If there are two independent exponential variables X and Y. $Pr(X < a)$ and $Pr(X < b)$ then what will be the probability $Pr(X*Y<c)$?
2026-03-30 00:19:31.1774829971
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Probability of product of two exponential variable
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Well, for strictly positive values $a,b$: $$\mathsf P(X<a) ~=~\int\limits_{\{x~:~0\leqslant x<a\}} f_X(x)\operatorname d x~=~F_X(a)\\\mathsf P(Y<b) ~=~\int\limits_{\{y~:~0\leqslant y<b\}} f_Y(y)\operatorname d y ~=~F_Y(b)$$
And since we have independence $$\mathsf P(XY<c) ~{= \mathop{\iint\qquad\qquad}\limits_{\{(x,y)~:~0\leqslant x~,~0\leqslant y<c/x\}} f_X(x)f_Y(y)\operatorname d (x,y)}$$
Where $f_X(\bullet),f_Y(\bullet)$ are the probability density functions of your independent exponential random variables (also $F_X(\bullet), F_Y(\bullet)$ are the cummulative distribution functions).
The rest is substitution and integrating over the indicated interval.
If $X$ and $Y$ have rate parameters $\mu$ and $\lambda$, so their pdf's are $f_X(x) = \mu e^{-\mu x}$ and $f_Y(y) = \lambda e^{-\lambda y}$ for $x,y \ge 0$, then for $c > 0$, $$ Pr(X Y < c) = \int_0^\infty dx \int_0^{c/x} dy\; \mu \lambda e^{-\mu x - \lambda y} = 1-2\,\sqrt { {c\mu\lambda}}{{ K}_{1}\left(2\,\sqrt { c\mu\lambda}\right)} $$ where $K_1$ is a modified Bessel function of the second kind.