Exponential of square of Normal Distribution

cooolge · #1 $Old$ November 9th, 2009, 06:46 PM

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)

matheagle · #2 $Old$ November 9th, 2009, 09:59 PM

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.

cooolge · #3 $Old$ November 10th, 2009, 08:01 PM

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!

matheagle · #4 $Old$ November 10th, 2009, 09:15 PM

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

$= {1\over \sqrt{3}} {\sqrt{3}\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-3x^2/2}dx= {1\over \sqrt{3}}$

cooolge · #5 $Old$ November 10th, 2009, 09:31 PM

Thank matheagle.
I got it!