Let $X$ and $Y$ be independent random variables for which $X \sim > \text{exp}(2)$ and $Y \sim \text{exp}(3)$. Find $P(\max(X, Y) - \min(X, Y) \gt 0.2) $.
1) I know this answer has already been asked here, but I'm interested in the "clever way" suggested by the question poster. Does anyone know what he had in mind?
2) Also, regarding his own convolution method: He sets $T= \max(X, Y) - \min(X, Y)$ which means that $T=|X-Y|$ why does he need two integrals when he finds $F_T(t)$?
3) In the comments for that particular question, there is a guy who says that the second integral shouls start at $t$. Why does it start at $t$ in the second and on $0$ in the first integral?
An interesting property of exponential distributions is memoryless property. You can see exponential random variables as a model for arrival time of a thing. The memoryless property says that if you have waited until time $t_0$ and you have not observed any arrival, i.e. you know that $X> t_0$, then the arrival time still follows the same exponential distribution namely: $$ P(X>t_0+s|X>t_0) = P(X>s). $$
Now suppose that $Y$ is an arbitrary independent random variable, then you can still see that $$ P(X>s+Y|X>Y)=P(X>s). $$ Therefore for independent $X\sim\exp(a)$ and $Y\sim\exp(b)$, we have $$ P(|X-Y|>s)=P(|X-Y|>s|X>Y)P(X>Y)+P(|X-Y|>s|Y\geq X)P(Y\geq X)\\ =P(X>Y)P(X>s)+P(Y\geq X)P(Y>s)\\ =P(X>Y)e^{-as}+P(Y\geq X)e^{-bs}. $$ It remains to find $P(X>Y)$. See that: $$ P(X>y|Y=y)=e^{-ay}\implies P(X>Y)=E(e^{-aY})=\int_{0}^\infty be^{-by}e^{-ay}\mathrm dy=\frac b{a+b}, $$ hence,