expectation of the number of empty cells

Question

You are given a random number, $N$, of balls, where $N$ has a Poisson distribution with parameter $\lambda > 0$. You then place these balls one by one among $r$ ($\geq 2$) cells according to the uniform distribution: each ball will be independently put in cell j with probability $\frac{1}{r}$ , for $j = 1, 2 , .. r$ (Assume that the random number of balls given to you and the way you place these balls among the cells are mutually independent.) Introduce the following indicator random variables: for $j = 1, 2, ... r$ $$I_j= \left\{ \begin{array}{lcc} 1 & \mbox{if cell j is empty}\ \\ 0, &\mbox{ otherwise,} \\ \\ \end{array} \right.$$ and denote $X$ to be the number of empty cells i.e $X = \sum_{j=1}^{r}I_j$

Qn(1) - Find $E[X]$

Qn(2) - Evaluate $P[I_i = 1; I_j = 1 | N = n]$, and then find the probability $P[I_i=1,I_j=1]$, where $1 \leq i < j \leq r$, and n is a non-negative integer.

My attempt -

$E[I_j|N=n] = P[I_j = 1|N=n] = (\frac{r-1}{r})^n $ $\mathbb{E}[X] = \mathbb{E}[\mathbb E[\sum_{i=1}^r I_j \mid N]] = \sum_{i=1}^r \sum_{n=0}^\infty \textbf P[I_j = 1 \mid N = n] \; \textbf P[N = n]$ $= \sum_{i=1}^r \sum_{n=0}^\infty \textbf (\frac{r-1}{r})^n \frac{e^{-\lambda} \lambda^n}{n!}$

For Qn1, I'm stuck here. I'm not sure how to simplify this expression.

As for Qn2), I don't really understand what the question is asking me. Shouldn't the probability for $P[I_i = 1; I_j = 1 | N = n]$ be the same as $P[I_i=1,I_j=1]$?

Answer 1

For (1) observe that firstly that $\sum_{i=1}^rc=rc$ and secondly that $\sum_{n=0}^{\infty}\frac{a^n}{n!}=e^a$.

Apply this on $a=\frac{r-1}{r}\lambda$.

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alternative route and making use of symmetry:

$P\left\{ I_{1}=1\right\} =\sum_{n=0}^{\infty}P\left\{ I_{1}=1\mid N=n\right\} P\left\{ N=n\right\} =\sum_{n=0}^{\infty}\left(1-\frac{1}{r}\right)^{n}e^{-\lambda}\frac{\lambda^{n}}{n!}=e^{-\frac{\lambda}{r}}$

hence

$\mathbb{E}X=\mathbb{E}\sum_{i=1}^{r}I_{r}=\sum_{i=1}^{r}\mathbb{E}I_{r}=r\mathbb{E}I_{1}=rP\left\{ I_{1}=1\right\} =re^{-\frac{\lambda}{r}}$

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On (2)

$P\left\{ I_{1}=1=I_{2}\mid N=n\right\} =\left(1-\frac{2}{r}\right)^{n}$ and $P\left\{ I_{1}=1=I_{2}\right\} =\sum_{n=0}^{\infty}P\left\{ I_{1}=1=I_{2}\mid N=n\right\} P\left\{ N=n\right\} $

I leave the rest to you.

Answer 2

Note that $\exp(-\lambda)$ is constant and can be pulled to the front of the sum. The remaining inner sum has the value $\exp\left(\frac{r - 1}{r} \lambda\right)$ (remember the representation of $\exp$ as power series).

The outer sum can easily be calculated when you see that the inner sum does not depend on the index of the outer sum.

To answer your second question: Clearly the probabilities are not equal. Note that if you have e.g. $r=2$ the probability is $1$ for $n = 1$ but $\frac{1}{2}$ for $n = 2$.