Law of total expectation

Let $A_1,\dots ,A_n$ be a partition of the sample space, and assume for all $A_i$ that $P(A_i)>0$. Then for any r.v. $Y$, the expectation can be expressed as

$$\mathbb{E}(Y)=\sum _{i=1}^n\mathbb{E}(Y\mid A_i)\cdot P(A_i)$$

Expressing LOTP using LOTE

Consider any event $B$, and let $I_B$ be an Indicator variable that equals 1 if $B$ occurs, and 0 otherwise. Using the same $A_1,\dots ,A_n$ from above, we can use LOTE to express the expectation of $I_B$ as

$$\mathbb{E}(I_B)=\sum _{i=1}^n\mathbb{E}(I_B\mid A_i)\cdot P(A_i)$$

Remember that $\mathbb{E}(I_B\mid A_i)$ simplifies to $0+1\cdot P(I_B\mid A_i)=P(I_B\mid A_i)$. Now we’re finding the probability that $I_B=1$, given $A_i$, which is equivalent to finding the probability that $B$ occurs, given $A_i$.

$$\mathbb{E}(I_B)=\sum _{i=1}^{n}P(I_B\mid A_i)\cdot P(A_i)=\sum _{i=1}^{n}P(B\mid A_i)\cdot P(A_i)=P(B)$$

We’ve just rediscovered LOTP using LOTE. In a sense, LOTP is a special case of LOTE when the random variable we’re finding the expectation of is an indicator variable that represents that event.