A limit question about Poisson process

Question

Assume$ \{ N(t)\} (t\ge 0)$ is a Poisson process，then how to show that $$P\left\{\lim_{t\rightarrow +\infty} N(t)＝+\infty \right\}＝1.$$

Answer 1

$\newcommand{\e}{\operatorname{E}}\newcommand{\v}{\operatorname{var}}$Since $N(t)$ is a nondecreasing function of $t,$ the only alternative to $\lim\limits_{t\,\to\,+\infty} N(t) = +\infty$ is that there is some number $L<\infty$ such that $\lim\limits_{t\,\to\,+\infty} N(t) = L,$ and then we have for all $t>0,$ $N(t)\le L.$

Recall that $\e N(t) = t\e N(1) = t\lambda$ and $\v N(t) = t \v N(1) = t\e N(1) = t\lambda$ (the variance of a Poisson distribution is the same as its expected value).

Recall Chebyshev's inequality: $\displaystyle \Pr( |X-\e X| > b) \le \frac{\v X}{b^2}.$

For every $0<L<+\infty$ and for $t$ large enough, we have \begin{align} & \Pr( N(t) \le L) \le \Pr(|N(t)-\lambda t| \ge 2\lambda t - L) \le \frac{t\lambda}{(\lambda t - L)^2} \\[10pt] & \Pr\left( \lim_{t\,\to\,+\infty} N(t)=+\infty \right) \ge 1 - \frac{\lambda t}{(\lambda t-L)^2}. \end{align} Thus $\displaystyle \Pr\left( \lim_{t\,\to\,+\infty} N(t)=+\infty \right)$ is $\ge$ every number less than $1.$