Cheng-Fu Chou, CMLab, CSIE, NTU Intro. to Stochastic Processes Cheng-Fu Chou.

Cheng-Fu Chou, CMLab, CSIE, NTUCheng-Fu Chou, CMLab, CSIE, NTU

Intro. to Stochastic Processes

Cheng-Fu Chou

Cheng-Fu Chou, CMLAB, CSIE, NTUCheng-Fu Chou, CMLAB, CSIE, NTU

P. 2

Outline

Stochastic Process Counting Process Poisson Process Markov Process Renewal Process


P. 3

Stochastic Process

A stochastic process N= {N(t), t T} is a collection of r.v., i.e., for each t in the index set T, N(t) is a random variable– t: time– N(t): state at time t– If T is a countable set, N is a discrete-time stochastic

process– If T is continuous, N is a continuous-time stoc. proc.


P. 4

Counting Process

A stochastic process {N(t) ,t 0} is said to be a counting process if N(t) is the total number of events that occurred up to time t. Hence, some properties of a counting process is– N(t) 0– N(t) is integer valued– If s < t, N(t) N(s)– For s < t, N(t) – N(s) equals number of events occurring in

the interval (s, t]


P. 5

Counting Process

Independent increments– If the number of events that occur in disjoint time intervals

are independent

Stationary increments– If the dist. of number of events that occur in any interval of

time depends only on the length of time interval


P. 6

Poisson Process

Def. A: the counting process {N(t), t0} is said to be Poisson process having rate l, l>0 if– N(0) = 0;– The process has independent-increments– Number of events in any interval of length t is Poisson dist.

with mean lt, that is for all s, t 0.

( )[ ( ) ( ) ]

! = 0,1,2,...

nt t

P N t s N s n en

n


P. 7

Poisson Process

Def. B: The counting process {N(t), t 0} is said to be a Poisson process with rate l, l>0, if – N(0) = 0– The process has stationary and independent increments– P[N(h) = 1] = lh +o(h)– P[N(h) 2] = o(h)– The func. f is said to be o(h) if – Def A Def B, i.e,. they are equivalent.– We show Def B Def A – Def A Def B is HW

0

( )lim 0h

f h

h


P. 8

Important Properties

Property 1: mean number of event for any t 0, E[N(t)]=lt.

Property 2: the inter-arrival time dist. of a Poisson process with rate l is an exponential dist. with parameter .l

Property 3: the superposition of two independent Poisson process with rate l1 and l2 is a Poisson process with rate l1+l2


P. 9

Properties (cont.)

Property 4: if we perform Bernoulli trials to make independent random erasures from a Poisson process, the remaining arrivals also form a Poisson process

Property 5: the time until rth arrival , i.e., tr is known as the rth order waiting time, is the sum of r independent experimental values of t and is described by Erlan pdf.


P. 10

Ex 1

Suppose that X1 and X2 are independent exponential random variables with respective means 1/l1 and 1/l2;What is P{X1 < X2}


P. 11

21

1

0

)(1

10

10 2

0 112121

}{

}|{}{

21

12

1

1

dxe

dxee

dxeXxP

dxexXXXPXXP

x

xx

x

x


P. 12

Conditional Dist. Of the Arrival Time

Suppose we are told that exactly one event of a Poisson process has taken place by time t, what is the distribution of the time at which the event occurred?


P. 13

t][0,over ddistributeuniformly isevent theof time theSo,

1} P{N(t)

t)}[s,in events 0 s)}P{[0,in event 1{

1} P{N(t)

t)}[s,in events 0 s),[0,in event 1{

}1)({

}1)(,{}1)(|{

)(

t

ste

ese

P

P

tNP

tNsxPtNsxP

t

sts


P. 14

Ex 2

Consider the failure of a link in a communication network. Failures occur according to a Poisson process with rate 4.8 per day. Find– P[time between failures 10 days]– P[5 failures in 20 days]– Expected time between 2 consecutive failures– P[0 failures in next day]– Suppose 12 hours have elapsed since last failure, find the

expected time to next failure


P. 15

hours 5 .5

4.

hours 5 .35!

20)*(4.8 .2

1 .1

8.4

20*8.45

10*8.4

e

e

e


Poisson, Markov, Renewal Processes

P. 16

Poisson Process:

Counting process

iid exponential times between

arrivals

Continuous Time Markov Chain:

Exponential times between transitions

Renewal Process:

Counting process

iid times between arrivals

Relax countingprocess

Relax exponentia

l interarrival

times


Markov Process


P. 18

Markov Process

P[X(tn+1) Xn+1| X(tn)= xn, X(tn-1) = xn-1,…X(t1)=x1] = P[X(tn+1) Xn+1| X(tn)=xn]– Probabilistic future of the process depends only on the

current state, not on the history– We are mostly concerned with discrete-space Markov

process, commonly referred to as Markov chains– Discrete-time Markov chains– Continuous-time Markov chains


P. 19

DTMC

Discrete Time Markov Chain: – P[Xn+1 = j | Xn= kn, Xn-1 = kn-1,…X0= k0]

= P[Xn+1 = j | Xn = kn] discrete time, discrete space A finite-state DTMC if its state space is finite A homogeneous DTMC if P[Xn+1 = j | Xn= i ] does not depend

on n for all i, j, i.e., Pij = P[Xn+1 = j | Xn= i ], where Pij is one step transition prob.


P. 20

Definition P = [ Pij] is the transition matrix

– A matrix that satisfies those conditions is called a stochastic matrix– n-step transition prob.

00 01 0

10 11 1

0

... ...

... ...

... ... ... ... ...

... ... ...

... ... ... ... ...

where 0 and 1

j

j

i ij

ij ijj

p p p

p p p

P

p p

p p

0[ | ]

, , 0, is the prob. of going from state

to in step

nij n

nij

p P x j x i

i j n p i

j n


P. 21

Chapman-Kolmogorov Eq.

Def.

Proof:

( )

For all n 0, m 0, i , j I

in matrix form where =[ ]

n m n mij ik kj

k I

n m n m n nij

p p p

P P P P p


P. 22

Question

We have only been dealing with conditional prob. but what we want is to compute the unconditional prob. that the system is in state j at time n, i.e.

0

0 0 0

0 0

0

( ) ( )

So, given the initial dist. of ,i.e.,

( ) ( ) and 1

we can get

[ ] ( | ) ( )

( )

n n

i I

n ni I

nij

i I

j p x j

x

i p x i

p x j p x j x i i

p i


P. 23

Result 1

For all n 1, pn = p0Pn, where pm = (pm(0),pm(1),…) for all m 0. From the above equ., we deduce that pn+1 = pnP. Assume that limn pn(i) exists for all i, and refer it as p(i). The remaining question is how to compute p

– Reachable: a state j is reachable from i. if

– Communicate: if j is reachable from i and if i is reachable form j, then we say that i and j communicate (i j)

0 for some 1nijp n


P. 24

Result 1 (cont.)

Irreducible:– A M.C. is irreducible if i j for all i,j I

Aperiodic:– For every state iI, define d(i) to be largest common divisor

of all integer n, s.t.,

0 if ( ) 1 then the state is aperiodicn

ijp d i


P. 25

Result 2

Invariant measure of a M.C., if a M.C. with transition matrix P is irreducible and aperiodic and if the system of equation p=pP and p1=1 has a strict positive solution then p(i) = limn pn(i) independently of initial dist.– Invariant equ. : p=pP – Invariant measure p


P. 26

Gambler’s Ruin Problem

Consider a gambler who at each play of game has probability p of winning one unit and probability q=1-p of losing one unit. Assuming that successive plays of the game are independent, what is the probability that, starting with i units, the gambler’s fortune will reach N before reaching 0?


P. 27

Ans

If we let Xn denote the player’s fortune at time n, then the process {Xn, n=0, 1,2,…} is a Markov chain with transition probabilities:– p00 =pNN =1– pi,i+1 = p = 1-pi,i-1

This Markov chain has 3 classes of states: {0},{1,2,…,N-1}, and {N}


P. 28

Let Pi, i=0,1,2,…,N, denote the prob. That, starting with i, the gambler’s fortune will eventually reach N.

By conditioning on the outcome of the initial play of the game we obtain– Pi = pPi+1 + qPi-1, i=1,2, …, N-1

Since p+q =1Pi+1 – Pi = q/p(Pi-Pi-1),

Also, P0 =0, so

P2 – P1 = q/p*(P1-P0) = q/p*P1

P3 - P2 =q/p*(P2-P1)= (q/p)2*P1


P. 29

1/2 p if 0

1/2p if p

q-1

N as that,Note

2/1p if 1

2/1p if )/(1

)/(1

obtain we,1 using Now,

1q

p if

1q

p if

)/(1

)/(1

i

1

1

1

i

N

N

i

i

P

N

pq

pq

P

P

iP

Ppq

pq

P


P. 30

If p > ½, there is a positive prob. that the gambler’s fortune will increase indefinitely

Otherwise, the gambler will, with prob. 1, go broke against an infinitely rich adversary.


P. 31

CTMC

Continuous-time Markov Chain– Continuous time, discrete state– P[X(t)= j | X(s)=i, X(sn-1)= in-1,…X(s0)= i0]

= P[X(t)= j | X(s)=i]– A continuous M.C. is homogeneous if

oP[X(t+u)= j | X(s+u)=i] = P[X(t)= j | X(s)=i] = Pij[t-s], where t > s

– Chapman-Kolmogorov equ.

For all t > 0, s > 0, i , j I

( ) ( ) ( ) ij ik kjk I

p t s p t p s


P. 32

CTMC (cont.)

p(t)=p(0)eQt

– Q is called the infinitesimal generator– Proof:


P. 33

Result 3

If a continuous M.C. with infinitesimal generator Q is irreducible and if the system of equations pQ = 0, and p1=1, has a strictly positive solution then p(i)= limt p(x(t)=i) for all iI, independently of the initial dist.


Renewal Process


Renewal Process

A counting process {N(t), t 0} is a renewal process if for each n, Xn is the time between the (n-1)st and nth arrivals and {Xn, n 1} are independent with the same distribution F.The time of the nth arrival iswith S0 = 0.

Can writeand if m = E[Xn], n 1, then the strong law of large numbers says that

1, 1,

n

n iiS X n

max : nN t n S t

as 1nSP n

n

Note: m is now atime interval, not a rate;

1/ m will be calledthe rate of the r. p.


Fundamental Relationship

It follows that

where Fn(t) is the n-fold convolution of F with itself.

The mean value of N(t) is

Condition on the time of the first renewal to get the renewal equation:

nN t n S t

1 1

1

n n n n

P N t n P N t n P N t n

P S t P S t F t F t

1 1 1

n nn n n

m t E N t P N t n P S t F t

0

tm t F t m t x f x dx


Exercise 1

Is it true that:

?nN t n S t

?nN t n S t

?nN t n S t


Exercise 2

If the mean-value function of the renewal process {N(t), t 0} is given byThen what is P{N(5) = 0} ?

2, 0m t t t


Exercise 3

Consider a renewal process {N(t), t 0} having a gamma (r,l) interarrival distribution with density

(a) Show that(b) Show that

Hint: use the relationship between the gamma (r,l) distribution and the sum of r independent exponentials with rate l to define N(t) in terms of a Poisson process with rate l.

1

, 01 !

rxe xf x x

r

!

it

i nr

e tP N t n

i

, where is the largest integer

!

it

i r

e tim t x x

r i


Limit Theorems

With probability 1,

Elementary renewal theorem:

Central limit theorem: For large t, N(t) is approximately normally distributed with mean t/ m and variancewhere s2 is the variance of the time between arrivals;in particular,

1 as t

N t

t

1 as t

m t

t

2 3t

2

3

Var as t

N t

t


Exercise 4

A machine in use is replaced by a new machine either when it fails or when it reaches the age of T years. If the lifetimes of successive machines are independent with a common distribution F with density f, show that(a) the long-run rate at which machines are replaced is

(b) the long-run rate at which machines in use fail equals

Hint: condition on the lifetime of the first machine

1

01

Txf x T F T

01

T

F T

xf x T F T


Renewal Reward Processes

Suppose that each time a renewal occurs we receive a reward. Assume Rn is the reward earned at the nth renewal and {Rn, n 1} are independent and identically distributed (Rn may depend on Xn). The total reward up to time t is

If then

and

and E R E X

as t 1E RR t

Pt E X

as tE R t E R

t E X

1

N t

nnR t R


Age & Excess Life of a Renewal Process

The age at time t is A(t) = the amount of time elapsed since the last renewal. The excess life Y(t) is the time until the next renewal:

tA(t) Y(t)

SN(t)

N tA t t S

What is the average value of the age

0lim

s

t

A t dt

s


Average Age of a Renewal Process

Imagine we receive payment at a rate equal to the current age of the renewal process. Our total reward up to time s isand the average reward up to time s is

If X is the length of a renewal cycle, then the total reward during the cycle is

So, the average age is

0

sA t dt

0reward during a renewal cycle

length of a renewal cycle

sA t dt E

s E

2

0 2

X Xtdt

2

2

E X

E X


Average Excess or ResidualNow imagine we receive payment at a rate equal to the current excess of the renewal process. Our total reward up to time s is

and the average reward up to time s is

If X is the length of a renewal cycle, then the total reward during the cycle is

So, the average excess is (also)

0

sY t dt

0reward during a renewal cycle

length of a renewal cycle

sY t dt E

s E

2

0 2

X XX t dt

2

2

E X

E X


Inspection Paradox

Suppose that the distribution of the time between renewals, F, is unknown. One way to estimate it is to choose some sampling times t1, t2, etc., and for each ti, record the total amount of time between the renewals just before and just after ti. This scheme will overestimate the inter-renewal times – Why?For each sampling time, t, we will recordFind its distribution by conditioning on the time of the last renewal prior to time t

1 1N t N t N tX S S


Inspection Paradox (cont.)

1 1N t N t N tP X x E P X x S t s

t

SN(t)SN(t)+1

t-s0

1

1 1 1

1

1

If then 1

If then

11

1

N t N t

N t N t N t N t

N t

N t

s x P X x S t s

s x P X x S t s P X x X s

P X x F xF x

F sP X s


Inspection Paradox (cont.)

t

SN(t)SN(t)+1

t-s0

For any s,so

1 1N t N tP X x S t s F x

1 1

1

N t N t N tP X x E P X x S t s

F x P X x

where X is an ordinary inter-renewal time.

Intuitively, by choosing “random” times, it is more likelywe will choose a time that falls in a long time interval.

Cheng-Fu Chou, CMLab, CSIE, NTU Intro. to Stochastic Processes Cheng-Fu Chou.

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