Arrival Counting Process

Let Ω be a sample space and P a probability measure on it. Throughout the following, by an

arrival process we will mean a stochastic process N = {N_t; t ≥ 0} defined on Ω such that for any ω

∈ Ω, the mapping t → N_t(ω) is non-decreasing, increases by jumps only, is right continuous, and has N₀(ω) = 0. See, for example, Figure 4.1.1.

Figure 4.1.1 The function Nt(ω) for the realization ω ∈ Ω for which the arrival times are T1(ω) = 0.4, T2(ω) = 0.6, T3(ω) = 1.8, T4(ω) = 2.1, T5(ω) = 2.4, . . . .

In the definition, axiom (b) expresses the independence of the number of arrivals in (t, t + s] from the past history of the process until time t. In particular, it implies that N_{t + s} − N_t is independent of N_t

< . . . < t_n in these arguments, and iterating, we reach the conclusion that N_t

1, N_t time-parameter version of the counting process for Bernoulli trials. Or somewhat more generally, what we

have is an example of a continuous time-parameter process with stationary and independent increments; here axiom (1.1c) gives the stationarity, and axiom (1.1b) gives the independent increments property.

Our definition of the Poisson process is completely qualitative. What is surprising is that these qualitative axioms completely specify the distribution of N_t and, thereby, the probability law of the whole process. The following results are directed at obtaining this distribution. Throughout this section N = {N_t; t ≥ 0} is a Poisson process.

(1.2) L^EMMA. For all t ≥ 0,

for some constant λ ≥ 0.

Proof. The number of arrivals in [0, t + s] is zero if and only if there are no arrivals in either [0, t]

or (t, t + s]; that is, the event {N_{t + s} = 0} is equal to the event {N_t = 0, N_{t + s} − N_t = 0}. Therefore, by the independence of N_{t + s} − N_t from N_t,

On the other hand, by the stationarity axiom (1.1c),

Hence, the function f(t) = P{N_t = 0} satisfies the equations

for all t, s ≥ 0.

Then it is known that either f(t) = 0 for all t ≥ 0 or else f(t) has the form

for some constant λ ≥ 0. This completes the proof once we show that f(t) does not vanish identically.

If f(t) were 0 for all t > 0, then for any t and almost all ω, N_{t + s}(ω) − N_t(ω) ≥ 1 no matter how small s almost all ω, thus contradicting the finiteness of N_t(ω).

The case λ = 0 corresponds to the degenerate case where N_t = 0 identically for all t. This case will be excluded from further consideration. Within the next proof we will need the fact that E[N_t] < ∞;

this can be proved using Proposition (2.2), to be given later, which will use only the definition and

(1.2).

(1.5) LEMMA. We have

Proof. Let h(t) = P{N_t ≥ 2}. Since N_t(ω) ≥ 2 implies N_{t + s}(ω) ≥ 2, by Proposition (1.1.8), h(t) ≤ h(t + s), that is, h is non-decreasing. Let ; then t ≤ 1/n_t and 1/t <

n_t + 1. Hence h(t) ≤ h(l/n_t), and

As t ↓ 0, n_t ↑ ∞ and (n_t + 1)/n_t ↓ 1. Therefore, to show that h(t)/t → 0 as t → 0, it is sufficient to show that nh(l/n)→ 0 as n → ∞.

Divide the unit interval [0, 1] into n subintervals of length 1/n each, and let S_n(ω) be the number of these subintervals during which there were two or more arrivals for the realization ω. We can think of S_n as the number of successes in n Bernoulli trials, where by “success at the kth trial” is meant having two or more arrivals during the kth subinterval. Then the probability of success at any one trial is p = h (l/n), and hence

By axiom (1.1a), for almost all ω ∈ Ω, the minimum time between two arrivals in [0, 1] is some number δ(ω) > 0. If n is large enough to have 1/n < δ(ω), then no subinterval can contain two arrivals and we must have S_n(ω) = 0. Hence, for almost all ω

Finally, S_n cannot be greater than the number of arrivals in [0, 1]; namely, the S_n are bounded by N₁, whose expected value E[N₁] is finite. Therefore the bounded convergence theorem (2.1.35) applies, and from (1.7) we get

In view of (1.6), this completes the proof.

(1.8) LEMMA. We have

where λ is the constant appearing in Lemma (1.2).

Proof. We have P{N_t = 1} = 1 − P{N_t = 0} − P{N_t ≥ 2}. Thus, by Lemmas (1.2) and (1.5),

The next theorem gives the distribution of N_t and, in view of axioms (1.1b) and (1.1c), specifies the joint distribution of N_t

1, N_t

2, . . ., N_t

n for any n ≥ 1 and .

(1.9) THEOREM. If {N_t; t ≥ 0} is a Poisson process, then for any t ≥ 0,

for some constant λ ≥ 0.

Proof. Let G(t) = E[α^Nt]. Writing N_{t + s} = N_t + (N_{t + s} − N_t), using the independence of N_{t + s} − N_t from N_t by axiom (1.1b) to apply Proposition (2.1.26), and using the stationarity axiom (1.1c), we get

Since G(t) = ∑ αⁿ P{N_t = n} ≥ P{N_t = 0} = e^−λt, G does not vanish for any t, and G(t + s) = G(t)G(s) can be satisfied only if

Note that g(α) is the derivative of G at t = 0, that is, since G(0) = 1 and

By Lemma (1.2), the first limit is −λ; by Lemma (1.8), the second limit is αλ; and by Lemma (1.5), for α ∈ [0, 1],

Putting these results in (1.11), we see that g(α) = −λ + λα, and (1.10) becomes

We thus have shown that for all α ∈ [0, 1],

The equality of these two power series in α implies that the corresponding coefficients are equal to each other, that is,

as was to be shown.

So far, λ is only a constant without any particular meaning attached to it. We now note that for any t

≥ 0,

Thus λ is the expected number of arrivals in an interval of unit length, or in other words, λ is the arrival rate. Incidentally, a computation similar to the one in Example (2.1.29) shows that

also. The distribution appearing in the preceding theorem is called the Poisson distribution with parameter λt.

The next corollary combines Theorem (1.9) with the axioms on stationarity and the independence of increments.

(1.15) C^OROLLARY. Let N = {N_t; t ≥ 0} be a Poisson process with rate λ. Then for any s, t ≥ 0, (1.15), they have the Poisson distributions with respective parameters 8 × 2.5 = 20, 8(3.7 – 2.5) = 9.6, 8(4.3 – 3.7) = 4.8. Hence, the probability desired is

From the practical point of view, checking to see if the axioms (1.1b) and (1.1c) hold for a particular process can be quite difficult. The following theorem reduces the checks involved considerably.

(1.17) THEOREM. N = {N_t; t ≥ 0} is a Poisson process with rate λ if and only if (a) for almost all ω each jump of N_t(ω) is of unit magnitude, and

(b) for any t, s ≥ 0,

Proof is omitted. The theorem states the following. Suppose that, as far as predicting the expected number of arrivals during (t, t + s] is concerned, the past history of arrivals before t has no value and that expectation is a constant λ times the length of the interval in question. Then that process is Poisson. Of course, if N is a Poisson process, then (1.17a) and (1.17b) are quite readily seen to hold.

What is surprising is that the seemingly mild condition (1.17b) implies the axioms (1.1b) and (1.1c).

The preceding theorem gives the simplest qualitative characterization of the Poisson process. It is easy to use, especially in situations where there is strong a priori evidence to suggest the independence of N_{t + s} − N_t from the past history of arrivals before t. The next theorem is useful in the opposite situation, where the independence axiom is hard to check but there are enough data to obtain the distribution of the number of arrivals falling in various time sets. Before stating this theorem we extend our notation somewhat.

Let N be a Poisson process with rate λ. For any subset B of we write N_B (ω) for the number of arrivals falling in B for the realization ω. For example, if B = (t, t + s], then N_B = N_{t + s}

− N_t; if B = [0, t], then N_B = N_t − N₀ = N_t; if B is the union of two disjoint intervals (t, u] and (s, υ], then N_B = N_{(t, u]} + N_{(s, υ]} = N_u − N_t + N_υ − N_s.

If B is an interval of length b, then by Theorem (1.9), the distribution of N_B is Poisson with parameter λb. Next let A and B be two disjoint intervals with respective lengths a and b, say, A = (t, t + a], B = (s, s + b]. Then N_A and N_B are independent random variables having the Poisson distribution with respective parameters λa and λb. From the stationarity axiom, N_B has the same distribution as N_C where C = (t + a, t + a + b], and N_A and N_C are still independent. Therefore, N_A + N_B has the same distribution as N_A + N_C. But the latter is the number of arrivals in (t, t + a + b] and hence has the Poisson distribution with parameter λ(a + b). Thus we have shown that N_A + N_B has the Poisson distribution with parameter λ(a + b). The same argument goes through for any number of intervals and shows the necessity part of the following

(1.18) T^HEOREM. N is a Poisson process with rate λ if and only if

for any subset B of which is the union of a finite number of disjoint intervals whose lengths sum up to b.

Proof of necessity was already outlined, and we omit the proof of sufficiency. As we mentioned before, the strength of the theorem lies in its avoidance of conditional expectations and distributions in characterizing the Poisson process. This is useful in situations where there are no intuitive reasons to expect axiom (1.1b) to hold. In words, this theorem states that an arrival process is Poisson with rate λ if the distribution of the number of arrivals occurring during any time set B is Poisson with parameter λ times the “length” of B.

Suppose next that the number of arrivals in an interval B = (t, t + b] is known to be k but the exact times of these k arrivals are not known. The next proposition shows that given N_B = k, each one of those k arrival times has the uniform distribution over the interval B independent of the times of the other k − 1 arrivals. In this sense, knowing the number of arrivals in an interval B yields no information on the times of those arrivals (other than the trivial information that they fall in B).

(1.19) P^ROPOSITION. Let A₁, . . ., A_n be disjoint intervals with union B, let a₁, . . ., a_n be their respective lengths, and set b = a₁ + · · · + a_n. Then for

,

Proof. The event {N_A

1 = k₁, . . ., N_A

n = k_n} implies {N_B = k}; therefore, the conditional probability in question is equal to

Since A₁, . . ., A_n are disjoint, N_A

1, . . ., N_A

n are independent, and

which becomes the desired result after rearrangement.

We close this section with a few remarks on the estimation of the parameter λ and the long-run behavior of N. We already computed E[N_t] to be λt. For any integer n, we can write N_n as the sum of n independent and identically distributed random variables (namely, N_n = N₁ + (N₂ − N₁) + · · · + (N_n

− N_n−1)), each one of which has the expected value λ. Thus, by the strong law of large numbers (3.4.10), for almost all ω,

From this it is easy to see that

for almost all ω. This is the strong law of large numbers for the Poisson process. By this result, we are justified in estimating λ by the average number of arrivals N_t(ω)/t we observed during the course of our observations.

A similar reasoning based on writing N_t as the sum of a large number of independent and identically distributed random variables also yields a central limit theorem:

for any . Namely, if λt is large enough, N_t has the normal distribution with mean λt and variance λt. For practical purposes, the approximation is quite good if λt ≥ 10.