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

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

A1:

L(y_{1},...,y_{n}|\theta)=f(y_{1},...,y_{n}|\theta )<br /> =\Pi_{i=1}^{n}a(\theta)b(y)e^{-[c(\theta)d(y)]}I_{[a,b]}(y)<br /> = [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 h(y_{1},...,y_{n})=\Pi_{i=1}^{n}I_{[a,b]}(y_{i}), which does not depend on \theta and 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 \theta interacts with the data y_{i} only through U=\sum_{i=1}^{n}d(y_{i}). So, by the factorization criterion, U=\sum_{i=1}^{n}d(y_{i}) is sufficient for \theta.

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

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

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

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