Generalizations of the Poisson Process

Nonhomogeneous Poisson Process

Definition

A counting process {N(t);t≥0} is said to be a nonhomogeneous Poisson process with intensity function λ(t), t >0 if

1. N(0)=0
   
2. N(t) has independent increments
   
3. P(N(t+h)-N(t)=1) = λ(t)h + o(h)
   
4. P(N(t+h)-N(t)≥2) = o(h)

Using induction it is then easy to show that

Example

say you run a hot dog stand that opens at 8am. From 8am to 11am customers arrive, on average, at a steadily increasing rate of that starts at 5 customers per hour and reaches 20 at 11am. From 11am to 1 pm that average rate is constant at 20 customers per hour. Finally from 1pm to 5pm the rate drops steadily from 20 to 12. If we assume that arrivals are independent, we have a nonhomogeneous Poisson process with

and λ(t)=λ(t-9) for t>9 (the next day)

What is the probability that now customers arrive between 8:30am and 9:30am?

The number of arrivals between 8:30am and 9:30am has a Poisson distribution with rate m(3/2)-m(1/2)

What is the expected number of customers between 8:30am and 9:30am?

It is m(3/2)-m(1/2) = 10

Compound Poisson Processes

Say {N(t),t>0} is a Poisson process, and {Y_i,i≥t} is a family of iid rv’s, independent also of N(t). Let

then X(t) is called a compound Poisson process.

Example

1. say Y_i=1, then X is a regular Poisson process

2. Suppose buses arrive at an event in accordance with a Poisson process, and suppose the number of people on each bus is are iid. Say X(t) is the total number of people who arrived at time t, the {X(t)} is a compound Poisson process

3. Suppose customers leave a supermarket in accordance to some Poisson process. If Y_i is the amount spent by customer i, and if the Y_i’s are iid then {X(t),t≥0} , the total spent up to time t, is a compound Poisson process.

Now

and

Marked Poisson

Example

say people arrive at an ATM machine at times given by a Poisson process with rate λ. At the machine they either deposit or withdraw money. The amount of money deposited or withdrawn by the i^th person is given by a rv Y_i, and we can assume the Y_i’s are independent. Say at 8am the machine is refilled with money. How much does it have to have so with a probability of 0.99 it still has money at 4pm, when it is scheduled to be filled again?

Let W_i be the waiting time until the i^th event in the Poisson process, then the sequence (W_i,Y_i) is called a marked Poisson process.

Let’s begin the analysis with a very simple version: Y_i~Ber(p). Let X₁(t) be the process that considers only the times when Y_i=1 and X₀(t) correspondingly. Then

Next we need to following

Theorem let N be a Poisson rv with rate λ. Conditional on N let X have a binomial distribution with parameters N and p. Then the unconditional distribution of X is Poisson with rate λp

proof:

Now back to our process X₁. By its definition it has independent and stationary increments, X₁(0)=0 and by the theorem above X₁(t) ~ Pois(λpt). Therefore {X₁(t),t≥0} is a Poisson process rate λp. Obviously also {X₀(t),t≥0} is a Poisson process rate λq. The surprising part is the following:

First note that X₁(t)|N(t)=n ~ Bin(n,p)

and now we find

and so we see that X₁(t) and X₀(t) are independent!

Example

Customers arrive at a store according to a Poisson process with rate of 2 per hour. Each customer is a “Buyer” with probability 0.3 or a “Window-Shopper” with probability q=0.7. What is the probability of at least 1 sale during a 2 hour period?

P(at least 1 sales) = P(N₁(t)≥1) = 1-P(N₁(t)=0) = 1-exp(-2*2*0.3) = 1-e^-1.2 =0.7

What is the probability that we have made a sale if by time t there where k customers in the store?

Let’s say the Y_i’s are discrete rv’s which take values 0, 1, .. with P(Y_i=k)=a_k

Now a straightforward extension of the calculations above shows that (X_k(t),t≥0} is Poisson process with rate λa_k

Example (Coupon Collection Problem)

There are m different coupons. Each time a person collects a coupon it is, independently of those previously obtained, of type j with probability p_j. Let N denote the number of coupons one needs in order to have a complete collection of at least one of each type. Find E[N]

Let N_j be the number of coupons needed until we have one of type j, then N=max_1≤j≤m N_j . It is easy to see that N_j ~ Geom(p_j), but they are not independent and so finding the distribution of their maximum is very difficult

Let’s assume that that coupons are collected according to a Poisson process with rate 1, and say an event is of type j if the coupon collected was of type j. If we let N_j(t) denote the number of type j coupons collected by time t, then it follows that {N_j(t),t≥0} are independent Poisson processes with rates p_j. Let X_j denote the time of the first event of type j, and let X=max_1≤j≤mX_j be the time when we have all the coupons. Now the X_j are the waiting times of independent Poisson processes, so they have an exponential distributions and are independent, so

Now let T_i be the i^th interarrival time, that is the time between finding the (i-1)^st and the i^th coupon. the X=∑T_i, but T_i~Exp(1), and they are independent, so E[X|N]=E[∑T_i|N]=NE[T₁|N]=N, so

E[X]=E{E[X|N]}=E[N]

For example, say p₁=..=p_n=p=1/m, then

m	2	3	4	5	6	7	8	9	10
E[N]	3	5.5	8.3	11.4	14.7	18.2	21.7	25.5	29.3

What if m-1 have the same probability, but one is rarer, say only half of the probability of the others? So (wlog) 2p₁==p₂=..=p_m=1, then p_i=1/(m-1/2) for i≤2≤m and p₁=1/[2(m-1/2)]

this integral has to be found numerically, using some numerical integration method. Then we get

m	2	3	4	5	6	7	8	9	10
E[N]	3.5	6.4	9.6	13.0	16.6	20.4	24.2	28.2	32.3

Proposition

If {N_i(t);t≥0} i=1,..,k represent the number of type i events occurring in (0,t] and if P_i(t) is the probability that an event occurring at time t is of type i, then

Example (HIV-Aids)

one of the difficulties in tracking the number of HIV infected people is its long incubation time, that is an infected person does not show any symptoms for a number of years, but is capable of infecting others.
Let us suppose that individuals contract HIV according to a Poisson process with unknown rate λ. Suppose that the incubation time until symptoms appear is a rv with cdf G, which is known, and suppose incubation times are independent. Let N₁(t) be the number of individuals that have shown symptoms at time t, and let N₂(t) be the number that have contracted HIV at time t but not yet shown symptoms. An individual that contracts HIV at time s will show symptoms at time t with probability G(t-s), so it follows from the above proposition that

say we know the number of individuals with system as time t is n₁, then

for example if t=16 years, μ=10 years and n₁=220,000, then n₂=219,00