Based on this: Equal in distribution but unequal almost everywhere?

My answer to questions like those is usually

Flip a fair coin. $1_H$ and $1_T$ have the same distribution but are never equal.

I didn't consider that independence could make a difference. Okay so if $X$ and $Y$ are identically distributed, independent in a finite sample space, what's a simple example to say that they aren't a.s. equal? What about for countably infinite sample space? What about uncountable sample space but RVs still discrete?

I have a feeling I'm missing some simple example to do with flipping a coin or rolling a die.


$X$ and $Y$ are defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$? Can they have different CDFs?

Random variables defined on the same probability space with different distributions

If two random variables have CDFs that have the same value for all x, can we assume the random variables are equal?

If X and Y are equal almost surely, then they have the same distribution


If $X$ and $Y$ are iid then the existence of some $x\in\mathbb R$ with $P(X=x)>0$ already excludes that $X\neq Y$ a.s.

This because: $$P(X=Y)\geq P(X=x\wedge Y=x)=P(X=x)P(Y=x)=P(X=x)^2>0$$


Suppose $X$ and $Y$ are i.i.d. Further suppose $X$ is not almost surely constant; then there exists an number $c$ so that $0<P(X\le c)<1$. Then $$ P(X\neq Y)\ge P(X\le c)P(Y>c)=P(X\le c)(1-P(X\le c))>0, $$ so that it is not true that $X=Y$ a.s.

  • $\begingroup$ I suspect the OP is looking for $X,Y$ iid together with $X\neq Y$ a.s. under specific conditions. $\endgroup$ – drhab May 16 '18 at 10:16

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