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Math Help - Normal Distribution Expected Value

  1. #1
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    Normal Distribution Expected Value

    I'm stuck with something that I think should be straightforward but can't see what to do.

    I'm trying to find E[e^X] where X~N( m, v^2). I know that this is  \int^\infty_{-\infty}{e^{x}\cdot\frac{1}{\sqrt{2{\pi}v^2}}\cdot\  * e^{\frac{-(x-m)^2}{(2v^2)}}dx} i.e. the integral of the inner product of e^{x} and the pdf of the Normal distribution. This is where it falls apart - how do I integrate this? I think it's done by substitution when dealing with E(X) with the standard normal but this has the extra exponential and I'm not sure how to do it.

    Help appreciated, thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    combine the exponents of e, you need to complete the square
    This can be found in many books.
    It's a bit messy but it works.
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  3. #3
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    Thanks for the reply.

    Combining e^{x} and e^{\frac{-(x-m)^2}{(2v^2)}} and then completing the square I get e^{(m-v^2+x)^2 -m -m^2 + 2mv^2 - v^2}.

    I think the idea is to now split this into e^{(m-v^2+x)^2} and e^{-m -m^2 + 2mv^2 - v^2} and then take the latter outside of the integral.

    This leaves me with \frac{e^{-m -m^2 + 2mv^2 - v^2}}{\sqrt{2{\pi}v^2}}\int^\infty_{-\infty}{e^{(m-v^2+x)^2}dx} but I still can't solve this even with a change of variable as it doesn't converge. Am I missing something or is this the wrong idea?
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  4. #4
    MHF Contributor matheagle's Avatar
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    You need the minus infront of the () in the exponent and where is the 2v^2

    By the way this is the MGF evaluated at t=1.
    I was looking for that online, but couldn't find it.
    I'm sure it's out there.
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  5. #5
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    Ah, silly mistake. I've now got it down to \frac{e^{\frac{v^2(2m+v^2)}{2v^2}}}{\sqrt{2{\pi}v^  2}}\int^\infty_{-\infty}{e^{-(m-v^2+x)^2}dx}  =  \frac{e^{\frac{v^2(2m+v^2)}{2v^2}}}{\sqrt{2{\pi}v^  2}}\cdot\sqrt{\pi}  =  \frac{e^{m+\frac{v^2}{2}}}{v\sqrt{2}}.

    This is close to the target solution, only the denominator v\sqrt{2} isn't included. Is this meant to cancel somewhere or is this actually the corrent solution?
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