I'm trying to find the PDF of the product of two random variables by first finding the CDF. I don't know where I'm going wrong.
Let $X\sim N(0,1)$ and $Y\sim Uniform\{-1,1\}$ and let $Z = XY$, then:
$F_Z(Z<z) = P(Z<z) = P(XY<z) = P(Y<\frac{z}{X})$
$\Rightarrow F_Z(Z<z) =
> \displaystyle\int_{-\infty}^{\infty}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}$$\Rightarrow F_Z(Z<z) =
> \displaystyle\int_{-\infty}^{-z}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}+\displaystyle\int_{-z}^{0}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}+\displaystyle\int_{-0}^{z}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}+\displaystyle\int_{z}^{\infty}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}$$\Rightarrow F_Z(Z<z) =
> \displaystyle\int_{-\infty}^{-z}{\left(\frac{\frac{z}{x}+1}{2}\right)f_X(x)dx}+\displaystyle\int_{0}^{z}{f_X(x)dx}+\displaystyle\int_{z}^{\infty}{\left(\frac{\frac{z}{x}+1}{2}\right)f_X(x)dx}$$\Rightarrow F_Z(Z<z) =
> \displaystyle\int_{-\infty}^{-z}{\left(\frac{z}{2x}\right)f_X(x)dx}+\displaystyle\int_{z}^{\infty}{\left(\frac{z}{2x}\right)f_X(x)dx}+\displaystyle\int_{0}^{z}{f_X(x)dx}+\displaystyle\int_{z}^{\infty}{\left(\frac{1}{2}\right)f_X(x)dx}+\displaystyle\int_{-\infty}^{-z}{\left(\frac{1}{2}\right)f_X(x)dx}$
Since $f_X(x)$ is an even function.
$F_Z(Z<z) = \displaystyle\int_{0}^{\infty}{f_X(x)dx}$
I want to know where am I going wrong.
Best Answer
I think your error is in the very first line.
$XY \lt z$ implies $Y \lt \frac zX$ if and only if $X$ is positive. You need to condition on the events $\lbrace X \gt 0 \rbrace$ and $\lbrace X \lt 0 \rbrace$. The case when $X=0$ is vacuous since $\lbrace X=0 \rbrace$ has measure zero.
You should also keep in mind that the support of $Y$ is $[-1,1]$.