# Poisson process time of arrival proposition

This is taken from a stochastic processes text.

Consider $$N$$ a Poisson process (number of arrivals at time $$t$$) with rate $$\lambda$$. Let $$T_n$$ denote an arrival time ($$n$$ a natural number). The probability that there are no arrivals in $$(t, t+s]$$ is $$e^{-\lambda s}$$, independent of the history of arrivals before $$t$$. This can be applied to arrival times: $$P(N_{T_{n+s}}-N_{T_n} = 0 | N_u; u \leq T_n) = e^{-\lambda s}$$

The following proposition is provided:

For any $$n$$ a natural number,

$$P(T_{n+1}-T_n \leq t | T_0, ..., T_n) = 1 - e^{-\lambda t}, t \geq 0$$

My question is about the proposition. It appears that the probability is the complement of no arrivals. So why is the interval $$\leq t$$? I'm getting confused about the exponential expression appearing at first with number of arrivals whereas it is appearing again as a complement with times of arrivals.

You are right about the fact that the event $$\{T_{n+1}-T_n \leq t\}$$ is the complement of the event $$\{N_{T_n+t}-N_{T_n}=0\}$$.
The way to understand the proposition, therefore, is to consider the probability, $$P(N_{T_n+t}-N_{T_n}=0|T_0,T_1,\ldots,T_n)$$, which is nothing but $$P(N_{T_n+t}-N_{T_n}=0|\{N_u:u\leq T_n\})$$. This is because the collection of random variables $$T_0,\ldots,T_n$$ describes precisely the arrival process $$\{N_u:u\leq T_n\}$$.
Now, we know from the Strong Markov property that $$P(N_{T_n+t}-N_{T_n}=0|\{N_u:u\leq T_n\})=e^{-\lambda t}$$, like you've mentioned. The required probability, $$P(T_{n+1}-T_n \leq t|\{N_u:u\leq T_n\}) = 1-e^{-\lambda t}$$, by the complement property.
• But I don't understand why $T_{n+1} - T_n \leq t$. It seems like it should be $> 0$, as it is the complement of $0$ arrivals. – Vahan May 21 at 17:15
• Could it be stated as follows: $P(T_{n+1}-T_n > t) = e^{-\lambda t}$, which would mean the probability of no arrivals between the two given arrival times within time $t$? – Vahan May 21 at 19:02
• Yes, that is correct. The event $\{T_{n+1}-T_n>t\}$ means that the next arrival, after the $n^{\text{th}}$, would occur only after a duration of $t$. Or in other words, there are no arrivals between $T_n$ and $T_n + t$. – arvram96 May 23 at 4:59