Results 1 to 2 of 2

Thread: Departing from the moment generating function

  1. #1
    Member
    Joined
    Dec 2007
    Posts
    86
    Awards
    1

    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:

    $\displaystyle
    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)
    $

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

    $\displaystyle
    E[x^{2n}]=\frac{n!}{\lambda^n}\ (n=0,1,2, ...)
    $

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

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

    Integrating by parts:

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

    $\displaystyle 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 =
    $

    $\displaystyle
    \left[-\frac{n}{\lambda}\ x^{2n-2}\ e^{-\lambda x^2}\right]_0^\infty+\ I_2$

    $\displaystyle 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 =
    $

    $\displaystyle
    \left[-\frac{n(n-1)}{\lambda^2}\ x^{2n-4}\ e^{-\lambda x^2}\right]_0^\infty\ +\ I_3$

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

    and, in general:

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

    The iteration stops when k=n with

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

    therefore:

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

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

    Thanks again for your patience.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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:

    $\displaystyle
    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)
    $

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

    $\displaystyle
    E[x^{2n}]=\frac{n!}{\lambda^n}\ (n=0,1,2, ...)
    $

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

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

    Integrating by parts:

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

    $\displaystyle 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 =
    $
    $\displaystyle I_{2n}=\int_0^\infty x^{2n}\ 2\lambda x\ e^{-\lambda x^2} dx = \left[-x^{2n}e^{-\lambda x^2}\right]_0^\infty+\frac{n}{\lambda}\ I_{2(n-1)}=\frac{n}{\lambda}\; I_{2(n-1)}$

    and as $\displaystyle I_0=1$ proves the result.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Moment Generating function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Apr 25th 2011, 01:49 PM
  2. Moment-Generating Function - Need help!!!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Mar 1st 2009, 05:40 PM
  3. moment-generating function
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: May 3rd 2008, 07:34 PM
  4. moment-generating function
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: Mar 26th 2008, 06:02 PM
  5. Moment generating function
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Feb 2nd 2008, 01:27 AM

Search Tags


/mathhelpforum @mathhelpforum