Originally Posted by

**mr fantastic** No.

$\displaystyle F(y) = \Pr(Y \leq y) = \Pr(XZ \leq y) =$

$\displaystyle \Pr\left(X \leq \frac{y}{Z} \right)$ when $\displaystyle Z > 0$

$\displaystyle \Pr\left(X \geq \frac{y}{Z} \right)$ when $\displaystyle Z < 0$

Therefore $\displaystyle F(y) =$

$\displaystyle \Pr(X \leq y)$ when $\displaystyle Z = 1$

$\displaystyle \Pr(X \geq -y)$ when $\displaystyle Z = -1$

Therefore:

$\displaystyle F(y) = \frac{1}{2} ( \, \Pr(X \leq y) + \Pr(X \geq -y) \, )$

$\displaystyle = \frac{1}{2} \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{y} e^{-t^2/2} \, dt + \frac{1}{\sqrt{2 \pi}} \int_{-y}^{+\infty} e^{-t^2/2} \, dt \right)$.

Therefore $\displaystyle f(y) = \frac{dF}{dy} = \frac{1}{2 \sqrt{2 \pi} } \left( e^{-y^2/2} + e^{-y^2/2} \right) = \frac{1}{\sqrt{2 \pi} } e^{-y^2/2}$.

This does not seem right but I don't have time now to find my mistake. I'll post again later.