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?

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

LHS is NOT a number; you can check Durrett Probability Theory and Examples that the second equality is actually correct. (version 4.1, page 191 at the end).

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.

  • $\begingroup$ So then $E[P(A|\mathcal{F})]=P(A)$? $\endgroup$ – asdf May 14 '18 at 18:00
  • $\begingroup$ Yes, exactly indeed. $\endgroup$ – E-A May 14 '18 at 18:18
  • $\begingroup$ 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? $\endgroup$ – jdods May 14 '18 at 23:45
  • 1
    $\begingroup$ 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. $\endgroup$ – E-A May 14 '18 at 23:47
  • $\begingroup$ @E-A, thanks for the verification. It was just a bit unclear from your answer, a slight edit might help. $\endgroup$ – jdods May 15 '18 at 0:55

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