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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?

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    $\begingroup$ "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
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    $\begingroup$ 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
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    $\begingroup$ 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

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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 $$ \boxed{\quad P(X_1\leq X_3, X_2\leq X_3)=\int_\mathbb R F^2(x)\,F'(x)\,dx=\frac{1}{3}\,.\quad} $$

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ +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

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