Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expectation of Normal Variable

  1. Sep 6, 2007 #1

    More of a general integration question, but I just saw the following proof for the derivation of the expectation of a normal variable:

    E[X] = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}{x exp\left( -\frac{1}{2\sigma^2}(x-\mu)^2 \right) dx}

    Set z=(x-mu):

    E[X] = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}{z exp\left( -\frac{1}{2\sigma^2}z^2 \right) dx} + \mu \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}{exp\left( -\frac{1}{2\sigma^2}z^2 \right)dx}


    Now, I dont really understand how this works: if z=x-mu, then I would assume that the term inside the integral becomes:

    (z+\mu) exp\left( -\frac{1}{2\sigma^2}z^2\right)

    = z \left( exp\left( -\frac{1}{2\sigma^2}z^2 \right) \right) + \mu \left( exp\left( -\frac{1}{2\sigma^2}z^2 \right) \right)

    However, I dont see how we get two separate integrals, as in the proof above. Can anyone help shed any light on this?

  2. jcsd
  3. Sep 6, 2007 #2
    It's a sum of two terms which need to be integrated, so you can distribute the integration over the summation?
  4. Sep 6, 2007 #3
    Hmm...I'm still a bit confused tho - I know that you can distribute integration as you have said, as it is a linear operation, but if this was the case, would you have the


    term in front of both integrals? I would have thought it was just be in front of the first one?
  5. Sep 6, 2007 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    a(b+c) = ab+ac

    you learnt that in kindergarten (albeit not in such an algebraic form).
  6. Sep 6, 2007 #5
    Got it, thanks :-)

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?