Introduction to Random ProcessesLecture 12Spring 2002Random Process• A random variable is a function X(e) that maps the set of ex-periment outcomes to the set of numbers.• A random process is a rule that maps every outcome e of anexperiment to a function X(t, e).• A random process is usually conceived of as a function of time,but there is no reason to not consider random processes that arefunctions of other independent variables, such as spatial coordi-nates.• The function X(u, v, e) would be a function whose value de-pended on the location (u, v) and the outcome e, and could beused in representing random variations in an image.Lecture 12 1Random Process• The domain of e is the set of outcomes of the experiment. Weassume that a probability distribution is known for this set.• The domain of t is a set, T , of real numbers.• If T is the real axis then X(t, e)isacontinuous-time randomprocess• If T is the set of integers then X(t, e)isadiscrete-time randomprocess• We will often suppress the display of the variable e and write X(t)for a continuous-time RP and X[n]orXnfor a discrete-time RP.Lecture 12 2Random Process• A RP is a family of functions, X(t, e). Imagine a giant strip chartrecording in which each pen is identified with a different e. Thisfamily of functions is traditionally called an ensemble.• A single function X(t, ek) is selected by the outcome ek. This isjust a time function that we could call Xk(t). Different outcomesgive us different time functions.• If t is fixed, say t = t1,then X(t1,e) is a random variable. Itsvalue depends on the outcome e.• If both t and e are given then X(t, e) is just a number.Lecture 12 3Random ProcessesLecture 12 4Moments and AveragesX(t1,e) is a random variable that represents the set of samples acrossthe ensemble at time t1If it has a probability density function fX(x; t1) then the momentsaremn(t1)=E[Xn(t1)] =∞−∞xnfX(x; t1)dxThe notation fX(x; t1) may be necessary because the probabilitydensity may depend upon the time the samples are taken.The mean value is µX= m1, which may be a function of time.The central moments areE[(X(t1) − µX(t1))n]=∞−∞(x − µX(t1))nfX(x; t1)dxLecture 12 5Pairs of SamplesThe numbers X(t1,e)andX(t2,e) are samples from the same timefunction at different times.They are a pair of random variables (X1,X2).They have a joint probability density function f(x1,x2; t1,t2).From the joint density function one can compute the marginal den-sities, conditional probabilities and other quantities that may be ofinterest.Lecture 12 6Covariance and CorrelationThe covariance of the samples isC(t1,t2)=E[(X1− µ1)(X2− µ2)∗]=∞−∞(x1− µ1)(x2− µ2)∗f(x1,x2; t1,t2)dx1dx2The correlation function isR(t1,t2)=E[X1X∗2]=∞−∞x1x∗2f(x1,x2; t1,t2)dx1dx2C(t1,t2)=R(t1,t2) − µ1µ2Note that both the covariance and correlation functions are conju-gate symmetric in t1and t2.C(t1,t2)=C∗(t2,t1)andR(t1,t2)=R∗(t2,t1)Lecture 12 7Mean-Squared ValueThe “average power” in the process at time t is represented byR(t, t)=E[|X(t)|2]and C(t, t) represents the power in the fluctuation about the meanvalue.Lecture 12 8Example: Poisson Random ProcessLet N(t1,t2) be the number of events produced by a Poisson processin the interval(t1,t2)when the average rate is λ events per second.The probability that N = n isP [N = n]=(λτ)ne−λτn!where τ = t2− t1. Then E[N(t1,t2)] = λτ.A random process can be defined as the number of events in theinterval (0,t). Thus, X(t)=N(0,t). The expected number of eventsin t is E[X(t)] = λt.Lecture 12 9Example-continuedFor a Poisson distribution we know that the variance isE[(X(t) − λt)2]=E[X2(t)] −(λt)2= λtThe “average power” in the function X(t)isE[X2(t)] = λt + λ2t2A graph of X(t) would show a function fluctuating about an averagetrend line with a slope λ.Lecture 12 10Program for Poisson Random ProcessFUNCTION PoissonProcess,t,lambda,p; S=PoissonProcess(t,lambda,p); divides the interval [0,t] into intervals of size; deltaT=p/lambda where p is sufficiently small so that; the Poisson assumptions are satisfied.;; The interval (0,t) is divided into n=t*lambda/p intervals; and the number of events in the interval (0,k*deltaT) is; returned in the array S. The maximum length of S is 10000.;; USAGE; S=PoissonProcess(10,1,0.1); Plot,S; FOR m=1,10 DO OPLOT,PoissonProcess(10,1,0.1)NP=N_PARAMS()IF NP LT 3 THEN p=0.1n=lambda*t/pu=RANDOMN(SEED,n,POISSON=p)s=INTARR(n+1)FOR k=1,n DO s[k]=s[k-1]+u[k-1]RETURN,sENDLecture 12 12Random Telegraph SignalConsider a random process that has the following properties:1. X(t)=±1,2. The number of zero crossings in the interval(0,t)is describedby a Poisson process3. X(0) = 1. (to be removed later)Find the expected value at time t.Lecture 12 14Random Telegraph SignalLet N(t) equal the number of zero crossings in the interval (0,t).with t ≥ 0.P (N = n)=(λt)ne−λtn!P [X(t)=1] = P[N = even number]= e−λt1+(λt)22!+(λt)44!+ ··· = e−λtcosh λtP [X(t)=−1] = P[N = odd number]= e−λtλt +(λt)33!+(λt)55!+ ··· = e−λtsinh λtLecture 12 15Random Telegraph SignalThe expected value isE[X(t)|X(0) = 1] = e−λtcosh λt − e−λtsinh λt = e−2λtNote that the expected value decays toward x = 0 for large t. Thathappens because the influence of knowing the value at t = 0 decaysexponentially.Lecture 12 16Random Telegraph SignalThe autocorrelation function is computed by finding R(t1,t2)=E[X(t1)X(t2)]. Let x0= −1andx1= 1 denote the two valuesthat X can attain. For the moment assume that t2≥ t1. ThenR(t1,t2)=1j=01k=0xjxkP [X(t1)=xk]P [X(t2)=xj|X(t1)=xk]The first term in each product is given above. To find the conditionalprobabilities we take note of the fact that the number of sign changesin t2− t1is a Poisson process. Hence, in a manner that is similar tothe analysis above,P [X(t2)=1|X(t1)=1] = P [X(t2)=−1|X(t1)=−1]= e−λ(t2−t1)cosh λ(t2− t1)Lecture 12 17Random Telegraph SignalP [X(t2)=−1|X(t1)=1] = P [X(t2)=1|X(t1)=−1]= e−λ(t2−t1)sinh λ(t2− t1)HenceR(t1,t2)= e−λt1cosh λt1e−λ(t2−t1)cosh λ(t2− t1) − e−λ(t2−t1)sinh λ(t2− t1)−e−λt1sinh λt1e−λ(t2−t1)cosh λ(t2− t1) − e−λ(t2−t1)sinh λ(t2− t1)After some algebra this reduces toR(t1,t2)=e−λ(t2−t1)for t2≥ t1A parallel analysis applies to the case t2≤ t1, so thatR(t1,t2)=e−λ|t2−t1|Lecture 12 18Random Telegraph SignalThe