3. Poisson Processes (12/10/12, see Adult and Exponential ...sman/courses/6761/6761-5... · Poisson...

3. Poisson Processes

3. Poisson Processes (12/10/12, see Adult and

Baby Ross)

Exponential Distribution

Poisson Processes

Poisson and Exponential Relationship

Generalizations

1


Exponential Distribution

Definition: The continuous RV X has the exponential

distribution with parameter λ if its p.d.f. is f(x) =

λe−λx, x ≥ 0.

Facts: F (x) = 1− e−λx, x ≥ 0.

E[X] = 1/λ, Var(X) = 1/λ2,

MX(t) = λ/(λ− t), t < λ.

2


Theorem: The exponential distribution has the mem-

oryless property . Namely, for s, t > 0,

Pr(X > s+ t|X > t) = Pr(X > s).

Example: If X ∼ Exp(1/10), then

Pr(X > 10|X > 5) = Pr(X > 5) = e−5/10 = 0.607.

Remark: The exponential is the only continuous dis-

tribution with this property.3


Definition: For a continuous distribution, the hazard

function (or failure rate) is

r(t) ≡f(t)

1− F (t).

The hazard function is the conditional p.d.f. that X

will fail at time t (given that X made it to t).

4


Why this interpretation?

r(t) dt =f(t) dt

1− F (t)

≈Pr(X ∈ (t, t+ dt))

Pr(X > t)

=Pr(X ∈ (t, t+ dt) and X > t)

Pr(X > t)= Pr(X ∈ (t, t+ dt)|X > t).

Example: X ∼ Exp(λ) implies that r(t) = λ. This

makes sense in light of the memoryless property. In

fact, the Exp(λ) is the only RV with constant r(t).

5


Remark: r(t) (uniquely) determines F (t).

“Proof:”

r(t) =f(t)

1− F (t)= = −

d

dt`n(1− F (t)),

so that

F (t) = 1− exp(−∫ t0r(s) ds

). ♦

6


Theorem: X1, X2, . . . , Xniid∼ Exp(λ) implies that

n∑i=1

Xi ∼ Erlangn(λ) ∼ Gamma(n, λ).

Many proofs — m.g.f.’s, induction, you name it. ♦

7


Poisson Processes

Definition: Consider discrete (non-fractional) events

occurring in continuous intervals (time, length, vol-

ume, etc.). The counting process N(t) ≡ the number

of events occurring in [0, t]. Let the rate λ > 0 be

the average number of occurrences per unit time (or

length or volume).

8


Examples: 1. Cars entering a shopping center (time);

λ = 5/min.

2. Defects on a wire (length); λ = 3/ft.

3. Raisins in cookie dough (volume); λ = 2.6/in3.

9


Definition: A counting process N(t) satisfying the fol-

lowing three assumptions is called a Poisson process

with rate λ (PP(λ)).

(A-1) Arrivals occur one-at-a-time,

(A-2) Independent increments, and

(A-3) Stationary increments.

Details follow. . .

10


(A-1) For any interval of sufficiently small length h,

(a) The prob of one arrival in that interval is

Pr(N(t+ h)−N(t) = 1) = λh+ o(h).

= λh.

(b) The prob of no arrivals is

Pr(N(t+ h)−N(t) = 0) = 1− λh+ o(h).

= 1− λh.

(c) The prob of more than one arrival is

Pr(N(t+ h)−N(t) ≥ 2) = o(h).

= 0,

where o(h) is a function that → 0 faster than h→ 0.11


(A-2) Independent Increments: The numbers of ar-

rivals in two disjoint intervals are independent. I.e., if

a < b < c < d, then N(d) − N(c) and N(b) − N(a) are

independent.

(A-3) Stationary Increments: The number of arrivals

in a time interval depends only on its length. Thus,

N(t+ s)−N(t) ∼ N(s)−N(0) for all t.

(A-4) (bonus assumption): N(0) = 0.

12


Example: People arriving to a restaurant is not a PP.

Arrivals occur in groups (violating A-1), and arrival

rates change throughout the day (violating A-3). ♦

Theorem: If N(t) is a PP(λ), then N(t) ∼ Pois(λt).

In particular, N(1) ∼ Pois(1). Further, E[N(t)] =

Var(N(t)) = λt.

Proof: Let px(t) = Pr(N(t) = x). Then for (small)

h > 0, we have

p0(t+ h) ≈ (1− λh)p0(t).13


Thus,p0(t+ h)− p0(t)

h≈ −λp0(t),

and

limh→0

p0(t+ h)− p0(t)

h=

d

dtp0(t) = −λp0(t).

Similarly, for x > 0,

px(t+ h) ≈ λhpx−1(t) + (1− λh)px(t),

so thatd

dtpx(t) = λpx−1(t) + λpx(t).

14


The solution to these equations is (check for yourself)

px(t) = (λt)xe−λt/x!, x = 0,1,2, . . .. ♦

Remark: Be careful with units of time. For example,

suppose that X ∼ Pois(3) is the number of phone calls

in a one-minute period.

Then the number of calls in a 3-minute period is

Pois(9) and the number in a 30-sec. period is Pois(1.5).

15


Poisson and Exponential Relationship

Definition: Let A1 denote the time until the first ar-

rival of a PP(λ).

For i ≥ 2, let Ai denote the time between the (i−1)st

and ith arrivals.

The Ai’s are called interarrival times.

We can get the distributions of the Ai’s. . .16


First of all, consider A1.

A1 > t iff no arrivals take place in [0, t]. Thus, for

t > 0,

Pr(A1 > t) = Pr(N(t) = 0) =e−λt(λt)0

0!= e−λt,

and so A1 ∼ Exp(λ).

17


Now A2. For 0 < s < t, we have

Pr(A2 > t|A1 = s)

= Pr(no arrivals in (s, s+ t]|A1 = s)

= Pr(no arrivals in (s, s+ t])

(by independent increments)

= Pr(no arrivals in (0, t])

(by stationary increments)

= e−λt (by previous arguments).

18


Thus, Pr(A2 > t|A1 = s) = e−λt for all s.

So Pr(A2 > t) = e−λt, i.e., A2 ∼ Exp(λ), independent

of the value of A1.

This can be generalized. . .

Theorem: A1, A2, . . .iid∼ Exp(λ).

19


Definition: The nth arrival time is Sn ≡∑ni=1Ai, n ≥ 1.

Theorem: Sn ∼ Gamma(n, λ) ∼ Erlangn(λ). Thus,

E[Sn] = n/λ and Var(Sn) = n/λ2.

Example: A UGA student can read X ∼ Pois(1) pages

of Dick and Jane a day. (a) The expected time until

the 10th page is completed is E[S10] = n/λ = 10 days.

(b) The prob that it takes > 2 days to read the 11th

page is Pr(A11 > 2) = e−2 .= 0.135. ♦

20


Theorem: Consider a PP(λ), N(t). Suppose arrivals

are either Type I or II (e.g., male or female), with

Pr(Type I) = p, Pr(Type II) = 1 − p. Let N1(t) and

N2(t) denote the numbers of Type I and II events

during [0, t]. Note that N(t) = N1(t) +N2(t).

Then N1(t) and N2(t) are independent PP’s with resp.

rates λp and λ(1− p).

21


Proof: Let’s find the joint pmf

Pr(N1(t) = n,N2(t) = m)

=∞∑k=0

Pr(N1(t) = n,N2(t) = m|N(t) = k)Pr(N(t) = k)

= Pr(N1(t) = n,N2(t) = m|N(t) = n+m)

×Pr(N(t) = n+m),

since the terms are 0 when k 6= n+m.

22


It’s obvious that the number of Type I arrivals is

N1(t) ∼ Bin(n + m, p) (and N2(t) = n + m − N1(t)).

So

Pr(N1(t) = n,N2(t) = m)

=(n+m

n

)pn(1− p)m

e−λλn+m

(n+m)!

=e−λp(λp)n

n!

e−λ(1−p)(λ(1− p))m

m!.

Since the joint pmf factors into two Poisson pmf’s,

we’re done. ♦23


Example: Suppose cars entering a parking lot follow a

PP(5/min). Further, suppose the prob that a driver is

female is 0.6. Find the prob that exactly 3 cars driven

by females will enter the lot in the next 2 minutes.

The # of cars entering in two min. ∼ Pois(λt = 10).

The # driven by females is Pois(λtp = 6).

So the desired prob is

Pr(Pois(6) = 3) =e−663

3!= 0.0892. ♦

24


Conditional Distribution of Arrival Times

Theorem: Pr(A1 < s|N(t) = 1) = st.

In other words, suppose we know that one arrival has

occurred by time t, i.e., N(t) = 1. What’s the (con-

ditional) distribution of the time of that arrival? An-

swer: Unif(0, t)!

25


Proof: We have

Pr(A1 < s|N(t) = 1)

=Pr(A1 < s and N(t) = 1)

Pr(N(t) = 1)

=Pr(1 arrival in [0, s]; none in (s, t])

Pr(N(t) = 1)

=Pr(1 arrival in [0, s]) Pr(none in (s, t])

Pr(N(t) = 1)(by independent increments)

=Pr(N(s) = 1) Pr(N(t)−N(s) = 0)

Pr(N(t) = 1)26


Proof (cont’d): Then

Pr(A1 < s|N(t) = 1)

=Pr(N(s) = 1)Pr(N(t− s) = 0)

Pr(N(t) = 1)(by stationary increments)

=e−λs(λs)1

1!

e−λ(t−s)(λ(t− s))0

0!

/e−λt(λt)1

1!

= s/t. ♦

27


Can generalize above result. . .

Theorem: Given that N(t) = n, the joint probability

distribution of the n arrival times S1, S2, . . . , Sn is the

same as the joint distribution of n i.i.d. Unif(0, t) RV’s.

Thus, if we know that n arrivals have occurred by

time t, the arrivals can be treated as if they were i.i.d.

Unif(0, t).

28


Aside: Before we prove the Theorem, we need a def-

inition. Suppose Y1, Y2, . . . , Yn are iid. The ordered

Yi’s, denoted Y(1) ≤ Y(2) ≤ · · · ≤ Y(n), are called order

statistics.

Fact: If the Yi’s are iid continuous RV’s with pdf f(y),

then the joint pdf of the order statistics is

fY(1),...,Y(n)(y1, . . . , yn) = n!

n∏i=1

f(yi), y1 ≤ · · · ≤ yn.

(The fact is proven via an easy permutation argument;

see any stats text.)29


Proof of Theorem: Let 0 < t1 < · · · < tn+1 = t, and

let hi be small enough so that ti + hi < ti+1 for all

i. Then, using the stationary and independent incre-

ments properties of the PP(λ),

Pr(ti ≤ Si ≤ ti + hi, i = 1, . . . , n |N(t) = n)

=Pr(1 arrival in each [ti, ti + hi] and none elsewhere)

Pr(N(t) = n)

=λh1e

−λh1 · · ·λhne−λhn × e−λ(t−h1−···−hn)

e−λt(λt)n/n!

=n!

tnh1h2 · · ·hn.

30


Now divide both sides by h1h2 · · ·hn and take the limits

as hi → 0 for all i. This gives the conditional pdf of

S1, . . . , Sn given that N(t) = n,

f(t1, . . . , tn) =n!

tn, 0 < t1 < · · · < tn.

The previous Fact indicates that this is also the joint

pdf of the order statistics from n iid Unif(0, t)’s, so

we are done. ♦

31


Example: Travelers arrive at a train station as a PP(λ).

If the train leaves at time t, let’s find the expected sum

of the waiting times of the customers arriving in [0, t].

First, condition on N(t) to get

E[N(t)∑i=1

(t− Si)∣∣∣∣N(t) = n

]

= E[ n∑i=1

(t− Si)∣∣∣∣N(t) = n

]

= nt− E[ n∑i=1

Si

∣∣∣∣N(t) = n].

32


If we let U1, U2, . . . , Un denote iid Unif(0, t) RV’s with

order statistics U(1) ≤ · · · ≤ U(n), then the previous

theorem implies

E[N(t)∑i=1

(t− Si)∣∣∣∣N(t) = n

]= nt− E

[ n∑i=1

U(i)

]

= nt− E[ n∑i=1

Ui

]

=nt

2.

33


Thus, we have the desired (unconditional) result,

E[N(t)∑i=1

(t− Si)]

=∞∑n=0

E[N(t)∑i=1

(t− Si)∣∣∣∣N(t) = n

]Pr(N(t) = n)

=∞∑n=0

nt

2Pr(N(t) = n)

=t

2E[N(t)] =

λt2

2. ♦

34


Example: A device is subject to PP(λ) shocks, and

the ith shock does damage Di. The Di’s are iid and

independent of their arrival times S1, S2, . . . and N(t).

Luckily, the damage due to a shock fades away ex-

ponentially in time; so if a shock has initial damage

D, then s time units later, the damage improves to

De−αs. On the other hand, damages are additive.

The total damage at time t is

D(t) =N(t)∑i=1

Die−α(t−Si).

35


In order to eventually find E[D(t)], we first get

E[D(t)|N(t) = n]

= E[N(t)∑i=1

Di e−α(t−Si)

∣∣∣∣N(t) = n]

=n∑i=1

E[Di e

−α(t−Si)∣∣∣∣N(t) = n

]

=n∑i=1

E[Di|N(t) = n]E[e−α(t−Si)

∣∣∣∣N(t) = n]

= E[D]e−αtn∑i=1

E[eαSi

∣∣∣∣N(t) = n]

(where the last two steps follow by independence).

36


If U(1) ≤ · · · ≤ U(n) denote iid Unif(0, t) order statistics,

the previous theorem implies

E[D(t)|N(t) = n]


E[eαU(i)

]


E[eαUi

]

= nE[D]e−αt(eαt − 1)

αt(Unif(0,t) mgf)

= nE[D](1− e−αt)

αt.

37


Thus,

E[D(t)] =∞∑n=0

E[D(t)|N(t) = n]Pr(N(t) = n)

= E[D](1− e−αt)

αt

∞∑n=0

nPr(N(t) = n)

= E[D](1− e−αt)

αtE[N(t)]

= E[D] (1− e−αt)λ

α. ♦

38


Bonus Theorem: Consider a PP(λ), N(t). Suppose

there are k possible types of arrivals. Further suppose

that the prob that an arrival is of Type i depends

on the time that it occurs — if it occurs at time t,

then the probability that it’s a Type i is Pi(t), where∑ki=1Pi(t) = 1. If Ni(t) denotes the number of Type i

arrivals by time t, then the Ni(t)’s, i = 1,2, . . . , k, are

independent Poisson RV’s with means

E[Ni(t)] = λ∫ t0Pi(s) ds.

Proof: Ross. ♦39


Example (infinite server queue): Customers arrive as a

PP(λ). They are immediately served, and the service

times are iid with cdf G(t). A Type 1 [2] customer

completes [doesn’t complete] service by time t. Let’s

get the joint distribution of the numbers N1(t) and

N2(t) of Type 1 and Type 2 customers.

If a customer enters at time s ≤ t and if its service

time is < t − s, then it will be a Type 1; and the

probability of this event is P1(s) = G(t− s).

40

Similarly, the probability of a Type 2 at time s ≤ t is

P2(s) = 1−G(t−s). The Bonus Theorem immediately

implies that N1(t) and N2(t) are independent Poisson

RV’s with

E[N1(t)] = λ∫ t0G(t− s) ds = λ

∫ t0G(s) ds,

and

E[N2(t)] = λ∫ t0G(s) ds. ♦


Generalizations

Definition: The counting process N(t) is a nonhomo-

geneous PP with intensity function λ(t) if it satisfies:

(A-1′) Arrivals occur one-at-a-time. In particular,

Pr(N(t+ h)−N(t) = 1) = λ(t)h+ o(h) and

Pr(N(t+ h)−N(t) = 0) = 1− λ(t)h+ o(h)

(A-2) Independent increments.

(A-4) N(0) = 0.

Note: We don’t require stationary increments.41


Fact: N(t+ s)−N(t) ∼ Pois(∫ t+st λ(x) dx

).

Remark: Arrivals may be more (or less) likely to oc-

cur as time progresses in a NHPP, since there is no

stationarity requirement.

42


Example: Distribution of cars arriving to a lot as the

day progresses is a NHPP. From 8:00–11:00 a.m.,

cars arrive at a steadily increasing rate: 5 cars/hr at

8:00 a.m. to 20 cars/hr at 11:00 a.m. I.e., λ(t) =

5 + 5t, 0 ≤ t ≤ 3 hrs.

Number of arrivals between 8:30–9:30 a.m. is

N(1.5)−N(0.5) ∼ Pois(∫ 1.5

0.5(5 + 5x) dx

)∼ Pois(10).

♦43


Definition: A stochastic process X(t) is a compound

PP if it can be written as

X(t) =N(t)∑i=1

Yi, t ≥ 0,

where N(t) is a PP(λ) and the Yi’s are i.i.d. and inde-

pendent of N(t).

Facts: E[X(t)] = E[N(t)]E[Y1] = λtE[Y1] and

Var(X(t)) = λtE[Y 21 ].

44


Example: If the Yi’s all equal 1, then X(t) = N(t),

the usual PP.

Example: Customers leave a market according to a

PP. Let Yi be the amount spent by customer i. If

X(t) is the total amount spent by time t, then X(t)

is a compound PP.

45


Example: A FB player makes Pois(λ = 2) scores/game.

Pr(score = x) =

1/6 if x = 11/3 if x = 31/2 if x = 6

Let Yi be the value of the ith score. E[Yi] = 256 ,

E[Y 2i ] = 127

6 . Let X(t) be the total points scored in t

games.

E[X(5)] = λtE[Y1] = 125/3,

Var(X(t)) = λtE[Y 21 ] = 635/3. ♦

46

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