Find the probability of $(X_1 < X_2)$

Question

If I have $2$ independent random variables $X_{1}$ and $X_{2}$, how can I prove that $Pr (X_{1} <X_{2} <t) = \int^{t}_{-\infty} F_{X_{1}}(x) f_{X_{2}}(x) dx$

Answer 1

Observe that $$ \mathbb{P}(X_1 < X_2 < t) = \int_{-\infty}^{t} \mathbb{P}(X_2 = x, X_1 < x) \text{d}x $$ and by the independence of $X_1, X_2$ you can split the above into

$$ \int_{-\infty}^{t} \mathbb{P}(X_2 = x) \cdot \mathbb{P}(X_1 < x) \text{d}x = \int_{-\infty}^t F_{X_1}(x) f_{X_2}(x) \text{d}x $$

The first equality holds because $$ \begin{align*} \{ X_1 < X_2 < t \} &= \bigcup_{x < y < t} \{X_1 = x, X_2 = y \} = \bigcup_{y < t} \{ X_1 \leq y, X_2 = y \} \end{align*} $$