}\vert\theta) = n(\log\log\theta-\log(\theta-1))+\sum x_{i}\log\theta$$

$$\sum\frac{\partial}{\partial\theta}\log f(x_{i}\vert\theta) = n\left(\frac{1}{\theta\log\theta}-\frac{1}{\theta-1}+\frac{\overline{x}}{\theta}\right)$$

$$= \frac{n}{\theta}\left(\frac{1}{\log\theta}-\frac{\theta}{\theta... ...t(\overline{x}-\left(\frac{\theta}{\theta-1}-\frac{1}{\log\theta}\right)\right)$$

So $\overline{X}$ is efficient for $\tau(\theta) = \frac{\theta}{\theta-1}-\frac{1}{\log\theta}$

7.39

Done in class.