# Time sampling an ordinary poisson process

My questions will be given at the end, let me just give some definitions first.

The counting process $\{ N(t), t \geq 0 \}$ is said to be a non homogenous Poisson process with intensity function $\lambda(t), t \geq 0$ if:

$N(0)= 0$

$\{ N(t),t \geq 0 \}$ has independent increments

$P\{ N(t+h) - N(t) \geq 2 \} = o(h)$

$P\{ N(t+h) - N(t) =1 \} =\lambda(t)h + o(h)$

Time sampling and ordinary poisson process generates a nonhomgenous poission process. Let $\{N(t),t\geq 0 \}$ be a poisson process with rate $\lambda$ and suppose that an event occurring at time $t$ is , independently of what has occurred prior to $t$, counted with probability $p(t)$,then the counting process $\{N_c(t), t \geq 0 \}$ is a nonhomogenous poisson process with intensity function $\lambda(t) = \lambda\cdot p(t)$

The book now shows that a time sampling satisfies the axioms above.verifying the last axiom we calculate

$P\{ N_c(t+h) - N_c(t) =1 \}=P\{ N_c(t,t+h)=1\} = \cdots =P\{N_c(t,t+h) = 1 \mid N(t,t+h) = 1 \}\lambda h + o(h) = p(t)\lambda h + o(h)$

Now, what i don't understand is how $P\{N_c(t,t+h) = 1 \mid N(t,t+h) = 1 \} = p(t)$, the way the book defined $p(t)$ (see above). First of all we are conditioning on that the regular poisson process is one, and also the above says that there is 1 event happening in the intervall $(t,t+h)$ , but $p(t)$ is the probability that an event occur at time $t$.

Could someone clarify this?

## 1 Answer

Events are occurring under the homogeneous process $N$ at rate $\lambda$. Each such event, occurring at time $t$ say, is then either counted in the $N_c$ process, with probability $p(t)$, or ignored with probability $1-p(t)$. So it's not correct to say that $p(t)$ is the probability that an event occurs at time $t$ in process $N_c$.

Mathematically, we can say:

$$p(t) = P(\text{Event at time t is counted}\mid\text{Event occurs under process N at time t}).$$

This is close to the line in the proof you refer to but there is a time interval of size $h$, so it is only an approximation

$$P\{N_c(t,t+h) = 1 \mid N(t,t+h) = 1 \} = p(t).$$

If the event occured at time $t_0 \in [t,t+h]$ then the true value for the RHS would be $p(t_0)$. But for small $h$, $p(t)$ is a reasonable approximation. Also, the "error" in this approximation is accounted for in the term $o(h)$ allowing the use of the equality sign.

• is this the right reasoning? : $P\{N_c(t,t+h) = 1 \mid N(t,t+h) = 1 \ \} = p(t) + g(h)$ where $g(h)$ is a function such that $lim_{h \rightarrow 0} g(h) = 0$? @Mick A, because as h get smaller and smaller then we finally get $p(t)$ – Danny Jun 17 '15 at 9:40
• @Danny Almost. The limit has to be stronger than that though. We need $\lim_{h\to 0} (g(h)/h) = 0$. So $g(h)$ has to approach $0$ much faster than $h$ itself does. For example, from the definition, we have $P\{N(t+h)-N(t)=1\}=\lambda(t)h+o(h)$. Here, as $h\to 0$, both $\lambda(t)h$ and $o(h)$ approach $0$ but $o(h)$ does so much faster than does $\lambda(t)h$. – Mick A Jun 17 '15 at 12:57
• Yes i know what you mean, but i think you forgot that $P\{N_c(t,t+h) = 1 \mid N(t,t+h) = 1 \} = p(t)$ above must be multiplied with $\lambda h$ (see above), so what I mean is that it should be $P\{N_c(t,t+h) = 1 \mid N(t,t+h) = 1 \}\lambda h = (p(t) + g(h))\lambda h$. and therefore $t(h) = g(h)\lambda h = o(h)$, then it is enough that $g(h)$ goes to zero. However when i think about it why should the added contribution to $p(t)$ be a function of 'h' so my reasoning seems not be right. @Mick A – Danny Jun 17 '15 at 16:09
• @Danny I don't think $P\{N_c(t,t+h)=1\mid N(t,t+h)=1\}=p(t)+g(h)$. It is just $P\{N_c(t,t+h)=1\mid N(t,t+h)=1\}=p(t)$ because you are given that there is a single event ($N(t,t+h)=1$) and we know that with probability $p(t)$ it is counted in process $N_c$. (If I understand you correctly, this is kind of what you're saying in your last sentence.) – Mick A Jun 17 '15 at 17:15
• .but $P\{N_c(t,t+h)=1\mid N(t,t+h)=1\}$is not $p(t)$, but as you said before, it is an approximation, so what is actually $P\{N_c(t,t+h)=1\mid N(t,t+h)=1\}$. @Mick A – Danny Jun 17 '15 at 17:20