I have trouble finding the following sufficient statistics.
How do you do this?
$$X\sim \Gamma(\alpha, \beta)$$ $$f(x;\alpha, \beta)=\frac{e^{-x/\beta}x^{\alpha-1}}{\Gamma(\alpha)\beta^\alpha}$$

Question: Is $log(X_1+X_2)$ a sufficient statistic for beta?

I am using the first way, conditional probability, but how to you plug it in and reduce? But there's a log in it.

I understand there are two ways to find it:

  1. Sufficiency principle: $$P(X_1=x_1, X_2=x_2,...X_n=x_n|T(X_1,...,X_n))$$ does not depend on $\theta$.
  2. Factorization theorem: $$P_\theta(x_1, x_2,...,x_n)=h(x_1, x_2,...x_n)g_\theta(T(x_1,x_2,...,x_n))$$ where $h$ only depends on the observation vector $(x_1,...x_n)$ and not $\theta$.
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    $\begingroup$ This looks like a standard textbook problem. If this is for some subject, or otherwise for the purpose of your own study, would you mind adding the self-study tag, please? $\endgroup$ – Glen_b Nov 22 '13 at 0:37

Sufficient for what? $\beta$ or $\alpha$?

If your sample size is 2, then you can quite easily show that $X_1+X_2$ is sufficient for $\beta$. Furthermore, any 1-1 function of a sufficient stats is itself a sufficient stats. therefore $\log(X_1+X_2)$ will be sufficient for $\beta$.

  • $\begingroup$ Yes, for $\beta$. How do you show? Way 1 or way 2? $\endgroup$ – user13985 Nov 22 '13 at 0:06
  • $\begingroup$ Factorization. In case this is hw. DIY the rest mate. $\endgroup$ – qoheleth Nov 22 '13 at 0:14
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    $\begingroup$ If the answer helped you solve your problem, you might consider indicating that it was helpful. $\endgroup$ – Glen_b Nov 22 '13 at 0:36
  • $\begingroup$ $X_1+X_2\sim\Gamma(2\alpha, \beta)$ Where is the sufficient stat? Like the sum of $X_i$ you usually see. $\endgroup$ – user13985 Nov 22 '13 at 0:46
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    $\begingroup$ I simply googled for some examples. I think the example in p.4 here will be helpful. stat.wisc.edu/courses/st312-rich/suff2.pdf $\endgroup$ – qoheleth Nov 22 '13 at 1:56

This is what I got, does it look correct? $$f(X_1, X_2|T=x_1+x_2)$$ $$=\frac{e^{\frac{-x_1}{\beta}}x_1^{\alpha-1}}{\Gamma(\alpha)\beta^\alpha} \frac{e^{\frac{-x_2}{\beta}}x_2^{\alpha-1}}{\Gamma(\alpha)\beta^\alpha}$$ $$=\frac{e^{\frac{-\sum{x_i}}{\beta}}(x_1x_2)^{\alpha-1}}{\Gamma(\alpha)^2\beta^{2\alpha}}$$

So, we have
$h(x_1,x_2)=(x_1x_2)^{\alpha-1}$, and $g_\beta(T(x_1,x_2))=\frac{1}{\Gamma(\alpha)^2\theta^{2\alpha}}e^{-\sum{x_i}/\beta}$

By the factorization theorem, $X_1+X_2$ is a sufficient statistic for $\beta$.

Since $log(\cdot)$ is a one to one function. We say $log(X_1+X_2)$ is a sufficient statistic.

  • $\begingroup$ The solution is incorrect as the first equation is the joint density of $(X_1,X_2)$ and not the conditional density. $\endgroup$ – Xi'an Dec 6 '17 at 19:05

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