# Gamma distribution from sample mean of Exponential distribution

1. Apr 11, 2014

### tmbrwlf730

1. The problem statement, all variables and given/known data
Let X1, X2,...,Xn be a random sample from the exponential distribution with mean θ and $\overline{X}$ = $\sum^{n}_{i = 1}X_i$

Show that $\overline{X}$ ~ Gamma(n, $\frac{n}{θ}$)

2. Relevant equations

θ = $\frac{1}{λ}$

MGF Exponential Distribution = $\frac{λ}{λ - t}$

MGF Gamma Distribution = ($\frac{β}{β - t}$)α

3. The attempt at a solution
I've tried using the generating function of the exponential distribution but I end up with

$\frac{(\frac{λ}{λ-t})^{n}}{n}$

I don't know what to do with the n in the denominator to get λ = $\frac{n}{θ}$

2. Apr 12, 2014

### Ray Vickson

We have $\bar{X} = S_n/n,$ where $S_n = \sum_{i=1}^n X_i$. The random variable $S_n$ has an n-Erlang distribution with mean $n \theta$. (Erlang is a special case of Gamma; its density is widely available in textbooks and on line.) To get the distribution pdf $f(x)$ of $\bar{X}$, use
$$f(x) =\frac{d}{dx} P (\bar{X} \leq x), \text{ and } P(\bar{X} \leq x) = P(S_n \leq nx).$$

3. Apr 12, 2014

### tmbrwlf730

I'm sorry but I'm still not getting it.

I understand about the erlang distribution and after plugging in what should be the parameters for it from

($\frac{λ}{λ - t}$)n

I get ∫$^{nx}_{0}$$\frac{s^{n-1} e^{-\frac{s}{θ}}}{θ^n \Gamma (n)}$

where $\Gamma(n)$ = ∫$^{∞}_{0}$sn-1 e-s ds

I'm guessing that the sn-1 cancel out but I feel like I'm missing a change of variables by a Jacobian.

Also in your response you mentioned we took $$\frac{d}{dx} P (\bar{X} \leq x)$$. Why do we take a derivative?

Thank you.

Last edited: Apr 12, 2014
4. Apr 12, 2014

### tmbrwlf730

So think I got it.

First to make things easier I'm just going to call $\frac{1}{θ}$ = λ

So..
∫$^{nx}_{0}$ $\frac{λ^{n} s^{n-1} e^{-λs}}{\Gamma (n)}$ ds

where the sn-1 cancel out right?

∫$^{nx}_{0}$ λn e-λs + s ds after working with the e-s in the denominator of the integrand.

∫$^{nx}_{0}$ $\frac{λ^{n} s^{n-1} e^{-λs}}{\Gamma (n)}$ ds

$\frac{-λ^{n}}{λ-1}$ e-s(λ-1)|$^{nx}_{0}$

After taking the derivative
$\frac{(-λ^{n})(-n)(λ-1)e^{-nx(λ-1)}}{λ-1}$

cancel out like terms and bring e^{-nx} back down to the denominator, and since $\frac{(nx)^{n-1}}{(nx)^{n-1}}$ = 1 we can put that back in so

$\frac{(nx)^{n-1}(λ^n)(n)e^{-nxλ}}{(nx)^{n-1}e^{-nx}}$

How do I get the integral sign back into the denominator? Also I still don't know why we integrated then took the derivative. Clarification on that would help. Thank you.

5. Apr 12, 2014

### Ray Vickson

If $g(w)$ is the density function of some random variable $W$, how do we find the density function of $W/n$? One way is to do it is to differentiate the cdf of $W/n$. However, if you prefer to use formulas for change-of-variables in probability you can do that instead. You should to it first for a general pdf $g(w)$, then specialize this to the case of the Erlang density.

I cannot figure out what you are doing with all your 'cancellations' and whatnot.

6. Apr 12, 2014

### tmbrwlf730

What I was thinking that something had to cancel so that I can integrate the function. I haven't worked with the gamma function so I'm not sure about some things.

So the $\Gamma$(n) = ∫ sn-1 e-s ds.

So we get $∫\frac{λ^{n}s^{n-1}e^{-λs}}{∫s^{n-1}e^{-s}ds}ds$

Would the sn-1 in the numerator cancel out with the one in the denominator?
Could you group e-λs and e-stogether to make e-s(λ+1) ?
How do I work with the gamma function, an integral, being a denominator of another integral?

Last edited: Apr 12, 2014