Identity of the Poisson distribution

Question

I have recently begun to study John Kingsman's "Poisson processes", and in the first chapter, the author defines $\mathbb{P}${$X$ = $n$} = $\pi_{n}$($\mu$) = $\mu^{n}e^{-\mu}/n!$

"Differentiating with respect to $\mu$, we have

$\frac{d\pi_{n}}{d\mu}$ = $\pi_{n-1} - \pi_{n}$ with the convention that $\pi_{-1}=0$.

Thus $\frac{d}{d\mu} \sum_{k=0}^n \pi_k = -\pi_n$.

Integrating from $0$ to $\mu$ gives the useful identity

$\sum_{k=0}^n \pi_k(\mu) = 1 - \int_0^{\mu} \pi_n(\lambda) d\lambda~ \qquad$(*) "

How does one get to the observation (*)? Integrating with respect to $\mu$ or $\lambda$? Where does the $1$ come from?

I am grateful for any tip or piece of advice

Answer 1

In general if $f'(x)=g(x)$ then - if $g$ is a suitable function - we may conclude at first hand that: $$f(x)=\int_0^xg(t)dt+c$$for some constant $c$.

Substituting $x=0$ reveals that $f(0)=c$ so that apparantly:$$f(x)=\int_0^xg(t)dt+f(0)$$

Something similar happens here.

Be aware that $\sum_{k=0}^n\pi_k(0)=1$.

Answer 2

Given $$\frac{d}{d\mu} \sum_{k=0}^n \pi_k = -\pi_n$$ we can integrate both sides and it's fairly simple to see that $$ \sum_{k=0}^n \pi_{k}(\mu) = c - \int_{0}^{\mu} \pi_n(\lambda) ~d\lambda. $$ To compute the constant $c$ plug-in $n = 0$ which implies $$ \pi_0(\mu) = c - \int_{0}^{\mu} \pi_0(\lambda) ~d\lambda \implies e^{-\mu} = c - \int_{0}^{\mu} e^{-\lambda} ~ d \lambda = c - e^{-\mu} - 1 \implies c = 1. $$