 Now, let us try to prove B implies C. This is a bit involved, but let us see if we can argue this. What we want now is, suppose we assume this B, then I want to show that for any tau, N tau is the Poisson distributed with rate tau lambda and further the conditional joint distribution of my count times is expressed in this format. So, now let us first focus on what does this event mean? So, given tau N tau equals to N means by the time tau N counts have happened. That means N tau equals to N means exactly N count might have happened before time t and N plus 1 count should have happened after tau. Then that is when I am going to call N tau equals to N. So, that is what I am writing here. So, that means like if I am going to look at my N of this random variable t1 all the way up to tN plus 1, it should be such that if you are going to look at the realization t1, t2 all the way up to tN should that is the times of this counts should have happened before or at least by the time tau including tau and the t1 count time should have happened after tau. This happens when this happens that is what you are going to call N tau equals to N. If this t plus N has happened before tau that means N plus count is already happened before time tau. So, that is why I have to express this completely I also have to include this tN plus 1. Suppose I just say this and leave it like this, it is not clear that this is going to capture this because I have not told you whether the tN plus 1 count has happened before tau or not. If that has happened before tau then this is not correct. So, for this to make sure I have to express this way that the N plus 1 count happened after tau. Now let us try to find out the conditional. So, how I am going to obtain this I want to find conditional PDF of this set of random variables given N tau equals to N. So, how I am going to leverage already the things that I know. I know that if you just forget the conditional part the joint distribution of this N random variable I have already computed it. I have this distribution. But now I have this conditioning here. What does this conditioning will imply to me? So, because of this conditioning if you go back and recall our conditional density property because of this conditional, then I will get an extra term which divides this PDF here. This probability is the one which divides over. So, let me make that clear here. So, if I want to, so I am just trying to use the definition of my conditional probability that we have defined earlier. So, because of this conditioning extra probability is showing up here in the denominator. Now if you just apply this definition here what I will have is this is nothing but lambda to the power. Now N is going to be replaced by what? In this formula, if I have to use that formula here I have to replace N with N plus 1. And then what? e to the power lambda times TN is going to be replaced by TN plus 1, right? Because I am looking at TN plus 1 here. And then this term is there and this is for the case where my T1 is greater than plus 2 T2 and is TN and then this is equals to tau and this is equals to TN plus 1 here and this is going to be 0 otherwise. Let us see that when I adopted this formula from here to this case that ordering, earlier the ordering did not include any tau here, right? It was just T1, T2, all the up to TN arranged in strictly increasing order. But now that I have this tau factor here, so this joint PDF is defined on this T such that this T is strictly increasing and such that this TN plus 1 is strictly larger than tau. That is where I could use the PDF in this format, okay? This is because I have to ensure that I have been conditioned on the fact that there are n counts till time tau, okay? Now, so we are there like we wanted to define this distribution, conditional distribution. We have this conditional distribution but the only thing that has thrown extra is we have ended up with this TN plus 1 variable here. But whereas what we wanted to show did not include TN plus 1, right? It was only talking about joint distribution of n random variable here. Now, how to derive, so I have this joint distribution which is now of n plus 1 random variables. Now, if I want to get a joint distribution of only in the first n variables here, how to get that? So just integrate this function over what TN plus 1 and what is the possible values of TN plus 1 tau to infinity, right? It could be any value. So if you just do that integration what you will end up is, so if you just do this integration T1, T2 all the way up to TN given n tau equals to n. So after doing this integration what you will end up is e to the power lambda, sorry, lambda to the power n is there anyway and then we have a lambda tau divided by probability that n tau equals to n. This is for all 0, T1, theta all the way up to TN and this is less than or equals to tau and this is going to be 0 otherwise. So I have just integrated this function between tau to infinity. If you just write like I am just ended up with this function. So now if you just look into this function, does this function depends on any of these variables here. Earlier when I have this function at least it depends on the variable TN plus 1 which was my argument. But now after doing this integrating out it is only lambda to the power e to the power lambda t tau and the denominator also is only my conditioned n tau equals to n. Now what is this? I have a distribution, I have pdf and this is a constant, a pdf which takes only constant value. It should then what this is? It should be kind of uniform distribution. Like I have ended up with a distribution which is uniform. And if it is a uniform distribution and also we know that a pdf should be such that if we integrate it over its region it should end up what? It should indicate what? So that is the case and if it is a uniform distribution or it takes a constant value throughout what should be the value of that pdf itself at every point that function itself. So let me refresh it. So if I have a pdf which is independent of my argument that means at every point it is going to take the same value. Like for example in this case you give me any point it is going to take the same value. Further if I going to integrate this over the entire region it should equals to 1. That means necessarily that at every point this function is nothing but the reciprocal of its area. Then only if I am going to integrate and if it is constant then I will end up with a one value. So in that case this value should whatever I got here should be reciprocal of the area of the region where this ordering holds. Do you understand this? So let us imagine for a two dimensional case. I am letting tau each one of this to go up to so I have this T1 I have this tau and this is I am only letting it go till tau tau because both T1 and T2 are random variables or variables which can take at most value tau. Now what is the region here where T1 is going to be greater than T2? So let us say I draw a 45 degree line in all of this region T1 will be greater than T2 or T2 will be greater than T1? T2 is going to be greater than if I am going to fix this and if I go all the way up T1 will be. So I have this region. So I am just let us focus on this. So if you just focus on this what is the area of this entire area tau square and now what is the range where T, this ordering in this case I am interested in T1 less than T2. So this and I mean I can ignore this part because that line itself do not have any area. The case where T1 equals to T2 I will what is the area of this region that is going to be exactly half right of T square by 2. In that case it is going to be tau square by 2. So similarly if you extend it to n dimension the area where this ordering holds if I am going to test it what that will be any guess tau raise to n divided by n factorial why that should be that is the result fine let us take it as a result. So then what do you expect this value to be equal to 1 by that quantity right tau to the power n divided by. So we expect this to be equals to reciprocal of that is n factorial t to the power n and you see that that is what I am going to have already mentioned here but this is just like one step of this right like I have to show that for each tau this is a Poisson random variable what I have just showed is this conditional distribution satisfies condition on this n tau equals to n is going to be this n factorial by tau to the power n. Now how to show this that for every tau greater than 1 n tau is a Poisson with rate lambda t whatever this relation I have you if I go ahead and manipulate it I am going to get probability that tau n tau equals to n is equals to what lambda n tau n e to the power lambda t lambda tau divided by n factorial. So what is this? So what is this what is this the random variable then n tau? So I have saying that the probability that this random variable n tau takes value n is expressed in terms of this but now what this looks similar to or what this exactly is Poisson with what so n tau is Poisson. So I could just write this as lambda tau to the power n and e to the power lambda tau divided by n square right. So this is my rate and this is Poisson with rate lambda tau and that is exactly our claim here. So in a way what this third point C point is telling us the Poisson process can also be thought of like an uniform random variables a collection of independent uniform random variables where that n points occurs can be thought of like they are going to happen uniformly in the in the interval tau that is each points I am going to draw that is going to happen uniformly with parameter tau but I have also this constraint that these points need not be arbitrary they have to be in increasing order right. So that is tau t 1 has to be greater than t 2 like this. So because of this I have to end up with this n factorial term here. Now quickly argue why if this is true this implies that my process is Poisson. So now what I need to show if I am going to assume this C property I need to show that all the three properties of the if I assume that some process is such that it is going to satisfy all it is satisfies point C then I have to show that this also satisfies this implies three properties of my Poisson process. Now let us let us try to say C equals to A. Suppose I am given this let us say t 1, t 2 and up to t k I am given this t k variables and also this n numbers n 1, n 2, n 3, n 2 all the way up to n k and then let us say I am going to define n to be n 1 plus n 2 plus all the way up to n k. So think of this I have been given as the time count times t 1, t 2 all the way up to t k. Now I am going to define and also imagine that this n 1, n 2, n k are nothing but the number of items you want to put in the interval 0 to t 1 is n 1 and that between t 1 to t 2 as n 2 like this. So I have total n items here n 1 which is the sum of all these items. Now I am going to define some ti here, ti is going to be defined as ti minus ti minus 1 divided by t k. So what is this suppose I have let us say t 1 here t 2 here and t 3 is bigger and t 4 is smaller and I have some t 5 here and this is like starting 0. So p 1 is going to be proportional to this interval p 2 is going to be proportional to this interval and p 3 is going to be proportional to this like this. So notice that ti, t k is constant that is the total length let us say this is t k. So this is t k and now for every pi I am just going to take the length of interval. So I have basically defined probability pi that is proportional to the length of the intervals. Pi I will just define now. So I will say where I am going to use it like pi is I have defined to be a number which is defined like this. You can think this as like a probability term here. So if you just add it over all i's it is going to add up to 1 right between 1 to k here. You have to convince yourself like if I am going to use assume this c holds and I think that just let me write this. So by c I am assuming c holds right by c the distribution of so what we are saying is if because of this uniform distribution as a interpretation we have from this c point we can assume that if you are going to define this p i's to be probabilities proportional to the interval n and if I have n numbers let us say given to me and if I throw this n of items on this interval probability that it falls in this interval is going to be proportional to the length of this interval. So that is what we are going to now use. So the distribution the number of counts in each interval is as if each of the n counts thrown into the sub interval at random following into the ith interval is going to be p i. So now, so if we say that we have n items and n i items are going to fall in the ith interval so what is this probability is going to be like if I have n objects and I throw them n 1 of them going to the first interval n 2 of them going to the second interval and n k of them going to the ith interval what is the probability is going to be like I am choosing each one of the object independently and I am just throwing every time I throw the probability that it is going to fall in the ith interval is p i which is independent of the other. So then what is the probability that n i counts going to the ith interval is it is going to be p 1 into n 1 p 2 into n 2 into p k into n k. So this is for a given n 1 n 2 all the way up to n k right. But when I have n total counts in how many ways I can partition them into n 1 n 2 all the way up to n k such that the sum is equals to n I have a multinomial distributions right in this. So the number of ways such that counts in the ith interval is simply n choose n 1 n 2 all the way up to n k which is defined as n factorial divided by n 1 factorial n 2 factorial all the way up to n k factorial right. So I have n things I am indifferent like where the count is happening I am just like throwing them right like I am just interested in how many of them going to the ith interval and then I am just looking at the probabilities in this fashion. So for a given n 1 n 2 like this is the probability but right now I do not know how this n total n gets split across this right. So I have this multinomial distribution. Now finally how to exploit this now. Now I have this. So suppose let us say n t i sorry n t i minus n t i minus 1 is equals to n i. Let us say for all i equals to 1 to n. So let us compute this probability I want to compute that okay in the interval of n 1 n 2 all the way up to n 1 n 2 all the way up to n 1 n 2 between t 1 and let us say t i and t i plus 1 or between t i and t i minus 1. So there should be i minus 1 there are n i objects okay. This you can write it as n t i n t k equals to n total number of objects and probability that n t i minus n t i minus 1 equals to n i between i to n and k given condition that n t k equals to 10. Now coming back to this c third property again here from the third property that point c here we already know that this distribution is what probability that n t equals to n this is a Poisson distributed with rate lambda t k right that is the length of the interval here. So this is going to be lambda t k and how many counts till then there are n counts e to the power lambda t k divided by n factorial and what is this. So I am basically asking there are n i counts between in this interval and there are n such intervals right. So how I am going to do I first going to look for a possible realization and then when I know the possible realization I have this is the probability associated with that. So I will consider it over all such possibilities right from there are these many ways of choosing these many ways of partitioning this n into n n 1 n 2 all the way up to n k when I have this kind of partition what is the associated probability the associated probability is e 1 to the power n 1 e to the power n k and all the way up to e to the power k n k. So if you just replace this guy by its n factorial divided by n 1 n 2 all the way up to n k factorial and further if you replace each of the probabilities as we defined earlier in terms of their intervals that is you will just replace this t i by what t i minus t i minus 1 divided by t k if you just replace what we are going to get t equals to i k equals to 1 for i equals to 1 to k then so you just verify that I am just directly writing this okay. So what is this now I am finally able to express this probability here as the product of k probability k terms here what is each term here if you just focus let us take one particular i here i is ranging from 1 to k right let us take one particular k. So if you just look into that what is this is going to be so that is going to be a Poisson with rate lambda t i minus t i minus 1 that is a lambda that is a random variable with Poisson distributed with rate and this is nothing but the probability that that random variable taking what value n i value and that is exactly this value right for particular i. But now what I have able to write it as it is now able to write it as all this as a product of all this k values so that means what we have done is this probability is nothing but so these are like increments right I am able to express these increments as the product probabilities each probability term here corresponds to a Poisson random variable so that means it I am already showing that the third part of the Poisson process that this independent increment so each of this random variable corresponds to Poisson random variable with rate which is going to be proportional to the length of the interval further we have shown that through this product that these increments are independent because each one of them I am able to write it as a product fashion right. So is it clear that like by expressing this I have been able to show both second and third properties that is independent increments and that each of this random variable is Poisson distributed with the rate corresponding to lambda times the length of the intervals. And the first property that counting process that is obvious right like we are doing a counting and you can just verify the definition of counting process on this. So with this we are able to show that C implies the three properties of a Poisson process so hence C also implies if C holds then it is clear that my process is a Poisson process. So any of this characterization you can check and that all applies the same Poisson process. So okay let us stop here if you have any doubts on this you can ask now.