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.

  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $S=\{i_1<i_2<\dots<i_k\}$. 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.


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.

  • $\begingroup$ Conditioned on $N=n$, the $X_i$ cannot be iid, because their sum is equal to $1$ almost surely. $\endgroup$ – Mike Earnest Feb 27 at 0:35
  • $\begingroup$ 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 $1$. $\endgroup$ – Goodarz Mehr Feb 28 at 2:51

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