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Thread: Problem with an exercise (moment generating function)

  1. #1
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    Problem with an exercise (moment generating function)

    Hi everybody,

    I attached a pdf file with an exercise (moment generating function). I dont understand two steps which are described in this file.

    I would be very glad if someone could help.

    Many thanks,
    Fab
    Attached Files Attached Files
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  2. #2
    MHF Contributor
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    Re: Problem with an exercise (moment generating function)

    You need to know these 2 facts to follow this answer:
    • $\displaystyle \int_0^\infty t^{z-1}e^{-t} dt = \Gamma(z)$
    • $\displaystyle \Gamma(r+1) = r!$ (if r is an integer)


    Answer
    $\displaystyle \int \frac{a^2 x_2^{r+1}}{r+1} e^{-ax_2} dx_2$

    move 1 power of "a" outside the fraction
    =$\displaystyle \int a\frac{a x_2^{r+1}}{r+1} e^{-ax_2} dx_2$

    multiply top and bottom by $\displaystyle a^r$
    =$\displaystyle \int a\frac{a^{r+1} x_2^{r+1}}{a^{r}(r+1)} e^{-ax_2} dx_2$
    =$\displaystyle \int a\frac{(ax_2)^{r+1}}{a^{r}(r+1)} e^{-ax_2} dx_2$


    substitute v=ax, noting that $\displaystyle dx = dv/a$


    =$\displaystyle \int \frac{a}{a} \frac{v^{r+1}}{a^{r}(r+1)} e^{-v} dv$
    =$\displaystyle \int 1 \frac{v^{r+1}}{a^{r}(r+1)} e^{-v} dv$

    move constants outside the integral
    =$\displaystyle \frac{1}{a^r(r+1)} \int \frac{v^{r+1}}{1} e^{-v} dv$

    The integral is now in the same form as the definition of the gamma function. i haven't written the limits in but they are still 0 and infinity, as required.


    =$\displaystyle \frac{1}{a^r(r+1)} \Gamma(r+2)$

    now (since r is in integer) we can use this property of the gamma function: $\displaystyle \Gamma(r+2) = (r+1)!$

    =>$\displaystyle \frac{1}{a^r(r+1)} (r+1)!$

    note that $\displaystyle \frac{(r+1)!}{r+1}=r!$
    =>$\displaystyle \frac{1}{a^r} r!$

    =$\displaystyle \frac{r!}{a^r}$
    Last edited by SpringFan25; Nov 20th 2011 at 05:07 PM.
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  3. #3
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    Re: Problem with an exercise (moment generating function)

    SpringFan25,

    This helped me a lot. Thank you very much for the comprehensive answer!!!

    Kind regards
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