Showing that sum of samples is sufficient statistic

Question

If $X_i$ where $i=1,2,\dots,n$ is a random independent sample from $\textrm{Beta(1,0)}$ having the pdf $f_X(x \mid \theta) = \theta x^{\theta-1}, 0 < x < 1$ and transform this random variable as $Y = -\ln{X}$, I'd like to show using purely the definition of sufficiency that $\sum_i^n Y_i$ is a sufficient statistic for $\theta$.

This is very easy to do using Fisher-Neymann Factorization, but I'd prefer to show that the ratio $\frac{\Pr(X=x)}{\Pr(T(X)=t)}$ does not depend on $\theta$.

Here is what I have so far:

I have verified that $X$ is a member of the single parameter exponential family by rewriting the pdf and using the transformation method, I've computed the pdf of $Y$ as $f_Y(y \mid \theta) = \theta e^{-\theta y}$.

I note that the above pdf signifies that $Y \sim \textrm{exp}(\theta)$. Setting $T = \sum_i^n Y$ and using the moment generating function of the exponential distribution, we have that $M_Z(t) = \prod_i^n M_{Y_i}$ and from this, we see that $T \sim \textrm{Gamma}(n, \theta)$. Now taking the ratio of the pdf's, we have:

$\frac{\Pr(X=x)}{\Pr(T(X)=t)} = \frac{\prod_i f_{X_i}}{f_T} = \frac{\theta^n (\prod_i^n x_i)^{\theta-1}}{\frac{\theta^n}{\Gamma(n)}t^{n-1} e^{-t/\theta}}$

Obviously, the $\theta^n$ terms cancel and I can use the fact that I know $t = -\sum_i \ln{x_i} = -\ln{(\prod_i x_i)}$ to get everything in terms of the same variable, but I can't seem to get rid of the other $\theta$ terms. Is this the right math to be doing?

Answer 1

You need to be careful about keeping your choice of parametrization consistent. If $Y \sim \operatorname{Exponential}(\theta)$, with $f_Y(y) = \theta e^{-\theta y}$, then that is a rate parametrization; thus, the sum $T$ is a gamma random variable that is also parametrized by rate; hence $$f_T(t) = \frac{\theta^n t^{n-1} e^{-\theta t}}{\Gamma(n)}$$ is the correct PDF. Then you have your cancellation, since $\prod_i x_i = e^{-t}$ implies the numerator is $$\Pr[X = x] = \theta^n e^{-t(\theta-1)}$$ hence $$\frac{\Pr[X = x]}{\Pr[T(X) = t]} = \frac{\theta^n e^{-t(\theta-1)}}{\theta^n t^{n-1} e^{-\theta t} / \Gamma(n)} = \ldots.$$