# Is $\log(X_1+X_2)$ a sufficient statistic for $\beta$?

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 $$\theta$$.
• 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? Nov 22, 2013 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 $\beta$.

• Yes, for $\beta$. How do you show? Way 1 or way 2? Nov 22, 2013 at 0:06
• Factorization. In case this is hw. DIY the rest mate. Nov 22, 2013 at 0:14
• If the answer helped you solve your problem, you might consider indicating that it was helpful. Nov 22, 2013 at 0:36
• $X_1+X_2\sim\Gamma(2\alpha, \beta)$ Where is the sufficient stat? Like the sum of $X_i$ you usually see. Nov 22, 2013 at 0:46
• 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 Nov 22, 2013 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.

• The solution is incorrect as the first equation is the joint density of $(X_1,X_2)$ and not the conditional density. Dec 6, 2017 at 19:05