Let X and Y be independent exponential random variables with mean 1.

I'm trying to derivative a convolution-like formula for $Z = XY$ but I'm having troubles getting started. The convolution formula is

$$\int_0^\infty fy(y)fx(z/y)dy$$

How can I start doing this problem. Any advice would be highly appreciate it.

  • $\begingroup$ I agree with the question posed by @Henry; convolution is for the sum of random variables, not the product. But then, you did say "convolution-like formula," so I'm assuming you do actually mean the product? $\endgroup$ – Math1000 Dec 6 '19 at 19:16
  • $\begingroup$ @Math1000 When I commented, the integrand was $fy(y)fx(z-y)$ but is now $fy(y)fx(z/y)$ so it seems the product was intended $\endgroup$ – Henry Dec 6 '19 at 19:44

In general, if $X$ and $Y$ are independent, continuous random variables with $f_X$ and $f_Y$, then we can find the density $f_z$ of $Z=XY$ by first considering the distribution function of $Z$; for $z\in\mathbb R$ we have

\begin{align} F_Z(z) :&= \mathbb P(Z\leqslant z)\\ &= \mathbb P(XY\leqslant z)\\ &= \mathbb P (XY\leqslant z,X\geqslant 0) + \mathbb P(XY\leqslant z,X\geqslant 0)\\ &= \mathbb P(Y\leqslant z/X,X\geqslant 0) + \mathbb P(Y\geqslant z/X, X\geqslant 0)\\ & \int_0^\infty f_x(x)\int_{-\infty}^{z/x}f_Y(y)\ \mathsf dy\ \mathsf dx + \int_{-\infty}^0f_X(x)\int_{z/x}^\infty f_Y(y)\ \mathsf dy\ \mathsf dx. \end{align} We find the density $f_z$ by differentiating with respect to $z$. Note that on the right-hand side, $z$ appears only in the integration limits, so the derivative follows from the fundamental theorem of calculus and the chain rule: \begin{align} f_Z(z) &= \frac{\mathsf d}{\mathsf dz} F_Z(z)\\ &= \int_0^\infty f_Xf_Y(z/x)\frac1x\ \mathsf dx - \int_{-\infty}^0 f_X(x)f_Y(z/x)\frac 1x\ \mathsf dx\\ &= \int_{-\infty}^\infty f_X(x)f_Y(z/x)\frac1{|x|}\ \mathsf dx. \end{align}

In the case where $X\sim\mathsf{Expo}(\lambda)$ and $Y\sim\mathsf{Expo}(\mu)$, this yields

\begin{align} f_Z(z) &=\int_{-\infty}^\infty f_X(x)f_Y(z/x)\frac1{|x|}\ \mathsf dx\\ &= \int_0^\infty \lambda e^{-\lambda x}\mu e^{-\mu z/x}\frac 1x\ \mathsf dx\\ &= 2 \lambda \mu K_0\left(\frac{2 \sqrt{\lambda }}{\sqrt{\frac{1}{z \mu }}}\right), \end{align} where $K_0$ is the modified Bessel function of the second kind with parameter $\nu=0$.

Now, when $\lambda=\mu=1$ we have $$ f_Z(z) = \int_0^\infty e^{-x}e^{-z/x}\frac1x\ \mathsf dx = 2 K_0\left(2 \sqrt{z}\right), $$ which is somewhat simpler, but still involves the use of Bessel functions.

(In the case that you actually mean $Z=X+Y$, the answer is simpler, and I have answered before on this site. I will find that answer if that is the case.)


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