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The Poisson Process: Definitions and Equivalent Characterizations · Tutorial 398 of 994

Independent Increments from Exponential Interarrivals

See how exponential memorylessness turns a sequence of iid gaps into independent Poisson counts across disjoint intervals.

Intermediate 12 min read

What You'll Learn

  • Define the arrival-count process from iid exponential interarrival times.
  • Use the exponential lack-of-memory property to identify the residual waiting time at a fixed boundary.
  • Prove that future gaps are independent of the arrival history up to that boundary.
  • Deduce independent counts on any finite collection of disjoint intervals.
  • Compute joint count probabilities and recognize the error in conditioning on a long wait.

Why Exponential Gaps Give Independent Counts

Knowing the count in one time interval does not, by itself, tell us how counts in other intervals behave. For a Poisson process, counts in disjoint intervals are independent. We will derive that property directly from the construction using independent exponential waiting times. The key step is at a deterministic interval boundary: conditional on waiting without an arrival up to that boundary, the remaining wait still has the original exponential distribution. The future gaps are also independent of the past.

Fix a rate \(\lambda>0\) and a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((\tau_k)_{k\ge1}\) be independent, identically distributed random variables with exponential distribution of rate \(\lambda\), so \(\mathbb{P}(\tau_k>x)=e^{-\lambda x}\) for \(x\ge0\). Set \(S_0=0\), \(S_n=\sum_{k=1}^n\tau_k\) for \(n\ge1\), and \(N_t=\#\{n\ge1:S_n\le t\}\) for \(t\ge0\). Thus \(S_n\) is the time of the \(n\)th arrival, and \(N_t\) is the number of arrivals by time \(t\).

Definition: The process just constructed is the exponential-gap arrival-count process. Counts in disjoint intervals are independent if, for every finite collection of pairwise disjoint intervals, the corresponding counts are mutually independent. We use intervals \((a,b]\); their endpoints are deterministic, and an arrival at any specified endpoint has probability zero.

The construction does not accumulate infinitely many arrivals in finite time. Indeed, the events \(\{\tau_k\ge1\}\) are independent and each has probability \(e^{-\lambda}>0\). For any fixed \(m\), the probability that none of these events occurs for \(k=m,\ldots,m+r\) is \((1-e^{-\lambda})^{r+1}\), which tends to zero as \(r\) tends to infinity. Hence the probability that only finitely many of the events occur is zero: that event is contained in the union, over \(m\ge1\), of the probability-zero events that none occur after \(m\). Almost surely infinitely many gaps are at least \(1\), so \(S_n\) tends to infinity. In particular, \(N_t\) is finite for every finite \(t\).

The Residual Wait at a Deterministic Boundary

For \(s\ge0\), let \(\mathcal{F}_s=\sigma(N_u:0\le u\le s)\), the information in the count process up to time \(s\). The residual waiting time is \(R_s=S_{N_s+1}-s\), the time from \(s\) to the next arrival. The defining exponential calculation, for \(r,z\ge0\), is

$$ \mathbb{P}(\tau_k-r>z\mid \tau_k>r) =\frac{\mathbb{P}(\tau_k>r+z)}{\mathbb{P}(\tau_k>r)} =\frac{e^{-\lambda(r+z)}}{e^{-\lambda r}} =e^{-\lambda z}. $$

The conditioning event has positive probability because \(e^{-\lambda r}>0\). This identity says more than that \(R_s\) has an exponential distribution: at a fixed boundary, the residual wait and the gaps after it are independent of the entire arrival history up to that boundary.

Lemma (fresh gaps after a fixed time): For every deterministic \(s\ge0\), the sequence \((R_s,\tau_{N_s+2},\tau_{N_s+3},\ldots)\) is independent of \(\mathcal{F}_s\), and its entries are independent exponential random variables of rate \(\lambda\).

Proof. Fix \(n\ge0\). On \(\{N_s=n\}\), the first \(n\) gaps determine the arrival history up to \(s\), and the condition on the next gap is \(\tau_{n+1}>s-S_n\). Given \(\tau_1,\ldots,\tau_n\) with \(S_n\le s\), this condition has probability \(e^{-\lambda(s-S_n)}>0\). By independence of the gaps and the displayed memoryless calculation, conditional on that condition, \(\tau_{n+1}-(s-S_n)\) has the exponential distribution of rate \(\lambda\), independently of \(\tau_{n+2},\tau_{n+3},\ldots\). This conditional law does not depend on the values of \(\tau_1,\ldots,\tau_n\). More explicitly, for any bounded measurable function \(h\) of the residual and finitely many following gaps, its conditional expectation on \(\{N_s=n\}\), given the first \(n\) gaps, is the same constant \(\mathbb{E}[h(\tau_1,\ldots,\tau_\ell)]\), where \(\ell\) is the number of arguments of \(h\). Any event in \(\mathcal{F}_s\), restricted to \(\{N_s=n\}\), is determined by those first \(n\) gaps. Multiplying by its indicator and summing the conditional-expectation identity over \(n\ge0\) proves independence from \(\mathcal{F}_s\), with the stated common joint law. The case \(n=0\) uses \(S_0=0\) and the same calculation. \(\square\)

A useful consequence is that the process of counts after \(s\), viewed from \(s\), has the same law as the original process and is independent of \(\mathcal{F}_s\). In fact, its first arrival after \(s\) occurs after \(R_s\), and its later gaps are \(\tau_{N_s+2},\tau_{N_s+3},\ldots\). The lemma says that these are fresh iid exponential gaps, independent of the past.

Independent Increments

We also need the count distribution on an interval. The exponential-gap construction gives, for \(t\ge0\) and integer \(k\ge0\),

$$ \mathbb{P}(N_t=k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}. $$

For completeness, when \(k=0\), this is \(\mathbb{P}(\tau_1>t)=e^{-\lambda t}\). When \(k\ge1\), the sum \(S_k\) has density \(\lambda^k u^{k-1}e^{-\lambda u}/(k-1)!\) for \(u>0\), the gamma density for a sum of \(k\) independent exponential gaps. The event \(N_t=k\) is \(S_k\le t<S_{k+1}\). Conditioning on \(S_k=u\), the probability of no further arrival by \(t\) is \(e^{-\lambda(t-u)}\). Therefore

$$ \begin{aligned} \mathbb{P}(N_t=k) &=\int_0^t \frac{\lambda^k u^{k-1}e^{-\lambda u}}{(k-1)!} e^{-\lambda(t-u)}\,du\\ &=e^{-\lambda t}\frac{\lambda^k}{(k-1)!}\int_0^t u^{k-1}\,du =e^{-\lambda t}\frac{(\lambda t)^k}{k!}. \end{aligned} $$
Theorem (Independent increments): For \(0\le t_0\le t_1\le\cdots\le t_m\), the counts \(N_{t_r}-N_{t_{r-1}}\), \(1\le r\le m\), are mutually independent. Each has the Poisson distribution with mean \(\lambda(t_r-t_{r-1})\). Intervals of length zero have count zero and are included in the statement.

Proof. At each deterministic boundary \(t_{r-1}\), the fresh-gaps lemma says that the arrival process after that time is independent of \(\mathcal{F}_{t_{r-1}}\) and has the original law. Consequently, the count over \((t_{r-1},t_r]\) is independent of \(\mathcal{F}_{t_{r-1}}\) and has the same distribution as \(N_{t_r-t_{r-1}}\). Every earlier interval count is measurable with respect to \(\mathcal{F}_{t_{r-1}}\). Thus the \(r\)th interval count is independent of all preceding interval counts. Applying this fact successively for \(r=2,\ldots,m\) proves mutual independence. The count formula just derived, with \(t=t_r-t_{r-1}\), gives the stated Poisson distribution; if the length is zero, the count is identically zero. \(\square\)

The argument uses the fresh-gap property at deterministic boundaries, not merely the fact that each individual count has a Poisson distribution. That distinction matters: marginal count distributions alone do not assert independence. Here the conditional law of the future given all the observed history is what supplies the independence.

Worked Example: One Arrival, Then No Arrival

Take rate \(\lambda=2\) and the intervals \((0,1]\) and \((1,2]\). Find the probability of exactly one arrival in the first interval and none in the second.

The lengths are both \(1\), so each count has a Poisson distribution with mean \(2\). The first interval count has probability \(e^{-2}2^1/1!=2e^{-2}\) of being one. The second has probability \(e^{-2}2^0/0!=e^{-2}\) of being zero. By independent increments, the joint probability is

$$ (2e^{-2})(e^{-2})=2e^{-4}. $$

Check: The factors are respectively the one-count and zero-count probabilities for intervals of length \(1\) at rate \(2\); their product is \(2e^{-4}\). In particular, conditional on exactly one arrival in \((0,1]\), the probability of no arrival in \((1,2]\) remains \(e^{-2}\), since the first event has probability \(2e^{-2}>0\).

Worked Example: Three Consecutive Intervals

Let \(\lambda=2\) and split \((0,2]\) into \((0,\tfrac12]\), \((\tfrac12,\tfrac32]\), and \((\tfrac32,2]\). Find the probability that the counts are respectively \(0,2,1\).

The interval lengths are \(\tfrac12,1,\tfrac12\), so their Poisson means are \(1,2,1\). The three probabilities are, in order, \(e^{-1}1^0/0!=e^{-1}\), \(e^{-2}2^2/2!=2e^{-2}\), and \(e^{-1}1^1/1!=e^{-1}\). The theorem permits multiplication, giving

$$ e^{-1}\cdot 2e^{-2}\cdot e^{-1}=2e^{-4}. $$

Check: Each factor uses the corresponding interval length and the specified count, and the product has exponent \(-1-2-1=-4\) and coefficient \(1\cdot2\cdot1=2\). The three counts are independent even though the intervals are consecutive and share boundary points under the chosen endpoint convention.

Worked Example: Separated Observation Windows

At rate \(\lambda=2\), find the probability of exactly one arrival in each of \((1,2]\) and \((4,5]\), with no restriction on arrivals between these windows.

Partition the relevant span into \((1,2]\), \((2,4]\), and \((4,5]\). Their lengths are \(1,2,1\), so the means are \(2,4,2\). The first and third counts are independent, and the middle count can be any nonnegative integer. Summing over its possible values gives probability \(1\), because its Poisson probabilities sum to \(1\). The requested probability is therefore

$$ \left(e^{-2}\frac{2^1}{1!}\right) \left(\sum_{j=0}^{\infty}e^{-4}\frac{4^j}{j!}\right) \left(e^{-2}\frac{2^1}{1!}\right) =(2e^{-2})(1)(2e^{-2})=4e^{-4}. $$

Check: The middle sum is \(e^{-4}e^4=1\) by the exponential series, and each requested window contributes \(2e^{-2}\). Thus leaving the intervening count unrestricted does not change the product for the two specified windows.

A Common Conditioning Error

A frequent mistake is to treat a long wait already observed as evidence that the next wait is longer. That reasoning is valid for many waiting-time distributions, but not for the exponential distribution. The memoryless calculation conditions on the event that the wait has exceeded the elapsed time and shows that the residual distribution is unchanged.

For a concrete correction, take rate \(\lambda=3\), and suppose the first gap has exceeded \(2\). The probability that it exceeds \(3\) as well, conditional on exceeding \(2\), is

$$ \mathbb{P}(\tau_1>3\mid\tau_1>2) =\frac{e^{-3\cdot3}}{e^{-3\cdot2}} =\frac{e^{-9}}{e^{-6}} =e^{-3}. $$

The incorrect calculation \(e^{-9}\) is the unconditional probability that the gap exceeds \(3\); it fails to divide by the positive conditioning probability \(e^{-6}\). The correct value \(e^{-3}\) is exactly the probability that a fresh rate-\(3\) exponential wait exceeds \(1\). At interval boundaries, this is why the future count law does not depend on how long the process has already waited.

Check Your Understanding

In each question, use the rate and interval lengths given, and distinguish a marginal count probability from a joint probability.

  1. For rate \(\lambda=1\), find the probability of no arrivals in \((0,2]\) and exactly one arrival in \((2,3]\).
  2. At rate \(\lambda=2\), what is the distribution of \(N_5-N_3\)? State its mean.
  3. Explain why the residual waiting time at a fixed time \(s\), conditional on the arrival history up to \(s\), still has an exponential distribution.
  4. For rate \(\lambda=3\), compute \(\mathbb{P}(\tau_1>5\mid\tau_1>4)\).
  5. Why does knowing the Poisson distribution of each interval count not alone prove that counts on different intervals are independent?

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