Exponential Interarrivals and the Poisson Special Case

From Exponential Waiting Times to Poisson Counts

Renewal processes can have quite different count behavior depending on their interarrival distribution. Exponential interarrivals are a special case: the number of renewals by time \(t\) has a Poisson distribution, and counts over disjoint time intervals are independent. The reason is not only that the exponential distribution has a convenient density. Its memoryless property means that, at any fixed time, the residual wait until the next renewal has the same distribution as a new interarrival time.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_k)_{k\ge1}\) be independent, identically distributed random variables with exponential distribution of rate \(\lambda>0\), so that \(\mathbb{P}(X_k>x)=e^{-\lambda x}\) for \(x\ge0\). Define the renewal epochs \(S_0=0\) and \(S_n=X_1+\cdots+X_n\) for \(n\ge1\). The renewal count is \(N(t)=\max\{n\ge0:S_n\le t\}\), for \(t\ge0\). It is a counting process with state space \(S=\{0,1,2,\ldots\}\). Since the interarrivals have continuous distributions, the probability of a renewal at any fixed \(t>0\) is zero; the convention \(S_n\le t\) therefore causes no ambiguity in fixed-time probabilities.

First derive the distribution at a single time. For \(n\ge1\), the event \(N(t)=n\) means \(S_n\le t<S_{n+1}\). Given the first \(n\) interarrivals, the probability that the next one lasts beyond \(t-S_n\) is \(e^{-\lambda(t-S_n)}\). Multiplying this survival probability by the joint density of \(X_1,\ldots,X_n\) gives a constant integrand on the region \(x_1,\ldots,x_n>0\), \(x_1+\cdots+x_n\le t\): the exponential factors combine to \(e^{-\lambda t}\), leaving the factor \(\lambda^n\).

Proof. For \(n=0\), \(N(t)=0\) exactly when \(X_1>t\), which has probability \(e^{-\lambda t}\). For \(n\ge1\), integrate over the possible values \(x_1,\ldots,x_n\) of the first \(n\) interarrivals. The joint density is \(\lambda^n e^{-\lambda(x_1+\cdots+x_n)}\), and the conditional probability that \(X_{n+1}>t-(x_1+\cdots+x_n)\) is \(e^{-\lambda(t-(x_1+\cdots+x_n))}\). Their product is \(\lambda^n e^{-\lambda t}\). The integration region is the \(n\)-dimensional simplex \(x_i>0\), \(\sum_i x_i\le t\), whose volume is \(t^n/n!\): its volume \(V_n(t)\) satisfies \(V_1(t)=t\) and, recursively, \(V_n(t)=\int_0^t V_{n-1}(t-x)\,dx=\int_0^t (t-x)^{n-1}/(n-1)!\,dx=t^n/n!\). Hence \(\mathbb{P}(N(t)=n)=\lambda^n e^{-\lambda t}t^n/n!\). At \(t=0\), \(N(0)=0\) with probability one, which agrees with the formula using \((\lambda t)^0=1\) and \((\lambda t)^n=0\) for \(n\ge1\). \(\square\)

Memorylessness Gives Independent Increments

For an exponential random variable \(X\), the memoryless identity is a direct calculation. If \(a,y\ge0\), then $$ \mathbb{P}(X>a+y\mid X>a)=\frac{e^{-\lambda(a+y)}}{e^{-\lambda a}}=e^{-\lambda y}. $$ The conditioning event has positive probability because \(e^{-\lambda a}>0\). Thus, after an interarrival has already lasted \(a\) units without ending, its residual duration has the same exponential distribution as a fresh interarrival.

To make the independence claim precise, use the natural filtration \(\mathcal{F}_s=\sigma(N(u):0\le u\le s)\), the information in the count history through time \(s\). The residual wait at \(s\) is \(R_s=S_{N(s)+1}-s\). On the event \(N(s)=n\), the history records the first \(n\) arrival times and that \(X_{n+1}>s-S_n\); it does not reveal the excess of \(X_{n+1}\) beyond that age. The memoryless calculation says that, conditional on this history, \(R_s\) has survival probability \(e^{-\lambda y}\) for every \(y\ge0\). The interarrivals after \(X_{n+1}\) are independent of this history and remain exponential.

Proof. Fix \(s\ge0\) and write \(n=N(s)\), so the age in the current interarrival is \(a=s-S_n\) (when \(n=0\), this is \(a=s\)). Conditional on the first \(n\) interarrivals and on \(X_{n+1}>a\), for any \(y\ge0\), $$ \mathbb{P}(R_s>y\mid X_1,\ldots,X_n,\ X_{n+1}>a) =\frac{\mathbb{P}(X_{n+1}>a+y)}{\mathbb{P}(X_{n+1}>a)} =\frac{e^{-\lambda(a+y)}}{e^{-\lambda a}} =e^{-\lambda y}. $$ The same ratio applies when conditioning on the entire history \(\mathcal{F}_s\): on each event \(N(s)=n\), the history gives the first \(n\) interarrivals and the condition \(X_{n+1}>a\), and the displayed conditional probability does not depend on their values. Also, \(X_{n+2},X_{n+3},\ldots\) are independent of \(\mathcal{F}_s\) and of \(R_s\). Consequently the waiting time from \(s\) to the first subsequent renewal, followed by all later interarrivals, has the same independent exponential sequence law as the original process, conditionally on \(\mathcal{F}_s\). Thus the process of counts after \(s\), measured relative to \(N(s)\), has the original law and is independent of \(\mathcal{F}_s\). Applying this argument successively at the endpoints of any finite collection of disjoint intervals proves independence of their counts. The single-time formula, applied to an interval of length \(h=t-s\) in the process restarted at \(s\), gives a Poisson count with mean \(\lambda h\). Its distribution depends on \(h\), not on \(s\), proving stationarity of the increments. \(\square\)

The Renewal Function

For this renewal process, define the renewal function by \(U(t)=\mathbb{E}[N(t)]\). The count distribution makes this expectation explicit. Since \(N(t)\) is nonnegative and has the probabilities just derived, summing its probability-weighted values gives $$ U(t)=\sum_{n=0}^{\infty}n e^{-\lambda t}\frac{(\lambda t)^n}{n!} =e^{-\lambda t}(\lambda t)\sum_{n=1}^{\infty}\frac{(\lambda t)^{n-1}}{(n-1)!} =\lambda t. $$ For \(t=0\), the sum is zero and the same formula holds. The exponential power series justifies the final equality. In particular, the expected number of renewals per unit time is constant and equals the interarrival rate \(\lambda\).

Common Pitfalls and Interpretation

A common error is to omit the factorial in the count formula. For example, at \(\lambda t=1\), writing \(\mathbb{P}(N(t)=n)=e^{-1}\) for every \(n\ge0\) would assign total probability \(\sum_{n=0}^{\infty}e^{-1}=\infty\), not \(1\). The factor \(1/n!\) comes from the volume of the simplex of possible interarrival times; it is essential.

Another error is to treat memorylessness as a statement that the elapsed waiting time is reset or erased. The elapsed time is known from the history. What memorylessness says is that, conditional on having waited that long without a renewal, the remaining wait has the fresh exponential distribution. It is this conditional law, together with independence of subsequent interarrivals, that supports the independent-increments argument.

Finally, the formula \(U(t)=\lambda t\) is an expectation, not a claim that exactly \(\lambda t\) renewals occur. A realized count is an integer-valued random variable; its mean can be noninteger. The linear renewal function and the full Poisson distribution are special consequences of exponential interarrivals, not properties to assume for a general renewal process.