# Proof of Poisson process distribution

I'm confused about part of a proof of this theorem.

Theorem: If {$$N_t; t \geq 0$$} is a Poisson process, then for any $$t \geq 0$$, $$P(N_t = k) = \frac{e^{\lambda t}(\lambda t)^k}{k!}$$ where $$k = 0, 1, ...$$ and some constant $$\lambda \geq 0$$.

Proof

Let $$G(t) = E(\alpha ^{N_t})$$. Writing $$N_{t+s} = N_t + (N_{t+s} - N_t)$$, using the independence of $$N_{t+s} - N_t$$ [and prior results] we get $$G(t+s) = E(\alpha ^{N_{t+s}})=E(\alpha ^{N_t}\alpha ^{N_{t+s}-N_t})=E(\alpha ^{N_t})E(\alpha ^{N_{t+s}-N_t})=G(t)(G(s)$$ Since $$G(t)=\sum \alpha ^n P(N_t = n) \geq P(N_t = 0)=e^{-\lambda t}, G$$ does not vanish for any $$t$$, and $$G(t+s) = G(t)G(s)$$ can be satisfied only if

$$G(t) = e^{tg(\alpha)}, t \geq 0$$

Note that $$g(\alpha)$$ is the derivative of $$G$$ at $$t=0$$

I'm confused about the previous two lines. Why does $$G(t)$$ need to have the form above? And how is it that $$g(\alpha)$$ is the derivative of $$G$$ at $$0$$? It seems like circular referencing, $$g$$ is the derivative, yet it appears in the expression of $$G$$. Strange.

• How do you define Poisson process? – Kavi Rama Murthy May 20 at 5:34
• Number of arrivals at time $t$ – Vahan May 20 at 6:20
• I think I understand why $G(t)$ needs to have an exponential. Since $G(t+s) = G(t)G(s)$, we need an exponential because exponentials have the property that $e^{t+s} = e^te^s$. I'm still lost on "$g(\alpha)$ is the derivative of $G$ at $t=0$". – Vahan May 20 at 19:32

## 1 Answer

There is the following general result

Let $$f:[0,\infty) \to \mathbb{C}$$ be a right-continuous function. If $$f$$ satisfies the Cauchy-Abel functional equation $$f(s+t) = f(s) f(t), \qquad s,t \geq 0, \tag{1}$$ then $$f(t) = f(1)^t$$ for all $$t \geq 0$$.

Applying this result we find that the function $$G(t) = \mathbb{E}\alpha^{N_t}$$ is of the form

$$G(t) = G(1)^t, \qquad t \geq 0,$$

which can be equivalently written as

$$G(t) = e^{t g(\alpha)}$$

where $$g(\alpha) := \log(G(1))$$ (note that it is well-defined since $$G(1)>0$$ for all $$\alpha \geq 0$$). Differentiating $$G$$ with respect to $$t$$ we find

$$G'(t) = g(\alpha) e^{t g(\alpha)},$$

and so

$$G'(0) = g(\alpha).$$

• Please provide a link about the Cauchy-Abel, I couldn't find a good one. And I was confused about the derivative because I thought the chain rule would complicate things, that is I thought there would be $g'(\alpha)$. Thanks for the answer. – Vahan May 21 at 17:37
• @Vahan See Theorem A.1 here – saz May 22 at 5:35
• Thank you but I don't understand that Theorem. But assuming that $G(t) = G(1)^t$, why is it that it can equivalently be written as $e^{tg(\alpha)}$? – Vahan May 22 at 6:06
• Actually I think I see it now. $G(1)$ is just a number, so let $g(\alpha)$ such that $e^{g(\alpha)} = G(1)$. Then $G(1)^t = e^{tg(\alpha)}$ – Vahan May 22 at 6:14
• @Vahan Well, $g(\alpha) = \log(G(1))$... as I wrote in my answer. And yes, $G(1)$ is just a deterministic number. – saz May 22 at 6:43