Product of two random variables that has exponential distribution

Question

I am trying to define the probability distribution of $Z$ such as $Z = X_1\cdot X_2$ where $X_1$ and $X_2$ are two independent and identically exponentially distributed variables.

$$P(X_1=x) = \lambda e^{-\lambda x}$$ $$P(X_2=x) = \lambda e^{-\lambda x}$$

I tried something like that….

$$P(Z=x) = P(X_1 = a) \cdot P(X_2 = x/a)$$

Is this first equation correct?

$$P(Z=x) = \lambda e^{-\lambda a} \cdot \lambda e^{-\lambda x/a} = \lambda^2 e^{-\lambda (a+x/a)}$$

I doesn't feel like I found the solution as the dummy variable $a$ remains in the final result. Can you please help me solving this problem?

Answer 1

See http://en.wikipedia.org/wiki/Product_distribution#Derivation_for_independent_random_variables where the formula for the product probability density is

$g(z) = \int_0^\infty f(x) f(z/x) \frac{1}{|x|} dx$

where $f$ is your exponential density.

Answer 2

Ok, here goes.

Assuming $X_1,X_2$ are independent and $Y=X_1X_2$ we have

\begin{multline} P(Y\leq y)= P(X_1X_2\leq y) = \int_0^\infty \lambda P(X_1\leq y/x) e^{-\lambda x} dx = \lambda \int_0^\infty (1-e^{-\lambda y/x}) e^{-\lambda x} dx =\\ 1- \lambda \int_0^\infty e^{-\{\lambda (x+y/x)\}} dx. \end{multline}

The density is the derivative of that with respect to $y$, i.e. \begin{equation} \lambda^2 \int_0^\infty \frac{1}{x} e^{-\{\lambda (x+y/x)\}} dx, \end{equation} for all $y>0$.

Answer 3

The density of $X_1 X_2$, obtained from that integral others have posted, is $f(z) = 2 \lambda^2 K_0(2 \lambda \sqrt{z})$ for $z > 0$, $0$ otherwise, where $K_0$ is a modified Bessel function of the second kind.