An exercise about Borel paradox

Question

If $X$ and $Y$ are independent standard normals, what is the conditional distribution of $Y$ given that $Z=1$, where $Z=I(X=Y)$?

Answer 1

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!

Answer 2

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:

1. Z=0, where Z=Y-X;
2. Z=1, where Z=Y/X;
3. 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.