# nonexistence of a sufficient statistic

Let $$X_1,X_2,\dots,X_n$$ be a random sample from a $$\Gamma(\theta,\theta)$$ distribution. Then $$\prod_{i=1}^n f(x_i;\theta) = \frac{1}{\Gamma(\theta)^n\theta^n}(\prod_{i=1}^n x_i)^{\theta-1}e^{-\frac{\sum_{i=1}^n x_i}{\theta}},$$ which cannot be factored as in Fisher–Neyman factorization theorem. In particular, neither $$\prod_i X_i$$ nor $$\sum_{i} X_i$$ is a sufficient statistic.

Can we conclude that a sufficient statistic does not exist in this case?

• The sample itself is always a sufficient statistic, so no. – jbowman Nov 17 '19 at 3:06

The sample $$(X_1,X_2,\ldots,X_n)$$ and the vector of order statistics $$(X_{(1)},X_{(2)},\ldots,X_{(n)})$$ are trivial sufficient statistics for any parametric distribution.
For the Gamma distribution in question, a sufficient statistic for $$\theta$$ is simply $$\left(\prod\limits_{i=1}^n X_i,\sum\limits_{i=1}^n X_i\right)$$ by the Factorization theorem. So in some cases where you make both parameters equal (or make one parameter a function of the other) in a two-parameter distribution, you will get a two-dimensional sufficient statistic for a one-dimensional parameter.