# Conditional distribution of the modulus of the output of AWGN channel given the modulus of the input

Hi everyone,

I will be too happy if anybody help me find a solution for the following problem. In fact, I have a big problem that I could not solve it for weeks.

Assume that we have we have two independent zero mean Gaussian random variables, $X$ and $Z$ and we define the new random variable $Y$ as $Y=X+Z$. I define the sign and the magnitude of this three random variables as $X_s$, $Y_s$ and $Z_s \in \{+1,-1\}$, $X_M$, $Y_M$ and $Z_M \in R^+$, respectively. It is clear that the pairs $X_s$ and $X_M$ are independent, this is the same for $Y_s$ and $Y_M$, $Y_s$ and $Y_M$.

My problem is how to find the conditional distribution $f(y_M|x_M)=?$.

I have a solution for this but I could not convince myself that my answer is true. We have, $$f(y_M|x_M)= f(y_M|x_M,x_s=+1)f(x_s=+1|x_M)+ f(y_M|x_M,x_s=-1)f(x_s=-1|x_M)$$ $$=f(y_m|x_M,x_s=+1)f(x_s=+1)+ f(y_M|x_M,x_s=-1)f(x_s=-1)$$ $$=0.5*(f(y_M|x_M,x_s=+1)+ f(y_M|x_M,x_s=-1))$$ then, since both $f(y_M|x_M,x_s=+1)$ and $f(y_M|x_M,x_s=-1)$ have the same distribution, i.e., "Folded normal distribution" then from above equation $f(y_M|x_M)$ follows "Folded normal distribution".

If it is correct it means that $f(y_M|x_M)= f(y_M|x_M,x_s)$ that implies $X_s$ and $Y_M$ condition on $X_M$ are independent!!!!!!!!!!!

But from $y_M = \lvert x+z \rvert$, $Y_M$ depends on both $X_M$ and $X_s$ !!!! I am really confused and I will be too grateful if anybody help me solve the problem. Thanks a lot in advance,

-

Yes, $Y_M$ is independent of $X_s$ - it doesn't matter whether or not you condition on $X_M$. You essentially give the answer yourself - conditioning on $X_M$, you get that $Y | X_M$ is a symmetric r.v., with $f(Y|X_M) = \frac{1}{2\sigma_Z}[\varphi(\frac{Y-X_M}{\sigma_z})+\varphi(\frac{Y+X_M}{\sigma_z})]$ where $\varphi$ is the standard normal density. Therefore, summing on $Y=\pm Y_M$ you get $f(Y_M|X_M) = \frac{1}{\sigma_Z}[\varphi(\frac{Y_M-X_M}{\sigma_z})+\varphi(\frac{Y_M+X_M}{\sigma_z})]$ (for any $Y_M \geq 0$) which is equal to $f(Y_M|X_M,X_s)$. It is true that $Y=X+Z$ depends on $X_s$ but you can write $f(Y | X_s=1) = f(-Y | X_s=-1)$ and when you take the absolute value $Y_M = |Y|$ becomes independent of $X_s$.
@ OrZuk, again thanks for your answer, but I did not tour 4th sentence, summing on $Y=\pm Y_M$ ..., I will be too happy if you explain me in more details? – Farzad May 30 '11 at 9:41
I meant that to get a particular value of $Y_M$ (which is non-negative), $Y$ can be either $Y_M$ or $-Y_M$. Therefore, to get the probability (actually density) of $Y_M$ attending a certain value, you should sum the density of $Y$ attaining this value or it's minus. – Or Zuk May 30 '11 at 13:03