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could anyone please indicate a general strategy (if there is any) to get the PDF (or CDF) of the product of two random variables, each having known distributions and limits? After having scanned related questions, I suspect there is no general strategy and the solution will depend on the distributions.

My particular need is the following:

$w=u \cdot v$

PDF of $u$: $\frac{1}{\pi u}\frac{1}{\sqrt{u^2-0.25}}$,for $z>0.5$

PDF of $v$: $\exp{[-\frac{v}{v_0}]}$, for $v_1<v<v_2$

what's the PDF of $w$?

Thank you

Sebastian

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2 Answers 2

up vote 1 down vote accepted

In case $U$ is a positive random variable with PDF $f_U$, and $V$ has a simple PDF $f_V$, so that the corresponding CDF, $F_V$, is simple too, it may be useful to use the following, assuming that $U$ and $V$ are independent. $$ {\rm P}(UV \le x) = \int {{\rm P}(UV \le x|U = u)f_U (u)\,du} = \int {{\rm P}\bigg(V \le \frac{x}{u}\bigg)f_U (u)\,du} = \int {F_V \bigg(\frac{x}{u}\bigg)f_U (u)\,du}. $$ You may then obtain the PDF of $UV$ upon differentiation.

EDIT: Here's a particularly simple example. Let $U$ and $V$ be independent uniform$(0,1)$ rv's. Here $f_U (u) = 1$, $0 < u <1$, $F_V (v) = v$, $0 < v < 1$, and $F_V (v) = 1$, $v \geq 1$. Thus, by the above formula, for any $0 < x \leq 1$, $$ {\rm P}(UV \le x) = \int_0^1 {F_V \bigg(\frac{x}{u}\bigg)du} = \int_0^x {F_V \bigg(\frac{x}{u}\bigg)du} + \int_x^1 {F_V \bigg(\frac{x}{u}\bigg)du} $$ $$ = \int_0^x {1\,du} + \int_x^1 {\frac{x}{u}\,du} = x - x\log x. $$ Hence the PDF of $UV$ is given, for $0 < x < 1$, by $$ f_{UV} (x) = \frac{d}{{dx}}(x - x\log x) = - \log x. $$

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Let $X$ and $Y$ be independent non-negative random variables, with density functions $f_X(x)$ and $f_Y(y)$. Let $Z=XY$.

Then $$P(Z\le z)=\iint_D f_X(x)f_Y(y)\,dx\,dy.$$ Here $D$ is the region in the first quadrant which is "below" the hyperbola $xy=z$.

Evaluate the integral, and differentiate the result with respect to $z$ to get the density function of $Z$. We can usually arrange to do the differentiation under the integral sign, but that still leaves one integral that may, like most integrals, not be expressible in terms of standard functions.

An alternative approach is to find the density functions of the random variables $\ln X$ and $\ln Y$, by using standard methods. Then use the usual "convolution" formula to find the density function of $U$, where $U=\ln X +\ln Y$. Finally, find the density function of $\exp(U)$.

Added: To my surprise, an integral not far from what is necessary for the first approach can be expressed in terms of modified Bessel functions and modified Struve functions (whatever those are). So says Wolfram Alpha. Thus there may be a sort of closed form for your density function. There would be, anyway, if what is called $v$ in the problem was a plain exponential. I would suggest also trying the second approach.

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