(Feller Vol.1, P.301) 4. Let $N$ have a Poisson distribution with mean $\lambda$, and let $N$ balls be placed randomly into $n$ cells. Show without calculation that the probability of finding exactly $m$ cells empty is ${n \choose m} e^{-\lambda m/n} [1- e^{-\lambda /n}]^{n-m}$.
Let $X_i$ be the number of balls in the $i$th cell. Then, $P(X_i =0)=\sum_j P(N=j)\frac{(n-1)^{j}}{n^j} = e^{-\lambda /n}$. So, the result follows from the binomial distribution with $e^{-\lambda/n}$.
- Continuation. Show that when a fixed number $r$ of balls is placed randomly into $n$ cells the probability of finding exactly $m$ cells empty equals the coefficient of $e^{-\lambda}\lambda^r/r!$ in the expression above. (a) Discuss the connection with moment generating function. (b) Use the result for the effortless derivation of ${n \choose m} \sum_{v=0}^{n-m} (-1)^v {n-m \choose v} (n-m-v)^r$.
I know that this probability is equal to ${n \choose m}{n-m-1 \choose r-1}/n^r$ since ${n-m-1 \choose r-1}$ represents the number of ways for $r$ balls to be placed in $n-m$ cells without making any cell empty. Then, I have the following generating function of this probability distribution $$\sum_{r=0}^\infty s^r P(N=r){n \choose m}{n-m-1 \choose r-1}/n^r = \sum_{r=0}^\infty s^r e^{-\lambda}\lambda^{r}/r!{n \choose m}{n-m-1 \choose r-1}/n^r.$$ I also know that The moment generating function looks like $\sum_{r=0}^\infty \frac{m_r}{r!} s^r$ for $m_r$ represents the $r$th moment.
I am not sure if this is correct because I can't see any connection between two generating functions. I would appreciate if you give some help.
Using the notation from the book, let $p_m(r,n)$ denote the probability of finding exactly $m$ cells empty when $r$ balls are randomly distributed among $n$ cells. Let $E_m(r,n)$ be the corresponding number of such distributions, so $E_m(r,n) = p_m(r,n)\cdot n^r$. Note that your formula for $p_m(r,n)$ is not correct because it assumes that the balls are indistinguishable and that all indistinguishable arrangements are equiprobable (Bose-Einstein statistics). For this problem, the balls are distinguishable and all of the $n^r$ possible distributions of the balls among the cells are equiprobable (Boltzmann-Maxwell statistics).
Let $M$ denote the number of empty cells in the process described in the problem. Then, \begin{align*} \dbinom{n}{m}e^{-\lambda m/n}\left(1-e^{-\lambda /n}\right)^{n-m}&=P(M=m) = \displaystyle\sum_{r=0}^\infty P(M=m,N=r)\\ &= \displaystyle\sum_{r=0}^\infty P(M=m|N=r)\cdot P(N=r)\\ & = \displaystyle\sum_{r=0}^\infty p_m(r,n)\cdot e^{-\lambda}\frac{\lambda^r}{r!}, \end{align*}
as desired.
(a) I am not entirely sure about the connection to moment-generating functions (MGFs) other than the fact that in the MGF, the $r^{th}$ moment is the coefficient of $\frac{s^r}{r!}$ and here, the corresponding coefficient is $p_m(r,n)\cdot e^{-\lambda}$. So I suppose you can say that $e^\lambda\cdot P(M=m)$ defines a MGF in $\lambda$ for a probability distribution whose $r^{th}$ moment is $p_m(r,n)$.
(b)
\begin{align*} \dbinom{n}{m}e^{-\lambda m/n}\left(1-e^{-\lambda /n}\right)^{n-m}&= \dbinom{n}{m}e^{-\lambda}e^{\left(\frac{n-m}{n}\right)\lambda}\left(1-e^{-\lambda /n}\right)^{n-m}\\ &= \dbinom{n}{m}e^{-\lambda}\left(e^{\frac{\lambda}{n}}\right)^{n-m}\left(1-e^{-\lambda /n}\right)^{n-m}\\ & = \dbinom{n}{m}e^{-\lambda}\left(e^{\lambda/n}-1\right)^{n-m}\\ & = \dbinom{n}{m}e^{-\lambda}\displaystyle\sum_{v=0}^{n-m}\dbinom{n-m}{v}(-1)^ve^{\lambda\left(\frac{n-m-v}{n}\right)} \end{align*}
Using the Taylor series $\displaystyle e^x = \displaystyle\sum_{r=0}^{\infty}\frac{x^r}{r!},$ we see that the coefficient of $\displaystyle e^{-\lambda}\frac{\lambda^r}{r!}$ in the last line above is given by $$p_m(r,n) = \dbinom{n}{m}\displaystyle\sum_{v=0}^{n-m}\dbinom{n-m}{v}(-1)^v\left(\frac{n-m-v}{n}\right)^r.$$
Multiplying by $n^r$ yields the desired formula for $E_m(r,n)$.