A random sample $X_{1},…,X_{n}$ is pulled from a gamma distribution. Are there jointly sufficient statistics based on these observations for the two unknown parameters?
The definition of a gamma distribution is that its probability density function $f$ is nonzero only for positive arguments, where it is given by
$$f(x;\alpha,\beta)=\frac{x^{\alpha-1 }}{\beta ^\alpha \Gamma(\alpha)}e^{-x/\beta}.$$
I kind of understand what a jointly sufficient statistic is; however, I am not sure what to do from here. Possibly taking the product $\prod_{i=1}^{n}$ in front of the distribution. Can anybody help? Thanks!
Best Answer
The sufficient statistics is $\left(\prod_{i=1}^n X_i, \sum_{i=1}^n X_i \right)$. To see this, note the likelihood can be expressed as \begin{eqnarray} &&L(X_1,\ldots X_n) \\ &=& \prod_{i=1}^n \frac{X_i^{\alpha-1}}{\beta^\alpha \Gamma(\alpha)} e^{-X_i/\beta} \\ &=& \frac{1}{\beta^{n\alpha} \Gamma^n(\alpha)} \left(\prod_{i=1}^nX_i\right)^{\alpha-1} e^{-\sum_{i=1}^nX_i/\beta} \end{eqnarray}
The last equation shows that you cannot factor out any terms just involving $\alpha$ and $\beta$. This means you need $\prod_{i=1}^n X_i$ and $\sum_{i=1}^n X_i$ in order to fully specify the likelihood. Knowing any additional information would not help you in defining your likelihood.
For example, suppose you have 3 observations and you know \begin{eqnarray} X_1 X_2 X_3&=& 4\\ X_1 + X_2 + X_3&=& 5 \end{eqnarray} There are infinitely many solutions to the above 2 equations as they have 3 variables. However, it doesn't matter what the individual values of your $X_i$ are, as long as they satisfy the above 2 equations you will end up with the same likelihood. That is why the term sufficient statistics is used as it is the minimal information you need for the model. Think about testing whether a coin is fair. You don't need to know whether it is a head or tail in each toss. All you need is the total number of heads.