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Thread: Normal Distribution

  1. #1
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    Normal Distribution

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

    $\displaystyle E(|X-\mu|)=\sqrt{\frac{2}{\pi}}\sigma$
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  2. #2
    MHF Contributor
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    aha!

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


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

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

    To get

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

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

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


    edit i did it to check:
    Evaluating the integral gives

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

    Which simplifies to the required answer.

    phew!


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