From Exponential Gaps to Poisson Count Probabilities

From Waiting Times to a Count

A process can be described through the times between its arrivals or through the number of arrivals observed by a given time. For exponential interarrival times, these two descriptions lead to a particularly simple connection: the probability of exactly \(n\) arrivals by time \(t\) is obtained by integrating a joint density over the possible arrival times. The integration region is an ordered simplex, whose volume is \(t^n/n!\).

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Fix a rate \(\lambda>0\), and let \((\tau_k)_{k\ge1}\) be independent random variables, each with exponential density \(f_\tau(x)=\lambda e^{-\lambda x}\) for \(x>0\), and zero otherwise. Define the arrival epochs by \(S_0=0\) and \(S_n=\sum_{k=1}^n\tau_k\) for \(n\ge1\). The count \(N(t)\) is the number of arrivals by time \(t\): when the sequence of arrival epochs is unbounded, \(N(t)=\max\{n\ge0:S_n\le t\}\). We will also show that infinitely many arrivals by a fixed finite \(t\) have probability zero, so this count is finite almost surely.

The Ordered Arrival-Time Density

For \(n\ge1\), the vector of the first \(n\) gaps has joint density \(\lambda^n e^{-\lambda(\tau_1+\cdots+\tau_n)}\) on positive coordinates, by independence. Change variables from gaps to epochs, \(s_k=\tau_1+\cdots+\tau_k\). The inverse transformation is \(\tau_1=s_1\) and \(\tau_k=s_k-s_{k-1}\) for \(k\ge2\). Its Jacobian determinant has absolute value \(1\), and positive gaps correspond to \(0<s_1<\cdots<s_n\). Since the sum of the first \(n\) gaps is \(s_n\), the arrival-epoch density is as follows.

Proof. The joint density of \((\tau_1,\ldots,\tau_n)\) at positive gaps is \(\prod_{k=1}^n\lambda e^{-\lambda\tau_k}=\lambda^n e^{-\lambda\sum_{k=1}^n\tau_k}\). The transformation from gaps to epochs is linear and triangular, with diagonal entries all equal to \(1\), so its determinant is \(1\). Thus the change-of-variables formula introduces no additional factor. The constraints \(\tau_1>0\) and \(\tau_k>0\) for \(k\ge2\) become \(s_1>0\) and \(s_k>s_{k-1}\); also \(\sum_{k=1}^n\tau_k=s_n\). Substitution gives the stated density on the ordered region, and the density is zero outside it. \(\square\)

To have exactly \(n\) arrivals by \(t\), the \(n\)th epoch must be at or before \(t\), and the next independent exponential gap must last longer than \(t-s_n\). For \(n\ge1\), the probability that this next gap exceeds \(t-s_n\) is \(e^{-\lambda(t-s_n)}\). Multiplying this survival probability by the density above cancels the dependence on the final epoch: \(\lambda^n e^{-\lambda t}\). It remains to find the volume of the ordered region.

Proof. Let \(V_n(t)\) denote this volume, and set \(V_0(t)=1\). For \(n\ge1\), fix the largest coordinate \(s_n=u\). The remaining coordinates range over \(0<s_1<\cdots<s_{n-1}<u\), a region of volume \(V_{n-1}(u)\). Integrating over \(0<u<t\) gives \(V_n(t)=\int_0^t V_{n-1}(u)\,du\). Starting from \(V_0(u)=1\), induction yields \(V_n(t)=\int_0^t u^{n-1}/(n-1)!\,du=t^n/n!\). Whether the boundary \(s_n=t\) is included does not change the volume, because that boundary has zero \(n\)-dimensional volume. The formula also gives zero when \(t=0\) and \(n\ge1\). \(\square\)

The Poisson Count Formula

Proof. First take \(n\ge1\). Using the arrival-epoch density and the independent next gap, the probability of exactly \(n\) arrivals is the integral over \(0<s_1<\cdots<s_n<t\) of \(\lambda^n e^{-\lambda s_n}e^{-\lambda(t-s_n)}\). The integrand is the constant \(\lambda^n e^{-\lambda t}\), so the simplex-volume lemma gives \(\mathbb{P}(N(t)=n)=\lambda^n e^{-\lambda t}t^n/n!\). For \(n=0\), there are no arrivals by time \(t\) precisely when \(\tau_1>t\), which has probability \(e^{-\lambda t}\); this agrees with the displayed formula at \(n=0\). To justify that these events describe a finite count almost surely, the joint density of the first \(m\) epochs is at most \(\lambda^m\). Therefore, for \(t>0\), \(\mathbb{P}(S_m\le t)\le\lambda^m V_m(t)=(\lambda t)^m/m!\), which tends to zero as \(m\to\infty\). The event of infinitely many arrivals by \(t\) is contained in \(\{S_m\le t\}\) for every \(m\), so its probability is at most \((\lambda t)^m/m!\) for every \(m\), and hence is zero. For \(t=0\), every gap is positive almost surely, so \(N(0)=0\). Thus \(N(t)\) is finite almost surely, and the formula gives its probabilities for every nonnegative integer \(n\). \(\square\)

The exponential factor in the result depends on both parts of the event “exactly \(n\)”: it accounts for the total waiting time up to the observation time, including the requirement that no additional arrival occurs before \(t\). The factorial comes from the volume of the ordered arrival-time region, not from an extra assumption about the count.

Why the “No Further Arrival” Condition Matters

A common mistake is to integrate only over the event \(S_n\le t\), as if that meant exactly \(n\) arrivals. It means at least \(n\) arrivals. For example, with rate \(\lambda=2\) and \(t=1\), the incorrect calculation for exactly one arrival might be \(\int_0^1 2e^{-2s}\,ds=1-e^{-2}\). This is actually \(\mathbb{P}(S_1\le1)\), the probability of at least one arrival. It does not exclude a second arrival before time \(1\). For exactly one arrival, the integrand must also include the probability that the second gap exceeds \(1-s\): \(2e^{-2s}e^{-2(1-s)}=2e^{-2}\). Integrating over \(0<s<1\) gives \(2e^{-2}\), the correct probability. Substitution in the count formula confirms \(e^{-2}(2\cdot1)^1/1!=2e^{-2}\).

This derivation explains how an arrival-time description determines a single-time count distribution. It uses independence of successive exponential gaps and their survival probabilities; it does not by itself establish independence of counts in disjoint time intervals. For a fixed observation time, the parameter \(\lambda t\) reflects the rate multiplied by the duration, while the ordered simplex accounts for the possible arrangements of the arrivals before that time.