Is it possible that $\mathbb P [ Y = 1 | X = x] >0$ whereas $\mathbb P [ X = x] = 0$?

Question

This question follows my previous one here, which is about the optimal classifier $g^*$ in case $X$ follows normal distribution.

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Let $X,Y$ be random variables in which

• $X$ follows normal distribution.

• $Y$ takes values in $\{-1,1\}$.

In measure-theoretic probability theory, $\mathbb P [ Y = 1 | X = x] := \mathbb E[\mathbf{1}_{\{Y=1\}} | \mathbf{1}_{\{X=x\}}]$ and $\mathbb P [ X = x] := \mathbb E[\mathbf{1}_{\{X=x\}}]$. Here $\mathbf{1}_{\{Y=1\}}$ and $\mathbf{1}_{\{X=x\}}$ are both integrable random variables. Then $\mathbb P [ Y = 1 | X = x]$ is well-defined even if $\mathbb P [ X = x] = 0$.

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I would like to ask if it's possible that $\mathbb P [ Y = 1 | X = x] >0$ whereas $\mathbb P [ X = x] = 0$?

Thank you so much for your clarification!

Answer 1

Here's an example to help you understand what's going on.

Suppose that $10$ people enter an elevator with capacity $2000$lbs. Denote $X$ as the combined weight of all $10$ people in the elevator. (I chose this example because weight is a random variable that classically possesses a normal distribution.) Now define an indicator random variable $Y$ such that $Y=1 \iff$ capacity is exceeded and $Y=0$ otherwise. Then $$P(Y=1|X=2200)=1 $$

Answer 2

Yes. Let $X,Y$ be independent. Using this for the conditional expectation with respect to the sigma field generated by $X$, we get: $$E[\mathbb{1}_{Y=1}|\sigma(X)] = E[\mathbb{1}_{Y=1}] = P(Y=1)$$