1 introduction to stochastic models gslm 54100. 2 outline counting process poisson process ...
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Introduction to Stochastic ModelsIntroduction to Stochastic ModelsGSLM 54100GSLM 54100
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OutlineOutline 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 ProcessCounting 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 ProcessPoisson Process
a Poisson process {N(t)} is a counting process of rate (per unit time) if the inter-arrival times are i.i.d. exponential of mean 1/
t
N(t)
0
1
S1
1
2
S2
2
S3
3
3
I ~ i.i.d. exp()
for each sample point, a Poisson
process is a graph
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Distribution of Arrival EpochsDistribution of Arrival Epochs
Sn = X1 + … + Xn, Xi ~ i.i.d. exp()
1( )( ) , 0
( 1)!n
x n
Se x
f x xn
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Distribution of Distribution of NN((tt))
P(N(t) = 0) =
P(N(t) = 1)
120 ( ) ( )tXP X t y f y dy
( )0t t y ye e dy
tte
t
N(t)
0S1
1
S2
2
3
S3
1( )P X t , 0te t
1 1 2( , )P X t X X t
1 1 2[ ( , )]E P X t X X t
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Distribution of Distribution of NN((tt))
1( )( ) , 0
( 1)!n
x n
Se x
f x xn
t
N(t)
0S1
1
S2
2
3
S3
) ( ) (P N t n
( )
!
nt t
en
1( , )n n nP S t S X t
1[ ( , | )]n n n nE P S t S X t S
10 ( ) ( )n
tn SP X t y f y dy
1( )
0( )
( 1)!
y nt t y e ye dy
n
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IncrementsIncrements
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 ProcessProperties of the Poisson Process
independent increments dependent increments
stationary 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: Property of the Poisson Process: Independent IncrementsIndependent Increments
number of increments in disjoint intervals are independent random variables
for t1 < t2 t3 < t4, N(t1, t2] = N(t2) - N(t1) and N(t3, t4] = N(t4) - N(t3) are independent random variables
dependent increments (Example 7.2.2)
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Property of the Poisson Process: Property of the Poisson Process: Stationary IncrementsStationary Increments
number of increments in an interval of length h ~ Poisson(h) a Poisson variable of mean h
N(s, t] ~ Poisson((ts)) for any s < t
non-stationary increments (Example 7.2.4)
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Property of the Poisson ProcessProperty 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 7.2.8Example 7.2.7 & Example 7.2.8
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Properties of the Poisson ProcessProperties 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 Summation of Independent Poisson Random VariablesPoisson Random Variables
X ~ Poisson(), Y ~ Poisson() , independent
Z = X + Y
distribution of Z?
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Summation of Independent Poisson Random Variables
Z ~ Poisson(+)
( )P Z n
( ) ( )
!
ne
n
( )P X Y n 0
( , )n
kP X n k Y k
0( ) ( )
n
kP X n k P Y k
0 ( )! !
n k kn
k
e e
n k k
( )
0 ( )! !
n k kn
ke
n k k
( )
0!
n n n k kk
k
eC
n
( )
0
!
!( )! !
n k kn
k
ne
k n k n
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Composition of Independent Composition of Independent Poisson Processes Poisson Processes
{X(t)} ~ Poisson process of rate
{Y(t)} ~ Poisson process of rate
Z(t) = X(t) +Y(t) distribution of
Z(t)? type of {Z(t)}?
t
X(t)
t
Y(t)
t
Z(t)
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Composition of Independent Composition of Independent Poisson Processes Poisson Processes
t
X(t)
t
Y(t)
X1
Y1Y2
t
Z(t)
Z1 = min(X1, Y1)Z2 = min(X’1, Y2)
X’1=(X1Y1|X1>Y1)
Xi ~ i.i.d. exp()
Yi ~ i.i.d. exp()
X’1 ~ i.i.d. exp()
Z1 ~ exp(+)
Z2 ~ exp(+), independent of Z1
by the same argument, Zi ~ exp(+), i.e., {Z(t)} is a
Poisson process of rate +
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Random Partitioning of Random Partitioning of Poisson Random VariablesPoisson 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 Random Partitioning of Poisson Random VariablesPoisson Random Variables
Y ~ Poisson(p)
[ ( | )]E P Y k X ( )P Y k ( | ) ( )n k
P Y k X n P X n
(1 )!
nn k n kk
n k
eC p p
n
(1 )( )
( )! !
n k n kk
n k
pe p
n k k
( ) (1 )
! ( )!
k n k n k
n k
e p p
k n k
0
( ) (1 )
! !
k m m
m
e p p
k m
(1 )( )
!
kpe p
ek
( )
!
p ke p
k
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Independent Partitioned Independent Partitioned Random VariablesRandom Variables
Y ~ Poisson(p), and XY ~ Poisson((1p))
surprising fact: Y and XY being independent
( , )P Y k X Y n k
( | ) ( )P Y k X n P X n
( , )P X n Y k
(1 )!
nn k n kk
eC p p
n
! (1 )
!( )! !
k n k nn p p e
k n k n
(1 )( ) [(1 ) ]
! ( )!
p k p n ke p e p
k n k
( ) ( )P Y k P X Y n k
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Random Partitioning Random Partitioning of Poisson Processesof Poisson Processes
{X(t)} ~ Poisson process of rate
each item is type 1 with probability p
distribution of Yi?
type of process of {Y(t)}
t
X(t)
X1
type 2
X2
type 2
X3
type 1
t
Y(t)
Y1
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Random Partitioning Random Partitioning of Poisson Processesof Poisson Processes
have argued that Y ~ exp(p) before, or
1, ~ i.i.d. exp( ), ~ ( )
M
i ii
Y X X M Geo p
( )Yf s1
( )M
ii
Xf s
1
1( ) ( )m
ii
XmP M m f s
11
1
( )(1 )
( 1)!
mm s
m
sp p e
m
1
1
[(1 ) ]
( 1)!
ms
m
p sp e
m
0
[(1 ) ]
!
ks
k
p sp e
k
(1 )s p sp e e
p sp e
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 Random Partitioning of Poisson Processesof Poisson Processes
{Y(t)} is a Poisson process of rate p {X(t)Y(t)} is a Poisson process of rate (1p) 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 DistributionsUniform Distributions
U, Ui ~ i.i.d. uniform[0, t] for 0 < s < t, P(U > s) = (ts)/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 1 22
2( )( )s s t s
t
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Conditional Distribution of Conditional Distribution of SSii
( S1|N(t) = 1) ~ uniform(0, t)
1 ( | ( ) 1)P S s N t
t s
t
1( ( , ] | ( ) 1)P X s t N t
( (0, ] 0, ( , ) 1)
( ( ) 1)
P N s N s t
P N t
( )( )
s t s
t
e t s e
te
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Conditional Distribution of Conditional Distribution of SSii
it can be shown that
( S1, S2 |N(t) = 2)
~ ( U[1], U[2] |N(t) = 2)
1 1 2 2 2 ( ( , ], ( , ] | ( ) 2)P S s s S s t N t
2 1 22
2( )( )s s t s
t
1 1 2 2( (0, ] 0, ( , ] 1, ( , ] 1)
( ( ) 2)
P N s N s s N s t
P N t
1 2 1 2
2 2
( ) ( )2 1 2
2!
( ) ( ) t
s s s t s
t e
e s s e t s e
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Conditional Distribution of Conditional Distribution of SSii
Given N(t) = n, S1, …, Sn distribute as the ordered statistics of i.i.d. U1, …, Un
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Example 7.2.15Example 7.2.15
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Equivalent Definition Equivalent Definition of the Poisson Process 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, ( )
( ( ) ( ) ) , 0,1,...!
nt t
P N t s N s n e nn
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Equivalent Definition Equivalent Definition of the Poisson Process 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