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Math Help - Moment generating function from p.d.f

  1. #1
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    Moment generating function from p.d.f

    X has chi square distribution with 4 degrees of freedom, with pdf

    f_{X}(x) = \frac{xe^-\frac{x}{2}}{4}, x>0

    (i) Find the moment generating function.

    Now I did moment generating function = E( e^{tx}) so for continuous r.v =

    \int e^{tx}\frac{xe^-\frac{x}{2}}{4}.

    Do I leave this here? I do I have to integrate and what happens with the t? Should I isolate the t outside of the integral first?

    (ii) If X1...Xn are independent, identically distributed random variables, with chi square distribution, with 4 degrees of freedom, show that the moment generating function Y = \sum X_{i} is (1-2t)^{-2n} and find the expecation and variance of Y.
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  2. #2
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    Re: Moment generating function from p.d.f

    Writing it out in full notation can provide hints on what the answer should look like:

    M_x(t) = \int_0^\infty e^{tx} \frac{ x e^{-0.5x}}{4} dx

    Written in full its clear that the next step is to complete the integration with respect to x. Its also clear from the LHS that the final answer should be a function of t (ie, you do not have to isolate or solve for t).
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  3. #3
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    Re: Moment generating function from p.d.f

    Quote Originally Posted by SpringFan25 View Post
    Writing it out in full notation can provide hints on what the answer should look like:

    M_x(t) = \int_0^\infty e^{tx} \frac{ x e^{-0.5x}}{4} dx

    Written in full its clear that the next step is to complete the integration with respect to x. Its also clear from the LHS that the final answer should be a function of t (ie, you do not have to isolate or solve for t).
    I used partical integration.

    I got:

    \frac{1}{4}(\frac{xe^{(t-\frac{1}{2})x}}{t-\frac{1}{2}}-\frac{e^{(t-\frac{1}{2})x}}{(t-\frac{1}{2})^2})

    is this correcT? I also have to specify the range of t. Would this just be t not equal 1/2? or is there another stipulation? also, moving onto the second part? any starters?

    (ii) If X1...Xn are independent, identically distributed random variables, with chi square distribution, with 4 degrees of freedom, show that the moment generating function Y = \sum X_{i} is (1-2t)^{-2n} and find the expecation and variance of Y.
    Last edited by Shizaru; November 30th 2011 at 03:08 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Re: Moment generating function from p.d.f

    this is a gamma density
    the idea is to combine the exponent of e and use the fact that a density integrates to one
    hence find your new BETA
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