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I've been asked to show the gamma distribution can be written in the form $p(x|\alpha, \beta) = f(x) g(\alpha, \beta) e^{h(\alpha,\beta)^T T(x)}$ where $T(x)$ is a sufficient statistic.

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I have done the following:

$p(x|\alpha,\beta) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} = \frac{\beta^\alpha}{\Gamma(\alpha)} exp[(\alpha-1)\log(x) - \beta x]$ and so identified $g(\alpha,\beta) = \frac{\beta^\alpha}{\Gamma(\alpha)}, f(x)=1, h(\alpha,\beta) = \begin{pmatrix} \alpha-1 \\ \beta \end{pmatrix}, T(x) = \begin{pmatrix} \log{x} \\ x \end{pmatrix}$ which I believe is correct.

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However, I do not understand why I cannot take the original form and identify $g(\alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)}, f(x) = x^{\alpha-1}, h(\alpha,\beta) = - \beta, T(x) = x$.

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My question is essentially in two parts:

1, Presumably this is because the question demands that $T(x)$ be a sufficient statistic. Now I understand that a sufficient statistic is basically all the information required so that we can compute the parameters $\alpha$ and $\beta$. Is that correct? Is there a better way to think about it?

2, Assuming that is correct, my problem reduces to determining $T(x)$ and justifying that my first approach is correct. I have read online that $T(x)$ is determined by finding the conditions necessary for $\frac{p(x|\alpha,\beta)}{p(y|\alpha,\beta)}$ to be independent of both $\alpha$ and $\beta$. I don;t really understand why though? Can someone explain?

Moving on, we get

$\frac{p(x|\alpha,\beta)}{p(y|\alpha,\beta)} = \frac{\frac{\beta^\alpha}{\Gamma(\alpha} x^{\alpha-1}e^{-\beta x}}{\frac{\beta^\alpha}{\Gamma(\alpha)} y^{\alpha-1}e^{-\beta y}} = \frac{x^{\alpha-1}}{y^{\alpha-1}} e^{-\beta(x-y)}$

At this point is seems to me that if I choose $x=y$ I can make this independent of both $\alpha$ and $\beta$ and so the sufficient statistic should just be $T(x)=x$ but this is definitely wrong.

I am just learning this by myself and haven't found many good resources online. If someone could fix my problem and also direct me to some good online notes or a good book I'd really appreciate it.

Thanks!

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I do not understand why I cannot take the original form and identify $$g(\alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)}, f(x) = x^{\alpha-1}, h(\alpha,\beta) = - \beta, T(x) = x$$

The reason for this decomposition to be incorrect is that $x^{\alpha-1}$ depends on the parameter $(\alpha,\beta)$, thus that there is no separation from the parameter.

That $T(\cdot)$ is a sufficient statistic proceeds from the factorisation theorem: $$p(x|\alpha, \beta) = f(x) g(\alpha, \beta) e^{h(\alpha,\beta)^\text{T} T(x)}$$shows that the density of $X$ factorises as a function of $x$, $f(x)$, and a function of $T(x)$ only, namely $g(\alpha, \beta) \exp\{h(\alpha,\beta)^\text{T} T(x)\}$

I have read online that $𝑇(𝑥)$ is determined by finding the conditions necessary for $𝑝(𝑥|𝛼,𝛽)/𝑝(𝑦|𝛼,𝛽)$ to be independent of both $𝛼$ and $𝛽$

This is incorrect: the ratio is independent of the parameter value when $T(x)=T(y)$, again by the factorisation theorem

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  • $\begingroup$ Hi Xi'an, thanks for your reply. I've added the tag. I can see that $x$ and $\alpha$ are connected from the $x^{\alpha-1}$ term and thus my $f(x)$ would actually be $f(x,\alpha)$ which is not allowed. However, I do not understand the rest of your post. This is probably a more serious misunderstanding of how to calculate sufficient statistics. Could you show me the details of how to find $T(x)$ or do you have a link I could read up on them that explains them clearly? Thanks. $\endgroup$ – user11128 Feb 13 at 13:04
  • $\begingroup$ I do not know how to explain the derivation by anything but an application of the factorisation theorem or of the exponential decomposition in the first equation of the question. It seems to me that every mathematical statistics book addresses this topic in a reasonable manner. $\endgroup$ – Xi'an Feb 13 at 13:37

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