Show that the sum of the observations of a random sample of size $n$ from a gamma distribution that has pdf $f(x,\theta)=(1/\theta)e^{-x/\theta}$, $0<x<\infty$, $0<\theta<\infty$, zero elsewhere, is a sufficient statistic for $\theta$.
This is what I did (but it's wrong, I don't know why):
$$Y=X_1 + X_2 + \cdots + X_n$$
$$f(y,\theta)= \left(\frac{1}{\theta}\right)e^{-y/(\theta)}$$
$$f(x_1;\theta)f(x_2; \theta)\cdots f(x_n,\theta) = \left(\frac{1}{\theta}\right)^n e^{\frac{-x_1-x_2-\cdots-x_n}{\theta}}$$
So…
$$\frac{f(x_1;\theta)f(x_2; \theta)\cdots f(x_n,\theta)}{f(y,\theta)} = \frac{(\frac{1}{\theta})^n e^{\frac{-x_1-x_2-\cdots-x_n}{\theta}}}{(\frac{1}{\theta})e^{-y/\theta}}$$
But that's equal to $(1/\theta)^{n-1}$, which does include $\theta$, so how is it sufficient? I'm sure I did a mistake but I don't know where…
Thanks in advance
Best Answer
\begin{eqnarray*} f \left( x_1, \ldots, x_n ; \theta \right) & = & \theta^{- n} \exp \left( - \frac{1}{\theta} \left( x_1 + \cdots + x_n \right) \right) \end{eqnarray*} From the last expression, you could see that this is an exponential family with parameter $- \frac{1}{\theta}$, sufficient statistic $x_1 + \cdots + x_n$ and normalizing constant $\theta^{- 1}$. You could appeal directly to Fisher-Neyman factorization theorem immediately here.