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# Chapter 2. Poisson Processes

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## Chapter 2. Poisson Processes

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### Outline

• Introduction to Poisson Processes

• Properties of Poisson processes – Inter-arrival time distribution – Waiting time distribution

– Superposition and decomposition

• Non-homogeneous Poisson processes (relaxing stationary)

• Compound Poisson processes (relaxing single arrival)

• Modulated Poisson processes (relaxing independent)

• Poisson Arrival See Average (PASTA)

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### Introduction

t

~1

x

~x2

~x3

0

S~ ~1

S ~2

S ~3

S ~4

S )

~ t(

n Inter-arrival time

arrival epoch

(i) nth arrival epoch ˜Sn is

S˜n = x˜1 + ˜x2 + . . . + ˜xn = ni=1 x˜i S˜0 = 0

(ii) Number of arrivals at time t is: ˜n(t). Notice that:

{˜n(t) ≥ n} if f⇔ { ˜Sn ≤ t}, {˜n(t) = n} if f⇔ { ˜Sn ≤ t and ˜Sn+1 > t}

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### Introduction

Arrival Process: X = {˜xi, i = 1, 2, . . .}; ˜xi’s can be any S = { ˜Si, i = 0, 1, 2, . . .}; ˜Si’s can be any

N = {˜n(t), t ≥ 0}; −→ called arrival process

Renewal Process: X = {˜xi, i = 1, 2, . . .}; ˜xi’s are i.i.d.

S = { ˜Si, i = 0, 1, 2, . . .}; ˜Si’s are general distributed N = {˜n(t), t ≥ 0}; −→ called renewal process

Poisson Process: X = {˜xi, i = 1, 2, . . .}; ˜xi’s are iid exponential distributed S = { ˜Si, i = 0, 1, 2, . . .}; ˜Si’s are Erlang distributed

N = {˜n(t), t ≥ 0}; −→ called Poisson process

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### Counting Processes

• A stochastic process N = {˜n(t), t ≥ 0} is said to be a counting process if ˜n(t) represents the total number of “events” that have occurred up to time t.

• From the deﬁnition we see that for a counting process ˜n(t) must satisfy:

1. ˜n(t) ≥ 0.

2. ˜n(t) is integer valued.

3. If s < t, then ˜n(s) ≤ ˜n(t).

4. For s < t, ˜n(t) − ˜n(s) equals the number of events that have occurred in the interval (s, t].

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### Definition 1: Poisson Processes

The counting process N = {˜n(t), t ≥ 0} is a Poisson process with rate λ (λ > 0), if:

1. ˜n(0) = 0

2. Independent increments relaxed ⇒ Modulated Poisson Process P [˜n(t) − ˜n(s) = k1|˜n(r) = k2, r ≤ s < t] = P [˜n(t) − ˜n(s) = k1] 3. Stationary increments relaxed ⇒ Non-homogeneous Poisson Process

P [˜n(t + s) − ˜n(t) = k] = P [˜n(l + s) − ˜n(l) = k]

4. Single arrival relaxed ⇒ Compound Poisson Process P [˜n(h) = 1] = λh + o(h)

P [˜n(h) ≥ 2] = o(h)

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### Definition 2: Poisson Processes

The counting process N = {˜n(t), t ≥ 0} is a Poisson process with rate λ (λ > 0), if:

1. ˜n(0) = 0

2. Independent increments

3. The number of events in any interval of length t is Poisson distributed with mean λt. That is, for all s, t ≥ 0

P [˜n(t + s) − ˜n(s) = n] = e−λt (λt)n

n! , n = 0, 1, . . .

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### Theorem: Definitions 1 and 2 are equivalent.

Proof. We show that Deﬁnition 1 implies Deﬁnition 2. To start, ﬁx u ≥ 0 and let

g(t) = E[e−u˜n(t)]

We derive a diﬀerential equation for g(t) as follows:

g(t + h) = E[e−u˜n(t+h)]

= E



e−u˜n(t)e−u[˜n(t+h)−˜n(t)]

= E



e−u˜n(t)



E



e−u[˜n(t+h)−˜n(t)]

by independent increments

= g(t)E



e−u˜n(h)



by stationary increments (1)

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### Theorem: Definitions 1 and 2 are equivalent.

Conditioning on whether ˜n(t) = 0 or ˜n(t) = 1 or ˜n(t) ≥ 2 yields E



e−u˜n(h)



= 1 − λh + o(h) + e−u(λh + o(h)) + o(h)

= 1 − λh + e−uλh + o(h) (2)

From (1) and (2), we obtain that

g(t + h) = g(t)(1 − λh + e−uλh) + o(h) implying that

g(t + h) − g(t)

h = g(t)λ(e−u − 1) + o(h) h

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### Theorem: Definitions 1 and 2 are equivalent.

Letting h → 0 gives

g(t) = g(t)λ(e−u − 1) or, equivalently,

g(t)

g(t) = λ(e−u − 1) Integrating, and using g(0) = 1, shows that

log(g(t)) = λt(e−u − 1) or

g(t) = eλt(e−u−1) → the Laplace transform of a Poisson r. v.

Since g(t) is also the Laplace transform of ˜n(t), ˜n(t) is a Poisson r. v.

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### Theorem: Definitions 1 and 2 are equivalent

Let Pn(t) = P [˜n(t) = n].

We derive a diﬀerential equation for P0(t) in the following manner:

P0(t + h) = P [˜n(t + h) = 0]

= P [˜n(t) = 0, ˜n(t + h) − ˜n(t) = 0]

= P [˜n(t) = 0]P [˜n(t + h) − ˜n(t) = 0]

= P0(t)[1 − λh + 0(h)]

Hence

P0(t + h) − P0(t)

h = −λP0(t) + o(h) h Letting h → 0 yields

P0(t) = −λP0(t) Since P0(0) = 1, then

P0(t) = e−λt

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### Theorem: Definitions 1 and 2 are equivalent

Similarly, for n ≥ 1

Pn(t + h) = P [˜n(t + h) = n]

= P [˜n(t) = n, ˜n(t + h) − ˜n(t) = 0]

+P [˜n(t) = n − 1, ˜n(t + h) − ˜n(t) = 1]

+P [˜n(t + h) = n, ˜n(t + h) − ˜n(t) ≥ 2]

= Pn(t)P0(h) + Pn−1(t)P1(h) + o(h)

= (1 − λh)Pn(t) + λhPn−1(t) + o(h) Thus Pn(t + h) − Pn(t)

h = −λPn(t) + λPn−1(t) + o(h) h Letting h → 0,

Pn (t) = −λPn(t) + λPn−1(t)

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### The Inter-Arrival Time Distribution

Theorem. Poisson Processes have exponential inter-arrival time distribution, i.e., {˜xn, n = 1, 2, . . .} are i.i.d and exponentially

distributed with parameter λ (i.e., mean inter-arrival time = 1/λ).

Proof.

˜

x1 : P (˜x1 > t) = P (˜n(t) = 0) = e−λt(λt)0

0! = e−λt

··

· x˜1 ∼ e(t; λ)

˜

x2 : P (˜x2 > t|˜x1 = s)

= P{0 arrivals in (s, s + t]|˜x1 = s}

= P{0 arrivals in (s, s + t]}(by independent increment)

= P{0 arrivals in (0, t]}(by stationary increment)

= e−λt ··· x˜2 is independent of ˜x1 and ˜x2 ∼ exp(t; λ).

⇒ The procedure repeats for the rest of ˜xi’s.

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### The Arrival Time Distribution of thenth Event

Theorem. The arrival time of the nth event, ˜Sn (also called the waiting time until the nth event), is Erlang distributed with parameter (n, λ).

Proof. Method 1 :

··· P [ ˜Sn ≤ t] = P [˜n(t) ≥ n] = 

k=n

e−λt(λt)k k!

··

· fS˜

n(t) = λe−λt(λt)n−1

(n − 1)! (exercise) Method 2 :

fS˜

n(t)dt = dFS˜

n(t) = P [t < ˜Sn < t + dt]

= P{n − 1 arrivals in (0, t] and 1 arrival in (t, t + dt)} + o(dt)

= P [˜n(t) = n − 1 and 1 arrival in (t, t + dt)] + o(dt)

= P [˜n(t) = n − 1]P [1 arrival in (t, t + dt)] + o(dt)(why?)

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### The Arrival Time Distribution of thenth Event

= e−λt(λt)n−1

(n − 1)! λdt + o(dt)

··

· lim

dt→0

fS˜

n(t)dt

dt = fS˜

n(t) = λe−λt(λt)n−1 (n − 1)!

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### Conditional Distribution of the Arrival Times

Theorem. Given that ˜n(t) = n, the n arrival times ˜S1, ˜S2, . . . , ˜Sn have the same distribution as the order statistics corresponding to n i.i.d.

uniformly distributed random variables from (0, t).

. . . . Order Statistics. Let ˜x1, ˜x2, . . . , ˜xn be n i.i.d. continuous random

variables having common pdf f . Deﬁne ˜x(k) as the kth smallest value among all ˜xi’s, i.e., ˜x(1) ≤ ˜x(2) ≤ ˜x(3) ≤ . . . ≤ ˜x(n), then ˜x(1), . . . , ˜x(n) are known as the “order statistics” corresponding to random variables

˜

x1, . . . , ˜xn. We have that the joint pdf of ˜x(1), ˜x(2), . . . , ˜x(n) is f˜x(1)x(2),...,˜x(n)(x1, x2, . . . , xn) = n!f (x1)f (x2) . . . f (xn), where x1 < x2 < . . . < xn (check the textbook [Ross]).

. . . .

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### Conditional Distribution of the Arrival Times

Proof. Let 0 < t1 < t2 < . . . < tn+1 = t and let hi be small enough so that ti + hi < ti+1 , i = 1, . . . , n.

··· P [ti < ˜Si < ti + hi, i = 1, . . . , n|˜n(t) = n]

=

P

exactly one arrival in each [ti, ti + hi]

i = 1, 2, . . . , n, and no arrival elsewhere in [0, t]

P [˜n(t) = n]

= (e−λh1λh1)(e−λh2λh2) . . . (e−λhnλhn)(e−λ(t−h1−h2...−hn)) e−λt(λt)n/n!

= n!(h1h2h3 . . . hn) tn

··

· P [ti < ˜Si < ti + hi, i = 1, . . . , n|˜n(t) = n]

h1h2 . . . hn = n!

tn

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### Conditional Distribution of the Arrival Times

Taking lim

hi→0,i=1,...,n( ), then fS˜

1, ˜S2,..., ˜Sn|˜n(t)(t1, t2, . . . , tn|n) = n!

tn , 0 < t1 < t2 < . . . < tn.

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### Conditional Distribution of the Arrival Times

Example (see Ref [Ross], Ex. 2.3(A) p.68). Suppose that travellers arrive at a train depot in accordance with a Poisson process with rate λ. If the train departs at time t, what is the expected sum of the

waiting times of travellers arriving in (0, t)? That is, E[i=1˜n(t)(t− ˜Si)] =?

. . . t

1

S~ ~2

S ~3

S ~~( )

t

Sn

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### Conditional Distribution of the Arrival Times

Answer. Conditioning on ˜n(t) = n yields E[

˜n(t)



i=1

(t − ˜Si)|˜n(t) = n] = nt − E[n

i=1

S˜i]

= nt − E[n

i=1

˜

u(i)] (by the theorem)

= nt − E[n

i=1

˜

ui] (···

n i=1

˜

u(i) =

n i=1

˜ ui)

= nt t

2 · n = nt

2 (··· E[˜ui] = t 2) To ﬁnd E[

˜n(t)



i=1

(t − ˜Si)], we should take another expectation

··

· E[

˜n(t)



i=1

(t − ˜Si)] = t

2 · E[˜n(t)]  

=λt

= λt2 2

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### Superposition of Independent Poisson Processes

Theorem. Superposition of independent Poisson Processes i, i = 1, . . . , N ), is also a Poisson process with rate

N 1

λi.

Poisson Poisson

Poisson

Poisson

λ1

λ2

λN rate =

N

1

λi

<Homework> Prove the theorem (note that a Poisson process must satisfy Deﬁnitions 1 or 2).

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### Decomposition of a Poisson Process

Theorem.

• Given a Poisson process N = {˜n(t), t ≥ 0};

• If ˜ni(t) represents the number of type-i events that occur by time t, i = 1, 2;

• Arrival occurring at time s is a type-1 arrival with probability p(s), and type-2 arrival with probability 1 − p(s)

⇓then

• ˜n1, ˜n2 are independent,

• ˜n1(t) ∼ P (k; λtp), and

• ˜n2(t) ∼ P (k; λt(1 − p)), where p = 1 t

 t

0 p(s)ds

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### Decomposition of a Poisson Process

Poisson Poisson (k;λ

p(s)ds)

### ∫

[1 ( )] )

;

(k λ p s ds

Poisson

λ p(s)

1-p(s)

special case: If p(s) = p is constant, then

Poisson Poisson rate

p

) 1

( − p

Poisson rate

p

1-p

(24)

### Decomposition of a Poisson Process

Proof. It is to prove that, for ﬁxed time t,

P [˜n1(t) = n, ˜n2(t) = m] = P [˜n1(t) = n]P [˜n2(t) = m]

= e−λpt(λpt)n

n! · e−λ(1−p)t[λ(1 − p)t]m m!

. . . .

P [˜n1(t) = n, ˜n2(t) = m]

=

 k=0

P [˜n1(t) = n, ˜n2(t) = m|˜n1(t) + ˜n2(t) = k] · P [˜n1(t) + ˜n2(t) = k]

= P [˜n1(t) = n, ˜n2(t) = m|˜n1(t) + ˜n2(t) = n + m] · P [˜n1(t) + ˜n2(t) = n + m]

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### Decomposition of a Poisson Process

• From the “condition distribution of the arrival times”, any event

occurs at some time that is uniformly distributed, and is independent of other events.

• Consider that only one arrival occurs in the interval [0, t]:

P [type - 1 arrival|˜n(t) = 1]

=

 t

0 P [type - 1 arrival|arrival time ˜S1 = s, ˜n(t) = 1]

×fS˜1|˜n(t)(s|˜n(t) = 1)ds

=

 t

0 P (s) · 1

t ds = 1 t

 t

0 P (s)ds = p

(26)

### Decomposition of a Poisson Process

··

· P [˜n1(t) = n, ˜n2(t) = m]

= P [˜n1(t) = n, ˜n2(t) = m|˜n1(t) + ˜n2(t) = n + m] · P [˜n1(t) + ˜n2(t) = n + m]

=

n + m n

pn(1 − p)m · e−λt(λt)n+m (n + m)!

= (n + m)!

n!m! pn(1 − p)m · e−λt(λt)n+m (n + m)!

= e−λpt(λpt)n

n! · e−λ(1−p)t[λ(1 − p)t]m m!

(27)

### Decomposition of a Poisson Process

Example (An Inﬁnite Server Queue, textbook [Ross]).

Poisson λ departure

G

• G˜s(t) = P ( ˜S ≤ t), where ˜S = service time

• G˜s(t) is independent of each other and of the arrival process

• ˜n1(t): the number of customers which have left before t;

• ˜n2(t): the number of customers which are still in the system at time t;

⇒ ˜n1(t) ∼? and ˜n2(t) ∼?

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### Decomposition of a Poisson Process

˜

n1(t): the number of type-1 customers

˜

n2(t): the number of type-2 customers

type-1: P (s) = P (ﬁnish before t)

= P ( ˜S ≤ t − s) = G˜s(t − s) type-2: 1 − P (s) = ¯G˜s(t − s)

··

· n˜1(t) ∼ P



k; λt · 1 t

 t

0 G˜s(t − s)ds



˜

n2(t) ∼ P



k; λt · 1 t

 t

0

G¯˜s(t − s)ds



(29)

### Decomposition of a Poisson Process

··

· E[˜n1(t)] = λt · 1 t

 t

0 G(t − s)ds

= λ

 0

t

G(y)(−dy) t − s = y s = t − y

= λ

 t

0 G(y)dy

As t → ∞, we have

t→∞lim E[˜n2(t)] = λ

 t

0

G(y)dy = λE[ ˜¯ S] (Little’s formula)

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### Non-homogeneous Poisson Processes

• The counting process N = {˜n(t), t ≥ 0} is said to be a non-stationary or non-homogeneous Poisson Process with time-varying intensity

function λ(t), t ≥ 0, if:

1. ˜n(0) = 0

2. N has independent increments 3. P [˜n(t + h) − ˜n(t) ≥ 2] = o(h)

4. P [˜n(t + h) − ˜n(t) = 1] = λ(t) · h + o(h)

• Deﬁne “integrated intensity function” m(t) =  t

0 λ(t)dt. Theorem.

P [˜n(t + s) − ˜n(t) = n] = e−[m(t+s)−m(t)][m(t + s) − m(t)]n n!

Proof. < Homework >.

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### Non-homogeneous Poisson Processes

Example. The “output process” of the M/G/∞ queue is a

non-homogeneous Poisson process having intensity function λ(t) = λG(t), where G is the service distribution.

Hint. Let D(s, s + r) denote the number of service completions in the interval (s, s + r] in (0, t]. If we can show that

• D(s, s + r) follows a Poisson distribution with mean λ ss+r G(y)dy, and

• the numbers of service completions in disjoint intervals are independent,

then we are ﬁnished by deﬁnition of a non-homogeneous Poisson process.

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### Non-homogeneous Poisson Processes

• An arrival at time y is called a type-1 arrival if its service completion occurs in (s, s + r].

• Consider three cases to ﬁnd the probability P (y) that an arrival at time y is a type-1 arrival:

s s+r t

Case 1 Case 2 Case 3

y y y

– Case 1: y ≤ s.

P (y) = P{s − y < ˜S < s + r − y} = G(s + r − y) − G(s − y) – Case 2: s < y ≤ s + r.

P (y) = P{ ˜S < s + r − y} = G(s + r − y)

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### Non-homogeneous Poisson Processes

– Case 3: s + r < y ≤ t.

P (y) = 0

• Based on the decomposition property of a Poisson process, we

conclude that D(s, s + r) follows a Poisson distribution with mean λpt, where p = (1/t)0t P (y)dy.

 t

0 P (y)dy =

 s

0 [G(s + r − y) − G(s − y)]dy +  s+r

s

G(s + r − y)dy +

 t

s+r

(0)dy

=

 s+r

0 G(s + r − y)dy −  s

0 G(s − y)dy

=

 s+r

0 G(z)dz  s

0 G(z)dz =

 s+r

s

G(z)dz

(34)

### Non-homogeneous Poisson Processes

• Because of

– the independent increment assumption of the Poisson arrival process, and

– the fact that there are always servers available for arrivals,

⇒ the departure process has independent increments

(35)

### Compound Poisson Processes

• A stochastic process {˜x(t), t ≥ 0} is said to be a compound Poisson process if

– it can be represented as

˜

x(t) =

˜n(t)



i=1

˜

yi, t ≥ 0 – {˜n(t), t ≥ 0} is a Poisson process

– {˜yi, i ≥ 1} is a family of independent and identically distributed random variables which are also independent of {˜n(t), t ≥ 0}

• The random variable ˜x(t) is said to be a compound Poisson random variable.

• E[˜x(t)] = λtE[˜yi] and V ar[˜x(t)] = λtE[˜yi2].

(36)

### Compound Poisson Processes

• Example (Batch Arrival Process). Consider a parallel-processing

system where each job arrival consists of a possibly random number of tasks. Then we can model the arrival process as a compound

Poisson process, which is also called a batch arrival process.

• Let ˜yi be a random variable that denotes the number of tasks comprising a job. We derive the probability generating function P˜x(t)(z) as follows:

P˜x(t)(z) = E



z˜x(t)



= E



E



z˜x(t)|˜n(t) = E



E



z˜y1+···+˜yn(t)˜ |˜n(t)

= E



E



z˜y1+···+˜yn(t)˜



(by independence of ˜n(t) and {˜yi})

= E E z˜y1 · · · E z˜yn(t)˜  (by independence of ˜y1, · · · , ˜y˜n(t))

= E



(P˜y(z))˜n(t)



= P˜n(t) (P˜y(z))

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### Modulated Poisson Processes

• Assume that there are two states, 0 and 1, for a “modulating process.”

0 1

• When the state of the modulating process equals 0 then the arrive rate of customers is given by λ0, and when it equals 1 then the arrival rate is λ1.

• The residence time in a particular modulating state is exponentially distributed with parameter μ and, after expiration of this time, the modulating process changes state.

• The initial state of the modulating process is randomly selected and is equally likely to be state 0 or 1.

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### Modulated Poisson Processes

• For a given period of time (0, t), let Υ be a random variable that indicates the total amount of time that the modulating process has been in state 0. Let ˜x(t) be the number of arrivals in (0, t).

• Then, given Υ, the value of ˜x(t) is distributed as a non-homogeneous Poisson process and thus

P [˜x(t) = n|Υ = τ] = 0τ + λ1(t − τ))ne−(λ0τ +λ1(t−τ)) n!

• As μ → 0, the probability that the modulating process makes no

transitions within t seconds converges to 1, and we expect for this case that

P [˜x(t) = n] = 1 2

0t)ne−λ0t

n! + 1t)ne−λ1t n!



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### Modulated Poisson Processes

• As μ → ∞, then the modulating process makes an inﬁnite number of transitions within t seconds, and we expect for this case that

P [˜x(t) = n] = (βt)ne−βt

n! , where β = λ0 + λ1 2

• Example (Modeling Voice).

– A basic feature of speech is that it comprises an alternation of silent periods and non-silent periods.

– The arrival rate of packets during a talk spurt period is Poisson

with rate λ1 and silent periods produce a Poisson rate with λ0 ≈ 0.

– The duration of times for talk and silent periods are exponentially distributed with parameters μ1 and μ0, respectively.

⇒ The model of the arrival stream of packets is given by a modulated Poisson process.

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### Poisson Arrivals See Time Averages (PASTA)

• PASTA says: as t → ∞

Fraction of arrivals who see the system in a given state

upon arrival (arrival average)

= Fraction of time the system is in a given state (time average)

= The system is in the given state at any random time

• Counter-example (textbook [Kao]: Example 2.7.1)

0 1 2 3 4 5

1 1

1/2 1/2

service time = 1/2 inter-arrival time = 1

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### Poisson Arrivals See Time Averages (PASTA)

– Arrival average that an arrival will see an idle system = 1 – Time average of system being idle = 1/2

• Mathematically,

– Let X = {˜x(t), t ≥ 0} be a stochastic process with state space S, and B ⊂ S

– Deﬁne an indicator random variable

˜

u(t) =

1, if ˜x(t) ∈ B 0, otherwise

– Let N = {˜n(t), t ≥ 0} be a Poisson process with rate λ denoting the arrival process

then,

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### Poisson Arrivals See Time Averages (PASTA)

t→∞lim

 t

0 u(s)d˜˜ n(s)

˜

n(t) = lim

t→∞

 t

0 u(s)ds˜ t

(arrival average) (time average)

• Condition – For PASTA to hold, we need the lack of anticipation assumption (LAA): for each t ≥ 0,

– the arrival process {˜n(t + u) − ˜n(t), u ≥ 0} is independent of {˜x(s), 0 ≤ s ≤ t} and {˜n(s), 0 ≤ s ≤ t}.

• Application:

– To ﬁnd the waiting time distribution of any arriving customer – Given: P[system is idle] = 1 − ρ; P[system is busy] = ρ

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### Poisson Arrivals See Time Averages (PASTA)

Poisson

Poisson

Case 1: system is idle

Case 2: system is busy

⇒ P ( ˜w ≤ t) = P ( ˜w ≤ t|idle) · P (idle upon arrival) + P ( ˜w ≤ t|busy) · P (busy upon arrival)

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### Memoryless Property of the Exponential Distribution

• A random variable ˜x is said to be without memory, or memoryless, if P [˜x > s + t|˜x > t] = P [˜x > s] for all s, t ≥ 0 (3)

• The condition in Equation (3) is equivalent to P [˜x > s + t, ˜x > t]

P [˜x > t] = P [˜x > s]

or

P [˜x > s + t] = P [˜x > s]P [˜x > t] (4)

• Since Equation (4) is satisﬁed when ˜x is exponentially distributed (for e−λ(s+t) = e−λse−λt), it follows that exponential random variable are memoryless.

• Not only is the exponential distribution “memoryless,” but it is the unique continuous distribution possessing this property.

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### Comparison of Two Exponential Random Variables

Suppose that ˜x1 and ˜x2 are independent exponential random variables with respective means 1/λ1 and 1/λ2. What is P [˜x1 < ˜x2]?

P [˜x1 < ˜x2] =



0 P [˜x1 < ˜x2|˜x1 = x]λ1e−λ1xdx

=



0 P [x < ˜x21e−λ1xdx

=



0 e−λ2xλ1e−λ1xdx

=



0 λ1e−(λ12)xdx

= λ1

λ1 + λ2

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### Minimum of Exponential Random Variables

Suppose that ˜x1, ˜x2,· · · , ˜xn are independent exponential random variables, with ˜xi having rate μi, i = 1, · · · , n. It turns out that the smallest of the ˜xi is exponential with a rate equal to the sum of the μi.

P [min(˜x1, ˜x2, · · · , ˜xn) > x] = P [˜xi > x for each i = 1, · · · , n]

=

n i=1

P [˜xi > x] (by independence)

=

n i=1

e−μix

= exp



 n



i=1

μi



x



How about max(˜x1, ˜x2, · · · , ˜xn)? (exercise)

In another word, the initial state description is the conjunct of the precondition of the task and the guard condition of the task’s method, and the state descriptions are

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