Proving Independent Increments Under Superposition

From Count Distributions to a Process Result

Knowing the distribution of the total number of arrivals in one time interval does not by itself establish that a process is Poisson. A Poisson process also has stationary independent increments: counts over disjoint intervals are independent, and the distribution of an increment depends only on the interval's length. For superposition, the first property requires a process-level argument, not just a calculation of a single count.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(N^{(1)},\ldots,N^{(m)}\) be a finite family of mutually independent Poisson processes, with respective rates \(\lambda_1,\ldots,\lambda_m\ge 0\). Mutual independence means the sigma-fields generated by the individual processes are independent. Each process starts at zero, has nondecreasing integer-valued paths, and has independent stationary increments; for \(0\le s\le t\), its increment \(N^{(r)}(t)-N^{(r)}(s)\) has a Poisson distribution with mean \(\lambda_r(t-s)\). Define the aggregate count, counting arrivals from all sources, by

When arrival times from different sources are distinct, this sum is also the count of the set-theoretic union of their arrivals. We will verify below that simultaneous arrivals have probability zero. The sum formulation remains useful even before that verification because it specifies exactly how counts are aggregated.

The Superposition Theorem

Proof. Fix a finite collection of pairwise disjoint intervals \((a_\ell,b_\ell]\), \(1\le\ell\le q\), with \(0\le a_\ell\le b_\ell\), and write \(h_\ell=b_\ell-a_\ell\). For source \(r\), let \(X_{r\ell}=N^{(r)}(b_\ell)-N^{(r)}(a_\ell)\). For each fixed source, these variables are independent because its intervals are disjoint. The vectors from different sources are independent by the assumed mutual independence of the processes. Consequently, all the variables \(X_{r\ell}\), across both indices, are independent, and \(X_{r\ell}\) has a Poisson distribution with mean \(\lambda_r h_\ell\). The aggregate increment on interval \(\ell\) is \(Y_\ell=\sum_{r=1}^{m}X_{r\ell}\).

For one interval of length \(h\), the sum of the source counts has probability, for \(k\ge0\),

The last equality is the multinomial expansion of \((\lambda_1+\cdots+\lambda_m)^k\). If \(\Lambda h=0\), every source count is zero with probability one, so the sum is zero with probability one, as required for a Poisson variable of mean zero.

It remains to show independence across intervals. For arbitrary nonnegative integers \(k_1,\ldots,k_q\), independence of all the source-interval counts gives

The factorization is the joint probability factorization for independent increments; the second line also gives each increment's distribution. Since that distribution depends on the interval only through its length, the increments are stationary. Finally, \(N(0)=0\), and the finite sum of the source counting paths is nondecreasing and integer-valued. These facts and the increment properties establish the stated definition of a Poisson process. \(\square\)

The proof separates two ingredients. Within a source, disjoint time intervals give independent counts. Across sources, the entire processes are independent. Together these imply independence of every source count used in the factorization. The one-interval convolution calculation gives the total rate, while the factorization across intervals supplies the additional process-level conclusion.

Why the Set-Theoretic Union Has No Ties

If “union” means the set of event times, rather than the sum of source counts, it is useful to know that independent sources almost surely do not arrive at the same time. Here is a direct argument for a finite family. Fix two sources with rates \(\lambda_r,\lambda_s\) and a finite horizon \(T>0\). Divide \([0,T]\) into \(M\) equal subintervals of length \(T/M\). For a fixed subinterval, the probability that both sources have at least one arrival in it is at most \(\lambda_r\lambda_s(T/M)^2\): the two counts are independent, and for a nonnegative integer-valued count \(Z\), \(\mathbb{P}(Z\ge1)\le\mathbb{E}[Z]\). Summing over the \(M\) subintervals bounds the probability of a shared subinterval containing arrivals from both sources by \(\lambda_r\lambda_s T^2/M\), which tends to zero.

A simultaneous arrival at a fixed partition endpoint has probability zero. Indeed, for a fixed time \(c>0\) and \(0<\varepsilon<c\), the probability of an arrival exactly at \(c\) is at most the probability of an arrival in \((c-\varepsilon,c]\), namely \(1-e^{-\lambda\varepsilon}\le\lambda\varepsilon\), which tends to zero. At time zero there is no arrival. Thus the probability of a tie between the two sources on \([0,T]\) is zero. There are finitely many source pairs, and applying the argument on each integer horizon \(T\) shows that, with probability one, no pair of sources ever has a simultaneous arrival. A zero-rate source has no arrivals and causes no exception.

Worked Count Calculations

A Common Mistake: Overlapping Intervals

Independent increments are asserted for disjoint intervals, not for every pair of increments. For example, take a Poisson process of rate \(1\), and define \(A=N(2)-N(0)\) and \(B=N(3)-N(1)\). These intervals overlap on \((1,2]\), so the theorem does not say that \(A\) and \(B\) are independent.

To calculate their dependence, set \(U=N(1)-N(0)\), \(V=N(2)-N(1)\), and \(W=N(3)-N(2)\). These are independent Poisson variables of mean \(1\), and \(A=U+V\), \(B=V+W\). Since \(U,V,W\) are independent, \(\operatorname{Cov}(U+V,V+W)=\operatorname{Var}(V)=1\). The covariance is nonzero, so the increments are not independent. A wrong calculation would apply the independent-increments rule to the overlapping intervals and conclude \(\operatorname{Cov}(A,B)=0\). The correction is to split the intervals into disjoint pieces and recognize the shared count \(V\).

For superposition, the corresponding safe procedure is to identify genuinely disjoint intervals first. For those intervals, source-wise independent increments and independence across sources give the needed joint factorization. Marginal Poisson distributions alone would not justify that step.