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CDF and expectation

  1. Oct 31, 2015 #1

    I have trouble with the following problem:

    Gaussian random variable is defined as follows

    [tex]\phi(t) = P(G \leq t)= 1/\sqrt{2\pi} \int^{t}_{-\infty} exp(-x^2/2)dx.[/tex]
    Calculate the expected value



    Because [itex]\phi[/itex] is a cumulative distribution function, [itex]\phi(+\infty) = 1[/itex].

    My attempt at solution:

    I start with:

    [tex] E(exp(G^2\lambda/2)) = \int^{\infty}_{-\infty}P(exp(G^2\lambda/2) \geq t)dt = \int^{\infty}_{-\infty}P(-\sqrt{2/\lambda*lnt}) \geq G \geq \sqrt{2/\lambda*lnt})dt[/tex]
    [tex]=1/\sqrt{2\pi} \int^{\infty}_{-\infty}(\int^{\sqrt{2/\lambda*lnt})}_{\sqrt{2/\lambda*lnt})}e^{-x^2/2}dx)dt.[/tex]

    Then my instinct would be to use Fubini theorem because I'd like to get rid of the integral of exp(-x^2/2) by [itex]\phi(+\infty) = 1[/itex].

    However, because both bounds are functions of t, it wouldn't work.

    Any help?
  2. jcsd
  3. Oct 31, 2015 #2


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    Science Advisor

    General formula: Let h(G) be a function of G. Let f(x) be the probability density function for G. Then:
    In your case [itex]h(G)=exp(G^2 \lambda /2)[/itex] and [itex]f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}[/itex].

    As you can see [itex]\lambda < 1[/itex] is necessary.
  4. Nov 2, 2015 #3
    Jaggis, I suggest you double-check your first equals sign.
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