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Math Help - Method of moment generating function with Gamma Distribution

  1. #1
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    Method of moment generating function with Gamma Distribution

    Assume that Y_1, \ Y_2, ... \ Y_n is a sample space of size n from a gamma distribution population with \alpha = 2 and \beta unknown.

    Us the method of moment generating function to show that 2 \sum_{i=1}^n \frac{Y_i}{\beta} is a pivotal quantity and has a \chi^2 distribution with 4n df.

    I know that the moment generating function of a Gamma Distribution is (1-\beta t)^{-\alpha}.

    I attempted to replicate http://www.mathhelpforum.com/math-he...-function.html, but got nowhere. I figure it has to like:

    (1-2t)^{-4n} since that's the mgf of a \chi^2 distribution with 4n df.
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    Quote Originally Posted by lllll View Post
    Assume that Y_1, \ Y_2, ... \ Y_n is a sample space of size n from a gamma distribution population with \alpha = 2 and \beta unknown.

    Us the method of moment generating function to show that 2 \sum_{i=1}^n \frac{Y_i}{\beta} is a pivotal quantity and has a \chi^2 distribution with 4n df.

    I know that the moment generating function of a Gamma Distribution is (1-\beta t)^{-\alpha}.

    I attempted to replicate http://www.mathhelpforum.com/math-he...-function.html, but got nowhere. I figure it has to like:

    (1-2t)^{-4n} since that's the mgf of a \chi^2 distribution with 4n df.
    Let X_i = \frac{2 Y_i}{\beta}.

    m_{Y_i}(t) = \frac{1}{(1 - \beta t)^2}.

    Therefore m_{X_i}(t) = \frac{1}{\left(1 - \beta \left( \frac{2}{\beta} \, t\right)\right)^2} = \frac{1}{(1 - 2t)^2}.

    Therefore the moment generating function of X = \sum_{i=1}^{n} X_i is equal to \left[\frac{1}{(1 - 2t)^2}\right]^n = \frac{1}{(1 - 2t)^{2n}}.

    Note that the moment generating function of the \chi^2 distribution with k degrees of freedom is \frac{1}{(1 - 2t)^{k/2}}.

    Compare moment generating functions: \frac{k}{2} = 2n \Rightarrow k = 4n.

    Therefore X = \sum_{i=1}^{n} X_i has a \chi^2 distribution with 4n degrees of freedom. Therefore the pdf is independent of \beta.

    Therefore .....
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    if I wanted to construct a confidence interval of 90% for \beta, would I integrate with respect to t which would give:

    \int_0^{\alpha} \frac{1}{(1-2t)^{2n}} dt = 0.05

    \int_\beta^{\infty} \frac{1}{(1-2t)^{2n}} dt = 0.05 ?
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    Quote Originally Posted by lllll View Post
    if I wanted to construct a confidence interval of 90% for \beta, would I integrate with respect to t which would give:

    \int_0^{\alpha} \frac{1}{(1-2t)^{2n}} dt = 0.05

    \int_\beta^{\infty} \frac{1}{(1-2t)^{2n}} dt = 0.05 ?
    NO! You certainly don't integrate the moment generating function!

    Read this thread: http://www.mathhelpforum.com/math-he...-interval.html.
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    Hopefully I got it, this is what I got for the confidence integral:

    \frac{1}{\Gamma{(\alpha)} \beta^{\alpha}} Y_i e^{-Y_i/\beta} now since \alpha = 2 and X_i = \frac{2Y_i}{\beta} where Y_i= \frac{\beta}{2}X_i and dY_i = \frac{\beta}{2}dX_i. Substituting these values in and taking the integral, I get:

    \int^{\alpha}_{0} \frac{1}{\Gamma{(2)} \beta^{2}}  \frac{\beta}{2}X_i e^{-\left( \frac{\beta}{2}X_i \right)/\beta} \frac{\beta}{2} \ dX_i

    = \int^{\alpha}_{0} \frac{X_i}{4} e^{-X_i/2} \  dX_i

    using integration by parts I get:

    \frac{1}{4} \bigg{[}-2X_i e^{-X_i/2} + 2e^{-X_i/2}\bigg{]}^{\alpha}_0 = 0.05

     = -2\alpha e^{-\alpha/2} + 2e^{-\alpha/2} -1 = 0.2

     = -\alpha e^{-\alpha/2} + e^{-\alpha/2} = 0.6

    now just isolate for \alpha (which doesn't look very solvable by hand), would this be correct?
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    Quote Originally Posted by lllll View Post
    Hopefully I got it, this is what I got for the confidence integral:

    \frac{1}{\Gamma{(\alpha)} \beta^{\alpha}} Y_i e^{-Y_i/\beta} now since \alpha = 2 and X_i = \frac{2Y_i}{\beta} where Y_i= \frac{\beta}{2}X_i and dY_i = \frac{\beta}{2}dX_i. Substituting these values in and taking the integral, I get:

    \int^{\alpha}_{0} \frac{1}{\Gamma{(2)} \beta^{2}} \frac{\beta}{2}X_i e^{-\left( \frac{\beta}{2}X_i \right)/\beta} \frac{\beta}{2} \ dX_i

    = \int^{\alpha}_{0} \frac{X_i}{4} e^{-X_i/2} \ dX_i

    using integration by parts I get:

    \frac{1}{4} \bigg{[}-2X_i e^{-X_i/2} + 2e^{-X_i/2}\bigg{]}^{\alpha}_0 = 0.05

     = -2\alpha e^{-\alpha/2} + 2e^{-\alpha/2} -1 = 0.2

     = -\alpha e^{-\alpha/2} + e^{-\alpha/2} = 0.6

    now just isolate for \alpha (which doesn't look very solvable by hand), would this be correct?
    No. Where has n gone? And I have pointed out a few times that the sort of integrals you're setting up are wrong.

    You need to closely follow the solution in the thread I refered you to.

    \Pr(a \leq X \leq b) = 0.9



     \Rightarrow \Pr\left( a \leq \frac{2}{\beta} \sum_{i=1}^{n} Y_i \leq b \right) = 0.9



    \Rightarrow \Pr\left( \frac{a}{2 \sum_{i=1}^{n} Y_i} \leq \frac{1}{\beta} \leq \frac{b}{2 \sum_{i=1}^{n} Y_i} \right) = 0.9



    \Rightarrow \Pr\left( \frac{2 \sum_{i=1}^{n} Y_i}{b} \leq \beta \leq \frac{2 \sum_{i=1}^{n} Y_i}{a} \right) = 0.9


    where \Pr(X \leq b) = \int_0^b \frac{(1/2)^{2n}}{\Gamma (2n)}\, u^{2n-1} e^{-u/2} \, du and \Pr(X \geq a) = \int^{+\infty}_a \frac{(1/2)^{2n}}{\Gamma (2n)}\, u^{2n-1} e^{-u/2} \, du.

    Nothing much can be done to get a value for a and b until the value of n is given. And when a value of n is given you should use technology to solve for the values of a and b.
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