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Math Help - Exponential of square of Normal Distribution

  1. #1
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    Exponential of square of Normal Distribution

    Dear All, I come across an equation when reading a paper. The author has skipped the steps to derive the equation. I am new to distributions. Could you please help me look into it? Thanks in advance!

    y~N(0,1+e/e)
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  2. #2
    MHF Contributor matheagle's Avatar
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    I cannot read the exponent on the right in front of the integral.
    But from what I can see, this does not make sense.
    On the RHS, y is a dummy variable of integration, but it's a variable on the LHS.
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  3. #3
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    Thank matheagle for your reply. My question can be simplified as following:

    Given x~N(0,1) , a standard normal variable.
    What is the expectation of exp(-x^2)?

    Thanks in advance!
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  4. #4
    MHF Contributor matheagle's Avatar
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    Just combine the exponentials and create a new normal rv.
    There's no reason to integrate.

    we know that {1\over b\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-(x-a)^2/(2b^2)}dx=1

    you want E(e^{-X^2}) = {1\over \sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-x^2} e^{-x^2/2}dx ={1\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-3x^2/2}dx

    = {1\over \sqrt{3}} {\sqrt{3}\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-3x^2/2}dx= {1\over \sqrt{3}}
    Last edited by matheagle; November 11th 2009 at 10:45 PM.
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  5. #5
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    Thank matheagle.
    I got it!
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