Expectation of product of indicator functions I was asked this question and somehow I cannot relate the expectation with the given probability. If $A,B$ are events and $1_A$,$1_B$ are the indicator variables for these events, then how do I show that
$$
E\big(1_{A}\times 1_{B}\big) = Pr\big(A\cap B\big).
$$
 A: $$\mathbb 1_A\cdot\mathbb 1_B=\mathbb 1_{A\cap B}\qquad\&\qquad E(\mathbb 1_C)=P(C)$$
A: It might help if you use the definition of indicator functions as well as expectation. The indicator function defines a Bernoulli random variable. Recall that 
\begin{align*}
1_{A} = \begin{cases}
& 1 & \text{ if $x\in A$}\\
& 0 & \text{ if $x\notin A$},
\end{cases}
\end{align*}
and we have for the expectation
\begin{align*}
E\big(1_{A}\big) &= \sum xPr(x) = 0\times Pr(x\notin A)+1\times Pr(x\in A)\\
&= Pr(x\in A).
\end{align*}
Similarly, 
\begin{align*}
1_{B} = \begin{cases}
& 1 & \text{ if $y\in B$}\\
& 0 & \text{ if $y\notin B$},
\end{cases}
\end{align*}
and $E(1_{B}) = Pr(y\in B)$. 
If we look at the event $1_{A}\times 1_{B}$, we get
\begin{align*}
1_{A}\times 1_{B} = \begin{cases}
& 1 & \text{ if $x\in A$ and $y\in B$ }\\
& 0 & \text{ if $x\notin A$ or $y\notin B$}.
\end{cases}
\end{align*}
The above in itself defines a Bernoulli variable, and so the expectation becomes:
\begin{align*}
E\big(1_A\times 1_B\big) &= \underbrace{0\times Pr(x\notin A \text{ or } y\notin B)}_{=0} +1\times Pr(x\in A \text{ and } y\in B) \\
&= Pr(x\in A \text{ and } y\in B) = Pr(A\cap B).
\end{align*}
