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Math Help - Departing from the moment generating function

  1. #1
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    Departing from the moment generating function

    Thanks for all your help, guys! Special thanks to Mr.F!

    The previous attempts were discouraging:
    http://www.mathhelpforum.com/math-he...-function.html

    Now I change approach totally, depart from the moment generating function and use the straight definition of moments. Please follow me...

    Bottom line, you will see that I'm not getting to the expected results yet. Therefore I would appreciate if you double check and see what I am doing wrong.

    Restating the problem - I need to analytically calculate the even moments of the following pdf:

    <br />
p(x;\lambda) =\left\{\begin{array}{cc}2 \lambda x\ e^{-\lambda{x^2}}& x\geq 0\\0 & x<0\end{array}\right. (\lambda>0)<br />

    I know that the even moments are these and I need to find a way to prove it:

    <br />
E[x^{2n}]=\frac{n!}{\lambda^n}\ (n=0,1,2, ...)<br />

    To do so, let's simply calculate the even-moments with the regular integration. By definition:

    E[X^{2n}]=\int_0^\infty x^{2n}\ 2\lambda x\ e^{-\lambda x^2} dx

    Integrating by parts:

    E[X^{2n}]=-\int_0^\infty x^{2n}\ (-2\lambda x)\ e^{-\lambda x^2} dx = \left[-x^{2n}e^{-\lambda x^2}\right]_0^\infty+\ I_1

    I_1=2n\int_0^\infty x^{2n-1}\ e^{-\lambda x^2} dx = -\frac{n}{\lambda}\int_0^\infty x^{2n-2}\ (-2\lambda x)\ e^{-\lambda x^2} dx =<br />

    <br />
\left[-\frac{n}{\lambda}\ x^{2n-2}\ e^{-\lambda x^2}\right]_0^\infty+\ I_2

    I_2=\frac{n}{\lambda}(2n-2)\int_0^\infty x^{2n-3}\ e^{-\lambda x^2} dx = -\frac{n(n-1)}{\lambda^2}\int_0^\infty x^{2n-4}\ (-2x\lambda)\ e^{-\lambda x^2} dx =<br />

    <br />
\left[-\frac{n(n-1)}{\lambda^2}\ x^{2n-4}\ e^{-\lambda x^2}\right]_0^\infty\ +\ I_3

    I_3\ =-\frac{n(n-1)(n-2)}{\lambda^3} \int_0^\infty x^{2n-6}(-2x\lambda)\ e^{-\lambda x^2} dx

    and, in general:

    I_k\ =-\frac{n!}{(n-k)!\ \lambda^k} \int_0^\infty x^{2(n-k)} d(e^{-\lambda x^2})=-\frac{n!}{(n-k)!\ \lambda^k}\left[x^{2(n-k)}\ e^{-\lambda x^2}\right]_0^\infty+\ I_{k+1}

    The iteration stops when k=n with

    I_n = (n!/\lambda^n)[e^{-\lambda x^2}]_0^\infty

    therefore:

    E[X^{2n}]=-\sum_{k=0}^n\frac{n!}{(n-k)!\ \lambda^k}\left[\frac{x^{2(n-k)}}{e^{\lambda x^2}}\right]_0^\infty\ =\ -\sum_{k=0}^n\frac{n!}{(n-k)!\ \lambda^k}\left[0-0\right]\ =\ 0

    And here I am... ZERO! That cannot be! What's wrong here?

    Thanks again for your patience.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by paolopiace View Post
    Thanks for all your help, guys! Special thanks to Mr.F!

    The previous attempts were discouraging:
    http://www.mathhelpforum.com/math-he...-function.html

    Now I change approach totally, depart from the moment generating function and use the straight definition of moments. Please follow me...

    Bottom line, you will see that I'm not getting to the expected results yet. Therefore I would appreciate if you double check and see what I am doing wrong.

    Restating the problem - I need to analytically calculate the even moments of the following pdf:

    <br />
p(x;\lambda) =\left\{\begin{array}{cc}2 \lambda x\ e^{-\lambda{x^2}}& x\geq 0\\0 & x<0\end{array}\right. (\lambda>0)<br />

    I know that the even moments are these and I need to find a way to prove it:

    <br />
E[x^{2n}]=\frac{n!}{\lambda^n}\ (n=0,1,2, ...)<br />

    To do so, let's simply calculate the even-moments with the regular integration. By definition:

    E[X^{2n}]=\int_0^\infty x^{2n}\ 2\lambda x\ e^{-\lambda x^2} dx

    Integrating by parts:

    E[X^{2n}]=-\int_0^\infty x^{2n}\ (-2\lambda x)\ e^{-\lambda x^2} dx = \left[-x^{2n}e^{-\lambda x^2}\right]_0^\infty+\ I_1

    I_1=2n\int_0^\infty x^{2n-1}\ e^{-\lambda x^2} dx = -\frac{n}{\lambda}\int_0^\infty x^{2n-2}\ (-2\lambda x)\ e^{-\lambda x^2} dx =<br />
    I_{2n}=\int_0^\infty x^{2n}\ 2\lambda x\ e^{-\lambda x^2} dx = \left[-x^{2n}e^{-\lambda x^2}\right]_0^\infty+\frac{n}{\lambda}\ I_{2(n-1)}=\frac{n}{\lambda}\; I_{2(n-1)}

    and as I_0=1 proves the result.

    RonL
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