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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.

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  • $\begingroup$ How do you define Poisson process? $\endgroup$ – Kavi Rama Murthy May 20 at 5:34
  • $\begingroup$ Number of arrivals at time $t$ $\endgroup$ – Vahan May 20 at 6:20
  • $\begingroup$ 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 $t=0$". $\endgroup$ – Vahan May 20 at 19:32
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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

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

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  • $\begingroup$ 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. $\endgroup$ – Vahan May 21 at 17:37
  • $\begingroup$ @Vahan See Theorem A.1 here $\endgroup$ – saz May 22 at 5:35
  • $\begingroup$ 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 $e^{tg(\alpha)}$? $\endgroup$ – Vahan May 22 at 6:06
  • $\begingroup$ 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 $G(1)^t = e^{tg(\alpha)}$ $\endgroup$ – Vahan May 22 at 6:14
  • $\begingroup$ @Vahan Well, $g(\alpha) = \log(G(1))$... as I wrote in my answer. And yes, $G(1)$ is just a deterministic number. $\endgroup$ – saz May 22 at 6:43

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