Joint Mass Function of a Poisson Process

Question

Let $N(t)$ be a Poisson process with parameter $\lambda$. We are asked to find the joint mass function of $N_s$ and $N_t$ with $s < t$. Because the process has stationary and independent increments, we have: \begin{align*} P(N_s = m, \ N_t = n) \ \ & = \ \ P(N_s = m, \ N_{t-s} = n-m) \\ & = \ \ P(N_s = m) \ P(N_{t-s} = n-m) \\ \\ & = \ \ e^{-\lambda s} \frac{(\lambda s)^m}{m!} \cdot e^{-\lambda(t-s)} \frac{(\lambda(t-s))^{n-m}}{(n-m)!} \\ & = \ \ e^{-\lambda t} \frac{(\lambda t)^n}{n!} \cdot \binom{n}{m} \left( \frac{s}{t}\right)^m \left( \frac{t-s}{t}\right)^{n-m} \end{align*} So it becomes a product of two mass functions. The fact that it factors out means that the Poisson and the binomial random variable are independent. Why is that? The binomial random variable uses $t$ and $n$ as constants.

Answer 1

You have found that the joint probability mass function is the product of the marginal probability mass function of the Poisson random variable, $N_t$ and the probability mass function of a Binomially distributed random variable. $$\mathsf P(N_s{=}m, N_t{=}n)~=~\dfrac{(\lambda t)^n e^{-\lambda t}}{n!}\cdot\binom n m \dfrac{s^m(t-s)^{n-m}}{t^n}$$

The binomial random variable is the count of Poisson point events that happen in interval $(0;s]$ when given that $n$ happen in interval $(0;t]$ (where $0{<}s{<}t$).   This is not independent of the count of point events that happen during $(0;t]$, it is conditional.

$$(N_{s}\mid N_{t}{=}n) ~\sim~\mathcal{Bin}(n, s/t)~~\iff~~ \mathsf P(N_s{=}m\mid N_t{=}n)~=~\binom n m \dfrac{s^m(t-s)^{n-m}}{t^n}$$

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$$\mathsf P(N_s{=}m, N_t{=}n)~=~\mathsf P(N_t{=}n)\,\mathsf P(N_s{=}m\mid N_t{=}n)$$