 Hello and welcome to the next lecture in the course on Introduction to Computer and Network Performance Analysis using queuing systems. I am Professor Varsha Apte, I am a faculty member in the Department of Computer Science and Engineering IIT Bombay. And today we are going to talk about memoryless distributions. So recall that when we were talking about the Kendall notation for queuing systems, there was one slightly strange thing that we talked about which is that the exponential distribution is denoted by the symbol m and that m stands for memoryless. It also actually there is one more word called Markovian that it stands for, you can say that m is either for Markovian or m is for memoryless and they both roughly mean the same thing. So what does this mean? And basically exponential distribution is a memoryless distribution and we will talk about what that means now. So again recall probability distributions. If x is a continuous random variable, then typically we describe it by a cumulative distribution function which is the denoted like this capital F sub x of t and that is a probability that x takes the value less than t. And then we have small f x t which is the derivative of the CDF and this is called the probability density function pdf. Remember this is not a probability, we will have a separate lecture explaining exactly what these things are and some maths about it. But right now we do not need to go into detail maths, we should try and understand all of these things intuitively. So these are some examples of probability distributions we talked about in the previous lecture. So denoted by m for memoryless like I said, so what is this memoryless? So memoryless property is this, I am going to state it and we are going to try and understand it. Suppose x is a random variable denoting the time a person talks on a time that a person talks on a cellular call turns out to be the, it is going to be equal to the holding time of the cellular channel. So let us assume that x, it is of course a random variable in the whole period of time that a cellular network operates, people talk for different amounts of time, it is nothing fixed. So we can denote it by a random variable and let x denote that time. Now suppose the person has been talking for some time t, imagine that there is a person talking on a cellular phone, that cellular phone has been, that call has been given a channel to do that conversation on in that particular cell and let us say the person has been talking for some time t. Now let us denote by y equal to x minus t, the remaining time of the call. If x is the random variable denoting the total time, denoting the total time and if t has already elapsed then x minus t is going to be the remaining time of the call. Now if the distribution of x is memoryless, if the distribution of this time of the call, time duration of the call is memoryless then the following holds. The probability that capital Y is less than small y given, remember this is a conditional probability given that x is greater than t is the same as probability x less than y. What does this mean? This is just, it is a very complicated equation. What does this mean that if I know that the person has been talking for some time t, then what I know is that the total time is going to be more than t, if the person has been talking for 5 minutes, then I already know that the call duration is either 5 minutes and 1 second or 5 minutes and 5 seconds or 10 seconds or 15 seconds or 20 seconds. One thing I know for sure is that it is going to be greater than 5 minutes. So given that the call is duration is going to be greater than 5 minutes, this is asking the probability as to is the remaining time less than say 10 minutes, let us say if y is equal to 10, then I am asking the question as to what is the probability that the remaining time is less than 10 minutes. That means I am asking actually the probability of the person going to be talking for 10 more minutes. Turns out this probability is the same as the total time being less than 10 minutes. What does that mean? As if the call just started that 5 minutes is like it did not even happen, it is as if the call just started and I am asking the question, asking this question is the same as asking the question as to is the call duration 10 minutes. So it is as if you forgot as if at any time when you are observing the call, you can just forget how much time has elapsed and the probability of the remaining time is the same. So that is what makes it memory less. So again I am going to state these things, so probability y less than y given x greater than t is equal to probability x less than y, which means now I am going to say this in a little more formal way, the conditional probability that the remaining time is less than y given that the total time is greater than t is the same as probability that total time is less than y. So one implication is that even if t seconds of the call have elapsed, the expected remaining time of the call will be the same as expected total time of the call. What do I mean by expected? Expected is one word that is often used instead of average. In a certain context it makes more sense to use the word expected. So let us say the cellular network had data of its cellular calls and average call duration is just hypothetically 5 minutes. So if 2 minutes have elapsed and if this is if memory less, if this call duration is memory less, then what is the expected time that the call will continue further even if 2 minutes have elapsed? We might want to say that average is 5 minutes, 2 minutes have elapsed, so 3 minutes must be left. So no, if it is memory less, even if 2 minutes have elapsed, then the average, the expected remaining time is still 5 minutes, it is as if this 2 minutes did not happen, we have no memory of it. So again even if 2 seconds of call have elapsed, the expected remaining time of the call will be the same as that expected total time of the call. It is as if there is no memory of the elapsed time, at any point it is as if the call is starting afresh and this is what is called a memory less property. In continuous distributions, only exponential distribution is memory less. Now it seems very weird, this memory lessness, so is it like, is it a completely non-intuitive thing, just a mathematical, artificial thing that somebody has cooked up? So no, I will give you an example if you consider discrete random variables. In continuous random variables, it is a little non-intuitive, it feels a little non-intuitive, but I will give you an example from discrete random variables and then it will start feeling intuitive. So let n be the number of coin tosses required to get till you get a head, standard example given probability and let head probability be P. So suppose you know, you had to do, you got tail, tail, tail, tail and then head, that means your n was equal to 5, this is you needed 5 tosses. So now we know that this we have done in your undergraduate mathematics that the probability that of needing i tosses, the probability of needing i tosses to get a head is that you should get i-1 tails and then one head. So this is to get for example probability n equal to 5 is equal to 1-P raised to 4 multiplied by 1 success, exactly this is the probability of needing exactly i tosses. And of course this i goes from 1, 2, 3 because you need at least one toss to get a head. Now suppose j tosses are done and they were all tails, they were all tails. Actually when you know this, what do you know? You know that you are going to need more than j, you are going to need at least one more toss to get a head. So in fact you know now that n is going to be strictly greater than j or n is going to be greater than or equal to j plus 1. So now let m denote the additional number of tosses required given that j tails are done. Now what is the probability that m equal to k? Suppose I am asking the question that I saw j tosses and now they were all tails, how many more am I going to need? A little bit of reasoning and thought will suggest to you that actually the probability m equal to k remains the same. It is also 1-P raised to k-1 to multiplied by P because why should it be different, right? Now take an example. Suppose I got 1, 2, 3, 4, 5, 6 tosses I did and all 6 were tail. Now if the question is probability of, probability needing 6 more tosses for head, does it depend on the fact that 6 tosses were done? Just think about it. Does it depend? Does anything change? It is the same coin, it is the same person doing tosses and just because 6 tosses are done is the probability of the next toss giving you a head anything different, okay? The coin is supposed to be such that every time the probability of getting the head is the same, it does not change. So there is no nothing changes even if you did a 100 tosses and you then the probability of needing 6 more tosses to get a head will continue to be 1-P raised to 5 multiplied by P. The 6 more does not change, okay? So it is the same as Pn equal to i, okay? It is the same as Pn equal to, sorry, k and the j successful tosses are just forgotten. So I hope that this was not as non-intuitive. Remember that I can be doing, you know, T, T, T, T, T, lots of, you know, I can have you know even 100 tosses that are tail, the probability of needing even 100 more given that 100 tails are done, okay, is actually still remains 1-P raised to 99 multiplied by P, okay? The fact that these 100 tosses were done is can be forgotten and so this is actually memoryless, this is a memoryless distribution, okay? So this is actually, you might remember, this is called the geometric distribution, right? This 1-P raised to k-1P, this is the geometric distribution and again to remind this is what you get when we are trying to have a situation like the number of coin tosses required till you get a head. And you can see here that a discrete equivalent of that same continuous property is satisfied here. What is the probability that m equal to k given that n is greater than j, okay? This we had the probability of capital Y less than y given x greater than t, right? We had that this probability was actually the same as x probability x less than y and that is what we get here, okay? So you should, you can think of exponential distribution as a continuous equivalent of geometric distribution that is just the best way to understand it. By itself it is difficult to accept the fact that if a cellular call duration for example is exponentially distributed, then if you are looking at a person talking on the cellular call and if the average was 10 minutes but it was a memory less distribution, then if you are observing the person for 8 minutes, you cannot say that now 2 minutes the expected remaining time is 2 minutes. Even if 8 minutes are passed, you have to say that expected remaining time is 10 minutes if it is memory less, okay? So this is actually a graph that shows the exponential distribution and the geometric distribution exponential is of course plotted as a continuous line and the geometric distribution is a specific probabilities for the geometric distribution taking values 1 and 2 and 3 and things like that. But you can say that for the, see that for the same average actually the exponential distribution is it looks very much like the geometric distribution, it really matches the actual values also if the average is the same. So they are kind of sister distributions and think of exponential as an intuitive distribution because in its discrete equivalent there does exist a distribution which is the geometric distribution which has a perfectly intuitive memory less behavior, right? Why are we talking about all of this in queuing systems? If you go back to the example of the cellular call time, I already said this that holding time is considered exponential which is memory less with an average of 3 minutes and we know that 1 minute has elapsed then the expected remaining call duration is still 3 minutes and this property will be used repeatedly in reasoning about queuing system metrics especially something like response time, waiting time, it becomes significantly easier to reason about this metrics if we can justifiably assume that sometimes some durations in the queuing system are exponentially distributed, right? So with that this part is over and now we will go on in the next lecture to observational or what are called operational laws in queuing systems, thank you.