# a confusion about independence related to random variables

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.

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

• "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. May 19 at 19:55
• Could you please explain why? this is very tricky. May 19 at 19:58
• 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 $1/3 \neq 1/4$. May 19 at 20:00
• 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 $X_2$.
– Karl
May 19 at 20:06

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 $$\boxed{\quad P(X_1\leq X_3, X_2\leq X_3)=\int_\mathbb R F^2(x)\,F'(x)\,dx=\frac{1}{3}\,.\quad}$$
• 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”) May 19 at 21:41