Probability of a Poisson Point Process

Question

If I have a poisson point process $N(t)$ with $\lambda=1$, and I am interested in $P(N(t) \geq 2t)$, how would I go about solving this? Would it just be a product of poisson CDF calculations?

Answer 1

For a fixed $t>0$, we have \begin{align} \mathbb P(N(t)\geqslant 2t) &= \mathbb P(N(t)\geqslant \lceil 2t\rceil)\\ &= \sum_{n=\lceil 2t \rceil}^\infty \mathbb P(N(t)=n)\\ &= \sum_{n=\lceil 2t \rceil}^\infty e^{-\lambda t}\frac{(\lambda t)^n}{n!}. \end{align} There isn't really a nice closed form for this expression, but Mathematica gives me this: $$ 1-\frac{\Gamma (\lceil 2 t\rceil ,\lambda t)}{\Gamma (\lceil 2 t\rceil )}, $$ where $\Gamma(z) = \int_0^\infty x^{z-1}e^{-x}\ \mathsf dx$ is the Gamma function and $\Gamma(s,x) = \int_x^\infty t^{s-1}e^{-t}\ \mathsf dt$ is the upper incomplete gamma function. The case where $\lambda=1$ does not particularly simplify things.