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Moment of a probability distribution

  1. Oct 27, 2004 #1
    when you calculate the Moment of the following equation

    [tex]

    p(x)=\left\{\begin{array}{cc}2Axe^{-Ax^2},&\mbox{ if }
    x\geq 0\\0, & \mbox{ if } x<0\end{array}\right.
    [/tex]

    We get

    [tex]
    Mn =2A \int_0^\infty x^{n+1}e^{-Ax^2}
    [/tex]

    solving it by parts I am getting

    [tex]
    Mn=(n+1)\int_0^\infty x^{n-1}e^{-Ax^2}
    [/tex]

    but, apparently, the right solution is

    [tex]
    Mn=n\int_0^\infty x^{n-1}e^{-Ax^2}
    [/tex]


    What am I doing wrong? What is the proper way to solve it? Could you please do it step by step?

    Thanks
     
  2. jcsd
  3. Oct 27, 2004 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    It looks like you didn't differentate properly when you integrated by parts:

    [tex]\frac {d x^n}{dx} = n x^{n-1}[/tex]
     
  4. Oct 28, 2004 #3
    sorry, still not following you.
    If we integrate by parts we have

    [tex]
    Mn =2A \int_0^\infty x^{n+1}e^{-Ax^2}
    [/tex]

    [tex]
    \int_0^\infty u\frac{dv}{dx} dx=uv-\int_0^\infty v\frac{du}{dx} dx
    [/tex]

    where

    [tex]
    u=x^{n+1} \ \ \ \ \ \frac{du}{dx}=(n+1)x^n
    [/tex]

    and

    [tex]
    \frac{dv}{dx}=e^{-Ax^2} \ \ \ \ \ v= -\frac{e^{-Ax^2}}{2xA}
    [/tex]

    so

    [tex]
    Mn=2A \ \{-x^{n+1} \ \frac{e^{-Ax^2}}{2xA} \ \ |_0^\infty +\int_0^\infty \frac{e^{-Ax^2}}{2xA}(n+1)x^n dx \ \}
    [/tex]

    [tex]
    Mn=0 +(n+1)\int_0^\infty x^{n-1} \ e^{-Ax^2} dx
    [/tex]

    So, How you get rid of the (n+1) term.
    Thanks
     
  5. Oct 28, 2004 #4

    Tide

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    Science Advisor
    Homework Helper

    This should get you where you want to go:

    [tex]\int_0^{\infty}x^{n+1} e^{-Ax^2} dx = - \frac {1}{2A}
    \int_0^{\infty}x^n \frac {d}{dx}e^{-Ax^2}dx[/tex]
     
  6. Oct 29, 2004 #5
    Thanks Tide,

    appreciate your patience
     
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