Solutions

8.5

a.

The likelihood can be written as

$$L(\theta,\nu) = \frac{\theta^{n}\nu^{n\theta}}{\prod x_{i}^{\theta+1}} 1_{[\nu,\infty)}(x_{(1)})$$

For fixed $\theta$, this increases in $\nu$ for $\nu \le x_{(1)}$ and is then zero. So $\widehat{\nu}=x_{(1)}$, and

$$L^{*}(\theta) = \max_{\nu} L(\theta,\nu) = \theta^{n}\prod\left(\frac{x_{(1)}}{x_{i}}\right)^{\theta} \frac{1}{\prod x_{i}}$$

$$\propto \theta^{n}e^{-\theta T}$$

So $\widehat{\theta} = n/T$.

b.

The likelihood ratio criterion is

$$\Lambda(x) = \frac{L^{*}(1)}{L^{*}(\widehat{\theta})} = \frac{e^{-T}}{\left(\frac{n}{T}\right)^{n}e^{-n}} = \times T^{n}e^{-T}$$

This is a unimodal function of $T$; it increases from zero to a maximum at $T = n$ and then decreases back to zero. Therefore for any $c > 0$

$$R = \{x: \Lambda(x) < c\} = \{x: T < c_{1} T > c_{2} \}$$

c.

The conditional density of $X_2, \dots, X_n$, given $X_1 = x$ and $X_i \ge X_1$, is

$$f({x}_{2},\ldots,{x}_{n}\vert x_{1}, x_{i} \ge x_{1}) = \frac{f(x_{1}) \cdots f(x_{n})} {f(x_{1})P(X_{2}>X_{1}\vert X_{1}=x_{1}) \cdots P(X_{n}>X_{1}\vert X_{1}=x_{1})}$$

$$= \frac{f(x_{2}) \cdots f(x_{n})}{P(X_{2}>x_{1}) \cdots P(X_{n}>x_{1})}$$

and

$$P(X_{i} > y) = \int_{y}^{\infty}\frac{\theta\nu^{\theta}}{x^{\theta+1}}dx = \frac{\nu^{\theta}}{y^{\theta}}$$

So

$$f({x}_{2},\ldots,{x}_{n}\vert x_{1}, x_{i} \ge x_{1}) = \theta^{n-1}\prod_{i=2}^{n}\frac{x_{1}^{\theta}}{x_{i}^{\theta+1}} 1_{\{x_{i} > x_{1}\}}$$

Let $Y_{1} = X_{i}/x_{1}$, $i = 2,\ldots,n$. Then

$$f_{Y}({y}_{2},\ldots,{y}_{n}\vert x_{1}, x_{i} > x_{1}) = x_{1}^{n-1}f(y_{2}x_{1},\ldots,y_{n}x_{1}\vert x_{1},x_{i}>x_{1})$$

$$= \frac{\theta^{n-1}}{y_{2}^{\theta+1},\ldots,y_{n}^{\theta+1}}$$

i.e. ${Y}_{2},\ldots,{Y}_{n}$ are $i.i.d.$ with density $\theta/y^{\theta+1}$, and $T = \log Y_{2}+ \cdots + \log Y_{n}$.

If $Z = \log Y$, then

$$f_{Z}(z) = f_{Y}(y)\frac{dy}{dz} = \frac{\theta}{e^{(\theta+1)z}}e^{z} = \theta e^{-\theta z}$$

and thus $T\vert\{X_{1}=x_{1}, X_{i} > X_{1}\} \sim$   Gamma$(n-1,1/\theta)$. By symmetry, this means that $T\vert X_{(1)} \sim$   Gamma$(n-1,1/\theta)$, which is indepentent of $X_{(1)}$, so $T$ has this distribution unconditionaly as well.

For $\theta = 1$,

$$T \sim (n-1,1)$$

$$2T \sim (n-1,2) = \chi^{2}_{n-1}$$

8.6

a.

The likelihood ratio criterion is

$$\Lambda = \frac{\left(\frac{n+m}{\sum X_{i}+\sum Y_{i}}\right)^{n+m}e^{-n... ...ac{n}{\sum X_{i}}\right)^{n}e^{-n} \left(\frac{n}{\sum Y_{i}}\right)^{m}e^{-m}}$$

$$= \frac{(n+m)^{n+m}}{n^{n}m^{m}} \frac{(\sum X_{i})^{n}(\sum Y_{i})^{m}}{(\sum X_{i}+\sum Y_{i})^{n+m}}$$

The test rejects if this is small.

b.

The likelihood ratio criterion is of the form $\Lambda =$   const$\times T^{n}(1-T)^{m}$. So the test rejects if $T$ is too small or too large.

c.

Under $H_{0}$, $T \sim$   Beta$(n,m)$.