How to show that a poisson distribution for a random variable becomes a binomial distribution, given the following conditional probability condition?

Question

For Poisson random variables $$N_1 \sim P(\lambda_1)$$ $$N_2 \sim P(\lambda_2)$$ $$N \sim P(\lambda)$$ where $N=N_1+N_2$ and $N_1$, $N_2$ are independent, how to show that $P(N_1=n_1)$ or $P(N_2=n_2)$ follows a binomial distribution, for a given value of $N$?

Answer 1

Let $N_1 \sim \mathcal{P}(\lambda_1)$ and $N_2 \sim \mathcal{P}(\lambda_2)$. Assume that $N_1$ and $N_2$ are independent. Let $N = N_1 + N_2$ and $\lambda = \lambda_1 + \lambda_2$. We then have $N = \mathcal{P}(\lambda)$.

From this page, we can then write: $$ \mathbb{P}[(N=n) \wedge (N_2 = n - n_1)] = \mathbb{P}[N_1 = n_1]\,\mathbb{P}[N_2 = n - n_1], $$ which implies: $$ \mathbb{P}[N_2 = n - n_1\,\vert\,N=n]\,\mathbb{P}[N=n] = \mathbb{P}[N_1 = n_1]\,\mathbb{P}[N_2 = n - n_1]. $$ We then have: $$ \mathbb{P}[N_2 = n - n_1\,\vert\,N=n] = \frac{\mathbb{P}[N_1 = n_1]\,\mathbb{P}[N_2 = n - n_1]}{\mathbb{P}[N=n]} = \frac{ \frac{\lambda_1^{n_1} e^{-\lambda_1}}{n_1!} \frac{\lambda_2^{n - n_1} e^{-\lambda_2}}{(n - n_1)!}}{\frac{\lambda^{n} e^{-\lambda}}{n!}} $$ $$ = \frac{n!}{n_1! (n - n_1)!} \frac{\lambda_1^{n_1}}{\lambda^{n_1}} \frac{\lambda_2^{n - n_1}}{\lambda^{n - n_1}} \frac{e^{-\lambda_1}e^{-\lambda_2}}{e^{-\lambda}}. $$ Since $\lambda_1 + \lambda_2 = \lambda$, we finally find: $$ \mathbb{P}[N_2 = n - n_1\,\vert\,N=n] = \left( \begin{array}{c} n \\ n_1 \end{array} \right) \left( \frac{\lambda_1}{\lambda} \right)^{n_1} \left(1 - \frac{\lambda_1}{\lambda} \right)^{n - n_1}. $$