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

Question

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 $\theta$.

Answer 1

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 $\beta$.

Answer 2

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.