Let $X$ and $Y$ be iid random variables for which $X \sim \text{Expo}(2)$ and $Y \sim \text{Expo}(3)$.
Find $ \mathbb P \{ \max(X, Y) - \min(X, Y) \gt 0.2 \} $.
SOLUTION:
There is a clever way to do this but, to convince myself, I tried to calculate it via "brute force," i.e., using convolution (or, rather, "cross-correlation," more specifically). However, I can't seem to get my answer to match with the given solution. Here's what I did:
Let $T \sim \{ \max(X, Y) - \min(X, Y) \} $. Then, $T \sim \lvert X - Y \rvert$, and so
$$ f_T(t) = \int_{0}^{\infty} f_X(t+y)f_Y(y)dy + \int_{0}^{\infty}f_X(y-t)f_Y(y)dy $$
$$ = \int_{0}^{\infty} 2e^{-2(t+y)} 3e^{-3y} dy + \int_{0}^{\infty}2e^{-2(y-t)} 3e^{-3y}dy $$
$$ = \frac{6}{5}\left(e^{-2t} + e^{2t}\right). $$
Now, for the survival function of $T$, we have
$$ \bar F_T(t) = 1 - \left[ \int_{0}^{t} \frac{6}{5}\left(e^{-2u} + e^{2u}\right) du \right] = 1 - \left[ \frac{6}{5} \sinh(2t) \right].$$
So, for our desired probability, we only need to evaluate $\bar F_T(0.2):$
$$\bar F_T(0.2) = 1 - \frac{6}{5} \sinh \left[ 2(0.2) \right] = 1 - \frac{6}{5} \sinh \left[ 2(0.2) \right] = 1 - \frac{6}{5} \sinh \left( 0.4 \right) \approx 0.5070972.$$
However, the provided solution states that
$$\bar F_T(t) = \frac{2}{5}e^{-3t} + \frac{3}{5}e^{-2t} $$
and so the solution is $ \approx 0.6217$.
Like I said, the provided solution goes about solving the problem in a much more clever (and quicker) way by conditioning on the cases in which $X \lt Y$ and $X \gt Y$, which I understand. My question here, rather, is why this brute force approach didn't work.