How does conditional expectation allow us to condition on zero-probability events?

Question

Can someone explain and provide example(s) of where we need to appeal/use conditional expectation to determine a conditional probability given a zero-probability event has occurred? I've seen books describe something like $E[X|Y]$ as allowing us to derive such conditional probabilities, which are not amenable to the usual $P(A|B)=\frac{P(A\cap B)}{P(B)}$ since $P(B)=0$....

Answer 1

In such situations it is good to recall that division is often defined as an inverse to multiplication: $x = b/a$ if $ax =b$. Whenever you are stack with some tricky division, you can just write it down in terms of equation involving the multplication. For example, you can define conditional probability as a solution of the following equation: $$ x= P(A|B) \quad \text{ if } \quad P(A\cap B) = x\cdot P(B) $$ or passing to the formula of total probability, given a finite partition $(H_i)$ of the probability space, the corresponding conditional probabilities $x_i = P(A|H_i)$ must satisfy $$ P(A) = \sum_i x_i(A) P(H_i) \qquad \forall A $$ Now, passing to a general case seems easy: just put a more general integral instead of a sum. That is, $$ P(A) = \int x_\omega(A) P(\mathrm d\omega) \qquad \forall A \tag{1} $$ is the (implicit) definition of (regular) conditional probability. Note that solution of $(1)$ is only unique almost surely, hence for any given null event you may change the conditional probability function $x_\omega(\cdot)$ and it will still solve this equation. For this reason, often you just need to guess what is a good candidate for the conditional probability, and then show that it solves the desired equation. E.g. Did's example looks very natural, although we can arbitrarily change it on rational numbers, and still get a valid conditional probability. In this case of course $E[X|Y]$ may differ from $\frac12 Y$ for rational numbers.

Disclaimer: $(1)$ is not perfectly formal, but it shall give an intuition.

Edit: regarding the question in the first comment - this is exactly what you do. Once you have defined conditional expectation, it is natural to say that conditional probability is defined by $P(A|\mathcal F) = E[1_A|\mathcal F]$. Now, the tower rule gives us exactly $(1)$: $$ P(A) = E[1_A] = E[E[1_A|\mathcal F]] = \int E[1_A|\mathcal F](\omega) P(\mathrm d\omega) $$ so that $E[1_A|\mathcal F](\omega)$ is a good candidate for $x_\omega(A)$. The only issue is that for some fixed $\omega$ it may happen that $E[1_A|\mathcal F](\omega)$ is not a measure as a function of $A$ - recall that conditional expectation is only defined $P$-a.s. uniquely, hence on some null set it may behave weirdly enough. For this reason, we distinguish regular conditional probabilities - those versions of $E[1_A|\mathcal F](\omega)$ that are probability measure as a function of $A$ for each fixed $\omega$. There are some measurable spaces and probability measures on them that do not admit regular versions, however for Borel spaces it works fine - you may wanna google about this if interested in details.

Answer 2

Consider $(X,Y)$ uniformly distributed in a triangle with vertices $(0,0)$, $(1,0)$, and $(0,1)$. Then $P(Y=y)=0$ for every $y$ and yet one wants to be able to declare that "when $Y=y$", $X$ is uniformly distributed on the interval $(0,y)$ hence the conditional expectation of $X$ should be $\frac12y$. This is rigorously stated as the fact that $E(X\mid Y)=\frac12Y$ almost surely.