Gamma distribution from sample mean of Exponential distribution

[tmbrwlf730]

Homework Statement

Let X₁, X₂,...,X_n 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})

Homework Equations

θ = \frac{1}{λ}

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

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

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}

 

[Ray Vickson]

Homework Statement

Let X₁, X₂,...,X_n 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})

Homework Equations

θ = \frac{1}{λ}

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

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

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}

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 }<br /> P(\bar{X} \leq x) = P(S_n \leq nx).

 

[tmbrwlf730]

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 }<br /> P(\bar{X} \leq x) = P(S_n \leq nx).

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}s^n-1 e^-s ds

I'm guessing that the s^n-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.

 

[tmbrwlf730]

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 }<br /> P(\bar{X} \leq x) = P(S_n \leq nx).

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 s^n-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.

 

[Ray Vickson]

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 s^n-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.

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.

 

[tmbrwlf730]

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.

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) = ∫ s^n-1 e^-s ds.

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

Would the s^n-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?