Poisson process at $t=0$

In the next definition of a Poisson process:

A counting process $$\left \{ N(t); t \geq 0 \right \}$$ is said to be a Poisson process with rate $$\lambda > 0$$ if

1) $$N(0)=0$$

2) $$N(t)$$ has stationary and independent increments

3) $$\mathbb{P}(N(h)=1) = \lambda h + o(h)$$

4) $$\mathbb{P}(N(h)\geq 2) = o(h)$$

Why can't we count an occurrence or event at the initial point $$t=0$$ in 1)?

Nothing can happen while $$t<0$$ since we don't consider the negative time. Now, $$N(0) \ne 0$$ means the event happens exactly at $$t=0$$.

What is the Probability that the event happens at $$t=0$$ exactly?

More generally, what is the probability, $$Pr(X=c)$$ when $$X$$ is a continuous random number and $$c$$ is a specific number?

Even though the process counts the number of occurrences, the process is the same as pointing occurrences on the timeline (as you mentioned). If the time is in between the time interval, the occurence is counted.

What is the probability that you point an occurrence on timeline, exactly at $$t=0$$?
This question is the same as What is the probability of N(0) = 1.

• The Poisson process is discrete random, but the time is continuous. $t$ is time. Therefore, when we talk about when the event will happen, we have to give a time interval, not a specific time. If you ask me what is the probability that an event happens at $t=0$ (or 1 or 2.35) exactly, the answer is 0 regardless of $\lambda$. – S. Phil Kim Jun 8 at 20:24
• To make it sure, Only way to have $N(0)=1$ is that an event happens at $t=0$. Therefore, $Pr(N(0)=1) = Pr($An event hapen at $t=0)$ – S. Phil Kim Jun 8 at 20:30
• Thanks for your response. I know that for continous random variables $Pr(X=c) is 0$. But in this case only the time is continous. The occurrences or events are discrete, points in time. They aren't a continous function. – roy212 Jun 8 at 20:33
• "To make it sure, Only way to have $N(0)=1$ is that an event happens at $t=0$. Therefore, $Pr(N(0)=1)=$Pr(An event hapen at $t=0$)". I don't quite understand it. In the first comment you mentioned, that "the probability that an event happens at $t=0$ (or $1$ or $2.35$) exactly, the answer is 0 regardless of λ." – roy212 Jun 8 at 20:39
• That means $Pr(N(0)=1) = 0$ regardless of $\lambda$. The same way, $Pr(N(0)=2) = 0$ regardless of $\lambda$. $Pr(N(0)=3) = 0$ regardless of $\lambda$. Therefore. $N(0) = 0$, (regradless of $\lambda$ – S. Phil Kim Jun 8 at 20:42