An exercise about Borel paradox If $X$ and $Y$ are independent standard normals, what is the conditional distribution of $Y$ given that $Z=1$, where $Z=I(X=Y)$? 
 A: It is undefined.  It doesn't make sense to condition on events of measure zero.
There are certain ways around this, for instance, if you are trying to condition on an event like $Z=a$, where $Z$ is a random variable, and for every $\epsilon$ we have $P(|Z-a| < \epsilon) > 0$.  That is not the case here: for any $\epsilon < 1$ we have $P(|Z-1| < \epsilon) = P(Z=1) = P(X=Y)=0$.  Indeed, up to almost sure equality, $Z$ is just the zero random variable.
Here is another way around it, however.  Let $V = X-Y$.  Then it is not hard to verify that the conditional distribution of $Y$ given $V$ is $N(-\frac{V}{2}, \frac{1}{2})$.  (Consider $U = X+Y$ and note that $U,V$ are iid $N(0,2)$.)  Based on this, you could argue that the conditional distribution of $Y$ given $V=0$ should be $N(0, \frac{1}{2})$.
The moral is that in cases like this, the answer you get may depend on how you phrase the question!
A: Thanks you very much for your answer, and I'm sorry to add comment here because I have not enough reputations.
I do the exercise while reading the "Statistical Inference", by Casella and Berger. In the text, in fact, there are three conditions:


*

*Z=0, where Z=Y-X;

*Z=1, where Z=Y/X;

*Z=1, where Z=I(Y=X).


Basically, I think that the conditional distribution in case3 is undefined, too. But the auther says each condition is a correct interpretation of the condition Y=X, and each leads to a different conditional distribution. Do I misunderstand the meaning of the above text?
Thank you for your comment. I never think it taht way.
