1. Intro. to Stochastic Processes
Cheng-Fu Chou

2. Outline Stochastic Process Counting Process Poisson Process
Markov Process
Renewal Process
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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.
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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]
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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
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6. Poisson Process
Def. A: the counting process {N(t), t0} 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.
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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
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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
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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.
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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}
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11. P. 11

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?
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13. P. 13

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
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15. P. 15

16. Poisson, Markov, Renewal Processes
Continuous Time Markov Chain: Exponential times between transitions
Relax counting
process
Poisson Process:
Counting process
iid exponential times between arrivals
Relax exponential interarrival times
Renewal Process:
Counting process
iid times between arrivals
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17. Markov Process

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
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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.
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20. Definition P = [ Pij] is the transition matrix
A matrix that satisfies those conditions is called a stochastic matrix
n-step transition prob.
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21. Chapman-Kolmogorov Eq.
Def.
Proof:
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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.
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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)
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24. Result 1 (cont.) Irreducible:
A M.C. is irreducible if i  j for all i,j I
Aperiodic:
For every state iI, define d(i) to be largest common divisor of all integer n, s.t.,
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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
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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?
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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}
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28. P3 - P2 =q/p*(P2-P1)= (q/p)2*P1
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 =1
Pi+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
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29. P. 29

30. If p > ½, there is a positive prob
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.
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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
P[X(t+u)= j | X(s+u)=i] = P[X(t)= j | X(s)=i] = Pij[t-s], where t > s
Chapman-Kolmogorov equ.
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32. CTMC (cont.) p(t)=p(0)eQt Q is called the infinitesimal generator
Proof:
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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 iI, independently of the initial dist.
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34. Renewal Process

35. time interval, not a rate;
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 is
with S0 = 0.
Can write
and if m = E[Xn], n  1, then the strong law of large numbers says that
Note: m is now a
time interval, not a rate;
1/ m will be called
the rate of the r. p.

36. 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:

37. Exercise 1
Is it true that:

38. Exercise 2
If the mean-value function of the renewal process {N(t), t  0} is given by
Then what is P{N(5) = 0} ?

39. Exercise 3
Consider a renewal process {N(t), t  0} having a gamma (r,l) interarrival distribution with density
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.

40. 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 variance
where s2 is the variance of the time between arrivals;
in particular,

41. 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
the long-run rate at which machines are replaced is
the long-run rate at which machines in use fail equals
Hint: condition on the lifetime of the first machine

42. 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

43. 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:
SN(t) t
A(t)
Y(t)
What is the average value of the age

44. 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 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 age is

45. Average Excess or Residual
Now 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)

46. 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 record
Find its distribution by conditioning on the time of the last renewal prior to time t

47. Inspection Paradox (cont.)
SN(t)
SN(t)+1
t-s t

48. Inspection Paradox (cont.)
SN(t)
SN(t)+1
t-s t
For any s, so
where X is an ordinary inter-renewal time.
Intuitively, by choosing “random” times, it is more likely
we will choose a time that falls in a long time interval.