## sufficient statistics and MVUE's

Q1: Suppose that $\displaystyle Y_{1},...,Y_{n}$ is a random sample from a probability density in the (one-parameter) exponential family so that

$\displaystyle f(y|\theta)=a(\theta)b(y)e^{-[c(\theta)d(y)]}I_{[a,b]}(y)$, where $\displaystyle I_{[a,b]}(y)$ is the indicator function and $\displaystyle a$ and $\displaystyle b$ do not depend on $\displaystyle \theta$. Show that $\displaystyle \sum_{i=1}^{n}d(Y_{i})$ is sufficient for $\displaystyle \theta$.

A1:

$\displaystyle L(y_{1},...,y_{n}|\theta)=f(y_{1},...,y_{n}|\theta )$$\displaystyle =\Pi_{i=1}^{n}a(\theta)b(y)e^{-[c(\theta)d(y)]}I_{[a,b]}(y)$=$\displaystyle [a(\theta)b(y)]^{n}exp[-c(\theta)\sum_{i=1}^{n}d(y_{i})]\Pi_{i=1}^{n}I_{[a,b]}(y_{i})$.

If I let $\displaystyle h(y_{1},...,y_{n})=\Pi_{i=1}^{n}I_{[a,b]}(y_{i})$, which does not depend on $\displaystyle \theta$ and $\displaystyle g(\sum_{i=1}^{n}d(y_{i}),\theta)=[a(\theta)b(y)]^{n}exp[-c(\theta)\sum_{i=1}^{n}d(y_{i})]$, where $\displaystyle \theta$ interacts with the data $\displaystyle y_{i}$ only through $\displaystyle U=\sum_{i=1}^{n}d(y_{i})$. So, by the factorization criterion, $\displaystyle U=\sum_{i=1}^{n}d(y_{i})$ is sufficient for $\displaystyle \theta$.

Q2: Let $\displaystyle Y_{1},...,Y_{n}$ denote a random sample from the probability density function

$\displaystyle f(y|\theta)=\theta\\y^{\theta\\-1}$ for $\displaystyle 0<y<1$, $\displaystyle \theta>0$ and $\displaystyle 0$ otherwise.

a) Show that this density function is in the (one-parameter) exponential family and that $\displaystyle \sum_{i=1}^{n}-ln(Y_{i})$ is sufficient for $\displaystyle \theta$.

b) If $\displaystyle W_{i}=-ln(Y_{i})$, show that $\displaystyle W_{i}$ has an exponential distribution with mean $\displaystyle \frac{1}{\theta}$.

Furthermore, in Q2, the book tells us to refrence back to Q1 as a hint.

A2:

I am really stuck on this one. I am not sure how to show something is in the one-parameter exponential family, as I have never seen such a thing before. So, some steps might be nice to see or even an explanation of the process. Any help would be greatly appreciated.