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Show that if two random variables X and Y are equal almost surely, then they have the same distribution. Show that the reverse direction is not correct.

If $2$ r.v are equal a.s. can we write $\mathbb P((X\in B)\triangle (Y\in B))=0$ (How to write this better ?)

then

$\mathbb P(X\in B)-\mathbb P(Y\in B)=\mathbb P(X\in B \setminus Y\in B)\le \mathbb P((X\in B)\triangle (Y\in B))=0$

$\Longrightarrow P(X\in B)=\mathbb P(Y\in B)$

but the other direction makes no sense for me, i don't know how this can be true.

BCLC
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derivative
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    FYI, the identity $\mathbb P(X\in B)-\mathbb P(Y\in B)=\mathbb P(X\in B \setminus Y\in B)$ in the question, is wrong in general (and wrong here). – Did Sep 14 '15 at 20:21
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    For the first direction you can use: If $A\cap R=A\cap S$ and $P(A)=1$ then: $P(R)=P(A\cap R)=P(A\cap S)=P(S)$. Now take $A={X=Y}$, $R={X\in B}$ and $S={Y\in B}$. If you throw exactly one fair coin and $X$ is the number of heads that fall and $Y$ the number of tails, then $X\neq Y$ but they have the same distribution. – drhab Sep 14 '15 at 20:35

3 Answers3

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Take $X$ and $Y$ with probabilities $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=0.5$ and which are independent. Then $$P(X=Y) = P(X=1, Y=1) + P(X=2,Y=2) =\\= P(X=1)P(Y=1)+P(Y=2)P(Y=2)=0.5,$$ meaning that $X=Y$ holds with probability $0.5$, not $1$.

5xum
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If $X$ is a random variable following uniform $\mathcal U(-1,1)$ distribution, then $X$ and $-X$ are identically distributed, but obviously $X$ and $-X$ are not almost surely equal, in fact $P(X=-X)=0$

user521337
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Consider flipping a fair coin. Then $1_H$ and $1_T$ have the same distribution.

However, they are more than just not equal almost surely ($P(X \ne Y)>0$): they are almost surely not equal ($P(X \ne Y)=1$)!

BCLC
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