Given is a Poisson process $(N_t)_{t \geq 0}$ with intensity $\lambda(t)$. I am interested in finding the distribution of the first arrival time $T$. I can use the fact that for any $t \geq 0$ it holds $T > t$ iff $N_t = 0$. By the definition of the process $N_t \sim Poi(\mu(t))$, where $\mu(t) = \int_0^t \lambda(z) dz$. I get
$$
\mathbb{P}[T > t] = \mathbb{P}[N_t = 0] = \exp(-\int_0^t \lambda(z)dz),
$$
and therefore the distribution is given by $\mathbb{P}[T \leq t] = 1 – \exp(-\int_0^t \lambda(z)dz)$.
If I consider the intensity $\lambda(z) = \frac1{1+z^2}$, I can calculate
$$
\lim_{t \to \infty} \mathbb{P}[T \leq t] = 1 – \exp(-\int_0^\infty \lambda(z)dz) = 1 – \exp(-\frac\pi2) \neq 1,
$$
so $\mathbb{P}[T < \infty] < 1$. Have I done something wrong or is it simply possible that $T$ attains infinity with positive probability, so the process never jumps?
Best Answer
A homogeneous Poisson process with intensity $\lambda > 0$ almost surely has a finite first arrival time, since $$\lim_{t \to \infty} \Pr[T \le t] = \lim_{t \to \infty} 1 - e^{-\lambda t} = 1.$$
However, as you have observed, this is not the case for an inhomogeneous process, since the cumulative intensity (aka cumulative hazard) need not be an unbounded function of time. That is to say, if $$\Lambda(t) = \int_{z = 0}^t \lambda(z) \, dz$$ obeys $$\lim_{t \to \infty} \Lambda(t) < \infty,$$ then the first arrival time may not be finite.