5.3.4 Further Properties of Poisson Processes
Consider a Poisson process 
having rate 
, and suppose that each time an event occurs it is classified as either a type I or a type II event. Suppose further that each event is classified as a type I event with probability 
or a type II event with probability 
, independently of all other events. For example, suppose that customers arrive at a store in accordance with a Poisson process having rate 
; and suppose that each arrival is male with probability 
and female with probability 
. Then a type I event would correspond to a male arrival and a type II event to a female arrival.
Let 
and 
denote respectively the number of type I and type II events occurring in 
. Note that 
.
Proposition 5.2
and 
are both Poisson processes having respective rates 
and 
. Furthermore, the two processes are independent.
Proof
It is easy to verify that 
is a Poisson process with rate 
by verifying that it satisfies Definition 5.3.
• 
follows from the fact that 
.
• It is easy to see that 
inherits the stationary and independent increment properties of the process 
. This is true because the distribution of the number of type I events in an interval can be obtained by conditioning on the number of events in that interval, and the distribution of this latter quantity depends only on the length of the interval and is independent of what has occurred in any nonoverlapping interval.
• 
Thus we see that 
is a Poisson process with rate 
and, by a similar argument, that 
is a Poisson process with rate 
. Because the probability of a type I event in the interval from 
to 
is independent of all that occurs in intervals that do not overlap 
, it is independent of knowledge of when type II events occur, showing that the two Poisson processes are independent. (For another way of proving independence, see Example 3.23.) 
Example 5.14
If immigrants to area 
arrive at a Poisson rate of ten per week, and if each immigrant is of English descent with probability 
, then what is the probability that no people of English descent will emigrate to area 
during the month of February?
Example 5.15
Suppose nonnegative offers to buy an item that you want to sell arrive according to a Poisson process with rate 
. Assume that each offer is the value of a continuous random variable having density function 
. Once the offer is presented to you, you must either accept it or reject it and wait for the next offer. We suppose that you incur costs at a rate 
per unit time until the item is sold, and that your objective is to maximize your expected total return, where the total return is equal to the amount received minus the total cost incurred. Suppose you employ the policy of accepting the first offer that is greater than some specified value 
. (Such a type of policy, which we call a 
-policy, can be shown to be optimal.) What is the best value of 
? What is the maximal expected net return?
Solution: Let us compute the expected total return when you use the 
-policy, and then choose the value of 
that maximizes this quantity. Let 
denote the value of a random offer, and let 
be its tail distribution function. Because each offer will be greater than 
with probability 
, it follows that such offers occur according to a Poisson process with rate 
. Hence, the time until an offer is accepted is an exponential random variable with rate 
. Letting 
denote the total return from the policy that accepts the first offer that is greater than 
, we have
(5.14)
Differentiation yields
Therefore, the optimal value of 
satisfies
or
or
It is not difficult to show that there is a unique value of 
that satisfies the preceding. Hence, the optimal policy is the one that accepts the first offer that is greater than 
, where 
is such that
Putting 
in Equation (5.14) shows that the maximal expected net return is
Thus, the optimal critical value is also the maximal expected net return. To understand why this is so, let 
be the maximal expected net return, and note that when an offer is rejected the problem basically starts anew and so the maximal expected additional net return from then on is 
. But this implies that it is optimal to accept an offer if and only if it is at least as large as 
, showing that 
is the optimal critical value. 
It follows from Proposition 5.2 that if each of a Poisson number of individuals is independently classified into one of two possible groups with respective probabilities 
and 
, then the number of individuals in each of the two groups will be independent Poisson random variables. Because this result easily generalizes to the case where the classification is into any one of 
possible groups, we have the following application to a model of employees moving about in an organization.
Example 5.16
Consider a system in which individuals at any time are classified as being in one of 
possible states, and assume that an individual changes states in accordance with a Markov chain having transition probabilities 
. That is, if an individual is in state 
during a time period then, independently of its previous states, it will be in state 
during the next time period with probability 
. The individuals are assumed to move through the system independently of each other. Suppose that the numbers of people initially in states 
are independent Poisson random variables with respective means 
. We are interested in determining the joint distribution of the numbers of individuals in states 
at some time 
.
Solution: For fixed 
, let 
denote the number of those individuals, initially in state 
, that are in state 
at time 
. Now each of the (Poisson distributed) number of people initially in state 
will, independently of each other, be in state 
at time 
with probability 
, where 
is the 
-stage transition probability for the Markov chain having transition probabilities 
. Hence, the 
will be independent Poisson random variables with respective means 
, 
. Because the sum of independent Poisson random variables is itself a Poisson random variable, it follows that the number of individuals in state 
at time 
—namely 
—will be independent Poisson random variables with respective means 
, for 
. 
Example 5.17
The Coupon Collecting Problem
There are 
different types of coupons. Each time a person collects a coupon it is, independently of ones previously obtained, a type 
coupon with probability 
. Let 
denote the number of coupons one needs to collect in order to have a complete collection of at least one of each type. Find 
.
Solution: If we let 
denote the number one must collect to obtain a type 
coupon, then we can express 
as
However, even though each 
is geometric with parameter 
, the foregoing representation of 
is not that useful, because the random variables 
are not independent.
We can, however, transform the problem into one of determining the expected value of the maximum of independent random variables. To do so, suppose that coupons are collected at times chosen according to a Poisson process with rate 
. Say that an event of this Poisson process is of type 
, if the coupon obtained at that time is a type 
coupon. If we now let 
denote the number of type 
coupons collected by time 
, then it follows from Proposition 5.2 that 
are independent Poisson processes with respective rates 
. Let 
denote the time of the first event of the 
th process, and let
denote the time at which a complete collection is amassed. Since the 
are independent exponential random variables with respective rates 
, it follows that
Therefore,
(5.15)
It remains to relate 
, the expected time until one has a complete set, to 
, the expected number of coupons it takes. This can be done by letting 
denote the 
th interarrival time of the Poisson process that counts the number of coupons obtained. Then it is easy to see that
Since the 
are independent exponentials with rate 1, and 
is independent of the 
, we see that
Therefore,
and so 
is as given in Equation (5.15).
Let us now compute the expected number of types that appear only once in the complete collection. Letting 
equal 
if there is only a single type 
coupon in the final set, and letting it equal 
otherwise, we thus want
Now there will be a single type 
coupon in the final set if a coupon of each type has appeared before the second coupon of type 
is obtained. Thus, letting 
denote the time at which the second type 
coupon is obtained, we have
Using that 
has a gamma distribution with parameters 
, this yields
Therefore, we have the result
The next probability calculation related to Poisson processes that we shall determine is the probability that 
events occur in one Poisson process before 
events have occurred in a second and independent Poisson process. More formally let 
and 
be two independent Poisson processes having respective rates 
and 
. Also, let 
denote the time of the 
th event of the first process, and 
the time of the 
th event of the second process. We seek
Before attempting to calculate this for general 
and 
, let us consider the special case 
. Since 
, the time of the first event of the 
process, and 
, the time of the first event of the 
process, are both exponentially distributed random variables (by Proposition 5.1) with respective means 
and 
, it follows from Section 5.2.3 that
(5.16)
Let us now consider the probability that two events occur in the 
process before a single event has occurred in the 
process. That is, 
. To calculate this we reason as follows: In order for the 
process to have two events before a single event occurs in the 
process, it is first necessary for the initial event that occurs to be an event of the 
process (and this occurs, by Equation (5.16), with probability 
). Now, given that the initial event is from the 
process, the next thing that must occur for 
to be less than 
is for the second event also to be an event of the 
process. However, when the first event occurs both processes start all over again (by the memoryless property of Poisson processes) and hence this conditional probability is also 
; thus, the desired probability is given by
In fact, this reasoning shows that each event that occurs is going to be an event of the 
process with probability 
or an event of the 
process with probability 
, independent of all that has previously occurred. In other words, the probability that the 
process reaches 
before the 
process reaches 
is just the probability that 
heads will appear before 
tails if one flips a coin having probability 
of a head appearing. But by noting that this event will occur if and only if the first 
tosses result in 
or more heads, we see that our desired probability is given by
5.3.5 Conditional Distribution of the Arrival Times
Suppose we are told that exactly one event of a Poisson process has taken place by time 
, and we are asked to determine the distribution of the time at which the event occurred. Now, since a Poisson process possesses stationary and independent increments it seems reasonable that each interval in 
of equal length should have the same probability of containing the event. In other words, the time of the event should be uniformly distributed over 
. This is easily checked since, for 
,
This result may be generalized, but before doing so we need to introduce the concept of order statistics.
Let 
be 
random variables. We say that 
are the order statistics corresponding to 
if 
is the 
th smallest value among 
, 
. For instance, if 
and 
, 
, 
then 
, 
. If the 
, are independent identically distributed continuous random variables with probability density 
, then the joint density of the order statistics 
is given by
(i) (
) will equal (
) if (
) is equal to any of the 
! permutations of (
);
(ii) the probability density that (
) is equal to 
is 
when 
is a permutation of 
.
If the 
, are uniformly distributed over 
, then we obtain from the preceding that the joint density function of the order statistics 
is
We are now ready for the following useful theorem.
Theorem 5.2
Given that 
, the 
arrival times 
have the same distribution as the order statistics corresponding to 
independent random variables uniformly distributed on the interval 
.
Proof
To obtain the conditional density of 
given that 
note that for 
the event that 
is equivalent to the event that the first 
interarrival times satisfy 
. Hence, using Proposition 5.1, we have that the conditional joint density of 
given that 
is as follows:
which proves the result. 
Remark
The preceding result is usually paraphrased as stating that, under the condition that 
events have occurred in 
, the times 
at which events occur, considered as unordered random variables, are distributed independently and uniformly in the interval 
.
Application of Theorem 5.2 (Sampling a Poisson Process)
In Proposition 5.2 we showed that if each event of a Poisson process is independently classified as a type I event with probability 
and as a type II event with probability 
then the counting processes of type I and type II events are independent Poisson processes with respective rates 
and 
. Suppose now, however, that there are 
possible types of events and that the probability that an event is classified as a type 
event, 
, depends on the time the event occurs. Specifically, suppose that if an event occurs at time 
then it will be classified as a type 
event, independently of anything that has previously occurred, with probability 
where 
. Upon using Theorem 5.2 we can prove the following useful proposition.
Proposition 5.3
If 
, represents the number of type 
events occurring by time 
then 
, are independent Poisson random variables having means
Before proving this proposition, let us first illustrate its use.
Example 5.18
An Infinite Server Queue
Suppose that customers arrive at a service station in accordance with a Poisson process with rate 
. Upon arrival the customer is immediately served by one of an infinite number of possible servers, and the service times are assumed to be independent with a common distribution 
. What is the distribution of 
, the number of customers that have completed service by time 
? What is the distribution of 
, the number of customers that are being served at time 
?
To answer the preceding questions let us agree to call an entering customer a type I customer if he completes his service by time 
and a type II customer if he does not complete his service by time 
. Now, if the customer enters at time 
, then he will be a type I customer if his service time is less than 
. Since the service time distribution is 
, the probability of this will be 
. Similarly, a customer entering at time 
, will be a type II customer with probability 
. Hence, from Proposition 5.3 we obtain that the distribution of 
, the number of customers that have completed service by time 
, is Poisson distributed with mean
(5.17)
Similarly, the distribution of 
, the number of customers being served at time 
is Poisson with mean
(5.18)
Furthermore, 
and 
are independent.
Suppose now that we are interested in computing the joint distribution of 
and 
—that is, the joint distribution of the number in the system at time 
and at time 
. To accomplish this, say that an arrival is
type 1: if he arrives before time 
and completes service between 
and 
,
type 2: if he arrives before 
and completes service after 
,
type 3: if he arrives between 
and 
and completes service after 
,
Hence, an arrival at time 
will be type 
with probability 
given by
Thus, if 
, denotes the number of type 
events that occur, then from Proposition 5.3, 
, are independent Poisson random variables with respective means
Because
it is now an easy matter to compute the joint distribution of 
and 
. For instance,
where the last equality follows since the variance of a Poisson random variable equals its mean, and from the substitution 
. Also, the joint distribution of 
and 
is as follows:
Example 5.19
A One Lane Road with No Overtaking
Consider a one lane road with a single entrance and a single exit point which are of distance 
from each other (See Figure 5.2). Suppose that cars enter this road according to a Poisson process with rate 
, and that each entering car has an attached random value 
which represents the velocity at which the car will travel, with the proviso that whenever the car encounters a slower moving car it must decrease its speed to that of the slower moving car. Let 
denote the velocity value of the 
th car to enter the road, and suppose that 
are independent and identically distributed and, in addition, are independent of the counting process of cars entering the road. Assuming that the road is empty at time 
, we are interested in determining
(a) the probability mass function of 
the number of cars on the road at time 
; and
(b) the distribution of the road traversal time of a car that enters the road at time 
.
Solution: Let 
denote the time it would take car 
to travel the road if it were empty when car 
arrived. Call 
the free travel time of car 
, and note that 
are independent with distribution function
Let us say that an event occurs each time that a car enters the road. Also, let 
be a fixed value, and say that an event that occurs at time 
is a type 
event if both 
and the free travel time of the car entering the road at time 
exceeds 
. In other words, a car entering the road is a type 
event if the car would be on the road at time 
even if the road were empty when it entered. Note that, independent of all that occurred prior to time 
an event occurring at time 
is a type 
event with probability
Letting 
denote the number of type 
events that occur by time 
, it follows from Proposition 5.3 that 
is, for 
a Poisson random variable with mean
Because there will be no cars on the road at time 
if and only if 
it follows that
To determine 
for 
we will condition on when the first type 
event occurs. With 
equal to the time of the first type 
event (or to 
if there are no type 
events), its distribution function is obtained by noting that
thus showing that
Differentiating gives the density function of 
:
To use the identity
(5.19)
note that if 
then the leading car that is on the road at time 
entered at time 
Because all other cars that arrive between 
and 
will also be on the road at time 
, it follows that, conditional on 
the number of cars on the road at time 
will be distributed as 
plus a Poisson random variable with mean 
Therefore, for 
Substituting this into Equation (5.19) yields
(b) Let 
be the free travel time of the car that enters the road at time 
and let 
be its actual travel time. To determine 
let 
and note that 
will be less than 
if and only if both 
and there have been no type 
events (using 
) before time 
. That is,
Because 
is independent of what has occurred prior to time 
, the preceding gives
Example 5.20
Tracking the Number of HIV Infections
There is a relatively long incubation period from the time when an individual becomes infected with the HIV virus, which causes AIDS, until the symptoms of the disease appear. As a result, it is difficult for public health officials to be certain of the number of members of the population that are infected at any given time. We will now present a first approximation model for this phenomenon, which can be used to obtain a rough estimate of the number of infected individuals.
Let us suppose that individuals contract the HIV virus in accordance with a Poisson process whose rate 
is unknown. Suppose that the time from when an individual becomes infected until symptoms of the disease appear is a random variable having a known distribution 
. Suppose also that the incubation times of different infected individuals are independent.
Let 
denote the number of individuals who have shown symptoms of the disease by time 
. Also, let 
denote the number who are HIV positive but have not yet shown any symptoms by time 
. Now, since an individual who contracts the virus at time 
will have symptoms by time 
with probability 
and will not with probability 
, it follows from Proposition 5.3 that 
and 
are independent Poisson random variables with respective means
and
Now, if we knew 
, then we could use it to estimate 
, the number of individuals infected but without any outward symptoms at time 
, by its mean value 
. However, since 
is unknown, we must first estimate it. Now, we will presumably know the value of 
, and so we can use its known value as an estimate of its mean 
. That is, if the number of individuals who have exhibited symptoms by time 
is 
, then we can estimate that
Therefore, we can estimate 
by the quantity 
given by
Using this estimate of 
, we can estimate the number of infected but symptomless individuals at time 
by
For example, suppose that 
is exponential with mean 
. Then 
, and a simple integration gives that
If we suppose that 
years, 
years, and 
thousand, then the estimate of the number of infected but symptomless individuals at time 16 is
That is, if we suppose that the foregoing model is approximately correct (and we should be aware that the assumption of a constant infection rate 
that is unchanging over time is almost certainly a weak point of the model), then if the incubation period is exponential with mean 10 years and if the total number of individuals who have exhibited AIDS symptoms during the first 16 years of the epidemic is 220 thousand, then we can expect that approximately 219 thousand individuals are HIV positive though symptomless at time 16. 
Proof of Proposition 5.3
Let us compute the joint probability 
. To do so note first that in order for there to have been 
type 
events for 
there must have been a total of 
events. Hence, conditioning on 
yields
Now consider an arbitrary event that occurred in the interval 
. If it had occurred at time 
, then the probability that it would be a type 
event would be 
. Hence, since by Theorem 5.2 this event will have occurred at some time uniformly distributed on 
, it follows that the probability that this event will be a type 
event is
independently of the other events. Hence,
will just equal the multinomial probability of 
type 
outcomes for 
when each of 
independent trials results in outcome 
with probability 
. That is,
Consequently,
and the proof is complete. 