Calculating $E(1_A|1_B)$

Question

I am trying to calculate $E(1_A|1_B)$, $A$ and $B$ being two events.

I know that $E(1_A) = P(A)$ and $P(B)=\frac{P(A\cap B)}{P(A|B)}$ and $E(X|B)=\frac{E(X1_B)}{P(B)}$ so

$E(X|B)=\frac{E(X1_B)}{\frac{P(A\cap B)}{P(A|B)}} = E(X1_B)\frac{P(A|B)}{P(A\cap B)}$

substituting $X$ with $1_A$ and knowing that $P(A\cap B) = E(1_A1_B)$ gives

$E(1_A|B)=E(1_A1_B)\frac{P(A|B)}{E(1_A1_B)}=P(A|B)$

I am trying to find a way to apply the same logic to $B$ in order to find $E(1_A|1_B)$ but to no avail, any help would be greatly appreciated!

Answer 1

Note that $\sigma(1_B)=\{\varnothing, B, B^c, \Omega\}$, so $$ \mathbb{E}[1_A\mid 1_B]=\mathbb{E}[1_A\mid B]1_B+\mathbb{E}[1_A\mid B^c]1_{B^c}=\mathbb{E}[1_A\mid B]1_B+\mathbb{E}[1_A\mid B^c](1-1_B) $$ Now you have shown $\mathbb{E}[1_A\mid B]=\mathbb{P}(A\mid B)$ and similarly $\mathbb{E}[1_A\mid B^c]$ is $\mathbb{P}(A\mid B^c)$ (assuming $B,B^c$ are not null).