I understand that $ p(\min(X_1, X_2) \geq \alpha) $ can be found by looking at the probability of each separate random variable being greater than or equal to $\alpha$, and similarly with $ \max $. However, how I'm in the middle of a problem where it requires me to solve $ p(\max(X_1, X_2) - \min(X_1, X_2) \geq \alpha) $. How would one go about finding this? Assume that $X_1$ and $X_2$ are independent and the PDFs are known.
Let the pdfs of $X$ and $Y$ be written $p(X)$ and $q(Y)$. Now consider the probability that $\max(X,Y)-\min(X,Y)\geq \alpha$.
There are two cases. Either $X>Y$ in which case $X - Y \geq \alpha$, or $X<Y$ in which case $Y-X \geq \alpha$. The conditional probability of the first case is $$ P_> = P\left(\max(X,Y)-\min(X,Y)\geq \alpha|X-Y>0\right)=\int \int p(x)q(y)H(x-y-\alpha) dx dy,$$ where $H$ is the Heaviside function, while the conditional probability of the second case is $$ P_< = P\left(\max(X,Y)-\min(X,Y)\geq \alpha|Y-X>0\right) = \int \int p(x)q(y)H(y-x-\alpha) dx dy.$$ Now using the probabilities $P(X>Y)$ and $P(Y>X)$, defined by integrals $$P(X>Y) = \int \int p(x)q(y) H(x-y)dx dy$$ $$P(X<Y) = \int \int p(x)q(y) H(y-x)dx dy,$$ we can write with the law of conditional probabilities $$ P(\max(X,Y)-\min(X,Y)\geq \alpha) = P_> P(X>Y) +P_< P(X<Y).$$