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)
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)
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}} $