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Consider tossing a coin with probability of heads p, n times. Let N_H and N_T denote the random variable for the number of heads and tails observed respectively. Clearly N_H and N_T are not independent. However if n \sim \mathtt{Poisson}(\mu), something beautiful happens. We get
Let N \sim \mathtt{Poisson}(\mu). Then N_H \sim \mathtt{Poisson}(p\mu) and N_T \sim \mathtt{Poisson}(q\mu), where q = 1 - p. Furthermore, N_H and N_T become independent.
Showing the p.m.f is as claimed
Let the number of coin flips be N \sim \mathtt{Poisson}(\mu). Then
if N=0 then \Pr{[N_H = 0]} = 1. If you do not have trials how can you have a heads.
for N = n for n \geq 1, say we have by independence of the two events
\begin{equation} \Pr{[N=n, N_H = k]} = \frac{e^{-\mu}\mu^n}{n!} \times \binom{n}{k}p^kq^{n-k} \end{equation}
Now if we marginalise over N_H we get
\begin{align} \Pr{[N_H = k]} &= \sum_{n=k}^{\infty} \Pr{[N=n, N_H = k]} \\ &= \sum_{n=k}^{\infty} \frac{e^{-\mu}\mu^n}{n!} \times \frac{n!}{(n-k)!k!}p^kq^{n-k}\\ &= \frac{e^{-\mu}p^k}{k!} \sum_{n=k}^{\infty} \frac{\mu^n}{(n-k)!}q^{n-k}\\ &= \frac{e^{-\mu}p^k}{k!} \sum_{j=0}^{\infty}\frac{\mu^n}{j!}q^{j} \\ &= \frac{e^{-\mu}p^k}{k!}\mu^k \sum_{j=0}^{\infty}\frac{q\mu^j}{j!} \\ &= e^{-\mu}\frac{(p\mu)^k}{k!} e^{q\mu} \\ & = e^{-p\mu}\frac{(p\mu)^k}{k!} \end{align}
Step (5) comes from a change of index n-k=j. Step (7) comes from the Taylor expansion of e^{(q\mu)}.
Proving independence
We showed that if N \sim \mathtt{Poisson}(\mu), then N_H \sim \mathtt{Poisson}(p\mu). Now all need to show is that \Pr{[N_H=h, N_T=t]} = \Pr{[N_T=t]}\Pr{[N_H=h]} where N_H + N_T = N. Let q = 1 - p
\begin{align} \Pr{[N_H=h, N_T=t]} & = \Pr{[N_H=h, N=h+t]} \\ &= \Pr{[N=h+t]}\Pr{[N_H=h]} \\ &= \Bigg(\frac{e^{-\mu}}{(h+t)!}\mu^{(h+t)}\Bigg)\frac{(h+t)!}{t!h!}p^hq^t \\ &= \Bigg(\frac{e^{-\mu(p+q)}}{1}\mu^h \mu^t\Bigg)\frac{1}{t!h!}p^hq^t \\ &= \Big(\frac{e^{-\mu p}}{h!}(p\mu)^{h}\Big) \Big(\frac{e^{-\mu q}}{t!}(q\mu)^{t}\Big)\\ &=\Pr{[N_T=t]}\Pr{[N_H=h]} \end{align}
\square
For the multinomials the same idea applies. Let \mathcal{P} \in \simplex{k}. The Poissonisation process to make analysis easier is works as follows:
Fix any \mathcal{P} \in \simplex{k}, and let (N_1, \dots, N_k) denote the vector of counts among the samples in the Poissonized sampling setting with parameter n. Then
For every j \in [k], N_j \overset{\text{i.i.d}}{\sim} \mathtt{Poisson}(np_j)
N_1, \dots, N_k are mutually independent.
The proof is identical to the binomial proof but for k \geq 2 \square
Let X \sim \mathtt{Poisson}(\lambda) and Y \sim \mathtt{Poisson}(\mu). Then X + Y \sim \mathtt{Poisson}(\lambda + \mu)
When exact binomial calculations are hard, one can use the Poisson distribution to approximate the binomial for large n and small p.
Let X \sim \mathtt{Poisson}(\lambda) and \mu > 0. Let Y = (X - \mu)^2 - X, then \E[Y] = (\lambda - \mu)^2 and \E[Y] = (\lambda - \mu)^4 + 4\lambda + 2\lambda(\lambda - \mu)^2