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Math Help - Normal Distribution

  1. #1
    Senior Member
    Feb 2008

    Normal Distribution

    If X is a normal random variable with mean \mu and variance \sigma^2, show that:

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  2. #2
    MHF Contributor
    May 2010

    its easier than i thought. make that substitution and you will get an integral you can actually evaluate directly.

    we started from
    2 \int_{\mu}^{\infty} (x-\mu) \frac{1}{\sqrt{2\pi\sigma^2}} \exp \left( \frac{-(X-\mu)^2}{2\sigma^2} \right) dx

    sub v=(X-\mu)^2
    so dx = \frac{dv}{2\sqrt{v}}

    To get

    2 \int_{0}^{\infty} \frac{\sqrt{v}}{2\sqrt{v}} * \frac{1}{\sqrt{2\pi\sigma^2}} * \exp(\frac{-v}{2\sigma^2}) dv

     = \frac{1}{\sqrt{2\pi\sigma^2}} \int_{0}^{\infty} \exp(\frac{-v}{2\sigma^2}) dv

    you can evaluate that integral using standard methods. hopefully the right answer drops out!
    (reminder: e^{-\infty} = 0 and e^0 = 1

    edit i did it to check:
    Evaluating the integral gives

    \frac{2\sigma^2}{\sqrt{2\pi\sigma^2}} \left( 0 - - 1 \right)

    Which simplifies to the required answer.


    not bad for an economist, if i do say so myself....even if it took about 10 tries
    Last edited by mr fantastic; June 5th 2010 at 03:22 PM. Reason: Merged posts. consequential editing
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