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,