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Math Help - probability generating functions- please help

  1. #1
    C.E
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    probability generating functions- please help

    A random variable Y is called gamma( \theta, n) for \theta>0 and natural n if it takes positive values and takes the following PDF:

    f(y)= \frac{1}{\theta (n-1)!}( \frac{y^{n-1}}{\theta^{n-1}}) exp \frac{-y}{\theta}

    Show how to find the moment generating function, expectation and variance of Y.

    my attempt:


     \int_{0}^{\infty} e^{\frac{-y}{\theta}}e^{ty} \frac{y^{n-1}}{(t\theta-1)^{n-1}}\frac{1}{\theta(n-1)!}dy


    = \int_{0}^{\infty} e^\frac{y(t\theta-1)}{\theta} \frac{y^{n-1}}{\theta^{n-1}}\frac{1}{\theta(n-1)!}dy


    let u=  {\frac{y(t\theta-1)}{\theta}}


    Then du=  \frac{(t\theta-1)}{\theta}dy


    =  \int_{0}^{\infty} \frac{1}{(t\theta-1)(n-1)!} \frac{u^{n-1}}{(t\theta-1)^{n-1}}e^u du


    = \frac{1}{(t\theta-1)^n} \int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!} du


    Is this right so far? Is there some reason why \int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!} du should be one? Could somebody show me how to integrate this by parts? (I know I should be able to do it but I am really struggling).
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by C.E View Post
    A random variable Y is called gamma( \theta, n) for \theta>0 and natural n if it takes positive values and takes the following PDF:

    f(y)= \frac{1}{\theta (n-1)!}( \frac{y^{n-1}}{\theta^{n-1}}) exp \frac{-y}{\theta}

    Show how to find the moment generating function, expectation and variance of Y.

    my attempt:


     \int_{0}^{\infty} e^{\frac{-y}{\theta}}e^{ty} \frac{y^{n-1}}{({\color{red}t\theta-1})^{n-1}}\frac{1}{\theta(n-1)!}dy
    I guess there's a little typo in red, though not important for your further working


    = \int_{0}^{\infty} e^\frac{y(t\theta-1)}{\theta} \frac{y^{n-1}}{\theta^{n-1}}\frac{1}{\theta(n-1)!}dy


    let u=  {\frac{y(t\theta-1)}{\theta}}


    Then du=  \frac{(t\theta-1)}{\theta}dy


    =  \int_{0}^{\infty} \frac{1}{(t\theta-1)(n-1)!} \frac{u^{n-1}}{(t\theta-1)^{n-1}}e^u du


    = \frac{1}{(t\theta-1)^n} \int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!} du


    Is this right so far? Is there some reason why \int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!} du should be one? Could somebody show me how to integrate this by parts? (I know I should be able to do it but I am really struggling).
    Well, actually, there's a problem, as for the convergence of the integral. But I'm a bit tired to study it ><

    Make the substitution u=\frac{y(1-t\theta)}{\theta}, assuming that t<\frac 1\theta . Otherwise, you'd have to change the boundaries of the integral, and it makes it more complicated.

    Then, you'll get something like \frac{1}{(1-t\theta)^n} \int_0^\infty e^{-u} \cdot \frac{u^{n-1}}{(n-1)!} ~du

    For this integral, see the Gamma function : Gamma function - Wikipedia, the free encyclopedia and it should be okay !



    I hope that what I babbled will help you in your process. Anyway, it's nice to see your working
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  3. #3
    C.E
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    Do you think I am ok to assume t< \frac{1}{\theta}? I doesn't say anything about the value of t in the question.
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  4. #4
    MHF Contributor matheagle's Avatar
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    It is important that your generating function exists in a neighborhood that contains zero.

    The trick to these is to make your generating function look like a density so you don't have to integrate.
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