Probability of One Random Variable Less than Another -- Why is this approach Wrong? This is not for homework. Just my curiosity. I am wondering why the following approach does not work.
Question: If $X_1, X_2$ have CDFs $F_{X_1},F_{X_2}$ respectively and are independent. What is $P(X_1\geq X_2)$? 
Proposed Answer: $P(X_1\geq X_2) = \int F_{X_2}(t)dF_{X_1}(t)$
I know what the correct answer is, but I guess I just want to understand why the above answer is wrong. My friend is very sure it is incorrect but can't explain why not adequately. I've seen an answer to this question that suggests the same thing on another post but it was down voted, so that gives me further confidence that it is wrong. That being said, it's not obvious to me why it is...
 A: Your expression is correct: Since $X_1 \ \bot \ X_2$ we can use a Riemann-Stieltjes integral to write the probability of interest as:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X_1 \geqslant X_2) 
&= \mathbb{P}(X_2 \leqslant X_1) \\[6pt]
&= \int \mathbb{P}(X_2 \leqslant X_1 | X_1 = t) \ d F_{X_1}(t) \\[6pt]
&= \int \mathbb{P}(X_2 \leqslant t) \ d F_{X_1}(t) \\[6pt]
&= \int F_{X_2}(x) \ d F_{X_1}(t). \\[6pt]
\end{aligned} \end{equation}$$
A: It seems correct to me, because what you write is actually $\int F_{X_2}(t)f_{X_1}(t)dt$ and can be interpreted as swiping $t$ values such that when $X_1$ is equal to $t$, we consider all $X_2$ smaller than $t$; applying the total probability law gives us $P(X_2\leq X_1)\approx\sum_t P(X_2\leq t)P(X_1=t)$, which makes sense.
An example: let $X_1\sim\exp(\lambda),X_2\sim\exp(\mu)\rightarrow F_{X_2}(t)=1-e^{-\mu t}$, $dF_{X_1}(t)=\lambda e^{-\lambda t}dt$. Then, the integral becomes:
$$P(X_2\leq X_1)=\int_0^\infty (1-e^{-\mu t})\lambda e^{-\lambda t}dt=1-\lambda\int_0^\infty e^{-(\lambda+\mu)t}dt=\frac{\mu}{\lambda+\mu}$$
which can be found via joint PDF directly as in @MichaelChernick's comment:
$$P(X_2\leq X_1)=\int_0^\infty\int_0^{x_1}\lambda\mu e^{-\mu x_2}e^{-\lambda x_1}dx_2dx_1=\underbrace{\int_0^\infty\lambda e^{-\lambda x_1}(1-e^{-\mu x_1})dx_{1}}_{\text{above integral}}$$
