Solutions

8.14

Use $R = \{x: \sum x_{i} > c\}$. $\alpha=0.01$ means

$$0.01 = P(\sum X_{i} > c\vert p=0.49) \approx P\left(Z > \frac{c-0.49}{\sqrt{0.49 \times 0.51}}\sqrt{n}\right)$$

So

$$\frac{c-0.49}{\sqrt{0.49 \times 0.51}}\sqrt{n} = 2.326$$

$\beta=0.99$ implies

$$0.99 = P(\sum X_{i} > c\vert p=0.51) \approx P\left(Z > \frac{c-0.51}{\sqrt{0.49 \times 0.51}}\sqrt{n}\right)$$

So

$$\frac{c-0.51}{\sqrt{0.49 \times 0.51}}\sqrt{n} = -2.326$$

So

$$c-0.49 = 2.326\sqrt{0.49 \times 0.51}\frac{1}{\sqrt{n}}$$

$$c-0.51 = -2.326\sqrt{0.49 \times 0.51}\frac{1}{\sqrt{n}}$$

or

$$\sqrt{n} \times 0.002 = 2 \times 2.326 \sqrt{0.49 \times 0.51}$$

$$n = (100)^{2} \times (2.326)^{2} \times 0.49 \times 0.51$$

$$= 13521$$

8.17

For ${X}_{1},\ldots,{X}_{n}$ $i.i.d.$ Beta($\mu,1$)

$$L(\mu\vert x) = \mu^{n} \prod x_{i}^{\mu-1}$$

$$= \mu^{n}e^{\mu\sum\log x_{i}} e^{-\sum\log x_{i}}$$

So

$$\widehat{\mu} = -\frac{n}{\sum \log x_{i}}$$

$$L(\widehat{\mu}\vert x) = \left(-\frac{n}{\sum\log x_{i}}\right)^{n}\exp\{-n-\sum\log x_{i}\}$$

and

$$\Lambda(x) = \frac{\left(-\frac{n+m}{\sum\log x_{i}+\sum\log y_{i}}\right)^{... ...og x_{i}\} \left(-\frac{m}{\sum\log y_{i}}\right)^{m}\exp\{-m-\sum\log y_{i}\}}$$

$$= \frac{(n+m)^{n+m}}{n^{n}m^{m}}T^{n}(1-T)^{m}$$

So

$$\{\Lambda < c\} = \{ T < c_{1} T > c_{2} \}$$

for suitable $c_{1}, c_{2}$.

Under $H_{0}$, $-\log X_{i},-\log Y_{i}$ are $i.i.d.$ exponential, so $T \sim$   Beta$(n,m)$.

To find $c_{1}$ and $c_{2}$ either cheat and use equal tail probabilities (the right thing to do by symmetry if $n=m$), or solve numerically.

8.15

$$L(\sigma^{2}\vert x) = \frac{1}{(\sigma^{2})^{n/2}} \exp\left\{-\frac{1}{2\sigma^{2}}\sum x_{i}^{2}\right\}$$

$$\frac{L(\sigma_{1}^{2})}{L(\sigma_{0}^{2})} = \left(\frac{\sigma_{0}}{\sigma_{1}}\right)^{n/2} \exp\left\{\fr... ..._{i}^{2} \left(\frac{1}{\sigma_{0}^{2}}-\frac{1}{\sigma_{1}^{2}}\right)\right\}$$

If $\sigma_{1} > \sigma_{2}$ then this is increasing in $\sum x_{i}^{2}$. So

$$L(\sigma_{1}^{2})/L(\sigma_{0}^{2}) > k$$

for some $k$ if and only if $\sum X_{i}^{2} > c$ for some $c$.

Under $H_{0}:\sigma=\sigma_{0}$, $\sum X_{i}^{2}/\sigma_{0}^{2} \sim \chi^{2}_{n}$, so

$$c = \sigma_{0}^{2} \chi^{2}_{n,\alpha}$$

8.25

a.

For $\theta_2 > \theta_1$

$$\frac{g(x\vert\theta_2)}{g(x\vert\theta_1)} = \frac{\exp\left\{-\... ...2}{2\sigma^2}\right\}} {\exp\left\{-\frac{(x-\theta_1)^2}{2\sigma^2}\right\}} = \times \exp\{x(\theta_2 - \theta_1)/\sigma^2\}$$

This in increasing in $x$ since $\theta_2 - \theta_1 > 0$.

b.

For $\theta_2 > \theta_1$

$$\frac{g(x\vert\theta_2)}{g(x\vert\theta_1)} = \frac{\theta_2^x e^{-\theta_2}/ x!}{\theta_1^x e^{-\theta_1}/ x!} = \times \left(\frac{\theta_2}{\theta_1}\right)^x$$

which is increasing $x$ since $\theta_2/\theta_2 > 1$.

c.

For $\theta_2 > \theta_1$

$$\frac{g(x\vert\theta_2)}{g(x\vert\theta_1)} = \frac{\binom{n}{x}\theta_2^s(1-\theta_2)^{n-x}} {\binom{n}{x}\theta_1^s(1-\theta_1)^{n-x}} = \times \left(\frac{\theta_2/(1-\theta_2)}{\theta_1/(1-\theta_1)}\right)^x$$

This is increasing in $x$ since $\theta/(1-\theta)$ is increasing in $\theta$.

8.28

a.

$$\frac{f(x\vert\theta_{2})}{f(x\vert\theta_{1})} = e^{\theta_{1}-\theta_{2}} \frac{(1+e^{x-\theta_{1}})^{2}}{(1+e^{x-\theta_{2}})^{2}} = \times \left(\frac{e^{\theta_{1}}+e^{x}}{e^{\theta_{2}}+e^{x}}\right)^{2}$$

Let

$$g(y) = \frac{A+y}{B+y}$$

$$g'(y) = \frac{B+y-A-y}{(B+y)^{2}} = \frac{B-A}{(B+y)^{2}}$$

Then $g'(y) \ge 0$ if $B \ge A$. So we have MLR in $x$.

b.

Since the ratio is increasing in $x$, the most powerful test is of the form $R = \{x > c \}$. Now

$$F_{\theta}(x) = 1-\frac{1}{1+e^{x-\theta}}$$

So for $H_{0}: \theta = 0$ and $\alpha=0.2 = 1/(1+e^{c})$, so

$$1+e^{c} = 5$$

$$e^{c} = 4$$

$$c = \log 4 = 1.386$$

The power is

$$\beta(1) = \frac{1}{1+e^{\log(4) -1}} = \frac{1}{1+4/e} = 0.405$$

c.

Since we have MLR, the test is UMP. This is true for any $\theta_{0}$. This only works for $n=1$; otherwise there is no one-dimensional sufficient statistic.

8.33

a.

$$P( Y_{1} > k Y_{n} > 1 \vert\theta=0) = P(Y_{1} > k\vert\theta=0) = (1-k)^{n} = \alpha$$

So $k = 1-\alpha^{1/n}$.

b.

$$\beta(\theta) = P( Y_{n} > 1 Y_{1} > k \vert\theta)$$

$$= P(Y_{n} > 1\vert\theta) + P( Y_{1} > k Y_{n} \le 1 )$$

$$= \begin{cases}1 & \theta > 1\\ 1-(1-\theta)^{n} + (1-\max\{k,\theta\})^{n} & \theta \le 1 \end{cases}$$

c.

$$f(x\vert\theta) = 1_{(\theta,\infty)}(Y_{1})1_{(-\infty,\theta+1)}(Y_{n})$$

Fix $\theta' > 0$. Suppose $k \le \theta'$. Then $\beta(\theta') = 1$.

Suppose $k >\theta'$. Take $k'=1$ in the NP lemma. Then

$$f(x\vert\theta') < f(x\vert\theta_{0}) \Rightarrow 0 < Y_{1} < \theta' < k x \not \in R$$

$$f(x\vert\theta') > f(x\vert\theta_{0}) \Rightarrow 1 < Y_{n} < \theta'+1 x \in R$$

So $R$ is a NP test for any $\theta'$.

So $R$ is UMP.

d.

The power is one for all $n$ if $\theta > 1$.