# Show the sample mean is a sufficient estimator of $\theta$ if the population is exponentially distributed with parameter $\theta$.

I used the Fisher-Neyman factorization theorem for this problem.

If $X$ is exponentially distributed, $f(x)=\theta e^{x \theta}$ for $x>0$.

If we have a random sample $X_1,\dots,X_n$, and $\bar{X}$ is the sample mean, their joint density is:

$$f(x_1,\dots,x_n)=\theta^ne^{n\bar{x}\theta}$$ Then if we take $h(x_1,\dots,x_n)=1$ and $g_\theta(x_1,\dots,x_n)=\theta^ne^{n\bar{x}\theta}$, by the Fisher-Neyman factorization theorem, $\bar{X}$ is a sufficient estimator of $\theta$.

I have two questions about this result. The first is that the Wikipedia article uses the sum of $X_1,\dots,X_n$ as an example of a sufficient statistic of $\theta$. Does this mean both of these estimators are sufficient, but not necessarily consistent? Also, by choosing $h(x_1,\dots,x_n)=1$ as in the article, I felt a little like I was cheating. What would be a situation in which I can take $h$ as above and still wind up with an insufficient statistic?

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