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If $X_1$, $X_2$, and $X_3$ are Continuous Random Variables that are i.i.d Uniform$([0, 1])$, what can deduce about $P(X_1 \le X_2)$ and $P(X_1 \le X_2 \le X_3)$?

My original thought is that $P(X_1 \le X_2) = 1/2$ because they are identically and uniformly distributed but I would love a more coherent/ mathematical argument.

Cheese Cake
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Anony mouse
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    Let $A={(x,y)\in\mathbb{R}^2\colon x\leq y}$. Since $X_1,X_2$ are i.i.d. we know that $(X_1,X_2)\overset{d}{=}(X_2,X_1)$. Hence, $P(X_1\leq X_2)=P((X_1,X_2)\in A)=P((X_2,X_1)\in A)=P(X_2\leq X_1)$. Combine this with the fact that $P(X_1=X_2)=0$ (because they are absolutely continuous) and you get $P(X_1\leq X_2)=1/2$. Can you do something similar for the other probability? – jakobdt Sep 15 '22 at 08:19
  • Yes thank you! I guess for the case with 3 variables it would just be 1/3! and you could generalise it for n variables to just 1/n!. – Anony mouse Sep 15 '22 at 19:42

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