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Exponential of Gaussian Distribution

  1. May 20, 2013 #1
    I'm looking for the expected value of an exponential Gaussian

    [itex]Y=\text{exp}(jX) \text{ where } X\text{~}N(\mu,\sigma^2) [/itex]

    From wolframalpha, http://www.wolframalpha.com/input/?i=expected+value+of+exp%28j*x%29+where+x+is+gaussian

    [itex]E[Y]=\text{exp}(j^2\sigma^2/2+j\mu)[/itex]

    If I were to use the expected value definition:
    [itex]E[Y]=\int_{-\infty}^\infty uf_Y(u)du[/itex]
    then I would have to figure out the pdf of Y.

    I'm having trouble remembering how to get the pdf of Y, is there a more explicit way to derive the expected value?
     
  2. jcsd
  3. May 20, 2013 #2

    mathman

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    You can do it directly by evaluating ∫exp(ix)f(x)dx where f(x) is the normal density function. Also note that this is simply φ(1) where φ(t) is the characteristic function of this particular normal distribution.
     
  4. May 24, 2013 #3
    What mathman said is right - I just wanted to add a couple of things:

    1) The Law of the Unconscious Statistician is a snarky name for the usual way to find the EV of a function of a random variable.

    ##
    E[y(X)] = \int y(x) f_X(x) dx
    ##
    ##
    E[e^{jX}] = \frac{1}{\sigma\sqrt{2\pi}} \int e^{jx} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} dx
    ##

    2) Exponentials of Gaussians show up often enough to have their own name. If ##Y = \exp(X)##, then ##Y## has a lognormal distribution. The name 'lognormal' is a reminder that ##\log(Y)## is normally distributed.
     
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