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Expected value of X*exp(X) for X normally distributed

  1. Oct 9, 2013 #1
    Assume we have [itex]X\sim\exp(\mu,\sigma^2)[/itex].
    How does one compute [itex]\mathbb{E}\left(Xe^X\right)[/itex] and/or what is the outcome value?
     
  2. jcsd
  3. Oct 9, 2013 #2

    CompuChip

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    The definition would say that

    $$\mathbb{E}[ X e^X ] = \int_{-\infty}^{\infty} x e^{x} e^{-(x - \mu)^2 / \sigma^2} \, dx$$

    That's a tricky one, I don't think any of the common integration strategies really work.
    WolframAlpha does give me an exact result for the standard distribution ##(\mu, \sigma) = (0, 1)##.

    Maybe this will also help.

    Sorry for the incomplete answer, hoping that this will get you started.
     
  4. Oct 9, 2013 #3

    D H

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    Rewrite ##\exp(x)\exp(-(x-\mu)^2/(2\sigma^2))## as ##\exp(x-(x-\mu)^2/(2\sigma^2)) = \exp((2\sigma^2x-(x-\mu)^2)/(2\sigma^2))##. Now complete the square so that you get ##a\exp(-(x-\mu')^2)/(2\sigma^2))##, where ##a## is the exponential of the nasty junk you had to add to complete the square and ##\mu'## has the effect of being a new mean.
     
  5. Oct 9, 2013 #4

    CompuChip

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    Very nice one, and congratulations on post number 13,000!
     
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