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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

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    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 :-)

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