Expectation of product of indicator functions

Question

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 $$ E\big(1_{A}\times 1_{B}\big) = Pr\big(A\cap B\big). $$

Answer 1

$$\mathbb 1_A\cdot\mathbb 1_B=\mathbb 1_{A\cap B}\qquad\&\qquad E(\mathbb 1_C)=P(C)$$

Answer 2

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*}