 So, now we have some understanding of how the this function mt, mt looks like but we have just defined one process mt based on our xi's and zk's we can actually define more random processes right. So, let us see what all we can discuss what can we can define. Before we do that some properties let us take zm of t. So, z subscripted by m of t. So, this m of t we know is a integer value right. So, what I am looking at is a renewal happening at the for the mt time or mt renewal. So, given a t you know that mt is going to denote the number of renewal that has happened in the 0t interval and if you look at z of mt. So, what is this this basically now I am looking. So, zt is a zz zk is a sequence of random variables right. Now, mt is a random variable I can treat this as a random time and now I am asking the value of z at this random time m of t right. Now, my claim is this quantity is going to be less than or equals to t. Why is that? So, what we are doing is m of t by our definition told me included all the renewals that has happened before time t right. So, naturally any renewal that has happened before t must have happened before time t right. So, because of that this property is natural. When you are going to look at m of t renewal of your process that then m of t renewal should have happened before time t. So, just go back and try to convince yourself all these properties. Now, similarly what about m of t plus 1 greater than or equals to t or strictly greater than it has to be greater than t right. Why is that? Because by definition m of t has included all the renewals that has happened till time t. If you are looking at the one more renewal that better have happened after time t right. So, that is why this property holds. And one more property is that is we are going to exploit is two events you may be let me write these two sets they are the same. Why is that? So, we are saying that suppose let us say this event implies let us first say this event is a subset of this. Suppose Zn is less than or equals to t. What does this mean? The nth renewal has happened before time t right. Now, if nth renewal has happened before time t then better be like mt should be larger than or at least equal to n right. Because if we already know Z nth renewal has happened before t maybe there could have been possibly more event renewal has happened just before t. So, because of that mt has to be at least n or more okay. So, we know that this event is included here. But what about the other case? Suppose if you know that m of t is going to be greater than or equals to n. That means in the interval 0 to t at least n renewal say happened okay. Then it must be the case that the nth renewal must have happened before time t right. nth renewal must have happened before time t because of that this is also included in this. So, we can now it is clear that these are the same sets. If not they are just again go and convince yourself okay. So, these are some property which we will exploit in coming up with some interesting results okay. Now, define another process I am called residual line time. So, I am going to define a process yt now capital Y of t that is defined for every t in this format Z of subscript Z of mt plus 1 minus t. So, what is the actually capturing? So, before we try to understand what this is actually capturing let us try to draw how this function looks like okay. This is t and let us see what is this yt. So, at t equals to 0 let us say take m of t to be 0. So, at t equals to 0 what is this? It is Z1 right minus t. So, t is 0 that means at time 0 this is Z1 but Z1 we know same as X1 by definition Z1 is same as X1. So, at time 0 so this quantity is Z1 or this is same as X1 okay and now let us increase t okay. As I increase t at some point the second interval will also be included right and then as I increase this t so as I increase this t till some point this guy m of t is going to remain 0 only right after that it is going to increase to index 1 okay. Let us say for time being it is not going to increase. So, in that case this guy is going to be Z1 and now there is a minus subtracting time here what is this going to happen as t increase it will be decreasing it will be decreasing till what time till first arrival can happen right and what is that can X1 right at X1 arrival has happened so because of that this guy suddenly becomes 1 now it was 0 so far and now it is Z2 right and what is this now Z2 and in the Z2 you are what is this t? t is already X1 right it has taken already 1. So, what is this jump in this case is it going to jump at this point? What is this jump? It is going to be the amount of X2 once it jumps to X2 after that this guy is not like immediately going to make 3 right it remains at 1 till some point but this guy is continue to increase right so it is going to pull it down and what this guy is going to pull down to it is going to get to 0. So, this guy at this point till this point this is already X1 now let us say I take it up to X2 here that time this guy is already so when I take it from okay let us say I increase it by amount X2 here. So, then this guy becomes X1 plus X2 right but X1 plus X2 is Z2 actually so this has made it become 0. So, this height here and this height they are the same so it is falling down to 0 at what right that you need negative right right. So, you can just keep on plotting like this so now based on this what you can see now again coming back to my battery when I charged it its battery is full whatever its inherent life was that time as the time process its life goes to 0. So, that is why we are going to at any time t you take an arbitrary time t this process is going to tell how much more time how much time is residual time is left in that before that guy dies. So, yt the function yt at any time t gives me that time remaining before our next renewal happens or in the battery example case the time before the battery gets discharged. So, till this point you are clear when X1 at when t equals to X1 you are not there so better to write this function. So, let us take t equals to X1 so t equals to X1 that is X1 is basically Z1 right. So, Z1 is definitely less than equals to Z1. So, k is included in that and m of t becomes 1 right at this point. So, at this point what is this Z of 1 plus 1. So, this is exactly Z2 at this point suddenly at this point it has become X2 why because this X is already X1 and this guy is. So, okay y of at t1 is equals to Z2 minus t1, but t1 we have taken to be X1 right. So, this is going to be Z2 I know is the X1 plus X2 minus X1 this is X2, but once I come here till my t increases by amount X2 my k value of m of t is not going to change right. So, till this time this guy is going to remain Z2, but this guy is increasing. So, it is going to pull it down right, but suddenly when you have increased till X2 then your t becomes X1 plus X2 which is Z2. So, this will make your m of t2 it will make. So, because of that now you become Z3 right. So, Z3, but t is. So, at time Z2 you have this becomes Z2 plus 1 minus Z2 right, but this is X1 plus X2 plus X3 minus Z1 X1 minus X2. So, this is suddenly here X3 jump and then it is going to go down to 0. So, like that. So, basically this function is y of t is going to capture. So, if you just say tell me what is the time if at this time this function defines how much more life is remaining that is why it is called residual lifetime. So, what is. So, if I am going to just take y to be Zk here at time t I am going to substitute Zk what we expect this value to be. So, when I had. So, this duration is what basically Z1 right Z1 and X1 are the same at this point this value happened to be X2 right. And so ok maybe just let us write this Z1 and what happens that y equals to Z2 that is at this point it has become X3. So, in general based on this can I write y of Zk is equals to Xk plus 1 and how what about y Zk minus just before the kth renewal. So, what is happening just before the jumps it is 0 right this point is going to be 0 this is going to be for all k it will be greater than or equals to 1 ok. So, now let us define another process this is called H process ut I am going to define as t minus Z of m of t ok. So, I have already argued that Z of mt is going to be less than or equals to t right because of this is this quantity is going to be positive or non-negative value it is going to be non-negative right because Z of mt is going to be always less than or equals to t. So, let us plot this. So, at time t equals to 0 m of t let us take it as 0 and Z of 0 let us take it 0 because Z of Z process is defined for k greater than or equals to 1 right. So, for Z of 0 let us take it as 0. So, at time t equals to 0 what is its value? It is going to be 0 and as I increase t what is happening right this guy is not going to suddenly change it is going to remain there for some time, but this guy t is increasing and it is increasing t increasing till what time? At some point when t I take it to be x1 suddenly this guy becomes z1 right at z1 at t is also z1. So, it falls to 0 what is this guy this guy here distance x1 and what is this value? y ut here right it has also increased by xt amount right because till when it is increasing this guy part was 0. So, it was like ut equals to t that was kind of linearly increasing this is also of the same height. Now, it has fallen 0 at t equals to x1. Now, after that if I go t to just beyond x1 this guy is not going to suddenly change it is going to still remain x1, but this guy t is now increasing from x1. So, what do you expect? It is increasing till what point x2 then? So, this is with slope 1 slope 1 and so was here. So, can you now see why it is called age process? So, age always increases right. Let us say you have replaced your battery at this time and you take some time at this point. So, what it basically tells you is basically if you just look at this time t and it has so happened that you have basically replaced your battery at this point your age has at this point is this much. See like when so the thing here the because I am subtracting this amount this is basically capturing that I have renewed replace the battery. So, even though I am looking at this time I am not counting age from this point I am counting the age from the point where it has been replaced or recharged right. So, and that part is trying to get rid of the things till that it only capturing what the time since the battery has been replaced. So, these are these processes this age process, residual process they try to help us understand how frequently maybe if I can characterize these processes for my system then they will tell us how frequently my component is going to fail even if I replace them and that will tell us how quickly I should take it for repair. So, if I know this is going to fail with this kind of behavior I can come up with a criteria okay before this guy fails I will make sure that I will do some maintenance work. So, that even that guy so then things fail you are in loss right maybe like if you have a company or something if some component there fails the machine is not working you are not basically earning money right. So, what you want is you are you do not want your machine or your process to stop. So, if you kind of analyze and anticipate when it is likely going to survive or what is the quality of my component then before it gets bad you can plan a scheduled repair process and make it continue to work without interruptions. Okay, this can be next one can be just this these are realizations right like you charged it and it worked for longer time for some reason and next time and it become bad and when you again charged it maybe it become bad again. Okay, so other basically the process the renewal process we have defined right. It was IID process for the point n greater than or equals to 2. So, to describe all the points after n greater than or equals to 2 I just have to specify one distribution and the first cycle could have different distribution. So, if I just specify you the distribution of the first cycle. So, I am going to interchangeably use cycles and renewals now like when a renewal happens I am going to call it as one cycle completed. Okay, so if I specify the distribution of the first cycle and the distribution of the second cycle that is enough right that completely specifies my renewal process here because that is that only two things I knew because these are independent process and for n greater than or equals to 2 this is all IID. But now suppose I have this information can I compute the quantities of my interests. For example, can I compute the distribution of my z i's can I compute my distribution of my m t right and can I compute the expected value of my m t okay. So, let us see how can we do that. Suppose a is cdf of x1 and f is cdf of x2. So, if I give these two cdfs is it enough for to characterize my renewal process right. So, the first cycle is distributed as x1 and the subsequence are distributed as per x2. Now, if I am going to capture what is the probability that z1 or equals to t what is this going to be. So, that can be computed this is nothing but the cdf of x1 right because z1 is same as x1. So, this is simply a of t okay. Now, what about probability of z2 greater than or equals to 2. So, z2 now is x1 plus x2 t which I can write as x2 less than or equals to 2 x1. So, you guys have already dealt with such cases right how to find the distribution of this. So, how you are going to find the distribution of this by using double integrations right. You first let this x1 take the values as per distribution and for each possible values then you compute x2 upper bounded by that value as per the distribution of x2. And you have when you do this you will end up with something like integration of okay basically we will end up with something like 0 to t f of t minus u and dv u. I am just writing this is just like simplification of this which you can basically I can think it as a convolution of a and f process at time t. So, this is then you can just like even just think of this as you can write express it as a convolution of your distributions a and f here. Now, you can repeat this process for any n now you can now think ask the question what is the probability that Zn will less than or equals to t. So, you can work out this I am just skipping the details it can be thought as a convolution of f and all these processes computed like this. So, these are basically convolution of a and n minus n one times the f functions convolution of this computed at a. Now, based on this now we are ready to kind of find the distribution of m of t why is that I know that probability of Zn less than or equals to t is equals to probability that m of t greater than or equals to n. We have already shown these two events are equivalent right. So, if these two events are the same then the probability be better at the same. So, you see that now we have this now how I am going to use this further information to find so fine I have this distribution. So, based on this I know just using a and f I will be able to find the distribution of my process m of t now how to count the expected value. So, we know that this expected value can also be found as probability that. So, I know that my m n m of t is going to take integer valued integer values. So, this is going to probability that it is equals to n equals to 1 to infinity it is correct. This also you know right the expectation of a random variable I can express it like this. So, here m of t is taking discrete valued outcomes that is why I am just writing a summation had it been a continuous value I had to replace it by integration fine. So, then I know this. So, now going back to this relation and this relation I have this is nothing but n equals to 1 of into. So, you just see that just by knowing these two CDFs of my renewal process I should be able to compute all the statistics of my related process this Z ions this m of t is and its mean values. This is equivalent to computing it CDF right it is a discrete valued right CDF should be enough for us. So, just use this you already have what more you want okay fine. So, let us conclude with the main result of this part just going to state it as a theorem this is called elementary renewal theorem lifetimes. So, it requires three hypothesis x2 almost surely here. So, what it says is okay as usual we are going to take a mutually independent not negative sequence of random variables such that they are IID distributed for n greater than or equals to 2 further let us say the probability that each of the renewal times being finite is 1 yeah. So, all my xk's are finite with probability 1 and I am also allowing but I am allowing this xk to take the value infinity okay. So, all I want is these expectation of xk to be any value but I want them to be strictly greater than 0 okay all these quantities for all k if this happens. So, notice that this expectation of k's they are same for all k when k is greater than or equals to 2 expectation of x1 could be different from expectation of x2 but expectation of x2, x3 and all others are going to be same. Now, it says that the mean number of renewals in that in in the interval 0 t as t goes to infinity is 1 by expectation of x2. So, what is m of t the number of renewals in the interval 0 t when you are dividing it by t that means basically mean number of renewals in the interval 0 t okay and you are letting t go to infinity. So, when you are going to let that go to infinity that rate is going to be like 1 upon expectation of x2 okay let us try to understand why expectation of x1 is not coming into picture why only expectation of x2 is coming into picture okay. So, see that for k greater than or equals to 1 we are assuming that expectation of sorry probability that xk is less than infinity right. So, that means I am going to hit state j whatever state I start from in some finite time right that is the why it is x of probability of x1 is less than infinity is probability you hope happens is probability one. So, at some point I am sure I am going to hit state j and after that I am looking at coming back to the state i j again and again right and now from state i to some state j I have come done that in the first cycle within some finite time and after that I am looking at again coming back to j again and again right. So, in the first cycle has some finite time but here I am looking at t letting go to infinity whatever the contribution that has happened from initially some state to coming to j for the first time that is going to be vanishing right it will not contribute. But what contributes is the sum of the other terms. So, this m of t is going to include all the renewals till time t right. The first renewal has happened sometime it is going to take some portion but most of the other time what happened the other renewals have happened in the other time. So, the time contributes to whatever like this that the first time contribution to m of t is going to be very small right and now because of that that does not going to affect expectation of X1 is not going to affect this. But why that then the the thing is that the others is happening at this rate fine. So, in m of t most of the renewals are like second, third right like other than first ones and now all of them are kind of identically distributed X2, X3 all of them and now when you are going to divide it by t this is basically we are asking number of returns to state j over a period of time t. So, in a way that has to be inversely proportional to. So, this you are basically acting on an average how many returns to state j are happening right. So, basically where if you are going to take some time t and then you are going to divide it by t. So, number of renewals by t are basically happening on a. So, how much time it took for one renewal to happen. So, one renewal happening time you expect it to be inversely proportional to the expected time of that renewal itself right. M of t is the number of renewals that has happened in time t. So, what are happening is basically per unit time how many renewals are happening right. So, this some number of m of t number of renewals are happening in the interval 0 t where dividing it by t. So, then how many renewals are happening per unit time that is the question you are asking right. So, we are saying that that number of renewals per unit time is 1 upon expectation of X2 right. If something is happening. So, you are repeating. So, X2 is basically telling you at what time you are repeating right. So, if you are repeating very fast. So, expectation of X2 is going to be very fast that means you are repeating very fast that means expectation of X2 is small because you are happening again and again very frequently. If this is small then you basically going to see lot of renewals in the given interval right. So, that is why if this expectation is going to be small you are going to see lot of renewals in a given duration. If this expectation is going to be large that means the renewals are happening after a long long time. So, you do not see more number of renewals in a given duration 0 t. So, that is why this number of renewals per unit time is going to be inversely related to the how frequently you are going to observe these renewals. So, and this is almost sure result because these m t are random quantities right. So, that is why this is random and now the last part of it says that if instead of m t you look at its expectation that is again converging to the same limit. So, you what was our law of large number said summation of Xi by n it went to mean value right. Is it necessary that always the average of random quantities has to converge to some random variable it can converge to a constant and that is exactly is happening. So, this is exactly saying that the number of renew average number of renewals is a constant and that constant is given by 1 upon expectation of X2 ok. So, this quantity is nothing this quantity here is nothing but the expectation of this, but as usual we cannot directly derive this quantity from this because in general we cannot just take expectation of this quantity here because expectation and limits cannot be interchanged. It needs some careful analysis. So, because of the lack of the time what we will do is we will not go into the proof of this. The proof is actually interesting, but in the last class we will just cover one more theorem called renewable reward theorem and just stop it there ok.