Solutions

6.3

The joint density for the data is

$$f({x}_{1},\ldots,{x}_{n}\vert\mu,\sigma) = \frac{1}{\sigma^{n}}\exp\{-\sum(x_{i}-\mu)/\sigma\} \prod 1_{(\mu,\infty)}(x_{i})$$

$$= \frac{1}{\sigma^{n}}\exp\left\{-\frac{n}{\sigma}\overline{x}+\frac{n\mu}{\sigma}\right\}1_{(\mu,\infty)}(x_{(1)})$$

So, by the factorization criterion, $(\overline{X},X_{(1)})$ is sufficient for $(\mu,\sigma)$.

6.6

The density of a single observation is

$$f(x\vert\alpha,\beta) = \frac{1}{\Gamma(\alpha)\beta^{\alpha}} x^{\alpha-1}e^{-x/\beta}$$

$$= \frac{1}{\Gamma(\alpha)\beta^{\alpha}} \exp\left\{(\alpha-1)\log x -\frac{1}{\beta}x\right\}$$

This is a two-parameter exponential family, so $(T_{1},T_{2}) = (\sum \log X_{i},\sum X_{i})$ is sufficient for $(\alpha,\beta)$. Since $\prod X_{i}$ is a one-to-one function of $\sum\log X_i$, the pair $(\prod X_{i},\sum X_{i})$ is also sufficient for $(\alpha,\beta)$.

6.9

a.

Done in class already.

b.

$\Theta=\mathbb{R}, \mathcal{X}=\mathbb{R}^{n}$, and

$$f(x\vert\theta)=e^{-\sum x_{i}+n\theta} 1_{(\theta,\infty)}(x_{(1)})$$

Suppose $x_{(1)} = y_{(1)}$. Then

$$f(x\vert\theta)=\exp\{\sum y_{i}-\sum x_{i}\}f(y\vert\theta)$$

for all $\theta$. So $k(x,y)=\exp\{\sum y_{i}-\sum x_{i}\}$ works.

Suppose $x_{(1)} \neq y_{(1)}$. Then for some $\theta$ one of $f(x\vert\theta), f(y\vert\theta)$ is zero and the other is not. So no $k(x,y)$ exists.

So $T(X)=X_{(1)}$ is minimal sufficient.

c.

$\Theta=\mathbb{R}, \mathcal{X}=\mathbb{R}^{n}$. The support does not depend on $\theta$ so we can work with ratios of densities. The ratio of densities for two samples $x$ and $y$ is

$$\frac{f(x\vert\theta)}{f(y\vert\theta)} = \frac{e^{-\sum(x_{i}-\theta)}}{e^{-\sum(y_{i}-\theta)}} \frac{\prod(1+e^{-(y_{i}-\theta)})^{2}}{\prod(1+e^{-(x_{i}-\theta)})^{2}}$$

$$= \frac{e^{-\sum x_{i}}}{e^{-\sum y_{i}}} \frac{\prod(1+e^{-y_{i}}e^{\theta})^{2}}{\prod(1+e^{-x_{i}}e^{\theta})^{2}}$$

If the two samples contain identical values, i.e. if they have the same order statistics, then this ratio is constant in $\theta$.

If the ratio is constant in $\theta$ then the ratio of the two product terms is constant. These terms are both polnomials of degree $2n$ in $e^{\theta}$. If two polynomials are equal on an open subset of the real line then they are equal on the entire real line. Hence they have the same roots. The roots are $\{-e^{x_{i}}\}$ and $\{-e^{y_{i}}\}$ (each of degree 2). If those sets are equal then the sets of sample values $\{x_i\}$ and $\{y_i\}$ are equal, i.e. the two samples must have the same order statistics.

So the order statistics $(X_{(1)}, \ldots,X_{(n)})$ are minimal sufficient.

d.

Same idea:

$$\frac{f(x\vert\theta)}{f(y\vert\theta)} = \frac{\prod(1+(y_{i}-\theta)^{2})}{\prod(1+(x_{i}-\theta)^{2})}$$

If the two samples have the same order statistics then the ratio is constant. If the ratio is constant for all real $\theta$ then two polynomials in $\theta$ are equal on the complex plane, and so the roots must be equal. The roots are the complex numbers

$$\theta = x_{j}\pm i, \theta = y_{j}\pm i$$

with $i = \sqrt{-1}$. So again the order statistics are minimal sufficient.

e.

$\Theta=\mathbb{R}, \mathcal{X}=\mathbb{R}^{n}$. The support does not depend on $\theta$ so we can work with ratios of densities. The ratio of densities for two samples $x$ and $y$ is

$$\frac{f(x\vert\theta)}{f(y\vert\theta)} = \exp\{\sum\vert y_{i}-\theta\vert-\sum\vert x_{i}-\theta\vert\}$$

If the order statistics are the same then the ratio is constant. Suppose the order statistics differ. Then there is some open interval $I$ containing no $x_{i}$ and no $y_{i}$ such that $\char93 \{x_{i} > I\} \ne \char93 \{y_{i} > I\}$. The slopes on $I$ of $\sum\vert x_{i}-\theta\vert$ and $\sum\vert y_{i}-\theta\vert$ as functions of $\theta$ are

$$n-2(\char93 \{x_{i} > I\}), n-2(\char93 \{y_{i} > I\})$$

So $\sum\vert y_{i}-\theta\vert-\sum\vert x_{i}-\theta\vert$ has slope

$$2(\char93 \{x_{i} > I\}-\char93 \{y_{i} > I\}) \ne 0$$

and so the ratio is not constant on $I$.

So again the order statistic is minimal sufficient.

6.10

To thos that the minimal sufficient statistic is not complete we need to fins a function $g$ that is not identically zero but has expected value zero for all $\theta$. Now

$$E[X_{(1)}] = \theta + \frac{1}{n+1}$$

$$E[X_{(n)}] = \theta + \frac{n}{n+1}$$

So $g(X_{(1)},X_{(n)})=X_{(n)}-X_{(1)}-\frac{n-1}{n+1}$ has expected value zero for all $\theta$ but is not identically zero for $n > 1$.

6.14

${X}_{1},\ldots,{X}_{n}$ are $i.i.d.$ from $f(x-\theta)$. This means $Z_{i}=X_{i}-\theta$ are $i.i.d.$ from $f(z)$. Now

$$\widetilde{X} = \widetilde{Z}+\theta$$

$$\overline{X} = \overline{Z}+\theta$$

So $\widetilde{X}-\overline{X} = \widetilde{Z}-\overline{Z}$ is ancillary.

6.20

a.

The joint density of the data is

$$f({x}_{1},\ldots,{x}_{n}\vert\theta)=2^{n}\left(\prod x_{i}\right) 1_{(0,\theta)}(x_{(n)})\frac{1}{\theta^{2n}}$$

$T=X_{(n)}$ is sufficient (and minimal sufficient). $X_{(n)}$ has density

$$f_{T}(t\vert\theta) = 2nt^{2n-1}\frac{1}{\theta^{2n}}1_{(0,\theta)}(t)$$

Thus

$$0 = \int_{0}^{\theta}g(t)\frac{2n}{\theta^{2n}} t^{2n-1}dt$$

for all $\theta > 0$ means

$$0 = \int_{0}^{\theta}g(t)t^{2n-1}dt$$

for almost all $\theta > 0$, and this in turn implies $g(t)t^{2n-1}=0$ and hence $g(t)=0$ for all $t > 0$. So $T=X_{(n)}$ is complete.

b.

Exponential family, $T(X)=\sum \log(1+X_{i})$,

$$\{w(\theta):\theta \in \Theta\} = (1,\infty)$$

which is an open interval.

c.

Exponential family, $T(X)=\sum X_{i}$,

$$\{w(\theta): \theta \in \Theta\} = \{\log \theta: \theta > 1\} = (0,\infty)$$

which is an open interval.

d.

Exponential family, $T(X)=\sum e^{-X_{i}}$,

$$\{w(\theta): \theta \in \Theta\} = \{-e^{\theta}: \theta \in \mathbb{R}\} = (-\infty,0)$$

which is an open interval.

e.

Exponential family, $T(X)=\sum X_{i}$,

$$\{w(\theta): \theta \in \Theta\} = \{\log \theta-\log(1-\theta): 0 \le \theta \le 1\} = [-\infty,\infty]$$

which contains an open interval.