 In the previous lecture, I introduced the idea of the tolerance intervals, the coverage, the coverage probabilities etcetera. And then towards the end I define what is known as empirical distribution function or the sample distribution function. If I have the sample x 1, x 2, x m and based on that the order statistics x 1, x 2, x m is defined. And if the observed values are taken, then based on the observed values we define a step function of this form. As I mentioned this is also the CDF of a discrete random variable which takes values x 1, x 2, x m each with probability 1 by m. So, we get the this as the function. Now in place of this small x i, suppose I put capital X i, then this will become a random variable. So, now let us consider that. Next we define empirical distribution function based on x 1, x 2, x m that is f m, x is equal to 0 for x less than x m sorry less than x 1. And it is equal to j by m, if x j is less than or equal to x less than x j plus 1 for j is equal to 1 to m minus 1. And it is equal to 1, if x m that means basically x is greater than or equal to x m. Now this has become a random quantity, but still we call it a empirical distribution function. So, for each x f m, x is a random variable. We can keep on changing x, but still in all the cases this will remain a random variable. Let us analyze this. If I consider say m times f m, x then what are the values? It is equal to 0 if x is less than x 1, it is equal to 1 if x is greater than or equal to x 1, but less than x 2. It is 2 if x is greater than or equal to x 2, but less than x 3 and so on. That means, it is exactly the number of x i's that is less than or equal to x. So, if I consider the distribution of m times f m, x then that is equal to probability of x j less than or equal to x less than x j plus 1. Now we can consider it in the following fashion. Here we can say this is same as probability that any of any j of x 1, x 2, x m are less than or equal to x and remaining n minus j of x 1, x 2, x m they are greater than x ok. So, if we consider the event x i less than or equal to x, suppose this is denoting a success and x i greater than x to be a failure, then basically what we are saying is that out of that means, because these are i i d this is a actual CDF here. Suppose I call it p then this is equal to simply probability of m f m, x is equal to j. This will simply become equal to m c j p to the power j 1 minus p to the power m minus j for j is equal to 0 1 to m, where p is nothing but f x here. So, we are actually able to derive the distribution of the empirical distribution function. That means, we are saying that m f m x is nothing but binomial m p that is f x basically. Now, based on the binomial distribution you have some simple properties. For example, what is expectation of m f m x that is equal to m times f x. So, this means f m x is unbiased for capital F x. So, that is the first property that is means we can say that the sample distribution function is an unbiased estimator of the CDF. So, you see the analog with the parametric inference. In the parametric inference we find out the unbiased estimator of some parametric function here. In the non-parametric case since the parameter is not there we are only having the form of the CDF. That means, basically we are saying that we the model is not known then we can actually estimate the CDF itself by using the empirical distribution function which is of course that means, you are taking m observations here x 1, x 2, x m and based on that you are constructing the estimate. And of course, this estimate that we have written it is based on this. So, of course, then observed values are there then it is becoming a estimate here and this is actually the estimator and that we are showing that it is an unbiased estimator. Of course, you may feel that see you are taking it as simply an step function and your CDF can be of any form. So, you may say that this may be too much different than the, but basically in the absence of any other information because the parametric form is not there therefore, this is the best that we can do ok. Based on this then we can have other properties also for example, what is variance of m f m x that is m f x into 1 minus f x that means, variance of f m x is nothing, but f x into 1 minus f x divided by m. And of course, you can see that this actually goes to 0 as m tends to infinity. So, we also have that f m x converges to f x in square mean in square mean this convergence will be there. And also since it is unbiased and the variance is going to 0 this is also becoming consistent here. That means, we are having f m x converging to f x in probability. In fact, you have a stronger so basically we can say that f m is consistent estimator of f x. In fact, one can prove stronger thing in fact, we can have f m x converging to f x almost surely. And we have even much more stronger result that is known by Glevenko cantelli lemma that is saying that probability of supremum of f m x minus f x greater than epsilon. This supremum is taken over all x on the real line even this probability goes to 0 as m tends to infinity. So, these are some of the very very you can say strong properties about the empirical distribution function. Now, we will develop a you can say theory which will be used for making useful inferences for the various two sample problems. For example, here I have told that we can actually talk about the median or the quantile. So, we have also discussed a test for the quantile, but many times we will be concerned about two populations. That means, we will be comparing like in the parametric case we had the testing about equality of the means of two normal populations equality of the variances and so on. So, similarly in the nonparametric case we may discuss the test about the equality of medians etcetera. So, based on the empirical distribution function I will construct some procedure which will help in this regard. So, let me develop this theory first. Let us now consider a random sample x 1 x 2 x m from c d f capital F x and say y 1 y 2 y n a random sample from say g y. So, this is different one and also we take the two samples to be independent of each other. Also the two samples are taken independently. Now based on this I consider the empirical distribution function here. So, u i I define to be f m y i for i is equal to 1 to n. Now this f m is the empirical distribution function I will use the term e d f actually e d f based on x 1 x 2 x m. All right, but in the argument I am substituting y i here. So, this is actually then becoming as we have seen what is the interpretation for the m times f m x, m times f m x was the number of x i's which are less than or equal to x. Therefore, m times f m y i will denote the number of x i's which are less than or equal to y i. So, and so this will become 1 by m times number of x j's which are less than or equal to y i. And of course, this notation is similar to the one which I used for f of x i, but now it is in a different context. So, these u i's are different from that that was the x i's were having c d f capital F. So, f of x 1 f of x 2 f of x n that was a random sample from uniform 0 1 that is the continuous distribution. Now these u i's I am defining based on the empirical distribution function and ok. So, the name u i is taken same, but this has some significance that will be clear a little later. So, if I consider u i as the 1 by m number of x j's less than or equal to y i then what will be u of i that will be 1 by m times well. So, that is equal to f m of y i that is equal to 1 by m times number of x j's which are less than or equal to y i ok. Since here it is number this is from 0 1 to m. So, what will be the values of u i's and u bracketed i's it will be 0 1 by m 2 by m etcetera. So, u i can take values 0 1 by m 2 by m and so on up to m minus 1 by m 1 and similarly u i can take values 0 1 by m 2 by m and so on m minus 1 by m 1 etcetera. What is the distribution of u i? Let us look at this distribution of u i ok. First of all what is the difficulty? Earlier the distribution of u i was simply uniform distribution because of the probability integral transform. Here I am actually doing an integral transform, but it is with respect to the CDF here. So, the CDF itself is having the random variables here and then this y i is coming here which is again having random variables. So, it is having a joint distribution here. So, let us derive this thing. What is the probability that u i is equal to say some number j by m where j can take values 0 1 to m ok. So, this is equal to probability of m times u i is equal to j that is equal to probability that j of x 1, x 2, x m are less than or equal to y i and m minus j of x 1, x 2, x m are greater than y i. So, this you can write as integral probability j of x 1, x 2, x m less than or equal to y m minus j of x 1, x 2, x m greater than y given that y i is equal to y say d g y where capital G was the distribution function of second sample. So, this then you can write as now what is happening that given y this becomes a fixed thing. So, this is simply coming from the binomial that is m c j f y to the power j 1 minus f y to the power m minus j and then d g y. So, actually we are able to determine the distribution of u i is here. Now, we take one particular case of course, I mean if I take for example, uniform distributions and here I take say some other distribution say normal distribution etcetera then this all can be easily written in a closed form. Now, let us take one particular case when both the samples are from the same population. If both the samples are from the same population then this probability of u i is equal to j by n that is minus infinity to infinity m c j f y to the power j 1 minus f y to the power m minus j d f y now. See earlier I had d g y, but if g is equal to f then I put it simply as d f y. Now, you can simply substitute say f y is equal to some u if that is happening then this is simply becoming 0 to 1 m c j u to the power i 1 minus u to the power m minus sorry u to the power j 1 minus u to the power m minus j d u which is simply a beta integrals this can be evaluated m factorial this m c j I write as m factorial by j factorial m minus j factorial. And this is becoming then j factorial m minus j factorial divided by m plus 1 factorial which is nothing but 1 by m plus 1. So, this is interesting what you are getting probability of u i is equal to j by m that is equal to 1 by m plus 1 where j is equal to 0 1 up to n this is nothing but a discrete uniform distribution. So, what we are saying it is something like this if I am considering say x 1 x 2 x m and y they are from the same distribution f. And if I consider f m of y where this is empirical distribution function based on x 1 x 2 x m then this is actually having discrete uniform on 0 1 by m 2 by m and so on. So, this if you see see if I have say x 1 x 2 x m from f then if I consider f of x 1 f of x 2 f of x m then this is iid from uniform 0 1. Here in place of f I am putting f m ok that means what I am saying is that if y 1 y 2 y n they are iid f and I define and if I define f m of y 1 f m of y n then they are discrete uniform, but they are not independent certainly they cannot be independent because all of them are based on the same x i is here. So, this can be considered as a sample analog of probability integral transform that is the first result which I gave in this particular section when we started the non-parametric methods we considered this probability integral transform that if x is having cdf f then f x is having uniform 0 1. So, if I have a sample x 1 x 2 x n then f of x 1 f of x 2 f of x n will be a sample from uniform 0 1, but here if I consider the sample distribution function here and then I define this then this is having discrete uniform, but it is not independent. So, that is the difference here. So, this is quite interesting result here. We will also be interested in the joint distributions of for example, 2 of them u i and u j. So, let me discuss this here f is equal to g let me take let us consider now the joint distribution of say 2 of them say u i u k of course, you are taking i not equal to k. So, what is the probability of say u i is equal to something like j by m and u k is equal to say l by m. Then you take here say j is less than l because there can be each of the u i's can take values 0 1 by m 2 by m up to m by m and similarly u k. So, there can be 3 cases j is equal to l j is less than l and j greater than l. So, let me take say j is less than l. If j is less than l then the advantage is that you are actually writing as the probability of m u i is equal to j and m u k is equal to l. That means, it is actually the probability of j of x i's let me say x s less than or equal to y i. This is I am talking about u i that is based on f of y i and then l minus j of x s are between y i and y k and then n minus l of x s are more than y k. So, this becomes basically a multinomial if we remember our effects there. So, we do the conditioning now. Now it will has to be conditioning on y i and y k here. So, we write probability of say j of x s less than or equal to say y 1, l minus j of x s between y 1 and y 2 and n minus l of x s greater than y 2. This is conditioned on y i is equal to y 1 and y j is equal to sorry y k is equal to y 2 d f y 1 d f y 2. See there could have been a general case where this will be g, but then of course, you will not be able to obtain a closed form expression for that. So, minus infinity to y 2 for y 1 and minus infinity to infinity for y 2 here. Now this is simple multinomial. So, we just write it here minus infinity to infinity minus infinity to y 2 n factorial actually there are m here. So, I have put wrongly this should be m this should be m here right. So, this will be m factorial divided by j l minus j and m minus l. Then you have f of y 1 to the power j f of y 2 minus f of y 1 to the power l minus j and 1 minus f of y 2 to the power m minus l d f y 1 d f y 2. Now in this one I can make the transformation say f of y 1 is equal to t 1 and f of y 2 is equal to t 2. So, then this will become d t 1 d t 2 and the range this will become minus infinity this will become 0 f of y 2 this will become t 2 this will become 0 to 1. So, this is becoming 0 to 1 0 to t 2 m factorial divided by j factorial l minus j factorial m minus l factorial here it will become t 1 to the power j t 2 minus t 1 to the power l minus j 1 minus t 2 to the power m minus l d t 1 d t 2. Now this can be simplified see you can do first time say something like t 1 is equal to u times t 2. Then if you do this then this is simplified this one will become equal to m factorial divided by j factorial l minus j factorial m minus l factorial integral 0 to 1 0 to 1 1 minus t 2 to the power m minus l t 2 to the power l plus 1 u to the power j 1 minus u to the power l minus j d u d t 2. So, this becomes simply a beta term and this is again becoming a beta term. So, all of this can be evaluated m factorial divided by j factorial l minus j factorial m minus l factorial this integral will give me j factorial l minus j factorial divided by l plus 1 plus 1 factorial and this integral will give us l plus 1 factorial m minus l factorial and m plus 2 factorial. So, you can see that these terms get cancelled out you are left with simply 1 by m plus 1 into m plus 2. Now the other case when j is greater than l will be similar. In fact, you can see here that this value is not dependent upon j and l here. So, only thing that we considered is j is less than l. So, if I consider j greater than l then in the expressions it will get reversed here it will become l factorial here it will become j minus l factorial and here it will become m minus j factorial that is all. So, all these things will be suitably interchanged, but the final value will still be the same. So, now the only other case that I need to consider is j is equal to l this case. If I consider this case then I am saying u i is equal to j by m and u k is also equal to j by m. Then this is equivalent to saying j of x s are less than or equal to y i. Now among i and k I have to assume something. So, it can be that y i is less than y k or y i can be greater than y k. So, let us take this and then m minus j of x s are greater than y k itself or it is the reverse of that. That is j of x s is less than or equal to y k is less than y i and then m minus j of x s they are greater than y i. So, basically you can do 2 times that thing. So, you can actually carry out the calculation it will become 2 by m plus 1 and 2 m plus 2 ok. Actually both the terms will have the same expression that is 1 by m plus 1 because basically what is happening? In between there is nothing, but since it was free from that choice it will be dependent only on this m here. So, we are able to derive the complete distribution or complete joint distribution of u i and u k. I have considered the case j less than l j greater than l will give the similar thing and j is equal to l. Next we consider say the moments of this. We consider the moment structure of u i's ok. So, what is expectation of u i for example, we derived the distribution of u i as simple discrete uniform distribution over the range 0 to m. So, when you consider the mean of this it is simply becoming equal to sigma j by m 1 by m plus 1 j is equal to 0 to m plus 1 sorry 0 to m. So, what is this term m into m plus 1 by 2 and this is m into m plus 1. So, the mean is half and you can of course, look at higher order moments also I am not going to write the full details here you can just check it. It will become equal to 2 m plus 1 by 6 m basically it is equal to 1 by 3 plus 1 by 6 m and therefore, variance of u i also you can see variance of u i is equal to 1 by 12 plus 1 by 6 m that is interesting. See if you remember the uniform 0 1 there the mean is 1 by 12 variance is 1 by 12 mean is half. So, here mean is still half, but the variance is 1 by 12 plus something 1 by 1 by 6 m. So, if m becomes large then this is approximately 1 by 12 ok. So, this is interesting thing to observe ok. So, basically what we are saying is that expectation of f m y i is equal to expectation of f of y i and expect variance of f m y i is actually greater than variance of f of y i, but variance of f m y i converges to variance of f of y i. So, this is the observation that we are having here based on these expressions that we derived here. We can also look at the covariance structure here. Let us look at for example, the product moment u i u k. So, that is equal to J L by m square 1 by m plus 1 into m plus 2. This is J is equal to 0 to m L is equal to 0 to m where J is not equal to L and when they are equal then it is becoming 2 times sigma J square by m square 1 by m plus 1 into m plus 2. Now, you can easily see that this denominator is common. So, I can keep it outside that is 1 by m square into m plus 1 into m plus 2 and we are left with summation J L J not equal to L plus summation J square which is actually coming out to be I can consider it as summation J whole square minus the cross product minus the terms which are J square here. So, this is then equal to 1 by m square m plus 1 m plus 2 that is equal to sigma J whole square plus sigma J square. I have made a mistake here this will become minus this here. So, after simplification this will turn out to be 1 by 4 plus 1 by 12 m. So, if I consider here covariance between u i and u k that will be equal to 1 by 12 m which actually goes to 0 as m tends to infinity and if I look at say correlation between u i and u k then that is equal to 1 by 12 m divided by m plus 2 by 12 m and this term is actually equal to 1 by m plus 2. So, that goes to 0 as m tends to infinity. That means, the amount of the correlation or correlatedness between i th and k th sample transformed value by of the order statistics that correlation becomes less and less as the sample size increases. Now, this is about the you can say sample analog of the order statistics of the any distribution. So, one we do with the original distribution function and second we do with the empirical distribution function. Now and we are actually able to analyze the distribution completely in this cases. Now on the right hand side in place of y i values if we consider ordered values then what will happen that is distribution of y i. So, that means, I consider u i's, u i's are actually f m of y i ok. In the case of f this was directly the order statistics from the uniform distribution and we were able to derive the distribution as a beta distribution, but here what it will give. So, let us consider this probability of u i is equal to j by m that is equal to probability of m times f m of y i is equal to j. So, that is equal to probability of m times f m y is equal to j given y i is equal to j d g y i y integral from minus infinity to infinity. Here the CDF of the ith order statistics from the second population is there and this one is of course, known here that is the binomial m c j f y to the power j f y to the power j 1 minus f y to the power m minus j and this distribution was known. So, if we substitute that there I get n factorial divided by i minus 1 factorial n minus i factorial g y to the power i minus 1 1 minus g y to the power n minus i. So, this is the d g y for j is equal to 0 1 to m because this is the ith order statistics from the second sample. So, that was a beta i n minus i plus 1 distribution for g. So, that we are able to get here. Let us so, this is the general form of the empirical distribution function transformed value of the ith order statistics from the second sample that is u y. So, this is a general expression. Now we can consider the particular case when g and f are the same then what happens f and g are same then this expression can be actually evaluated. So, I will show that thing m c j f y to the power j 1 minus f y to the power j 1 minus f y to the power m minus j n factorial divided by i minus 1 factorial n minus i factorial then this is becoming f y to the power i minus 1 1 minus f y to the power n minus i d f y for j is equal to 0 1 to m. Here all this powers get added up further you can substitute say f y is equal to something like u then this is becoming integral from 0 to 1 m c j then this is all this term will come here that is n factorial divided by i minus 1 factorial n minus i factorial. Then f y this power will get added up i plus j minus 1 and then 1 minus u to the power m plus n minus i minus j d u that is equal to m factorial divided by j factorial m minus j factorial n factorial i minus 1 factorial n minus i factorial then this is becoming i plus j factorial m plus n minus i minus j factorial and here it will become m plus n plus 1 factorial. So, many of the terms can get adjusted here and you can write it as I am not sure whether all the terms are ok here this is m plus sorry this will become m plus n factorial right because m plus n plus i plus j minus i minus j. So, this will get cancelled out minus 1 plus 1. So, it will become m plus n factorial right now I think this is correct. So, then if you have this then this is becoming m plus n minus i minus j c m minus j then you have i plus j plus 1 sorry minus 1. So, I made one more mistake here yes I am sorry this will become i plus j minus 1 factorial right now it is ok. So, i plus j minus 1 c j divided by m plus n c m which is actually hyper geometric distribution. So, this is quite interesting we are able to obtain the distribution of this transforms using the empirical distribution function. Let me show it again for the convenience. We got the u i as a discrete uniform distribution on the points 0, 1 to m and the corresponding ordered one that means, the empirical distribution function transform value of the i th order statistics that is y i is coming out to be a hyper geometric distribution here. That means, here the two samples they are added up actually m and n are the respective sample sizes. So, it is becoming m plus n where m or x's and y are n, n y i's are there. So, you can see here out of that if you are choosing m things here then it is becoming. So, it is dependent upon how many things I am saying i values that is j values of x's are less than or equal to y i that is i th one ok. So, that is simply playing a role here. So, this is very very interesting here and we can also talk about the joint distribution of u i and say u j kind of thing that is when we are considering two different things. So, let us look at this. So, these are leading to very very interesting observations and we will show later on that we will use it for the inference purpose. That means, when we do the testing about the locations for the two sample problems this will be really used here. So, let us consider now joint distribution of say two of them say u p and u q of course, I am taking the case when f and g are same and let us take say 0 p is less than q here. So, probability that u p is equal to j by m and u q is equal to l by m that is equal to probability of m times u p is equal to j m times u p q is equal to l. This will be equal to probability j of x s are less than or equal to y p l minus j of x s they are less than or equal to they are between actually y p and y q and m minus l of x s are greater than y q. So, that is equal to double integral probability of j of x s are less than or equal to s l minus j of x s are between s and t and m minus l of x s are greater than t d f y p y q s. Remember here I am not writing separately this. The reason is that the distributions of y p and y q are not independent unlike the case that I discussed a little earlier where the distributions were independent. So, there I was able to write separately you see here here I was able to write d f y 1 d f y 2 because this y i and y k they were taken to be independent, but here they are ordered. So, they are not independent. So, I had to write the joint one here, but this is not a problem because we actually know this. So, we can write minus infinity to infinity minus infinity to t this is m fact that factorial divided by j factorial l minus j factorial m minus l factorial then you are having the CDF things. So, f of s is to the power j f of t minus f of s to the power l minus j and 1 minus f of t to the power m minus l. Now, here you are to write the joint distribution of y p and y q, but that we know. So, we substitute here that is n factorial divided by p minus 1 factorial q minus p minus 1 factorial n minus q factorial f s to the power p minus 1 f t minus f s to the power q minus p minus 1 then 1 minus f t to the power n minus q d f s d f t where s is from minus infinity to t and s is from minus infinity to plus infinity. So, now for purpose of evaluation this can be transformed you can consider say f s is equal to u and f t is equal to v. In that case you can see here this will become 0 to v and this will become from 0 to 1 and all other things will be added up. So, the powers will get added. So, this can be evaluated in a closed form let me show you this here. So, this is equal to 0 to 1, 0 to v and all this coefficients will become 0 to v. So, this is coming here let me write it here m factorial n factorial divided by j factorial l minus j factorial m minus l factorial p minus 1 factorial q minus p minus 1 factorial n minus q factorial. So, all these terms will be coming there and then you have the powers let me write here. You will get u to the power that is f s f s to the power j plus p minus 1. So, that is becoming u to the power j plus p minus 1 then we have v minus u to the power l plus q minus j minus p minus 1. Then you have 1 minus v to the power m plus n minus p minus l minus q d u d v. So, this is actually a bivariate beta integral and this can be easily evaluated. In fact, I can show you the method of calculation. Let us put say u is equal to v w that is d u is equal to v times d w in the inner integral. In the inner integral if we put this then when u is equal to 0, w is 0 and when u is equal to v, w will become equal to 1. So, basically what is happening is that both are becoming beta integral 0 to 1 m factorial n factorial j factorial l minus j factorial m minus l factorial p minus 1 factorial q minus p minus 1 factorial n minus q factorial. Then you have u to the power j plus p. So, that is becoming v to the power j plus p minus 1 w to the power j plus p minus 1 and then 1 v is coming here also. So, I will put here this then here you get v minus v w. So, another v is coming out l plus q minus j minus p minus 1 and then you have here v w is coming out so it will become 1 minus w that is 1 minus w to the power l plus q plus minus j minus p minus 1 and then you have 1 minus v to the power m plus n minus l minus q d w d v. So, now, this is becoming m factorial n factorial divided by j factorial l minus j factorial m factorial m minus l factorial p minus 1 factorial q minus p minus 1 factorial n minus q factorial. Now, when you evaluate these beta integrals you have w to the power j plus p minus 1. So, you will get j plus p minus 1 factorial then 1 minus w to the power l plus q minus j minus p minus 1 factorial and then you add when you add this j plus p will get cancelled out you will get l plus q minus 1 factorial and now let us look at the powers of v and 1 minus v. So, for v it is l plus q minus 1. So, l plus q minus 1 factorial 1 minus v is m plus n minus l minus q then I will get m plus n factorial. So, here you can see some of the terms may get cancelled out. For example, I can see here l plus q minus 1 factorial and l plus q minus 1 factorial gets cancelled out here and the other terms we adjust here. So, this becomes j plus p minus 1 c j then this is m minus l plus n minus q c m minus l l minus j plus q minus p plus 1 c l minus j and this is m plus n c n where j and l are from 0 to m j is less than or equal to l. So, this is something like a bivariate hypergeometric distribution. So, again let us compare with the one when ordering was not taken then what we had we got the distribution as a bivariate discrete uniform. Distribution here this was the distribution of u i and u k here. So, I had got it as a bivariate discrete uniform and now you can see here that we are getting it as a bivariate hypergeometric. So, the comparison you can see you have discrete uniform well let us consider the sample thing firstly x 1 x 2 x n random sample from capital F then f of x 1 f of x 2 f of x n is a random sample from uniform 0 1. So, the corresponding if I take ordered observations then it becomes corresponding order statistics from a uniform random sample from 0 to 1. If I replace capital F by the empirical distribution function and I consider two things f and g that means, two samples x 1 x 2 x m and y 1 y 2 y n, but if I am taking f is equal to g then what interesting thing that I am observing that if I take unordered one that is f m y 1 f m y 2 f m y n then they are identically discrete uniform distributions on 0 1 by m 0 2 by m up to 1, but they are not necessarily independent. Now you consider again two of them suppose I consider x i x j then the corresponding thing for f of x i f of x j they are independent ok. If I consider f of x bracketed i f bracketed x j that is the ordered ones then they are order jointly distributed order statistics from uniform distribution and therefore, their distributions are derived as something like a bivariate beta kind of thing. If I am considering the empirical version of that in that case in the first case when I am taking unordered one I am obtaining a bivariate hyper geometric distribution bivariate discrete uniform distribution and now I am obtaining it as a bivariate hyper geometric distribution. So, bivariate beta bivariate discrete uniform and now a bivariate hyper geometric distribution that we are getting here. So, we have discussed in detail this applications of the empirical distribution function for some two sample cases. In the next class I will do a few more properties and then we will move over to the we will for example, look at the moment structure of this also and then we will look at the application to the testing problems ok. So, in the next class I will be trying to cover that.