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Thread: 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 $\displaystyle E[e^X]$ where $\displaystyle X$~N($\displaystyle m$,$\displaystyle v^2$). I know that this is $\displaystyle \int^\infty_{-\infty}{e^{x}\cdot\frac{1}{\sqrt{2{\pi}v^2}}\cdot\ * e^{\frac{-(x-m)^2}{(2v^2)}}dx}$ i.e. the integral of the inner product of $\displaystyle e^{x}$ 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 $\displaystyle E(X)$ 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 $\displaystyle e^{x}$ and $\displaystyle e^{\frac{-(x-m)^2}{(2v^2)}}$ and then completing the square I get $\displaystyle e^{(m-v^2+x)^2 -m -m^2 + 2mv^2 - v^2}$.

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

    This leaves me with $\displaystyle \frac{e^{-m -m^2 + 2mv^2 - v^2}}{\sqrt{2{\pi}v^2}}\int^\infty_{-\infty}{e^{(m-v^2+x)^2}dx}$ 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 $\displaystyle 2v^2$

    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 $\displaystyle \frac{e^{\frac{v^2(2m+v^2)}{2v^2}}}{\sqrt{2{\pi}v^ 2}}\int^\infty_{-\infty}{e^{-(m-v^2+x)^2}dx} = \frac{e^{\frac{v^2(2m+v^2)}{2v^2}}}{\sqrt{2{\pi}v^ 2}}\cdot\sqrt{\pi} = \frac{e^{m+\frac{v^2}{2}}}{v\sqrt{2}}$.

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