Let N be a Poisson process with rate λ For almost all ω ∈ Ω, the function t → Nt(ω) is non-decreasing and right continuous and increases by jumps of size one. Such a function is completely determined by the times of its jumps; let these be T1(ω), T2(ω), . . . . Then T1, T2, . . . are the successive instants of arrivals.
We have seen in Lemma (1.2) that if t is an arbitrary point in time, the probability that there are no arrivals in (t, t + s] is e−λs, independent of the history of arrivals before t. The same result happens to hold even when the arbitrary time point t is replaced by a time of arrival, say Tn; namely,
where {Nu; u ≤ Tn} is the history of the process until the time Tn of the nth arrival. Noting that the event {NT
n+s − NT
n = 0} is equal to the event {Tn + 1 − Tn > s}, and that the information contained in the history {Nu; u ≤ Tn} is the same as that contained in {T1, . . ., Tn}, what we have can be restated as follows (we write T0 = 0 for convenience):
(2.2) PROPOSITION. For any n ≥ 0,
In other words, the interarrival times T1, T2 − T1, T3 − T2, . . . are independent and identically distributed random variables with the common distribution being
The distribution (2.3) above is called the exponential distribution with parameter λ. It is differentiable, and the probability density function corresponding to it is
Note that this density function is monotone decreasing. As a result, an inter-arrival time is more likely to have a length in [0, s] than a length in [t, t + s] for any t. Thus, a Poisson process has more short intervals than long ones. Therefore, a plot of the times of arrivals on a line looks, to the naive eye, as if the arrivals occur in clusters.
Another property, usually referred to as the memorylessness of the exponential distribution, is the prime reason for the pleasant independence properties of the Poisson process: if a random variable X has an exponential distribution, then
and conversely, if X has this property, then it must have an exponential distribution. That is, knowing that an interarrival time has already lasted t units does not alter the probability of its lasting another s units.
In certain applied situations it is more efficient to work with the times of arrivals. Then it is of value to find conditions on interarrival times which guarantee that the counting process will be
Poisson. The answer turns out to be that the conditions of (2.2) above are sufficient. We omit the proof of this result :
(2.5) THEOREM. Let T1, T2, . . . be the successive times of arrivals of an arrival process N = {Nt; t ≥ 0}. Then N is a Poisson process with rate λ if and only if the interarrival times T1, T2 − T1, . . . are independent and identically distributed exponential random variables with parameter λ.
(2.6) EXAMPLE. An item has a random lifetime whose distribution is exponential with parameter λ.
When it fails, it is immediately replaced by an identical item; and when that fails, it is replaced by another identical item; etc. What this means is that the lifetimes X1, X2, . . . of the successive items in use are independent and identically distributed random variables with distribution
If T1, T2, . . . are the times of successive failures, then T1 = X1, T2 = X1 + X2, . . .; and we see that the conditions of Theorem (2.5) are satisfied by the sequence T1, T2, . . . . Hence, if Nt is the number of failures in [0, t], we may conclude by Theorem (2.5) that the process of failures {Nt; t ≥ 0} is Poisson with rate λ.
It follows from the fact that Tn + 1 − Tn has an exponential distribution with parameter λ that the expected value of an interarrival time in a Poisson process is (we are using Theorem (2.1.9))
and the variance is (see Example (2.1.30) for an illustration)
By writing Tn = T1 + (T2 − T1) + · · · + (Tn − Tn−1) and using the fact that the expectation of a sum is the sum of the expectations, we obtain from (2.7) that
Using the independence of T1, T2 − T1, . . ., Tn − Tn−1. proved in (2.2) to apply (2.3.9) (namely, that the variance of a sum of independent random variables is the sum of the variances), we obtain from (2.8) that
Similarly, various higher moments of Tn may be obtained by utilizing Proposition (2.2) once such moments are obtained for one interarrival time.
(2.11) EXAMPLE. In Example (2.6), suppose λ = 0.0002 (time is measured in hours). Then the expected lifetime of an item is E[Xn] = 1/λ = 5,000 hours, and the variance is Var(Xn) = 1/λ2 = 25 × 106 hours2. Hence, if a piece of equipment contains 3 such items connected so that the second starts functioning as soon as the first fails, and the third starts functioning as soon as the second fails, then the lifetime of this equipment is T3 = X1 + X2 + X3. It has mean E[T3] = 3/λ = 15,000 hours and the variance Var(T3) = 3/λ2 = 75 × 106 hours2.
(2.12) EXAMPLE. In Example (2.6), again suppose that the items are replaced as soon as they fail.
Suppose the cost of a replacement is β dollars, and suppose the discount rate of money is a α > 0, so that one dollar spent at time t has the present value e−αt (typically, α is the interest rate). Then for the realization ω ∈ Ω, time of the nth failure is Tn(ω), and the present value of the cost of replacement is β exp(−αTn(ω)), Summing this over all n, we obtain the present value of all future replacement costs;
this is
We are interested in the expected value of C. Since the expected value of a sum is the sum of the expectations,
For fixed n we can write Tn = T1 + (T2 − T1) + · · · + (Tn − Tn−1) where T1, T2 − T1, . . ., Tn − Tn−1 are independent and identically distributed. Hence, using Proposition (2.1.26)
Since the distribution of T1 is exponential with parameter λ,
Hence, remembering that 1 + x + x2 + · · · = 1/(1 − x) for x ∈ [0, 1),
In particular, if the mean lifetime is 5,000 hours, replacement cost 800 dollars, and the interest rate 24 percent per year, then β = 800, λ = 1/5,000, α = 0.24/(365 × 24) = 0.01/365, and hence E[C] = 800 × 36,500/5,000 = 5840 dollars.
Though in many computations we can do without the distribution of Tn, there are situations when it is helpful to know it. One way of obtaining the distribution of Tn would be to invert the Laplace transform of Tn, which is E[e−αTn] = [λ/(α + λ)]n. However, it is easier (and smarter) to observe first that for any realization ω, the time T5(ω) of the fifth arrival is before t if and only if the number of arrivals Nt(ω) in [0, t] is equal to five or more. This reasoning gives, for any n and t,
From this equality between two events we obtain, by (1.9), (2.14) PROPOSITION. For any n ∈ {1, 2, . . . },
The distribution of Tn is called the Erlang-n distribution. It is a member of the gamma family of distributions. It is differentiable, and the derivative is
It is important to distinguish this probability density function t → λ(λt)n−1 e−λt/(n − 1)! of the random variable Tn taking values in from the distribution n → e−λt(λt)n/n! of the discrete random variable Nt.
(2.16) EXAMPLE. Taking a certain fixed point on a highway, let U1, U2, · · · be the successive interarrival times of vehicles at this point. Suppose U1, U2,· · · are independent and identically distributed random variables with the distribution
We are interested in the distribution of the number Mt of vehicles crossing this fixed point during [0, t].
First we observe that the distribution of Uk given is the Erlang-2 distribution. Thus, we may think
of each Uk as being the sum of two interarrival times in a Poisson process with rate λ. That is, the times vehicles cross the given point may be thought of as U1 = T2, U1 + U2 = T4, U1 + U2 + U3 = T6, . . . where T1, T2, . . . are the times of arrivals in a Poisson process {Nt} with rate λ. Then the number Mt(ω) of vehicles crossing is equal to 6 if and only if the Poisson process N has, for that realization ω, 12 or 13 arrivals in [0, t]. So for any t ≥ 0,
Next is a computational result which extends that of Example (2.12) to situations where the interest rate, instead of remaining constant over time as in (2.12), may vary as a function of time. Again we assume that T1, T2, . . . are the times of arrivals of a Poisson process with rate λ.
(2.17) PROPOSITION. For any non-negative function f on ,
Proof. Using Proposition (2.1.23), noting that the probability density function of Tn is given by (2.15),
Hence,
We can now re-do the computations of Example (2.12): E[C] = λ ∫ f, where f(t) = βe−αt. So ∫ f = β/
α, and E[C] = λβ/α.
We close this section by giving a more general version of the property (2.1) with which we started.
In certain situations, one is interested in the number of arrivals occurring in an interval (T, T + s]
where T is a random variable instead of a fixed number. It turns out that for a certain class of random
times T, the independence of Nt + s − NT from the past history {Nu; u ≤ T] until T is still preserved, and furthermore, the distribution of Nt + s − NT is again Poisson with parameter λs. Such “good”
random times T are characterized by the property that for any number t ≥ 0, one can determine whether the event {T ≤ t} has occurred or not by knowing the history {Nu; u ≤ t} of the arrival process until the time t. Such random times are called stopping times. For example, the time T6 of the sixth arrival is a stopping time, since we can tell whether T6 ≤ 3.7 or not by knowing what the arrival process looked like until the time t = 3.7 (and similarly for any other t). The first time T when an interarrival time exceeds a certain critical value c is a stopping time (this is, for example, the random time when a pedestrian completes crossing a highway if he needs a gap of c time units in traffic in order to cross safely).
(2.18) THEOREM. For any number s ≥ 0 and any stopping time T,
Proof is omitted. Note that Proposition (2.2) is a special case, where T = Tn and k = 0.
(2.19) EXAMPLE. The times vehicles cross a fixed point on a highway form a Poisson process with rate λ vehicles per second. A pedestrian arrives at this point at time 0 wanting to cross the highway.
He needs 4 seconds to cross it. Therefore, for the realization ω, if T1(ω) ≤ 4, T2(ω) − T1(ω) ≤ 4, . . ., T6(ω) − T5(ω) ≤ 4, and T7(ω) − T6(ω) > 4, then the time T(ω) he completes crossing is T6(ω) + 4; if the first interval greater than 4 were T13(ω) − T12(ω), then T(ω) would have been T(ω) = T12((ω) + 4.
The time T is a stopping time (because the picture of N in the interval [0, t] is sufficient for us to tell whether T ≤ t or not).
After he completes crossing, he will have escaped his doom by less than 2 seconds with probability
(2.20) EXAMPLE. Arrivals of buses at a certain stop form a Poisson process with rate λ = 0.2 buses per minute. (Though the arrival times are scheduled, when the average time between arrivals is short as is the case here, traffic delays, etc., introduce enough variation to make the arrival process become a Poisson process.) An inspector from the bus company arrives at the stop with the fifth bus after 9:00
A.M. We are interested in the distribution of the number of buses he will count during the one hour he plans to stay there. If 9:00 A.M. corresponds to time t0 = 120 (i.e., if the time origin is taken to be 7:00
A.M.), then the time the inspector arrives is
This random time T is a stopping time; therefore, the distribution of the number of buses he will count in the next 60 minutes is