Transcript Introduction to Stochastic Models GSLM 54100 Outline   counting process Poisson process definition: interarrival ~ i.i.d.

Introduction to Stochastic Models
GSLM 54100
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Outline


counting process
Poisson process
definition: interarrival ~ i.i.d. exp
 properties

independent increments
 stationary increments
 P(N(t+h)  N(t) = 1)  h for small h
 P(N(t+h)  N(t)  2)  0 for small h
 composition of independent Poisson processes
 random partitioning of Poisson process
 conditional distribution of (Si|N(t) = n)

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Counting Process
 {N(t)}
is a counting process if N(t) = the total
number of occurrences of events on or
before t
 N(0)
=0
 N(t)
is a non-negative integer
 N(t)
is increasing (i.e., non-decreasing) in t
 for
s < t, N(s, t] = the number of events
occurring in the interval (s, t]
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Poisson Process
a
Poisson process {N(t)} is a counting
process of rate  (per unit time) if the interarrival times are i.i.d. exponential of mean
1/
I ~ i.i.d. exp()
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for each sample
point, a Poisson
process is a graph
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N(t)
1

S1
0
1

S2
2

S3
t
3
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Distribution of Arrival Epochs
 Sn
= X1 + … + Xn, Xi ~ i.i.d. exp()
ex (x)n1
f Sn ( x ) 
, x0
(n  1)!
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Distribution of N(t)
 P(N(t)
= 0) = P( X1  t )  et , t  0
 P(N(t)
= 1)  P( X1  t , X1  X 2  t )
t

 E[ P( X1  t , X1  X 2  t )] 0 P( X 2  t  y) f X1 ( y)dy

 tet
t  (t  y ) y
e dy
0 e
3
2
N(t)
1
0

S1

S2
t

S3
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Distribution of N(t)
P( N (t )  n)
 P( Sn  t , Sn  X n1  t )
 E[ P(Sn  t , Sn  X n1  t | Sn )]
 0t P( X n1  t  y) f Sn ( y)dy
y
(y )n1
t  (t  y ) e
 0 e
dy
(n  1)!
n
(

t
)
 e t
3
n!
ex (x)n1
f Sn ( x ) 
, x0
(n  1)!
2
N(t)
1
0

S1

S2
t

S3
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Increments
 N(s,
t] = N(t)  N(s)
= number of arrivals in (s, t]
= increments in (s, t]
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Properties of the Poisson Process
 independent
 dependent
 stationary
increments
increments
increments
 non-stationary
increments
 P(N(t+h)
 N(t) = 1)  h for small h
 P(N(t+h)
 N(t)  2)  0 for small h
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Property of the Poisson Process:
Independent Increments
 number
of increments in disjoint intervals
are independent random variables
< t2  t3 < t4, N(t1, t2] = N(t2) - N(t1) and
N(t3, t4] = N(t4) - N(t3) are independent random
variables
 for t1
 dependent
increments (Example 7.2.2)
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Property of the Poisson Process:
Stationary Increments
 number
of increments in an interval of
length h ~ Poisson(h)
a
Poisson variable of mean h
 N(s,
t] ~ Poisson((ts)) for any s < t
 non-stationary
increments (Example 7.2.4)
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Property of the Poisson Process
 P(N(t+h)
 N(t) = 1)  h for small h
 P(N(t+h)
 N(t)  2)  0 for small h
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Example 7.2.7 & Example 7.2.8
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Properties of the Poisson Process

composition of independent Poisson processes
 summation
of independent Poisson random
variables

random partitioning of Poisson process
 random
partitioning of Poisson random variables
 Example

7.2.12
conditional distribution of (Si|N(t) = n)
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Summation of Independent
Poisson Random Variables
X
~ Poisson(), Y ~ Poisson() , independent
Z
=X+Y
 distribution
of Z?
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Summation of Independent
Poisson Random Variables
n
  P( X  n  k , Y  k )
P ( Z  n)  P ( X  Y  n)
k 0
  P( X  n  k ) P (Y  k )
e  nk e k
 
k!
k 0 (n  k )!
nk
k


 e() 
k 0 (n  k )! k !
e() n n nk k

 Ck  
n ! k 0
n!
 nk  k
e

n!
k 0 k !(n  k )!
e( ) (  ) n

n!
n
n
k 0
n
 ( )

n
Z ~ Poisson(+)
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Composition of Independent
Poisson Processes
 {X(t)}
~ Poisson
process of rate 
X(t)
t
Y(t)
 {Y(t)}
~ Poisson
process of rate 
 Z(t)
t
= X(t) +Y(t)
 distribution
of
Z(t)?
Z(t)
 type
t
of {Z(t)}?
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Composition of Independent
Poisson Processes
Xi ~ i.i.d. exp()
X(t)
Yi ~ i.i.d. exp()
t
X1
Y(t)
X’1 ~ i.i.d. exp()
X’1=(X1Y1|X1>Y1)
t
Y1
Y2
Z1 ~ exp(+)
Z(t)
Z2 ~ exp(+),
independent of Z1
t
Z2 = min(X’1, Y2)
Z1 = min(X1, Y1)
by the same argument, Zi ~
exp(+), i.e., {Z(t)} is a
Poisson process of rate +
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Random Partitioning of
Poisson Random Variables
X
items, X ~ Poisson()
 each
item, if available, is type 1 with
probability p, 0 < p < 1
Y
= # of type 1 items in X
 distribution
of Y?
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Random Partitioning of
Poisson Random Variables
P(Y  k )  E[ P(Y  k | X )]

  Ckn p k (1  p)nk
nk
e  n
n!

  P(Y  k | X  n) P ( X  n)
nk
nk n k
(1

p
)

 e ( p)k 
(n  k )!k !
nk

e ( p)k  (1  p)nk  nk


k!
(n  k )!
nk
e ( p)k  (1  p)m  m


k!
m!
m 0
e ( p)k (1 p )

e
k!
e p ( p) k

k!
Y
~ Poisson(p)
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Independent Partitioned
Random Variables
Y
~ Poisson(p), and XY ~ Poisson((1p))
 surprising
fact: Y and XY being
independent
P(Y  k , X  Y  n  k )  P( X  n, Y  k )
e  n
p)
 P(Y  k | X  n) P( X  n)
n!
e p ( p)k e(1 p ) [(1  p)]nk
n ! p k (1  p)nk e  n


k!
(n  k )!
k !(n  k )!
n!
 Ckn p k (1 
n k
 P(Y  k ) P( X  Y  n  k )
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Random Partitioning
of Poisson Processes
 {X(t)}
~ Poisson
process of rate 
type 1
type 2
X(t)
type 2
 each
t
X1
X2
item is type 1
with probability p
X3
Y(t)
 distribution
t
Y1
of Yi?
 type
of process of
{Y(t)}
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Random Partitioning
of Poisson Processes
M
Y   X i , X i ~ i.i.d. exp(), M ~ Geo( p)
i 1
 have
argued that Y ~ exp(p) before, or
fY ( s)
 fM
 Xi

(s)
i 1

  (1  p)m1 pes
m1
  P( M  m) f m
(s)m1
(m  1)!
k
[(1

p
)

s
]
 pes 
k!
k 0

 Xi
m1
 pe
(s)
i 1
m1
[(1

p
)

s
]
 pes 
(m  1)!
m1

s (1 p )s
e
 pe ps
can argue that Yi ~ i.i.d. exp(p), i.e.,
{Y(t)} is a Poisson process of rate p
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Random Partitioning
of Poisson Processes
 {Y(t)}
is a Poisson process of rate p
 {X(t)Y(t)}
is a Poisson process of rate
(1p)
 no
dependence of interarrival times among
{Y(t)} and {X(t)Y(t)}
 {Y(t)}
and {X(t)Y(t)} are independent
Poisson processes
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Uniform Distributions
U, Ui ~ i.i.d. uniform[0, t]
 for 0 < s < t, P(U > s) = (ts)/t
 for 0 < s1 < s2 < t, one of U1 and U2 in (s1, s2] and
the other in (s2, t]

 P(one
of U1, U2 in (s1, s2] & the other in (s2, t] )
= P(U1  (s1, s2], U2  (s2, t])
+ P(U2  (s1, s2], U1  (s2, t])
2( s2  s1 )(t  s2 )
=
t2
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Conditional Distribution of Si
P(S 1  s | N (t )  1)
 P( X 1 ( s, t ] | N (t )  1)
(
S1|N(t) = 1) ~
uniform(0, t)
P( N (0, s]  0, N ( s, t )  1)

P( N (t )  1)
s
 (t  s )
e (t  s)e

tet
ts

t
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Conditional Distribution of Si
P(S1  (s1, s2 ], S2  (s2 , t ] | N (t )  2)
P( N (0, s1 ]  0, N ( s1, s2 ]  1, N ( s2 , t ]  1)

P( N (t )  2)

es1( s2  s1 )e( s2 s1 )(t  s2 )e(t s2 )

2( s2  s1 )(t  s2 )
t2
 2t 2et
2!
it can be shown that
( S1, S2 |N(t) = 2)
~ ( U[1], U[2] |N(t) = 2)
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Conditional Distribution of Si
N(t) = n, S1, …, Sn distribute as the
ordered statistics of i.i.d. U1, …, Un
 Given
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Example 7.2.15
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Equivalent Definition
of the Poisson Process
a
counting process {N( t)} is a Poisson
process of rate  (> 0) if
 (i)
N(0) = 0
 (ii)
{N( t)} has independent increments
 (iii)
for any s, t  0,
P( N (t  s )  N ( s)  n)  e t
( t ) n
, n  0,1,...
n!
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Equivalent Definition
of the Poisson Process
a
counting process {N( t)} is a Poisson
process of rate  (> 0) if
 (i)
N(0) = 0
 (ii)
{N( t)} has stationary and independent
increments
 (iii)
P(N(h) = 1)  h for small h
 (iv)
P(N(h)  2)  0 for small h
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