Finding the probability distribution of the sample mean drawn from an exponential population.

Question

How can I find the probability distribution $f_{\overline{x}}(n, \mu)$ of the sample mean from a sample of size n drawn from an exponential population with mean $\mu$?

I have no idea where to begin on this. I know the same problem is easy with the normal distribution substituted for the exponential, since $$Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$$

yields

$$ f_{\overline{x}} = \sqrt{\frac{\sigma}{2\pi n}} e^{\frac{(\overline{x}-\mu)^2}{2 {\sigma}^2}}+\mu $$

But I don't think there is any similar identity with the exponential distribution, relating the sample mean to a well known distribution. Any help is appreciated.

Answer 1

You want to find the distribution of

$$\overline{X} = \frac{1}{n}\sum_{i=1}^{n}X_i,$$

where $X_i$ are exponential r.v. with expected value $\mu$.

While it is well known that the sum of Gaussian r.v.s produces another Gaussian r.v., for the exponential case we are not so lucky.

For the exponential case, it turns out that the distribution of:

$$Y = \frac{1}{n}\sum_{i=1}^n X_i,$$

follows a Gamma distribution with parameters $n$ and $\mu$. Check this link for further details.

Specifically, the gamma distribution of $Y$ has the following form:

$$f(y) = \frac{1}{(n-1)!\mu^n}y^{n-1}e^{- \frac{y}{\mu}}.$$

Since $\overline{X} = \frac{1}{n}Y$, then its distribution is:

$$f(\overline{x}) = \frac{n}{(n-1)!\mu^n}(n\overline{x})^{n-1}e^{-\frac{ n\overline{x}}{\mu}} \\ = \frac{n^n}{(n-1)!\mu^n}\overline{x}^{n-1}e^{-\frac{n\overline{x}}{\mu}}$$

since $f(y)dy = f(n\overline{x})nd\overline{x}.$

Addition

Since $n \in \mathbb{N}$, then the gamma distribution with parameters $n$ and $\frac{1}{\mu}$ is also known as Erlang distribution.