Negative values in Poisson process

Question

I'm trying to make some sense of the following definition:

A collection of random variables $\{ N_t \}$$_{t \geq 0}$ is called a Poisson process with rate parameter $\lambda > 0$ if

• $N_0 = 0$
• $N_{t_2} - N_{t_1}$ and $N_{t_4} - N_{t_3}$ are independent for all $0 \leq t_1 < t_2 \leq t_3 < t_4$
• For all $t \geq 0$ and $h > 0$, we have $P(N_{t+h} - N_t = 1) = \lambda h + o(h)$ and $P(N_{t+h} - N_t \geq 2) = o(h)$

Now my question is, does this definition imply that $N_t \geq 0$ for all $t$? I know that this must obviously be true since they represent counts, but how do you actually get this from the definition?

Answer 1

I suspect the definition is incomplete and you may need an extra condition, for example that $P(N_{t+h} - N_t \lt 0) = 0$ or perhaps at most $o(h)$ in the third bullet, or that $P(N_{t+h} - N_t =0) = 1-\lambda h +o(h)$.

As an example, suppose $L_t$ is a conventional Poisson process with parameter $\lambda$ and independently $M_t$ a conventional Poisson process with parameter $\mu$, with $N_t=L_t-M_t$.

Then the first two conditions are satisfied. So too is the third, since when $h \to 0^+$:

• $P(N_{t+h} - N_t = 1) = \lambda h + o(h)$
• $P(N_{t+h} - N_t \ge 2) = o(h)$
• $P(N_{t+h} - N_t = 0) = 1-\lambda h-\mu h + o(h)$
• $P(N_{t+h} - N_t = -1) = \mu h + o(h)$
• $P(N_{t+h} - N_t \le -2) = o(h)$

In this case $N_t$ would have a Skellam distribution with parameters $\lambda t$ and $\mu t$, and would have a positive probability of being negative for any $t>0$, so would not be a conventional Poisson process.