# Interarrival Time Distribution of a Poisson Process

For a Poisson Process with parameter $$\lambda$$ restricted to the interval $$[0, 1]$$, what is the probability that at least one of the interarrival times (including the time between $$0$$ and the first arrival time and between the last arrival time and $$1$$) is greater than or equal to $$d$$, where $$d$$ is a given parameter?

In other words, if $$T_{1}, T_{2}, \ldots, T_{N}$$ are the arrival times in the interval $$[0, 1]$$, where $$N$$ is a Poisson random variable with parameter $$\lambda$$, and $$X_{0}, X_{1}, \ldots, X_{n}$$ are the interarrival times, what is the probability of $$P[\exists i: X_{i} \ge d] = 1 - P[X_{i} < d\,\,\,\forall\, 0 \le i \le n]$$.

I did some numerical simulations in MATLAB and it seems that the probability is Guassian as a function of $$\lambda$$ and $$d$$ individually, but I may be wrong.

• Hint: you are in fact asking to find $1-\mathbb{P}\left(\color{blue}{\max\{X_1,\ldots,X_N\} < d}\right)$. To do this, try and condition on the value of $N$ and use the law of total probability. – Minus One-Twelfth Feb 23 at 19:55

Conditioned on $$N=n$$, the set of arrival times is distributed like $$\{U_i:i=1,2,\dots,n\},$$ where $$U_i$$ are iid uniformly distributed on $$[0,1]$$ (see here for proof). Given $$n$$ uniform samples, we want the probability that they divide $$[0,1]$$ into pieces whose lengths are all at most $$d$$.

I claim this probability is $$P({\textstyle\max_{i=0}^N} X_i\le d\mid N=n)=\sum_{k=0}^{\lfloor1/d\rfloor}(-1)^k\binom{n+1}{k}(1-dk)^n\tag{1}$$ This follows by a sort of inclusion-exclusion argument. Let $$E_i$$ be the event that $$X_i>d$$. We want the the probability of the intersection $$E_0^c\cap E_1^c\cap \dots\cap E_n^c$$. This is equal to the sum of $$(-1)^{|S|}P\left(\bigcap_{i\in S}E_i\right)$$ where $$S$$ ranges over subsets of $$\{0,1,\dots,n\}$$. The event $$\bigcap_{i\in S}E_i$$ is a certain region of the hypercube $$[0,1]^n$$. Let $$S=\{i_1. I claim that the following is a volume preserving bijection from $$\bigcap_{i\in S}E_i$$ to the hypercube $$[0,1-dk]^n$$. Namely, if $$T_1 is the $$U_i$$ in sorted order, then

• Take all points $$T_j$$ for which $$j\ge i_1+1$$, and decrease their values by $$d$$.
• Take all points $$T_j$$ for which $$j\ge i_2+1$$, and decrease their values by $$d$$.
• $$\vdots$$
• Take all points $$T_j$$ for which $$j\ge i_k+1$$, and decrease their values by $$d$$.

Points can have their values decreased multiple times in this procedure. For example, $$T_n$$ is decreased once for each $$i\in S$$ for which $$i, so $$|S\setminus \{n\}|$$ times.

Since the volume of this hypercube is $$(1-dk)^n$$ when $$dk\le 1$$, the probability of $$\bigcap_{i\in S}E_i$$ is $$(1-dk)^n$$. We only need to sum up to $$|S|=\lfloor1/d\rfloor$$, because for larger sets the probability is zero. Putting this all together proves $$(1)$$. Combining this with $$P({\textstyle\max_{i=0}^N} X_i\le d)=\sum_{n=0}^\infty P({\textstyle\max_{i=0}^N} X_i\le d\mid N=n)\cdot e^{-\lambda}\frac{\lambda^n}{n!}$$ answers your question.

Update. The solution below is incorrect.

Original Answer. Here is the way I figured it out (which is similar to the answer above).

Noting that $$X_{i}, 0 \le i \le n$$ form a set of iid exponential random variables (this assumption is not correct), we can write

$$P[\max_{i = 0} ^ {N}(X_{i}) < d] = \sum_{n = 0} ^ {\infty} P[\max_{i = 0} ^ {N}(X_{i}) < d \vert N = n]P[N = n] = \sum_{n = 0} ^ {\infty} \prod_{i = 0} ^ {n} P[X_{i} < d]P[N = n] = \sum_{n = 0} ^ {\infty} (1 - e ^ {- \lambda d}) ^ {n + 1} e ^ {- \lambda} \frac{\lambda ^ {n}}{n!} = (1 - e ^ {- \lambda d})e ^ {- \lambda e ^{- \lambda d}}.$$

This expression seems to agree with the numerical simulations.

• Conditioned on $N=n$, the $X_i$ cannot be iid, because their sum is equal to $1$ almost surely. – Mike Earnest Feb 27 at 0:35
• I see, but what if we define $X_{N}$ as the interarrival time between the $N$-th and $N + 1$-th arrivals, as opposed to the interarrival time between the $N$-th arrival and $1$? Conditioned on $N = n$, $X_{i}, 0 \le i \le n - 1$ are still iid's, but I am not sure if that extends to $X_{n}$, since we require that the arrival happen after $1$. – Goodarz Mehr Feb 28 at 2:51