Could someone explain to me why the following equation holds? It is related to the law of total probability , but I don't get it. I'm confused because it's two random variables on the right side and therefore it should also be a joint distribution, right ?

$$\begin{align} \int \limits_0^\infty f_X(x) \ \mathbb{P}(Y<x) \ dx &= \mathbb{P}(Y<X) \end{align}$$

I understand that $$\begin{align} \mathbb{P}(X \in A, Y \in B) = \int_B f_Y(y) \int_A f_X(x|Y = y) \ dx \ dy . \end{align}$$

But how do I put this together?


1 Answer 1


Consider the continuous random variables $W$ and $V$ and their joint distribution $p(w,v)$. By the law of total probability, note that $$ p(w) = \int p(w,v) \ \text{d}v $$ and by Bayes' rule \begin{align} p(w) &= \int p(w,v) \ \text{d}v \\ &= \int p(w \mid v) \cdot p(v) \ \text{d}v \end{align} Going back to your random variables, note that $$ p(Y < X) = \int p(Y < X \mid x) \cdot p(x) \ \text{d}x $$ Since $X$ is given in $p(Y < X \mid x)$, then we can let $$ p(Y < X \mid x) = p(Y < x) $$ such that $$ p(Y < X) = \int p(Y < x) \cdot p(x) \ \text{d}x $$ Another way to see this is to let $$ Z = \begin{cases} 1 \quad \text{if} \quad Y < X \\ 0 \quad \text{otherwise} \end{cases} $$ We are therefore interested in $p(Z = 1)$. By the law of total probability and Bayes' rule, $$ p(Z = 1) = \int p(Z = 1 \mid x) \cdot p(x) \ \text{d}x $$

  • $\begingroup$ Hi, thank you for your reply! Could you be more specific with the Integral limits? Do you mean integrating x out? Why is it then still a random variable... For example, in your first equations (law of total prob.), we are integrating v out, therefore we get the marginal. But dont hold anymore, by consindering Y < X ? $\endgroup$
    – Camilo
    Commented Nov 28, 2021 at 19:00
  • $\begingroup$ Yes we are integrating $v$ out. Not sure what you are asking about... $\endgroup$
    – mhdadk
    Commented Nov 28, 2021 at 19:11

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