Limit of Poisson process probability

Question

$X(t)$ is a Poisson process with intensity $\lambda = 3$.

I need to compute a limit like:

$$\lim_{\Delta t\to 0}\frac{P(X(\Delta t)=1)}{P(X(\Delta t)\ge1)}$$

Any hints?

Answer 1

By the very definition of a Poisson process, we have $X_t \sim \text{Poi}(\lambda t)$ for any $t \geq 0$; in particular

$$\mathbb{P}(X_t \geq 1) = 1- \mathbb{P}(X_t = 0) = 1- e^{-\lambda t}$$

and

$$\mathbb{P}(X_t = 1) = e^{-\lambda t} \lambda t.$$

Consequently,

$$\begin{align*} \frac{\mathbb{P}(X_t =1)}{\mathbb{P}(X_t \geq 1)} &= e^{-\lambda t} \left( \frac{-e^{-\lambda t}+1}{\lambda t} \right)^{-1} \stackrel{t \to 0}{\to} 1 \cdot \left( \frac{d}{ds} (-e^{-s}) \bigg|_{s=0} \right)^{-1} = 1. \end{align*}$$

Answer 2

One definition of the Poisson process is that, in your notation, for $t \geq 0$ and sufficiently small $\nabla t$, $$P[X(t+\nabla t)- X(t) = 1] = \lambda \cdot \nabla t + o( \nabla t)$$ $$P[X(t+ \nabla t) - X(t) > 1] = o( \nabla t)$$ where $o$ is the so-called little-o notation and it basically refers to any function for which it is true that: $$\underset{\nabla t \rightarrow 0}{\lim} \frac{o(\nabla t)}{\nabla t} = 0$$

From this it should be easy to evaluate your limit. Do you see how?

It may also be worth reiterating that $o$ isn't a specific function so, informally, we usually write things like $o(h) + o(h) = o(h)$ without any trouble.