Solutions

9.1

Since $L \le U$,

$$\{L \le \theta \le U\}^{c} = \{L > \theta\} \cup \{U < \theta\}$$

Also

$$\{L > \theta\} \cap \{U < \theta\} = \emptyset$$

So

$$P(\{L \le \theta \le U\}^{c}) = P(L > \theta) + P(U < \theta) = \alpha_{1} + \alpha_{2}$$

and thus

$$P(L \le \theta \le U) = 1 - \alpha_{1} - \alpha_{2}$$

9.2

This can be interpreted conditionally or unconditionally. Given ${X}_{1},\ldots,{X}_{n}$,

$$P(X_{n+1} \in \overline{x} \pm 1.96/\sqrt{n}\vert\overline{x}) = P(Z \in \overline{x}-\theta \pm 1.96/\sqrt{n})$$

$$\le P(Z \in 0 \pm 1.96/\sqrt{n})$$

$$< 0.95 n > 1$$

$$\le 0.95 n=1$$

Equality holds only if $n=1$ and $\overline{x}=\theta$.

Unconditionally,

$$P(X_{n+1} \in \overline{X} \pm 1.96/\sqrt{n}) = P(Z-\overline{Z} \in 0 \pm 1.96/\sqrt{n}$$

$$= P\left(Z' \in 0 \pm \frac{1.96}{\sqrt{n}\sqrt{1+1/n}}\right)$$

$$= P(Z' \in 0 \pm 1.96/\sqrt{n+1})$$

$$< 0.95 n \ge 1$$

9.4

In this problem

$$X_{i} i.i.d. N(0,\sigma_{X}^{2})$$

$$Y_{i}/\sqrt{\lambda_{0}} i.i.d. N(0,\sigma_{X}^{2})$$

a.

$$\Lambda = \frac{\left(\frac{n+m}{\sum X_{i}^{2}+\sum Y_{i}^{2}/\lambda_{0... ...X_{i}^{2}}\right)^{n/2}\left(\frac{m}{\sum Y_{i}^{2}/\lambda_{0}}\right)^{m/2}}$$

$$= C T^{n/2}(1-T)^{m/2}$$

with

$$T = \frac{1}{1+\frac{\sum Y_{i}^{2}}{\lambda_{0}\sum X_{i}^{2}}} = \frac{1}{1+\frac{m}{n}F}$$

So $\Lambda < k$ if and only if $F < c_{1}$ or $F > c_{2}$.

b.

$F/\lambda_{0} \sim F_{m,n}$. Choose $c_{1}, c_{2}$ so $c_{1} = F_{m,n,1-\alpha_{1}}$, $c_{2} = F_{m,n,\alpha_{2}}$, $\alpha_{1}+\alpha_{2}=\alpha$, and $f(c_{1})=f(c_{2})$ for

$$f(t) = \left(\frac{1}{1+\frac{m}{n}t}\right)^{n/2} \left(1-\frac{1}{1+\frac{m}{n}t}\right)^{m/2}$$

c.

$$A(\lambda) = \{X,Y: c_{1} \le F \le c_{2}\}$$

$$= \left\{X,Y: c_{1} \le \frac{\sum Y_{i}^{2}/m}{\lambda\sum X_{i}^{2}/n} \le c_{2}\right\}$$

$$C(\lambda) = \left\{\lambda: \frac{\sum Y_{i}^{2}/m}{c_{2}\sum X_{i}^{2}/n} \le \lambda \le \frac{\sum Y_{i}^{2}/m}{c_{1}\sum X_{i}^{2}/n}\right\}$$

$$= \left[\frac{\sum Y_{i}^{2}/m}{c_{2}\sum X_{i}^{2}/n},\frac{\sum Y_{i}^{2}/m}{c_{1}\sum X_{i}^{2}/n}\right]$$

This is a $1-\alpha$ level CI.

9.12

All of the following are possible choices:

1. $\sqrt{n}(\overline{X}-\theta)/S \sim t_{n-1}$
2. $(n-1)S^{2}/\theta \sim \chi^{2}_{n-1}$
3. $\sqrt{n}(\overline{X}-\theta)/\sqrt{\theta} \sim N(0,1)$

The first two produce the obvious intervals.

For the third, look for those $\theta$ with

$$-z_{\alpha/2} < Q(X,\theta) = \frac{\overline{X}-\theta}{\sqrt{\theta/n}} < z_{\alpha/2}$$

If $\overline{x} \ge 0$, then $Q(x,\cdot)$ is decreasing, and the confidence set is an interval.

If $\overline{x} < 0$, then $Q(X,\theta)$ is negative with a single mode. If $\overline{x}$ is large enough (close enough to zero, then the confidence set is an interval, corresponding to the two solutions to $Q(x,\theta) = -z_{\alpha/2}$.

If $\overline{x}$ is too small, then there are no solutions and the confidence set is empty.

9.13

a.

$U = \log X \sim \exp(1/\theta)$, $\theta U \sim \exp(1)$. So

$$P(Y/2 < \theta < Y) = P(1/2 < \theta U < 1)$$

$$= e^{-1}-e^{-1/2} = 0.239$$

b.

$\theta U \sim \exp(1)$.

$$P(-\log(1-\alpha/2) < \theta U < -\log(\alpha/2)) = 1-\alpha$$

$$P\left(\frac{-\log(1-\alpha/2)}{U} < \theta < \frac{-\log(\alpha/2)}{U}\right) = 1-\alpha$$

$$[-\log(1-\alpha/2)Y,-\log(\alpha/2)Y] = [0.479 Y, 0.966 Y]$$

c.

The interval in b. is a little shorter,

$$\frac{b}{a} = \frac{0.487}{0.5}$$

though it is not of optimal length.