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If this question has been asked, I apologize but I could not find it.

I was wondering if it was possible construct $X$, $Y$ two iid rv's such that they equal in distribution, i.e. $P_X(B) = P(X^{-1}(B)) = P(Y^{-1}(B)) = P_Y(B), B \in Borel(R)$, but differ every a.s., so that $X(\omega) \neq Y(\omega), \forall \omega \in \Omega$, except on a set of measure zero. I realize that likely one has to use the Axiom of Choice, but I am unsure about how to proceed.

I see here " If X and Y are equal almost surely, then they have the same distribution, but the reverse direction is not correct " that equality in distribution does not mean that $P(X = Y) = 1$ even if $X,Y$ are iid rv's.

Any advice/links would be greatly appreciated!

Community
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user3083171
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2 Answers2

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If rv $U$ is uniformly distributed on $(-1,1)$ then so is $-U$.

They have the same distribution but $U=-U\iff U=0$ and $P(U=0)=0$.

If you want this under the extra condition that the random variables are independent, then let $(X,Y)$ be uniformly distributed on $(0,1)^2$. Then $X$ and $Y$ are independent and both uniformly distributed on $(0,1)$ (so they are iid). However: $P(X=Y)=0$.

drhab
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Without independence: Flip a fair coin. $1_H$ and $1_T$ have the same distribution but are never equal.

With independence: Pick any identically distributed continuous RVs $X$ and $Y$. $P(X=Y)=0$, independent or not. Consider in Excel/Google Sheets:

A1=RAND()

B1=RAND()

A1 will never be equal to B1.

However if C1=A1

then

C1 will always be equal to A1 by definition.

BCLC
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