21 The Exponential Distribution

1 21 The Exponential DistributionFrom Discrete-Time to Continuous-Time:In Chapter 6 of the text we will be considering Markov processes in con-tinuous time. In a sense, we already have a very good understanding ofcontinuous-time Markov chains based on our theory for discrete-timeMarkov chains. For example, one way to describe a continuous-timeMarkov chain is to say that it is a discrete-time Markov chain, exceptthat we explicitly model the times between transitions with contin-uous, positive-valued random variables and we explicity consider theprocess at any timet, not just at transition single most important continuous Distribution for building andunderstanding continuous-time Markov chains is the Exponential dis-tribution.

2 For reasons which we shall explore in this THE Exponential DISTRIBUTIONThe Exponential Distribution :A continuous random variableXis said to have an Exponential ( ) Distribution if it has probability density functionfX(x| ) ={ e xforx >00forx 0,where >0is called therateof the the study of continuous-time stochastic processes, the exponentialdistribution is usually used to model thetime until something hap-pens in the process. The mean of the Exponential ( ) Distribution iscalculated using integration by parts asE[X] = 0x e xdx= [ xe x 0+1 0e xdx]= [0 +1 e x 0]= 1 2=1.}

3 So one can see that as gets larger, the thing in the process we rewaiting for to happen tends to happen more quickly, hence we thinkof as a an exercise, you may wish to verify that by applying integration byparts twice, the second moment of the Exponential ( ) Distribution isgiven byE[X2] = 0x2 e x=..=2 the first and second moments we can compute the variance asVar(X) =E[X2] E[X]2=2 2 1 2=1 Memoryless Property:The following plot illustrates a key property of the Exponential distri-bution. The graph after the pointsis an exact copy of the originalfunction.

4 The important consequence of this is that the distributionofXconditioned on{X > s}isagain Exponential Functionxexp( - x) : The Exponential Functione x18021. THE Exponential DISTRIBUTIONTo see how this works, imagine that at time 0 we start an alarm clockwhich will ring after a timeXthat is exponentially distributed withrate . Let us callXthelifetimeof the clock. For anyt >0, wehave thatP(X > t) = t e xdx= e x t=e we go away and come back at timesto discover that the alarmhas not yet gone off. That is, we have observed the event{X > s}.

5 If we letYdenote theremaininglifetime of the clock given that{X > s}, thenP(Y > t|X > s) =P(X > s+t|X > s)=P(X > s+t,X > s)P(X > s)=P(X > s+t)P(X > s)=e (s+t)e s=e this implies that the remaining lifetime after we observe the alarmhas not yet gone off at timeshas the same Distribution as the originallifetimeX. The really important thing to note, though, is that thisimplies that the Distribution of the remaining lifetimedoes not dependons. In fact, if you try settingXto haveany othercontinuousdistribution, then ask what would be the Distribution of the remaininglifetime after you observe{X > s}, the Distribution will depend property is called thememorylessproperty of the exponentialdistribution because I don t need to remember when I started theclock.

6 If the Distribution of the lifetimeXis Exponential ( ), then ifI come back to the clock at any time and observe that the clock hasnot yet gone off, regardless of when the clock started I can assert thatthe Distribution of the time till it goes off, starting at the time I startobserving it again, is Exponential ( ). Put another way, given that theclock has currently not yet gone off, I can forget the past and stillknow the Distribution of the time from my current time to the timethe alarm will go off. The resemblance of this property to the Markovproperty should not be lost on is a rather amazing, and perhaps unfortunate, fact that the exponen-tial Distribution is the only one for which this works.

7 The memorylessproperty is like enabling technology for the construction of continuous-time Markov chains. We will see this more clearly in Chapter 6. Butthe Exponential Distribution is even more special than just the memo-ryless property because it has a second enabling type of Important Property of the Exponential :LetX1,..,Xnbe independent random variables, withXihaving anExponential( i) Distribution . Then the Distribution ofmin(X1,..,Xn)is Exponential ( 1+..+ n), and the probability that the minimumisXiis i/( 1+..+ n).

8 Proof:P(min(X1,..,Xn)> t) =P(X1> t,..,Xn> t)=P(X1> t)..P(Xn> t)=e nt=e ( 1+..+ n) THE Exponential DISTRIBUTIONThe preceding shows that the CDF ofmin(X1,..,Xn)is that of anExponential( 1+..+ n) Distribution . The probability thatXiis theminimum can be obtained by conditioning:P(Xiis the minimum)=P(Xi< Xjforj6=i)= 0P(Xi< Xjforj6=i|Xi=t) ie itdt= 0P(t < Xjforj6=i) ie itdt= 0 ie it j6=iP(Xj> t)dt= 0 ie it j6=ie jtdt= i 0e ( 1+..+ n)tdt= i e ( 1+..+ n)t 1+..+ n 0= i 1+..+ n,as required. To see how this works together with the the memoryless property,consider the following :(Ross, #20).

9 Consider a two-server system inwhich a customer is served first by server 1, then by server 2, andthen departs. The service times at serveriare Exponential randomvariables with rates i,i= 1,2. When you arrive, you find server1 free and two customers at server 2 customerAin service andcustomerBwaiting in line.(a) FindPA, the probability thatAis still in service when you moveover to server 2.(b) FindPB, the probability thatBis still in the system when youmove over to 2.(c) FindE[T], whereTis the time that you spend in the :(a)Awill still be in service when you move to server 2 if your service atserver 1 ends beforeA s remaining service at server 2 ends.

10 NowAis currently in service at server 2 when you arrive, but becauseof memorylessness,A s remaining service is Exponential ( 2), andyou start service at server 1 that is Exponential ( 1). Therefore,PAis the probability that an Exponential ( 1) random variable isless than an Exponential ( 2) random variable, which isPA= 1 1+ 2.(b)Bwill still be in the system when you move over to server 2 ifyour service time is less than the sum ofA s remaining servicetime andB s service time. Let us condition on the first thing tohappen, eitherAfinishes service or you finish service:18421.