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.

  • $\begingroup$ Do you mean that you're trying to find the PDF... or CDF... or something of the product of $X$ and $Y$? $\endgroup$ Feb 11, 2014 at 17:09
  • $\begingroup$ By the way, I think I know your mistake. It is in the second line, where you use convolution. According to the equation I have, you have to divide by $|x|$. $\endgroup$ Feb 11, 2014 at 17:12
  • 1
    $\begingroup$ @Clarinetist I think you divide by $|x|$ when using pdf of two random variables. Here I'm using cdf of one and pdf of other. Also I want to first find cdf and then get pdf from it. $\endgroup$
    – probMath
    Feb 11, 2014 at 17:25
  • $\begingroup$ Have you tried using moment generating function or characteristic function? $\endgroup$
    – Jlamprong
    Feb 11, 2014 at 18:05
  • $\begingroup$ I am not sure if you are actually looking for (a) the solution to your problem or (b) someone to tell you where you are going wrong. Generally, and perhaps unfortunately, people don't have the time nor inclination to work through other people's mistakes ... which may perhaps explain why this has not attracted any answers. If you are actually seeking (a) the answer ... then I think you have posed your question in an unattractive manner by asking others to work through your calculations. $\endgroup$
    – wolfies
    Feb 18, 2014 at 16:59

2 Answers 2


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 $[-1,1]$.


As pointed out in other answer, you need to be careful with passing terms in inequalities when sign can be negative. I would write:

$$P(Z \le z)= \int f_X(x) P(XY \le z \mid X=x)\,dx=\\ = \int_{x<0} f_X(x) P(Y \ge \frac{z}{x}) \,dx + \int_{x\ge 0} f_X(x) P(Y\le \frac{z}{x})\,dx $$

Can you go on from here?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.