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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- Nov 9th 2009, 05:46 PMcooolgeExponential 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) - Nov 9th 2009, 08:59 PMmatheagle
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. - Nov 10th 2009, 07:01 PMcooolge
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! - Nov 10th 2009, 08:15 PMmatheagle
Just combine the exponentials and create a new normal rv.

There's no reason to integrate.

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

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

$\displaystyle = {1\over \sqrt{3}} {\sqrt{3}\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-3x^2/2}dx= {1\over \sqrt{3}} $ - Nov 10th 2009, 08:31 PMcooolge
Thank matheagle.

I got it!