# Product of two random variables

• Nov 21st 2011, 01:13 AM
james121515
Product of two random variables
Hi guys,

Need a hint with the following problem:

Let $X$ and $Y$ be independent random variables, with probability distribution functions $f_{X}(x)$ and $f_{Y}(y)$. Let $Z= XY$ What is the probability distribution function for $Z$?

$F_{Z}(x) =P(Z \leq z) = P(XY \leq z) = P (Y \leq \frac{z}{x}) = \int_{-\infty}^{\infty}\int_{-\infty}^{\frac{z}{x}}f_{XY}(x, y)\,dy\,dx$
Since $X$ and $Y$ are independent, this gives:

$F_{Z}(z) = \int_{-\infty}^{\infty}\int_{-\infty}^{\frac{z}{x}}f_{X}(x)f_{Y}(y)\,dy\,dx$

Thus, we can obtain $f_{Z}(z)$ by differentiating both sides w/ respect to $z$ and using the fundamental theorem of calculus, which gives

$f_{Z}(z) = \int_{-\infty}^{\infty}\frac{1}{x}f_{X}(x) g_{Y}(\frac{x}{z})\,dx$

Thanks any help!

James
• Nov 21st 2011, 03:19 AM
mr fantastic
Re: Product of two random variables
Quote:

Originally Posted by james121515
Hi guys,

Need a hint with the following problem:

Let $X$ and $Y$ be independent random variables, with probability distribution functions $f_{X}(x)$ and $f_{Y}(y)$. Let $Z= XY$ What is the probability distribution function for $Z$?

$F_{Z}(x) =P(Z \leq z) = P(XY \leq z) = P (Y \leq \frac{z}{x}) = \int_{-\infty}^{\infty}\int_{-\infty}^{\frac{z}{x}}f_{XY}(x, y)\,dy\,dx$
Since $X$ and $Y$ are independent, this gives:

$F_{Z}(z) = \int_{-\infty}^{\infty}\int_{-\infty}^{\frac{z}{x}}f_{X}(x)f_{Y}(y)\,dy\,dx$

Thus, we can obtain $f_{Z}(z)$ by differentiating both sides w/ respect to $z$ and using the fundamental theorem of calculus, which gives

$f_{Z}(z) = \int_{-\infty}^{\infty}\frac{1}{x}f_{X}(x) g_{Y}(\frac{x}{z})\,dx$

Thanks any help!

James

Product distribution - Wikipedia, the free encyclopedia
• Nov 21st 2011, 04:49 AM
basovic88
Re: Product of two random variables
Hey,

Interesting, you have the same question as I have!
http://www.mathhelpforum.com/math-he...es-192173.html

I'm specifically looking for the case where X is exponentially distributed, and Y follows a normal distribution.

A closed-form solution seems impossible, hence I tried to estimate the pdf of Z.

Via Monte Carlo simulation, it appears that Z=XY, follows again an exponential distribution for z>0.