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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

Michael Hardy
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1 Answers1

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\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.

Learner
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  • Thanks a lot...but I'm sorry I think I coppied the wrong question. The pdf is $(1/\theta)e^{-x/\theta}$ instaed of $\theta^n(1-\theta)^x$. I edited my question. –  Feb 26 '13 at 12:49
  • @Artus That´s even simpler. – Learner Feb 26 '13 at 13:00
  • I think you dont need any algebra here. It is obvious that $h(x)=1$ and the other function is not not a function of any single part of $x$ – Seyhmus Güngören Feb 26 '13 at 13:11
  • @SeyhmusGüngören That's why it is a single line! – Learner Feb 26 '13 at 13:13
  • @Learner Thanks. I actually did try to use the Neyman theorem. Did you look at my answer? I was just wondering where I went wrong. I was trying to show that if we divide $f(x_1; \theta)...f(x_n,\theta)$ by $f(y,\theta)$, then we get a function that does not inlcude theta in it. But when I actually did divide them, the result was $(1/\theta)^{n-1}$ which does include $\theta$. So is it ok if you tell me where I went wrong? –  Feb 26 '13 at 13:14
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    @Artus This is because $f(y,\theta)$ is not a density (normalizing constant must be raised to the power $n$ to get a density for $y$). – Learner Feb 26 '13 at 13:18