CS 547 Lecture 9: Conditional Probabilities and the ...

1 CS 547 Lecture 9: Conditional Probabilities and the memoryless PropertyDaniel MyersJoint ProbabilitiesFor two events,EandF, thejoint probability, writtenP(EF), is the the probability that both events example, letEbe the probability that a die roll is even andFbe the probability that a die roll isgreater than 3 . We have the following sets to describe each event:E={2,4,6}F={4,5,6}E F={4,6}The probability that the joint event occurs is the probability that the outcome is inE F, which ProbabilitiesFrequently, we are interested in analyzing the probability of an event with respect to some known events, theconditional probability ofEgivenFisP(E|F) =P(EF)P(F)Intuitively,P(EF) is the portion of the sample space that is in bothEandF.

2 If we know that a pointof the sample space is definitely inF, the probability that it is also inEis given by the portion ofFthatoverlaps withE, which is exactly what the formula rearranging the formula gives us a way to calculate joint Probabilities in terms of Conditional proba-bilitiesP(EF) =P(E|F)P(F)IndependenceTwo events areindependentifP(E|F) =P(E). That is, knowing thatFhas occurred tells us nothing aboutwhetherEalso occurs. Using the formula for joint probability, we getP(EF) =P(E|F)P(F) =P(E)P(F)The joint probability of independent events is just the product of their individual this knowledge, we re now able to interpret some earlier results.

3 Recall the pmf of the geometricrandom variable,f(k) = (1 p)k 1pWe defined this to be the probability that a series of independent Bernoulli trials yields its first success ontrialk. That is, we are interested in the probability of a sequence ofk 1 failures, each of which occurwith probability 1 p, and one success, which occurs with probabilityp. Because the Bernoulli trials areindependent, the probability of this sequence is simply the product of the Probabilities of its individualevents, which gives pmf of the Minimum of ExponentialsSuppose we have a set of independent and identically distributed (iid) variables,X1,X2.

4 ,Xk, which areexponentially distributed with the same parameter . What is the distribution of the random variableX= min(X1,X2,..,Xk)?Let s reason aboutP(X>t).Xis defined as the minimum of all theXi, so the only way thatXcan begreater than a certain valuetis if all of its components are also greater thant. Therefore,P(X>t) =P(X1>tandX2> >t)The individualXiare all independent, so we can re-write the joint probability as the product of theirindividual (X>t) =P(X1>t)P(X2>t)..P(Xk>t)To find the final distribution, we need to knowP(Xi>t) whenXi exp( ). This is given by thecomplementary CDF (CCDF) of the random variable, which is defined in terms of the CDF,F(t), asF(t) =P(Xi>t)= 1 P(Xi t)= 1 F(t)For the exponential distribution,F(t) = 1 F(t)= 1 (1 e t)=e tNow that we have the exponential CCDF, we can find the distribution (X>t) =P(X1>t)P(X2>t).

5 P(Xk>t)=e te t=e k tThe distribution of the minimum of a set ofkiid exponential random variables is also exponentially dis-tributed with parameterk . This result generalizes to the case where the variables are still independent,but have different memoryless PropertyThe memoryless proeprty tells us about the Conditional behavior of exponential random variables. It s oneof our key results, which we ll use in deriving the solution of queueing exponentially distributed with parameter . Suppose we knowX>t. What is the probabilitythatXis also greater than some values+t? That is, we want to knowP(X>s+t|X>t)This type of problem shows up frequently in queueing systems where we re interested in the time betweenevents.

6 For example, suppose that jobs in our system have exponentially distributed service times. If wehave a job that s been running for one hour, what s the probability that it will continue to run for more thantwo hours?2 Using the definition of Conditional probability, we haveP(X>s+t|X>t) =P(X>s+tandX>t)P(X>t)IfX>s+t, thenX>tis redundant, so we can simplify the (X>s+t|X>t) =P(X>s+t)P(X>t)Using the CCDF of the exponential distribution,P(X>s+t|X>t) =P(X>s+t)P(X>t)=e (s+t)e tThee tterms cancel, giving the surprising resultP(X>s+t|X>t) =e sIt turns out that the Conditional probability does not depend ont! The probability of an exponentialrandom variable exceeding the values+tgiventis the same as the variable originally exceeding that values, regardless oft.

7 In our job example, the probability that a job runs for one additional hour is the same asthe probability that it ran for one hour originally, regardless of how long it s been exponential distribution ismemorylessbecause the past has no bearing on its future behavior. Everyinstant is like the beginning of a new random period, which has the same distribution regardless of howmuch time has already exponential is the only memoryless continuous random of the memoryless PropertyThe memoryless property makes it easy to reason about the average behavior of exponentially distributeditems in queuing we re observing a stream of events with exponentially distributed interarrival times.

8 Because of thememoryless property, the expected time until the next event is always1 , no matter how long we ve beenwaiting for a new arrival to behavior is a bit counterintuitive. We might expect that arrivals getmorelikely the longer we example, if the bus is supposed to come every ten minutes, and we have been waiting for nine minuteswithout seeing a bus, we expect that the next bus should be along very soon. If the time between bus arrivalsis exponentially distributed, however, the memoryless property tells us that our waiting time no matterhow long it s been is of no use in predicting when the next bus will we have a queue with exponentially distributed service times.

9 If a new customer arrives to thequeue to find someone in service, theresidual service timeis the time until the currently running customerfinishes service and departs the queue. Because of the memoryless property, the distribution of the residualservice times does not depend on how long the customer has been in service. The probability that thecurrent customer runs for an additional minute and then departs is the same as the probability that a newcustomer just entering service runs for one minute. Likewise, the average remaining service time is simplys, the expected time for a new customer just entering course, the time between real bus arrivals is never exponentially distributed.

10 People want to know exactly when theirbuses will come, so bus schedules are nearly deterministic, which justifies our intuition about longer waiting times increasingthe probability of an RatesIn the field of reliability theory, it s common to use a random variable to represent the lifespan of a of the main problems in this area is predicting the likelihood that a component fails in the very nearfuture given its current age. This likelihood is summarized by thefailure rate(also called the hazard rate)of the most components, the failure rate changes with time. There are three possible relationships. increasing failure rate decreasing failure rate constant failure rateIf the failure rate isincreasing, then failures becomes more likely as the component ages.