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Given a Poisson process $N_t$. If we set $$f_t(n):=P(N_t=n)=\frac{e^{-t}t^n}{n!}.$$ We know $\sum_{n\ge 0} f_t(n)=1$.
My question is that how to conclude that $$\sum_{n>et} f_t(n)<e^{-t}?$$
This is equivalent as to show that $$\sum_{n>et}\frac{t^n}{n!}<1$$
Note that $n!\approx\sqrt{2\pi n}n^ne^{-n}$, then $$\sum_{n>et}\frac{t^n}{n!}\le \sum_{n>et}\frac{t^ne^n}{\sqrt{2\pi n} n^n}$$
The mean value form of the remainder $\sum_{n>m} \frac{t^n}{n!}$ of the $m$th-degree Taylor approximation to $e^t$ is $\frac{1}{(m+1)!} \xi^{m+1}$ where $\xi$ is between $0$ and $t$.
When $m \approx et$ (hand-waving a bit here, you can be more rigorous with floor/ceiling functions), we have $\xi^{m+1} \le t^{et + 1}$ and $$\frac{1}{(m+1)!} \overset{*}{\le} \left(\frac{e}{m+1}\right)^{m+1} \le \frac{e^{et+1}}{(et)^{et+1}} = \frac{1}{ t^{et+1}}$$
where the starred inequality is due to $e^k = \sum_{n \ge 0} \frac{k^n}{n!} \ge \frac{k^k}{k!}$.
