0
$\begingroup$

How would I go about obtaining the probability density function of a random variable that results in a product of a discrete variable and a continuous variable? I know that if $X$ and $Y$ are both continuous, then the probability density function of $Z=XY$ is given by: $$f_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y\left(\frac{z}{x}\right)\frac{1}{|x|}dx$$ but what is the method of obtaining $f_Z(z)$ if $X$ is a discrete variable and $Y$ is continuous?

More specific to my case, I have $X$ following a Rademacher distribution, that is $$\begin{align*} P(X=x) = \begin{cases} \frac{1}{2} & x = -1\\ \frac{1}{2} & x = 1\\ 0 & otherwise \end{cases} \end{align*}$$ and $Y$ following a Rayleigh distribution. What would be the product of these two random variables?

$\endgroup$
  • $\begingroup$ Use $f_X(x)=\sum_k P(X=k)\delta (x-k)$ (see en.wikipedia.org/wiki/Dirac_delta_function). $\endgroup$ – J.G. Dec 25 '18 at 13:13
  • $\begingroup$ @J.G. but does that $f_Z(z)$ formula still apply to this case? Is $f_X(x)$ considered continuous? $\endgroup$ – rea Dec 25 '18 at 13:35
  • $\begingroup$ Yes, it applies. $\endgroup$ – J.G. Dec 25 '18 at 14:26
0
$\begingroup$

We can try from the definition: $$ \begin{split} F_Z(z) &= \mathbb{P}[Z \le z] \\ &= \mathbb{P}[XY \le z] \\ &= \sum_{k=1}^\infty \mathbb{P}[XY \le z, X = k] \\ &= \sum_{k=1}^\infty \mathbb{P}[Y \le z/k] \mathbb{P}[X = k] \\ &= \sum_{k=1}^\infty F_Y(z/k) f_X(k). \end{split} $$

Note I did not handle the case where $k \le 0$...

One can alternatively condtion on $Y$ similarly and end up with an integral instead of a sum: $$ \begin{split} F_Z(z) &= \mathbb{P}[Z \le z] \\ &= \mathbb{P}[XY \le z] \\ &= \int_\mathbb{R} \lim_{\epsilon \to 0} \mathbb{P}[XY \le z, |Y-y| < \epsilon] dy \\ &= \int_\mathbb{R} F_X(z/y) f_Y(y) dy. \end{split} $$

$\endgroup$
0
$\begingroup$

For the special case of Rademacher $X$, $$P(Z\le z)=P(X=1)P(Y\le z)+P(X=-1)P(Y\ge -z)=\frac{F_Y(z)+1-F_Y(-z)}{2}.$$with $F_Y$ the cdf of $Y$.

$\endgroup$

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.