Presentation is loading. Please wait.

Presentation is loading. Please wait.

Introduction to Stochastic Models GSLM 54100

Similar presentations


Presentation on theme: "Introduction to Stochastic Models GSLM 54100"— Presentation transcript:

1 Introduction to Stochastic Models GSLM 54100

2 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) 2

3 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] 3

4 for each sample point, a Poisson process is a graph
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/ I ~ i.i.d. exp() t N(t) 3 2 for each sample point, a Poisson process is a graph 1 S1 S2 S3 3 1 2 4

5 Distribution of Arrival Epochs
Sn = X1 + … + Xn, Xi ~ i.i.d. exp() 5

6 Distribution of N(t) P(N(t) = 0) = P(N(t) = 1) t N(t) S1  S2 S3 3 2 1
S1 1 S2 2 3 S3 6

7 Distribution of N(t) N(t) ~ Poisson (t) E[N(t)] = t 7 t N(t) S1  1
S1 1 S2 2 3 S3 7

8 Increments N(s, t] = N(t)  N(s) = number of arrivals in (s, t]
= increments in (s, t] 8

9 Properties 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 9

10 Property of the Poisson Process: Independent 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) 10

11 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) 11

12 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 12

13 Example & Example 7.2.8 13

14 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 conditional distribution of (Si|N(t) = n) 14

15 Summation of Independent Poisson Random Variables
X ~ Poisson(), Y ~ Poisson() , independent Z = X + Y distribution of Z? 15

16 Summation of Independent Poisson Random Variables
Z ~ Poisson(+) 16

17 Composition of Independent 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) 17

18 Composition of Independent Poisson Processes
X(t) Xi ~ i.i.d. exp() Yi ~ i.i.d. exp() X1 X’1 ~ i.i.d. exp() t Y(t) X’1=(X1Y1|X1>Y1) Y1 Y2 t Z(t) Z1 ~ exp(+) Z2 ~ exp(+), independent of Z1 by the same argument, Zi ~ exp(+), i.e., {Z(t)} is a Poisson process of rate + Z2 = min(X’1, Y2) Z1 = min(X1, Y1) 18

19 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? 19

20 Random Partitioning of Poisson Random Variables
Y ~ Poisson(p) 20

21 Independent Partitioned Random Variables
Y ~ Poisson(p), and XY ~ Poisson((1p)) surprising fact: Y and XY being independent 21

22 Random Partitioning of 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)} X3 type 1 t X(t) X2 type 2 type 2 X1 t Y(t) Y1 22

23 Random Partitioning of Poisson Processes
have argued that Y ~ exp(p) before, or can argue that Yi ~ i.i.d. exp(p), i.e., {Y(t)} is a Poisson process of rate p 23

24 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 24

25 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]) = 25

26 Conditional Distribution of Si
( S1|N(t) = 1) ~ uniform(0, t) 26

27 Conditional Distribution of Si
it can be shown that ( S1, S2 |N(t) = 2) ~ ( U[1], U[2] |N(t) = 2) 27

28 Conditional Distribution of Si
Given N(t) = n, S1, …, Sn distribute as the ordered statistics of i.i.d. U1, …, Un 28

29 Example 29

30 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, 30

31 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 31


Download ppt "Introduction to Stochastic Models GSLM 54100"

Similar presentations


Ads by Google