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!


2 Answers 2


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.

  • $\begingroup$ Ok, so my question here is WHY are those the sufficient statistics? $\endgroup$
    – EhBabay
    Commented Oct 8, 2014 at 18:39
  • 1
    $\begingroup$ Peter has given a fine explanation (+1). What additional explanations are you looking for? $\endgroup$
    – whuber
    Commented Oct 8, 2014 at 19:32
  • $\begingroup$ Ok, so after taking the product of the gamma distribution I to have gotten those sufficient statistics. Now I am going to assume that these are the sufficient statistics BECAUSE they do not depend on $\alpha$ and $\beta$. Or do I have wrong? $\endgroup$
    – EhBabay
    Commented Oct 8, 2014 at 22:36
  • $\begingroup$ The other answer throws some additional light on this. $\endgroup$
    – Glen_b
    Commented Oct 8, 2014 at 22:58

Sufficient statistics are defined in terms of parameters. Jointly sufficient statistics, are together sufficient for one or more parameters.

Recall that sufficiency is defined typically in terms of the joint distribution of the sample conditioned on the sufficient statistic; therefore joint sufficiency is obtained by conditioning the joint distribution of the sample on the joint distribution of the sufficient statistics. I would advise you to revisit the factorization theorem and the sufficient statistics for $\alpha$ and $\beta$ then come back with any questions you have.

The trivial solution to this problem is the sample; but these are probably not the sufficient statistics you are looking for.


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