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?

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    $\begingroup$ The sample itself is always a sufficient statistic, so no. $\endgroup$ – 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.

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  • $\begingroup$ I should add that the first sentence considering order statistics is true only for continuous distributions. $\endgroup$ – StubbornAtom Jun 29 at 15:09

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