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Math Help - Sum of independent gamma samples

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    Sum of independent gamma samples ~Solved

    Y has gamma distribution with parameters α, λ > 0.

    pdf is f(y)= {(λ^α)(y^(α-1))(e^(-λy))}/Γ(α)


    If Y1, Y2, ..., Yn are independent random samples from such a gamma distribution, show that their sum also has a gamma distribution and give the parameters.







    I can find E(ΣYi) and Var(ΣYi) and this leads me to believe the parameters of ΣYi are nα, λ, but I don't understand how to write down a more thorough arguement than just my first guesses.
    Last edited by Thomas154321; November 9th 2008 at 01:31 AM.
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  2. #2
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    Quote Originally Posted by Thomas154321 View Post
    Y has gamma distribution with parameters α, λ > 0.

    pdf is f(y)= {(λ^α)(y^(α-1))(e^(-λy))}/Γ(α)

    If Y1, Y2, ..., Yn are independent random samples from such a gamma distribution, show that their sum also has a gamma distribution and give the parameters.

    I can find E(ΣYi) and Var(ΣYi) and this leads me to believe the parameters of ΣYi are nα, λ, but I don't understand how to write down a more thorough arguement than just my first guesses.
    By induction, it suffices to prove it for two random variables X of parameters (a,\lambda) and Y of parameters (b,\lambda).

    The probability density function of the sum of two independent random variables (with densities) is the convolution of their density functions: in your case, this is:
    f_{X+Y}(z)=\int_0^z f_X(z-y)f_Y(y)dy = \int_0^z \lambda^a (z-y)^{a-1}e^{-\lambda (z-y)}\frac{1}{\Gamma(a)}\lambda^b y^{b-1}e^{-\lambda y}\frac{1}{\Gamma(b)}dy.
    There are a few simplifications, and you get:
    f_{X+Y}(z)=\frac{\lambda^{a+b}}{\Gamma(a)\Gamma(b)  }e^{-\lambda z} \int_0^z (z-y)^{a-1} y^{b-1}dy =\frac{\lambda^{a+b}}{\Gamma(a)\Gamma(b)}e^{-\lambda z}z^{a+b-1}\int_0^1 (1-u)^{a-1}u^{b-1}du
    (after the change of variable y=zu). Call the last integral B(a,b). Then we have found that the density function of X+Y is \frac{\lambda^{a+b}B(a,b)}{\Gamma(a)\Gamma(b)}z^{a  +b-1}e^{-\lambda z}. This is a multiple of the density function of a Gamma distribution with parameters (a+b,\lambda), and this is a density function, hence this is the density function of a Gamma distribution with parameters (a+b,\lambda).
    As a consequence this shows as well that B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, which you already knew, perhaps.


    Why is the convolution involved? This is because, for every positive measurable function f, we have, using Fubini:
    E[f(X+Y)]=\int_0^\infty \left(\int_0^\infty f(x+y)f_Y(y)dy \right)f_X(x)dx =\int_0^\infty \left(\int_x^\infty f(z)f_Y(z-x)dz\right)dx =\int_0^\infty f(z)\left(\int_0^z f_X(x)f_Y(z-x)dx\right) dz=\int_0^\infty f(z)f_{X+Y}(z)dz,
    and because these integrals characterize the distribution of X+Y.
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    Fascinating! Really enjoyed reading that. But I haven't done anything on convolution yet... Do you think there is a way of deducing the distribution using Moment-Generating functions of Y, using that all Yi have the same gamma(α,λ) distribution?
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  4. #4
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    Quote Originally Posted by Thomas154321 View Post
    Fascinating! Really enjoyed reading that. But I haven't done anything on convolution yet... Do you think there is a way of deducing the distribution using Moment-Generating functions of Y, using that all Yi have the same gamma(α,λ) distribution?
    That's exactly what I was going to suggest:

    1. Read this: PlanetMath: moment generating function of the sum of independent random variables

    2. Now get the moment generating function of Y (find it at Gamma distribution - Wikipedia, the free encyclopedia)

    3. Now put it all together and get the moment generating function of the sum.

    4. Now recognise the result as being the moment generating function of a gamma distribution with the given parameters (note that when a moment generating function exists there is a unique distribution corresponding to that moment generating function).

    Can you do these four things?
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  5. #5
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    Yes thanks, all straight forward using that. Since each integral is identical, it is just the moment generating function of Y to the power n, and nα obviously pops out.

    Cheers.
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