Distribution of the sample mean of an exponential distribution

[buggy418]

Let's say you have a random sample of 5 values that are drawn from an exponential distribution with a mean of 8.

How do I find the distribution of Ybar, which is the sample mean of the 5 random variables? [Note: Ybar = 1/5(Y₁+Y₂+Y₃+Y₄+Y₅)]

I know that for an exponential distribution with mean 8 (i.e. Y~exp(8)), the variance would be 64.
So it seems like the distribution of Ybar can't also be exponential, since the variance is supposed to be the mean squared. I figure the mean of Ybar will be 8, but the variance must be something other than 64.
I don't know what approach to take...this seems harder than the approach for a normal distribution.

 

[Stephen Tashi]

You need to compute the five fold convolution of the exponential distribution. Have you studied convolutions?

 

[SW VandeCarr]

How do I find the distribution of Ybar, which is the sample mean of the 5 random variables? [Note: Ybar = 1/5(Y₁+Y₂+Y₃+Y₄+Y₅)]


So it seems like the distribution of Ybar can't also be exponential

True. What distribution would you expect for the sample means?

 

[buggy418]

I think it may have a gamma distribution of some sort. We've learned the method of distribution functions. So maybe if I let V=5Y, and then found the distribution of V/5 that would work? But it seems like that's just going to make me end up at the exponential distribution again, since V/5 = 5Y/5 = Y.

 

[Stephen Tashi]

So maybe if I let V=5Y

No, that would imply you had 5 copies of the same outcome, not 5 independent outcomes. You're going to have to do a convolution or look up the answer in a reference. Generating functions might help with a convolution.

 

[mattjw]

Sorry to dig up a two-month old thread, but in the course of solving a different problem, I may have found a straight-forward answer to the problem presented by the OP. Here it is, for future reference.

The OP was on the right lines with the Gamma distribution. In particular, it is the Erlang distribution, which is a special case of the Gamma distribution, that is appropriate in this case.

Recall that the distribution of the sum of k iid exponential distributions is described by the Erlang distribution. That is,
Erl(k,r) ~ Exp(r) + … + Exp(r). (Sum of k exponential distros.)
Here we use r to denote the rate parameter (more commonly denoted by lamda), where r = 1/mean.

The pdf of the Erlang distro is given by
f( x; k, r ) = ( r^k * x^(k-1) * e^(-r*x) ) / (k-1)! .
(Wiki page for more info on the Erlang distro: http://en.wikipedia.org/wiki/Erlang_distribution )

So, to find the distribution of the sample mean of k values drawn from k iid exponential distributions we simply need to find
1/k * Erl(k,r).
This is a scalar multiple of a random variable. The transformation of the random variable yields a distribution with pdf
f'( x; k, r ) = f( x*k; k, r ) = ( r^k * (x*k)^(k-1) * e^(-r*x*k) ) / (k-1)!.

In the OP's case we have k=5; plugging this into the pdf gives
f'( x; 5, r ) = (625/24) * e^(-5*x*r) * r^5 * x^4.

This is my first post here. I may have made a mistake. A quick numerical test gives similar results. Also the convolution method for k=2 gives the same result.

In case anyone's interested, my own problem is to find the distribution of the sample *variance* of k iid exponential distributions. I have yet to find the solution.

 

[bpet]

Sorry to dig up a two-month old thread, but in the course of solving a different problem, I may have found a straight-forward answer to the problem presented by the OP. Here it is, for future reference.

The OP was on the right lines with the Gamma distribution. In particular, it is the Erlang distribution, which is a special case of the Gamma distribution, that is appropriate in this case.

Recall that the distribution of the sum of k iid exponential distributions is described by the Erlang distribution. That is,
Erl(k,r) ~ Exp(r) + … + Exp(r). (Sum of k exponential distros.)
Here we use r to denote the rate parameter (more commonly denoted by lamda), where r = 1/mean.

The pdf of the Erlang distro is given by
f( x; k, r ) = ( r^k * x^(k-1) * e^(-r*x) ) / (k-1)! .
(Wiki page for more info on the Erlang distro: http://en.wikipedia.org/wiki/Erlang_distribution )

So, to find the distribution of the sample mean of k values drawn from k iid exponential distributions we simply need to find
1/k * Erl(k,r).
This is a scalar multiple of a random variable. The transformation of the random variable yields a distribution with pdf
f'( x; k, r ) = f( x*k; k, r ) = ( r^k * (x*k)^(k-1) * e^(-r*x*k) ) / (k-1)!.

In the OP's case we have k=5; plugging this into the pdf gives
f'( x; 5, r ) = (625/24) * e^(-5*x*r) * r^5 * x^4.

This is my first post here. I may have made a mistake. A quick numerical test gives similar results. Also the convolution method for k=2 gives the same result.

In case anyone's interested, my own problem is to find the distribution of the sample *variance* of k iid exponential distributions. I have yet to find the solution.

A factor of k seems to be missing from your f', but anyway if you just need to find the sample variance for sample size k, it's just (1/k) times the variance of the exponential distribution.