Results 1 to 2 of 2

Thread: 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:

    Follow Math Help Forum on Facebook and Google+

  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; Jun 5th 2010 at 03:22 PM. Reason: Merged posts. consequential editing
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Dec 27th 2011, 01:08 PM
  2. normal distribution prior and posterior distribution proof
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Mar 9th 2011, 06:12 PM
  3. Replies: 2
    Last Post: Mar 29th 2010, 02:05 PM
  4. Replies: 2
    Last Post: Aug 25th 2009, 10:39 PM
  5. Replies: 1
    Last Post: Apr 20th 2008, 06:35 PM

Search Tags

/mathhelpforum @mathhelpforum