Showing the sample mean is a sufficient statistics from an exponential distribution

Question

Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($\theta$), where $\theta>0$ is unknown. If we observe $\overline x = 5.2$ for a sample of $20$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?

The likelihood is pretty straightforward to find: $$L(\theta \mid x_1,\ldots,x_{20})=\prod_{i=1}^{20}\theta e^{-\theta \overline x} = \theta^{20}e^{-20 \theta \overline x}$$

I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $h(\overline x)g_\theta (T(\overline x))$

Answer 1

You've already done what you think you haven't done.

You should write it as $h(x_1,\ldots,x_{20}) g_\theta( (x_1 + \cdots + x_{20})/20),$ or in other words as $h(x_1,\ldots,x_{20}) g_\theta(T(x_1,\ldots, x_{20})).$

What's going on is camouflaged by the fact that $h(x_1,\ldots,x_{20}) = 1$ regardless of the value of $(x_1,\ldots,x_{20}).$

That often happens. (One situation where it does not happen is with an i.i.d. sample from the family of Poisson distributions.)