How do I compute the distribution of averages of an exponential distribution?

Question

I'm given an exponential distribution and told the mean is $0.88$. I believe that means the distribution is notated $Exp(\lambda)$ where $\lambda = \frac{1}{0.88} $; is that correct? Then I'm told that groups of 25 samples of this distribution are averaged and I'm asked to find the distribution of the averages, notated $\bar{X}$.

So my understanding is that this is an application of the central limit theorem. The distribution of $\bar{X}$ will be normal, and I'm asked to find the mean and standard deviation.

I think the way to look at this is that the exponential distribution is the population with a mean of $0.88$ and also a standard deviation of $0.88$ (I believe that the standard deviation of an exponential distribution is equal to the mean). My sample mean will match the population mean, that is, $0.88$.

What I'm iffy about is the next step. Is it true that I can apply this?

$$s = \frac{\sigma}{\sqrt{N}} = \frac{0.88}{5} = 0.176 $$

That is, the sample standard deviation will be the population std deviation, $\sigma = 0.88$ divided by square root of the sample size, $25$?

Does this formula work with any original population for which $\sigma$ can be computed? That's pretty amazing if so.

Answer 1

I am not quite sure you can use CLT. Still, let $X_i$ be an exponential. It's not hard to show that $E(X_i) = \frac{1}{\lambda} $ and $ Var(X_i) = \frac{1}{\lambda^2}$. Then:

$$ P\left( \frac{\sum_{i=1}^{n} X_i}{n} \leq a \right) = P\left( \frac{\sum_{i=1}^{n} X_i - n\mu}{n} \leq a - \mu \right) $$

$$P\left( \frac{\sum_{i=1}^{n} X_i - n\mu}{\sigma\sqrt{n}} \leq \frac{(a - \mu) \sqrt{n}}{\sigma}\right) = P\left(Z \leq \frac{(a - \mu) \sqrt{n}}{\sigma} \right)$$ where $Z \sim N(0,1)$.

Notice that the left side of the inequality is precisely the equation of the central limit theorem. In your case, $n=25$, $\mu =1/\lambda$ and $ \sigma = 1/\lambda^2 $.

Now you can say it's approximatelly normal.

$$ P\left( Z \leq \frac{5(a-0.88)}{0.88} \right) $$.

That doesn't give you an answer about the distribution, though...

Yes, you can use the equality above. Because since all $X_i$ are iid, you have:

$$ Var\left(\frac{\sum X_i}{n}\right) = \sum_{i=1}^{n} \frac{1}{n^2} Var(X_i) = \frac{\sigma^2}{n}$$

Now for sd: $$ SD\left(\frac{\sum X_i}{n} \right) = \frac{\sigma}{\sqrt{n}}$$

Note that this is not the standard deviation of your sample, but the standard deviation of your estimator to the populational mean!

Hope I helped :)

Answer 2

Yet the rigorous answer is as it follows:

Note that the exponential distribution is given by: $$ f(x|p) = p e^{-px} $$ where $p$ is your parameter. Note that it is also a gamma $\Gamma(1,p)$distribution. The general gamma $\Gamma(\alpha,\beta)$ is given by: $$ f(x|p) = \frac{1}{\Gamma(\alpha)}\beta^{\alpha} x^{\alpha-1} e^{-\frac{x}{\beta}}$$ so it's not difficult to see that it's a gamma.

We can evaluate:

$$ P\left( \frac{X_i}{n} \leq a \right) = P(X_i \leq na) = \int_{0}^{na} p e^{-px} dx = 1 - e^{-np a}$$

Hence, for $Y=\frac{X}{n}$: $$ f(y|p) = np e^{-np y}$$

This happens to be a gamma $\Gamma(1, np)$. So:

$$ \frac{\sum_{i=1}^{n} X_i}{n} \sim \Gamma(n,np)$$

Finally we evaluate the distribution of that random variable, which is:

$$ f(x|p) = \frac{(np)^n}{\Gamma(n)} x^{n-1} e^{-np x}$$