Interarrival Time Distribution of a Poisson Process

Question

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.

Answer 1

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 $S=\{i_1<i_2<\dots<i_k\}$. 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 $T_1<T_2<\dots<T_n$ 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 $i<n$, 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.

Answer 2

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.