I'm working with a problem known in some places as the Elevator Problem. It goes as follows:
Suppose the of people entering on the ground floor of a building is modeled by a Poisson random variable with mean $\lambda$. There are $N$ floors above the ground floor and each person is equally likely to get off at any one of the $N$ floors, independently of others. We define the indicator random variable $I_i$ as $$I_i = \begin{cases} 1 & \text{if elevator stops at floor }i\\0 & \text{o/w.}\end{cases}$$Let $X$ represent the number of people that entered on the ground floor. Then
- Compute $\mathbb{E}(I_i\ | X = k)$ for $i=1,\ldots, N$, $k\geq 0$, and
- Compute $\mathbb{E}\left(\sum^N_{i=1} I_i\right)$.
So far, I've managed to compute both expectations, but I'm seeking further simplication for the second question. My answers:
- for the first one $$\mathbb{E}(I_i | X=k) = 1-\left(\frac{N-1}{N}\right)^k$$
- and the second \begin{align*} \mathbb{E}\left(\sum^N_{i=1} I_i\right) &= \sum^N_{i=1} \mathbb{E}(I_i)\\ &= \sum^N_{i=1} \sum^{\infty}_{k=1} \mathbb{E}(I_i | X = k)\cdot P(X = k)\\ &= \sum^N_{i=1} \sum^{\infty}_{k=0} \left[1 - \left(\frac{N-1}{N}\right)^k \right]\cdot \frac{10^k\cdot e^{-10}}{k!} \end{align*}
Supposing I am allowed to treat $k$ as a constant (such that summing it out is not required), are there any simplifications I could make to the double summation? All suggestions would be appreciated.
Since each conditional expectation of $I_i$ given $X = k$ is the same regardless of $i$, your double sum simplifies to $$\begin{align} \operatorname{E}\left[\sum_{i=1}^N I_i\right] &= N \sum_{k=0}^\infty \left(1 - (1 - 1/N)^k \right) e^{-\lambda} \frac{\lambda^k}{k!} \\ &= N \left( \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^k}{k!} - \sum_{k=0}^\infty e^{-\lambda} \frac{((1 - 1/N)\lambda)^k}{k!} \right) \\ &= N \left( 1 - \frac{e^{-\lambda}}{e^{-(1 - 1/N)\lambda}} \sum_{k=0}^\infty e^{-(1-1/N)\lambda} \frac{((1-1/N)\lambda)^k}{k!} \right) \\ &= N \left(1 - e^{-\lambda} e^{\lambda - \lambda/N} \right) \\ &= N (1 - e^{-\lambda/N}). \end{align}$$