I have a question: Determine the conditional expectation $\mathrm{E}(A|B)$ for:
The number $B(\geq 0)$ of bats that leave a cave at the time of a nuclear explosion has a geometric distribution with mean $\lambda$. Suppose the bats each contribute independent exponentially distributed amounts of wind movement with mean $\frac{1}{\mu}$. Let $A$ be the total amount of wind movement at the time of a nuclear explosion(with no wind movement accumulated if no bats are flying.)
From the accepted answer: I take number $B$ of faults and multiply it with $\frac{1}{\mu}$ to yield $\mathrm{E}(A|B)$ this is because ~ $exp(\frac{1}{\mu})$ applied B times goes to $\frac{B}{\mu}$ due to large numbers right?
This is known as a random or conditional sum, and it's quite straight-forward once you wrap your head around the question. The total amount of movement is $A=\sum_{i=1}^b a_i$ where $a_i \sim \text{Exp}(1/\mu)$ and $b$ is the realization of $B$ you condition upon. You want the conditional expectation, so you are using $b$, a fixed constant and not random, and no expectation of $B$ is needed. Since it is a fixed constant, you get that $$ E(A|B=b)=E\left(\sum_{i=1}^ba_i\right)=\sum_{i=1}^bE(a_i)=\sum_{i=1}^b\frac{1}{\mu}=\frac{b}{\mu}. $$ Note that this is a number. The conditional expectation $E(A|B)$ is a random variable, and is: $$ E(A|B)=E\left(\sum_{i=1}^Ba_i\right)=\sum_{i=1}^BE(a_i)=\sum_{i=1}^B\frac{1}{\mu}=\frac{B}{\mu}. $$