Conditional probability and relationship to conditional expectation

It is well-known that $$\mathbb{P}(A)=\mathbb{E}[\mathbb{1}_{A}]$$

for an event $A \in \Omega$.

However, if we have a $\sigma$-algebra $\mathcal{F}$ then it certainly it is not true that $$\mathbb{P}(A|\mathcal{F})=\mathbb{E}[\mathbb{1}_{A}|\mathcal{F}]$$

since the LHS is a real number, while the RHS is a function.

My question is: Do we have a similar relationship between conditional probability (on a $\sigma$-algebra) and the expectation of an indicator function?

• You wrote LHS twice "since the LHS is a real number, while the LHS is a function." – john May 14 '18 at 21:47

The way that people sometimes use $\mathbb{P}(A \mid B)$ is simply the value of this random variable on the event B, i.e. then it becomes a number.
• So then $E[P(A|\mathcal{F})]=P(A)$? – asdf May 14 '18 at 18:00
• Doesn't the value of $P(A|\sigma(1_B))$ depend on further information, but $P(A|B)$ does not and is fixed? $P(A|\sigma(1_B))=P(A|\{\emptyset,B,B^c,\Omega\})$ can equal several values, including $P(A|B)$ or $P(A|B^c)$, yes? – jdods May 14 '18 at 23:45
• Right, so P(A | B) is somewhat of an abuse of a notation in my opinion; you are essentially evaluating P(A | $\sigma(1_B)$) on the set B. – E-A May 14 '18 at 23:47