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Gamma distribution from sample mean of Exponential distribution

  1. Apr 11, 2014 #1
    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 [itex]\overline{X}[/itex] = [itex]\sum^{n}_{i = 1}X_i[/itex]

    Show that [itex]\overline{X}[/itex] ~ Gamma(n, [itex]\frac{n}{θ}[/itex])

    2. Relevant equations

    θ = [itex]\frac{1}{λ}[/itex]

    MGF Exponential Distribution = [itex]\frac{λ}{λ - t}[/itex]

    MGF Gamma Distribution = ([itex]\frac{β}{β - t}[/itex])α



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

    [itex]\frac{(\frac{λ}{λ-t})^{n}}{n}[/itex]

    I don't know what to do with the n in the denominator to get λ = [itex]\frac{n}{θ}[/itex]
     
  2. jcsd
  3. Apr 12, 2014 #2

    Ray Vickson

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    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
    [tex] f(x) =\frac{d}{dx} P (\bar{X} \leq x), \text{ and }
    P(\bar{X} \leq x) = P(S_n \leq nx). [/tex]
     
  4. Apr 12, 2014 #3
    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

    ([itex]\frac{λ}{λ - t}[/itex])n

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

    where [itex]\Gamma(n)[/itex] = ∫[itex]^{∞}_{0} [/itex]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 [tex] \frac{d}{dx} P (\bar{X} \leq x)[/tex]. Why do we take a derivative?

    Thank you.
     
    Last edited: Apr 12, 2014
  5. Apr 12, 2014 #4
    So think I got it.

    First to make things easier I'm just going to call [itex]\frac{1}{θ}[/itex] = λ

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

    where the sn-1 cancel out right?

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

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

    [itex]\frac{-λ^{n}}{λ-1}[/itex] e-s(λ-1)|[itex]^{nx}_{0}[/itex]

    After taking the derivative
    [itex]\frac{(-λ^{n})(-n)(λ-1)e^{-nx(λ-1)}}{λ-1}[/itex]

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

    [itex]\frac{(nx)^{n-1}(λ^n)(n)e^{-nxλ}}{(nx)^{n-1}e^{-nx}}[/itex]

    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.
     
  6. Apr 12, 2014 #5

    Ray Vickson

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    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.
     
  7. Apr 12, 2014 #6


    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 [itex]\Gamma[/itex](n) = ∫ sn-1 e-s ds.

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

    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
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