I would like to compute $E[N|D]$, expectation of some random variable $N$ given that event $D$ occurs. In the outcome space, there are a set of disjoint events $B_1, B_2, B_3$.
How can I apply the law of total expectation to this $E[N|D]$?
I don't think it makes sense to write $$ E[N|D] = E[N|D,B_1]P(B_1) + E[N|D,B_2]P(B_2) + E[N|D,B_3]P(B_3) $$
and I also don't think it makes sense to write (in part because it looks weird) $$ E[N|D] = E[N|D|B_1]P(B_1) + E[N|D|B_2]P(B_2) + E[N|D|B_3]P(B_3) $$
The one thing I can think of doing is defining a random variable $M$ to be $N|D$. Then I can write it in a reasonable looking way: $$ E[M] = E[M|B_1]P(B_1) + E[M|B_2]P(B_2) + E[M|B_3]P(B_3) $$
But how can I write this in terms of the original $N|D$?
The formula you are looking for is: \begin{align*} \mathbb{E}[N | D] = \mathbb{E}[N | D,B_1]\mathbb{P}(B_1 | D) + \mathbb{E}[N | D,B_2]\mathbb{P}(B_2 | D) + \mathbb{E}[N | D,B_3)\mathbb{P}(B_3 | D) \end{align*} This is because by definition of conditional expectation and the fact that $B_1,B_2,B_3$ are disjoint and together form the state space it holds that: \begin{align*} \mathbb{E}[N | D] = \frac{\mathbb{E}[N1_{D}]}{\mathbb{P}(D)} = \frac{\mathbb{E}[N1_{D}1_{B_1}] + \mathbb{E}[N1_{D}1_{B_2}] + \mathbb{E}[N1_{D}1_{B_3}]}{\mathbb{P}(D)}. \end{align*} Filling in the following expression for $j = 1,2,3$ yields the desired result: \begin{align*} \mathbb{E}[N1_D1_{B_j}] = \mathbb{E}[N1_{D,B_j}] = \mathbb{E}[N | D,B_j]\mathbb{P}(D,B_j) = \mathbb{E}[N | D,B_j]\mathbb{P}(B_j | D)\mathbb{P}(D) \end{align*}