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Problem

Suppose random variables $X$ and $Y$ are independent, and $X$ has a half-Gaussian distribution with $\mu=0$ and $\sigma^2=1$, $Y$ has a Rayleigh distribution with unknown $b$. Then what is the distribution of $Z=XY$

What I have Done

This question seems to be pretty straightforward but I got stuck halfway.

Obviously $f_X(x)=\frac{2}{{\sqrt{2 \pi}}}\exp(-\frac{x^2}{2})\ (x>0)$ and $f_Y(y)=\frac{y^2}{b^2}\exp(-\frac{y^2}{2b^2})$. Then $$ f_Z(z)=f_{XY}(x,y)=f_X(x)f_Y(y)=\frac{2y^2}{\sqrt{2 \pi}b^2}\exp(-\frac{1}{2}(\frac{y^2}{b^2}+x^2)) $$

I got stuck right here. I do not know what to do next since it seems that I could not transform the form with both $x$ and $y$ into the one with merely $z$.

Could anyone help me, thank you in advance.

Mr.Robot
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1 Answers1

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It is in general not true that $f_Z(z) = f_X(x) f_Y(y) $ for $Z=XY$ even if $X$ and $Y$ are independent.

Let $F(\cdot) $ denote the CDF of a random variable. We first write down the CDF of $Z$. Let $z\geq 0$: \begin{align} F_Z(z)=P(Z\leq z) = P\left( XY\leq z\right) = \int^\infty_0P(X\leq z/y) f_Y(y) dy \end{align} Differentiating under the integral sign yields: \begin{align} f_Z(z) = \int^\infty_0\frac{1} {y} f_X(z/y) f_Y(y) dy \end{align} So you have the CDF of $Z$. I don't know if that can be found exactly. I'll leave the calculations for you.

Shashi
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