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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?

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  • $\begingroup$ $P(X=x)=0$ for all $x\in\mathbb R$ for any continuous random variable $X$. So $P(X_1=x)=0$ not $\lambda e^{-\lambda x}$. There's a difference between $P(X=x)$ and density function $f(x)$. The first equation is correct, but not useful since $0=P(Z=x)=P(X_1=a)\cdot P(X_2=x/a)=0\cdot0$. $\endgroup$ – Cm7F7Bb Mar 27 '14 at 15:28
  • $\begingroup$ This might be useful: Product distribution. $\endgroup$ – Cm7F7Bb Mar 27 '14 at 15:35
  • $\begingroup$ @V.C. yes they are independent (I edited my question to make this clear). I think I understood why $P(X_1=x) = 0$ for all $x$. My technic would be meaningful for probability mass function only, is it correct? So how can we calculate the distribution of $Z$? Can you given me a hint? Using the characteristic equation (inverse Fourrier transform)? $\endgroup$ – Remi.b Mar 27 '14 at 15:36
  • $\begingroup$ Yes, you would use the pmfs in the discrete case, but still your approach lacks some steps since $a$ remains in the final result. You can calculate the density function of $Z$ as described here. $\endgroup$ – Cm7F7Bb Mar 27 '14 at 15:53
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See http://en.wikipedia.org/wiki/Product_distribution#Derivation 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.

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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 $y>0$.

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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.

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