Solutions

8.31

a.

The joint PMF of the data is

$$f(x\vert\lambda) = \frac{\lambda^{\sum x_{i}} e^{-n\lambda}}{\prod x_{i}!}$$

For $\lambda_{2} > \lambda_{1}$,

$$\frac{f(x\vert\lambda_{2})}{f(x\vert\lambda_{1})} = \left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{\sum x_{i}} e^{n(\lambda_{1}-\lambda_{2})}$$

is increasing in $\sum x_{i}$, so has MLR. So a test which rejects the null hypothesis if $\overline{X} > c$ is UMP of its size.

b.

If $\lambda=1$, then $\overline{X} \sim$   AN$(1,1/n)$, so $c \approx 1+z_{\alpha}/\sqrt{n}$.

If $\lambda=2$, then $\overline{X} \sim$   AN$(2,2/n)$, so

$$P(\overline{X} > 1+z_{\alpha}/\sqrt{n}\vert\lambda=2) \approx P\left(Z > \left(\frac{z_{\alpha}}{\sqrt{n}}-1\right)\sqrt{\frac{n}{2}}\right)$$

$$= P\left(Z > \frac{z_{\alpha}}{\sqrt{2}}-\sqrt{\frac{n}{2}}\right)$$

For $\alpha = 0.05$, $z_{\alpha} = 1.645$.

For $\beta(2)=0.9$,

$$\frac{z_{\alpha}}{\sqrt{2}}-\sqrt{\frac{n}{2}} = -z_{0.1} = -1.282$$

so

$$n = (z_{\alpha}+\sqrt{2}z_{1-\beta})^{2} = (1.645+\sqrt{2} \times 1.282)^{2} = 11.27$$

so this suggest using $n=12$.

It might be better to use the variance-stabilizing transformation $\sqrt{\overline{X}}$. Either way, use of the CLT is a bit questionable.

8.34

a.

$T \sim f(t-\theta)$, $T_{0} \sim f(t)$. So

$$P(T > c\vert\theta) = P(T_{0}+\theta > c)$$

This is increasing in $\theta$.

b.

First approach: MLR implies that $T > c$ is UMP of $\theta_{1}$ against $\theta_{2}$ for size $\alpha=P(T>c\vert\theta_{1})$. Since $\phi'(t) \equiv \alpha$ is size $\alpha$ and $\beta'(t)=E[\phi'(T)\vert\theta] = \alpha$ for all $\theta$, UMP implies that $\beta(\theta_{2}) \ge \beta'(\theta_{2}) = \alpha = \beta(\theta_{1})$.

Second approach: Let $\alpha=P(T>c\vert\theta_{1})$, $h(t) = g(t\vert\theta_{2})/g(t\vert\theta_{1})$. Then

$$P(T>c\vert\theta_{2})-\alpha = E[\phi(T)-\alpha\vert\theta_{2}]$$

$$= \frac{E[(\phi(T)-\alpha)h(T)\vert\theta_{1}]}{E[h(T)\vert\theta_{1}]}$$

$$\ge \frac{h(c)E[\phi(T)-\alpha\vert\theta_{1}]}{E[h(T)\vert\theta_{1}]}$$

$$= \frac{h(c)(\alpha-\alpha)}{E[h(T)\vert\theta_{1}]}$$

$$= 0$$

Can also go back to the NP proof.

8.49

L^ATEX2Html is giving me grief about this one-it is available in the pdf version.

a.

The $p$-value is $P(X \ge 7\vert\theta=0.5) = 1 - P(X \le 6\vert\theta = 0.5) = 0.171875$. This can be computed in R with

> 1 - pbinom(6,10,0.5)
[1] 0.171875

b.

The $p$-value is $P(X \ge 3\vert\lambda=1) = 1 - P(X \le 2\vert\lambda=1) = 0.0803014$. This can be computed in R with

> 1 - ppois(2, 1)     
[1] 0.0803014

c.

If $\lambda=1$ then the sufficient statistic $T = X_1 + X_2 + X_3$ has a Poisson distribution with mean $\lambda_T = 3$. The observed value of $T$ is $t = 9$, so the $p$-value is $P(T \ge 9\vert\lambda_T = 3) = 1 - P(T \le 8\vert\lambda_T = 3) = 0.001102488$.

8.54

a.

From problem 7.22 the posterior distribution of $\theta\vert x$ is normal with mean and variance

$$E[\theta\vert X=x] = \frac{\tau^2}{\tau^2+\sigma^2/n} \overline{x}$$

$$(\theta\vert X=x) = \frac{\tau}{1+n\tau^2/\sigma^2}$$

So

$$P(\theta \le 0\vert x) = P\left(Z \le - \frac{(\tau^2/(\tau^2+\si... ...ft(Z \ge \frac{\tau}{\sqrt{(\sigma^2/n)(\tau^2+\sigma^2/n)}}\overline{x}\right)$$

b.

The $p$-value is

$$P(\overline{X} \ge \overline{x}\vert\theta = 0) = P\left(Z \ge \frac{1}{\sigma/\sqrt{n}}\overline{x}\right)$$

c.

For $\tau = \sigma = 1$ the Bayes probability is larger than the $p$-value for $\overline{x} > 0$ since

$$\frac{1}{\sqrt{(\sigma^2/n)(1+\sigma^2/n)}} < \frac{1}{\sqrt{1/n}}$$

d

. As $n \to \infty$,

$$\frac{\tau}{\sqrt{(\sigma^2/n)(\tau^2+\sigma^2/n)}}\overline{x} =... ...}{\sqrt{(\sigma^2/n)(1+\sigma^2/(\tau^2n))}}\overline{x} \to \frac{1}{\sigma/n}$$

and therefore $P(\theta \le 0\vert x)$ converges to the $p$-value.

8.55

L^ATEX2Html is giving me grief about this one-it is available in the pdf version.

8.56

L^ATEX2Html is giving me grief about this one-it is available in the pdf version.