Let $X_1, X_2, X_3, X_4$ be independent and identically distributed continuous random variables. Then we know that for any $s,t$ real numbers $$P(X_1\leq s, X_2\leq t)=P(X_1\leq s)P(X_2\leq t)$$.

Then, can I claim that

  1. $P(X_1\leq X_3, X_2\leq X_3)=P(X_1\leq X_3)P(X_2\leq X_3)$?

If this is true, then I can also claim that $$P(\max(X_1,X_2)<\min(X_3,X_4))=P(X_1<\min(X_3,X_4))P(X_2<\min(X_3,X_4))$$ and then iterating this I can get $$P(\max(X_1,X_2)<\min(X_3,X_4))=P(X_1<X_3)^4$$.

This is not correct because my simulation result contradicts. What is wrong here?

  • 1
    $\begingroup$ "Then, can I claim that $P(X_1\leq X_3, X_2\leq X_3)=P(X_1\leq X_3)P(X_2\leq X_3)$?" -- No. $\endgroup$ May 19 at 19:55
  • $\begingroup$ Could you please explain why? this is very tricky. $\endgroup$ May 19 at 19:58
  • 3
    $\begingroup$ In the context you've given, you can evaluate $$P(X_1 \leq X_3, X_2 \leq X_3) = 1/3 \\ P(X_1 \leq X_3) = 1/2 \\ P(X_2 \leq X_3) = 1/2$$ so $1/3 \neq 1/4$. $\endgroup$ May 19 at 20:00
  • 2
    $\begingroup$ The intuition is that the event $X_1\le X_3$ tells you something about $X_3$, so it tells you something about the event $X_2\le X_3$ even though it tells you nothing about $X_2$. $\endgroup$
    – Karl
    May 19 at 20:06

1 Answer 1


The claim $$ P(X_1\leq X_3, X_2\leq X_3)=P(X_1\leq X_3)P(X_2\leq X_3) $$ is not true because the events $\{X_1\leq X_3\}$ and $\{X_2\leq X_3\}$ are not independent. The following should be true: $$ P(X_1\leq X_3, X_2\leq X_3)=\int_{\mathbb R}P(X_1\leq x)P(X_2\leq x)\,p(x)\,dx $$ where $p(x)$ is the PDF of the $X_i\,.$

Since $X_i$ are i.i.d. we can write this integral using the cumulative distribution function $F$ of $X_i$ as $$ \int_\mathbb R F^2(x)\,p(x)\,dx=\int_\mathbb R F^2(x)\,F'(x)\,dx. $$ Integration by parts now gives $$ -\underbrace{\int_\mathbb R 2F(x)\,F'(x)\,F(x)\,dx}_{2\,P(X_1\leq X_3, X_2\leq X_3)}+\underbrace{F^3(x)\Big|_{x=-\infty}^{x=+\infty}}_{1} $$ It follows -as Brian Moehring and Milten have pointed out- that $$ \boxed{\quad P(X_1\leq X_3, X_2\leq X_3)=\int_\mathbb R F^2(x)\,F'(x)\,dx=\frac{1}{3}\,.\quad} $$

  • $\begingroup$ I think these types of questions are helpful for students, this is why I asked this question. $\endgroup$ May 19 at 20:13
  • $\begingroup$ what a wonderful community we have :) $\endgroup$ May 19 at 20:16
  • $\begingroup$ I would like to take your attention to the following question as well: math.stackexchange.com/q/3642494/377953 $\endgroup$ May 19 at 20:22
  • 1
    $\begingroup$ And as Brian Moehring pointed out, the integral will always evaluate to $1/3$ because of the symmetry. (The event is “$X_3$ is the largest”) $\endgroup$
    – Milten
    May 19 at 21:41
  • 1
    $\begingroup$ +1 with a technical caveat: if the distribution has a nonzero singular part, then there's no PDF and then your steps don't work as written. However, each statement you wrote has an sister statement using Riemann-Stieltjes integrals that works in full generality and gives the same conclusion. $\endgroup$ May 21 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.