 Yesterday I introduced you to the topic of distribution free methods or what we call as non-parametric statistics, non-parametric statistical inference. I mentioned that a primary quantity or you can say the primary variable or primary statistic of interest in non-parametric is the order statistics. That means, when we have the observations from the sample then we consider their ordered versions. So, they are called order statistics. Yesterday we have seen how to derive the distribution the joint distribution of the order statistics, distribution of one order statistics or the joint distribution of a couple of order statistics and a general methodology of taking any subset of the collection of order statistics how to derive the distribution of that. So, certainly as I mentioned yesterday that in various problems we are interested in different kind of order statistics. In particular we may be interested in the minimum of the observations or the maximum of the observations or the middle of the observations or a particular position. To give you an example some examples of distributions of say maximum, minimum etcetera. Let us consider say first example, let us take say x 1, x 2, x n follow a exponential distribution with parameters. Let me take the standard exponential distribution lambda. That means, I am considering the density function f x is equal to lambda e to the power minus lambda x, where x is greater than 0, lambda is greater than 0, 0 otherwise. Let us consider the distribution of say here let us consider say CDF. So, what is the CDF here? CDF here is equal to 1 minus e to the power minus lambda x for x positive and of course, it is 0 for x less than or equal to 0. If I consider say minimum of the observations that is x 1, then yesterday we have seen the distribution of the minimum can be written separately. It can be obtained in a simple way as n times 1 minus f y 1 to the power n minus 1 f y 1 if the common density is small f, where capital F is the CDF. So, if I apply this formula here, then this will have density it will be equal to n times 1 minus f x. So, this will become e to the power minus lambda y 1 to the power n minus 1 into lambda e to the power minus lambda y 1. Now, you can see here this powers will get added up. So, you are getting the density of this minimum as n lambda e to the power minus n lambda y 1. Now, this is interesting because it is of the same form. So, what you are getting that is if I am having x i is exponential lambda, then this is following exponential n lambda. So, a generalization of this could be that if I consider say x 1, x 2, x n following exponential mu lambda, then you will have x 1 following exponential mu n lambda. So, this is interesting the distribution of the minimum when we are considering the observations from a exponential distribution is again an exponentially distributed random variable. Let us take some more interesting example. Let us take say x 1, x 2, x n follow Pareto distribution with parameters say alpha and beta. That means, I am considering the density as say beta alpha to the power beta divided by x to the power beta plus 1 x greater than or equal to alpha, where alpha and beta are positive. So, I am considering this density as the Pareto density and let us consider here c d f. So, c d f here is equal to alpha to t. So, let me take 1 minus c d f here. So, that will become equal to x 2 infinity beta alpha to the power beta divided by t to the power beta plus 1 d t. So, this then simply becomes if I integrate t to the power minus beta minus 1. So, I will get t to the power minus beta divided by minus beta minus beta minus beta will cancel out and then t to the power minus beta will give me x. So, that means, I will get alpha by x to the power beta for x greater than or equal to alpha. Therefore, if I consider now the density function of the smallest then it is n times 1 minus f of y 1 to the power n minus 1 into f of y 1. So, this is equal to n times and you consider alpha by x to the power sorry y 1 to the power n minus 1 beta and then you have the density here that is beta alpha to the power beta divided by y 1 to the power beta plus 1. So, if you combine the terms you will get n beta alpha to the power n beta divided by y 1 to the power n beta plus 1 for y 1 greater than alpha. So, you compare this with this density beta has changed to n beta alpha remains the same that is we are saying x 1 follows Pareto alpha n beta. So, this is another interesting example where the distribution of the order statistics is in the same family of distributions. Let me give one example of say maximum here say x 1, x 2, x n follows say uniform 0 theta and let us consider the distribution of x n. Then by the same logic here the density function of x will be 1 by theta between 0 to theta. So, the cdf will be x by theta and if I apply this formula for the maximum here which is derived yesterday as n f to the power n minus 1 into f of y n. If I apply this formula then I will get the density of this as n y n to the power n minus 1 divided by theta to the power n 0 less than y less than or equal to theta. So, these are some of the densities where the forms of this distributions of the order statistics can be derived in a simple fashion or you can say in a closed form sometimes they are recognizable sometimes they may be in a different form also. Now, I mentioned yesterday that one of the particular order statistics that will be in use or useful is the median, median of the sampled observations. So, we also may be interested in the distribution of the sample median. Let us look at that distribution of the sample median. So, in fact, I should also mention that what is the further use of this properties. For example, here if I have shown that if x 1, x 2, x n follows exponential lambda then x 1 follows exponential n lambda. Now, if we are interested in the averages here then the average or you can say if they are denoting the lifetimes then the average life here is 1 by lambda for each of the observations, but if I am looking at the minimum then the average life is becoming 1 by n times lambda. That means, it is 1 by n th of the original lifetime. Now, this can be an eye opener because in the say in a certain industrial situation if we are interested to use any of the components then it is all right we can say that the average life is say 1 by lambda. Suppose, I am saying lambda is 6 months. So, if I put say 1 by 6 say if I put for example, lambda is equal to half suppose we are measuring in the years then it is becoming 2 years, but if I have say 10 observations and then I consider the one which fails the first for example, it may be a series system. If it is a series system then the life of the minimum will determine the whole thing and here it will become 1 by n lambda that means, it will become 2 by 10 that means, it is only. So, it is coming a few months only because 24 by 10 if I put it is turning out to be only 2 months 2.4 months which is much smaller which you are thinking that it will be 2 years, but it is not so. So, actual lifetime of the entire system will become very small compared to in a similar way you can think of the Pareto. Here this is beta here and here it has become n beta. So, in certain case when you are using beta as a parameter then it is becoming actually n times. So, that is the difference if we are making use of the order statistics. These facts provide interesting information about the nature of the means are. So, basically various characteristics of this order statistics we are able to find out. So, we will explore this further that is why the distributions and means where n says and other things asymptotic distribution etcetera will be of much interest for order statistics. So, let us now look at the distribution of the sample median. So, if the sample size n is odd. So, odd means generally we put say something like 2 k plus 1 then the sample median is the middle of the observation. Let me call it median as m then that is equal to x k plus 1. So, we have already derived the form of a general order statistics which is of the rth term. So, let me just show it again where we will substitute the value. So, if you look at this term here the distribution of the rth order statistics is obtained as n factorial divided by r minus 1 factorial n minus r factorial f to the power r minus 1 and 1 minus f to the power n minus r multiplied by small f where capital F is the CDF. So, now if I take r is equal to m plus 1 here where n is 2 r is equal to k plus 1 where n is equal to 2 k plus 1. So, let us substitute here that will give me the distribution of the sample median for this particular case. So, the pdf is let me call it x f m and I will use the notation is small m is sometimes used for the sample size etc. So, let me not confuse let me use it something like say u here then it is equal to 2 k plus 1 factorial r minus 1 that will give me k factorial then n minus r will again give me k factorial. So, it is k factorial square f of u to the power m 1 minus f to the power m small f of u. So, you are able to obtain the distribution of the sample median if it is if the sample size is odd it is much simpler here. You can also look at a special case that is say f is that means, original distributions are uniform distribution on the interval 0 to 1. In that case this will become 2 k plus 1 factorial by k factorial square u to the power m 1 minus u to the power m 0 less than u less than 1. Now if you are looking at this one this is nothing but a beta m plus 1 m plus 1 distribution. So, we know the moment structure and other things also for a beta distribution for example, the mean of a mean of this it becomes simply half because it is alpha by alpha plus beta and similarly other moments of this can be obtained. Let us take the other case let us consider the case when n is even say n is equal to 2 k then the median is equal to median is x k plus x k plus 1 by 2. Now if that is so I firstly use the joint distribution of x k and x k plus 1 and again let us go back to the formula which I derived for the joint distribution of rth and sth order statistics that was obtained as n factorial by r minus 1 factorial, n minus s factorial, s minus r minus 1 factorial, f to the power r minus 1 at y r 1 minus f at y s is to the power n minus s then f of y s minus f of y r to the power s minus r minus 1 multiplied by the densities at y r and y s. So, now I will take r to be k and s to be k plus 1. So, if we do that then what do we get? The joint probability density function of x k and x k plus 1 this is given by f of use a notation y k y k plus 1 points that is equal to so it is 2 k factorial that is n, then you have r minus 1 that is becoming k minus 1 factorial, then you have k plus 1 minus k minus 1 that is simply becoming 0. So, 0 factorial that we take as 1 and then you have 2 k minus k plus 1 that is again k minus 1 factorial. So, it becomes 2 k factorial divided by k minus 1 factorial square, then I get f of y k to the power r minus 1 then again the term which is corresponding to f of y k plus 1 minus f of y k to the power s minus r minus 1. So, since s minus r minus 1 is becoming equal to 0 therefore, that power and that term will give me simply 1. So, I will get f of y k plus 1 to the power k minus 1 and then I get f of y k f of y k plus 1 and here of course, I have to write y m y k is less than y k plus 1. In order to derive the distribution of m that is half of this I have to define a transformation. So, let me write m is equal to half of x k plus x k plus 1 and I define another variable let me call it say v that is equal to x k plus 1. So, the inverse transformation we consider the inverse transformation is this will turn out to be x k plus 1. So, basically you are having x k plus 1 is equal to v. So, if I put it here I get 2 m minus v. So, the Jacobian of the transformation. So, that will give me 2 minus 1 0 1. So, that is becoming equal to 2. So, if I look at the joint pdf of u and v then the joint probability density function of m and v that is f of 2 k factorial divided by k minus 1 factorial square and it will be multiplied by 2 because the Jacobian was 2 then I get f of y k is equal to 2 m minus v to the power k minus 1 1 minus f of v to the power k minus 1 f of 2 m minus v f of v and you are having the region here x k is less than x k plus 1 that will give me 2 m minus v less than v that is m is less than v. So, m is less than v that region I will get. So, the marginal pdf of m is obtained from integrating the joint density divided by k minus 1 factorial square f of 2 m minus v to the power k minus 1 1 minus f v to the power k minus 1 f of 2 m minus v f of v d v and v is integrated from m to infinity. So, this gives me the formula for the derivation of the distribution of the sample median when the sample size is even. Again as I saw the case for the uniform distribution when it was odd we are getting simply a beta distribution. Let us see what we get the original distribution is say uniform 0 1 then this will become equal to 2 2 k factorial divided by k minus 1 factorial square m to infinity. Now, this will go only up to when I am considering the uniform distribution the range of the random variables original random variables is from 0 to 1. So, if I consider this then I will get here see 0 less than say y k less than y k plus 1 this is less than 1. So, this is same as saying 0 less than 2 m minus v less than v less than 1 which is same as saying u is less than v that is less than minimum of 2 u and 1. So, if I consider the region here of integration see this is 1 if I consider this line as u m is equal to v this is m here and this is m. So, m is equal to v and I consider m is equal to 2 v then we are actually getting this region here basically on this side I am having m and on this side I am showing v here. So, this is then becoming equal to m to minimum of 2 m 1 2 u minus sorry 2 m minus v to the power k minus 1 1 minus v to the power k minus 1 d v. So, when you look at this integration I will get 2 parts here this will become equal to 2 times 2 k factorial divided by k minus 1 factorial square integral from m to 2 m to m minus v to the power k minus 1 1 minus v to the power k minus 1 d v where m is between v to the power k minus 1 d v where m is between 0 and half and it is equal to 2 times 2 k factorial divided by k minus 1 factorial square m to 1 2 m minus v to the power k minus 1 1 minus v to the power k minus 1 1 minus v to the power k minus 1 d v if m is between half and 1. For various values of k for example, if I consider k is equal to 5 or k is equal to here k is even. So, suppose I take k is equal to 4 or k is equal to 2 then this expressions can be directly evaluated otherwise also they can be evaluated, but you have to do several integration by parts here. The distribution of the sample median we are able to derive for both the cases when the sample size is odd then in that case it is a straight forward form, but in the case of even it is in the form of an integral, but that integral may be evaluated for a specific choices of the distributions. In every case it may not be possible to do. For example, if I consider exponential distribution then I can then one of them will become a proper term and then the other one will become a finite expansion which can be easily done and we can really evaluate. Similarly, if I consider say Pareto distribution etcetera, so this can be obtained, but in some other distributions say normal etcetera this may be much more complicated. Another quantity of interest in the case of non-parametric statistics is the range of the observations. That means, the difference between the maximum and the minimum. So, in the real life situation also you can see the range is generally taken as an important quantity and therefore, the distribution of the range is also of quite importance. When we are considering the parametric situation then we generally consider the variance or the standard deviation, but in case of non-parametric when we do not have the exact form of the CDF or the PDF it is difficult to study that. Therefore, a more useful quantity would be the distribution of the range. So, we look at that thing now. So, range is defined as xn let me call it capital R. So, xn minus x1. So, we look at the distribution of that. Now, once again if you look at this we will require the distribution of the minimum the joint distribution of the minimum and the maximum. We have already derived the separate distributions of the minimum and the maximum, but that will not be helpful. So, we again make use of the formula for the joint distribution of the order statistics yr and ys. So, here you take r is equal to 1 and s is equal to n. Now, that has some advantage here if you take this as 1 then this factorial will vanish this factorial will vanish because this will be giving you simply n minus 1 and here you will get sorry n minus n that is becoming 0 1 minus 1 that is becoming 0. So, you will get only n minus 2 here. So, this term can be simplified here and then we can write this in a closed form as the joint probability density function of x1 and xn that is equal to n into n minus 1 f of yn minus f of y1 to the power n minus 2 f of y1 into f of yn and certainly you will have that y1 is less than yn. In order to obtain the distribution of this we define the transformation r is equal to yn minus y1 and then you take say some new variable say s is equal to say yn or y1 for example. So, if you take the reverse of this you will get y1 is equal to s minus r and yn is equal to s. So, if I consider the Jacobian I will get minus 1 1 0 1. So, that is equal to minus 1. So, if I take the absolute value of this that is becoming equal to 1. So, we can easily write down the joint distribution of r and s now. Therefore, the joint pdf of r and s n into n minus 1 f of yn that is s minus f of y1 that is s minus r to the power n minus 2 f of s minus r f of s. Also you look at the region here s minus r is less than s. So, r is greater than basically 0 which is true r will be greater than 0. So, because you are subtracting the maximum minus the minimum and s is anything that is means it is having the full region. So, the desired pdf of r is obtained as n into n minus 1 integral and when you integrate with respect to s this will be from minus infinity to infinity f of s minus f of s minus r to the power n minus 2 f of s minus r f s ds. One important thing that you notice in the distribution of the sample median when the sample size was odd it was quite difficult to write down the integral form there, but in this case since the range is the full the derivation of the distribution of the sample range will not be that much complicated. In fact, we can see for example, if I consider say uniform distribution suppose we have uniform 0 1 then this is becoming s minus s minus r which is nothing, but simply becoming r to the power n minus 2 and this is simply 1. So, this is and then the integral will become from r to 1 because s is actually here you will have 0 less than y 1 less than y n less than 1. So, this will give me 0 less than s minus r less than s less than 1 because both are positive. So, this region will come here now this will give me 0 less than r less than s less than 1. So, there is a small error here because if I put the lower limit here we are actually getting here r is less than s that is coming here and at the same time from the second one r is greater than 0 also. So, when we do this one this should be from r. So, this was an error here. So, if we do that in the case of uniform then this is becoming n into n minus 1 r to 1 and this will become r to the power n minus 2 d s. So, that is becoming n into n minus 1 r to the power n minus 2 1 minus r 0 less than r less than 1 which is again simply a beta distribution. Actually, this is a beta n minus 1 2 distribution. In the general the distribution of the sample range is not so, straightforward. However, statisticians like Hartley etcetera they have tabulated the cumulative CDF of this for the normal population for n like 10 11 up to 20 and the asymptotic distribution has been studied by Gumbel in 1944. So, friends what we have done is we have discussed the distribution of the order statistics some special order statistics and certain special functions like median or the range etcetera. In a similar way we can also study say quantile for example, a particular position what is the distribution of x 4 or the one which is coming at the 20 say 1 by 4 position n by 4 or 3 n by 4 or 1 by 5 or 4 by 5 and so on. Different positions can be the distributions of all of them can be obtained sometimes they will be unique sometimes they will be average of the two values, but in all the cases the distributions can be derived. Now, next thing is that we can talk about the moments of the order statistics like we discussed the moments of the random variables. Now, when we have the distribution of the order statistics what happens to that? Now, why is that important as I mentioned in one of the examples of exponential distribution each of the x i's had the mean 1 by lambda, but the mean of x 1 that is the minimum was becoming 1 by n lambda that is 1 by n th of the original value. So, that is a drastic change. So, in general we will be interested in the distributions of the or the moments of the order statistics. So, in the next section I study this. So, we consider moments of order statistics. Now, you can see here that the general distribution of the order statistics is complicated as you have seen here the form is like this. So, if I consider say for example, what is expectation of y r to the power k then you cannot say anything about this except that this is an integral y r to the power k into all these things from minus infinity to infinity certainly the evaluation of this integral is not possible except that you may consider some sort of special cases etcetera. So, that part we will see that we will consider approximations of this expression. Another thing is that if I actually take a specific form for example, if I am taking exponential distribution or a Pareto distribution etcetera then in some cases the forms can be written. I am doing the simplest one which is the uniform distribution and uniform distribution is of course, quite important as we have seen that if original random variables are from any distribution then f of that becomes from uniform distribution. I actually started the topic of non parametric methods with that. So, if that is so then and since capital F is an increasing function therefore, it makes sense to consider the order statistics of uniform random variables. So, if I do that then I will show you that the forms can be easily calculated because we have derived the form of the distribution of order statistics of the uniform distribution as a beta distribution. Now, moments of beta distribution are known so let us consider when the sample is from say uniform 0 1. So, in this case f of y r that is the rth order statistics it was given by basically it was n factorial divided by r minus 1 factorial n minus r factorial y r to the power r minus 1 1 minus y r to the power n minus r 0 less than y r less than 1. So, if I consider say mu k prime that is expectation of y r to the power k then it is equal to n factorial divided by r minus 1 factorial n minus r factorial integral 0 to 1 y r to the power k plus r minus 1 1 minus y r to the power n minus r d y r which you can again recognize as a beta. So, this is n factorial divided by r minus 1 factorial n minus r factorial. So, this will actually give me beta of k plus r and n minus r plus 1. So, we expand this it becomes n factorial divided by r minus 1 factorial of course, you can also say it as beta k plus r n minus r plus 1 divided by beta r n minus r plus 1. So, it is simply a ratio of 2 beta, but if you want expressions in the form of fractions then we can write it as gamma k plus r that is k plus r minus 1 factorial then gamma n minus r plus 1 that is n minus r factorial and divided by k plus n and plus 1. So, it is becoming k plus n factorial. Now, this you can expand in very systematic way like this you can cancel out this term you can cancel out and sorry this will get cancelled out. So, you will get here it is equal to I adjust this term with this and this with this. So, I will get k plus r minus 1 and so on k plus r minus 2 and so on up to r plus 1 r divided by n plus k n plus k minus 1 up to n plus 2 n plus 1. In particular if I consider mu 1 prime that is expectation of x 1 sorry x r that is expectation of y r then that is becoming r by n plus 1. Similarly, you can consider expectation of x r square that will become r into r plus 1 divided by n plus 1 into n plus 2. So, you can also talk about the variance here the variance of x r that is turning out to be r into n minus r plus 1 divided by n plus 1 square into n plus 2. So, this is quite interesting here we are able to obtain the moments of the order statistics from the uniform distribution. Now, as before if I consider the joint distribution between r th and s th then from uniform what I will get that I will get from here by substituting the f y r as y r and 1 minus f y as 1 minus y s and this will become y s minus y r. So, for uniform distribution the form that I will obtain will be similarly the joint p d f of say y r and y s are less than s from uniform 0 1. Then that will be f of y r y s that is equal to n factorial r minus 1 factorial s minus r minus 1 factorial n minus s factorial y r to the power r minus 1 y s minus y r to the power s minus r minus 1 1 minus y s to the power n minus s 0 less than y r less than y s less than 1. If I want to find out say correlation between y r and y s in that case I need the product moment also. We have already obtained the form of the expectation of x r and expectation of x s. So, similarly I can calculate the product moment. So, for example, what will be expectation of y r y s. Now, in order to evaluate this kind of integrals we can make certain transformations here. So, I will not get too much into the detailed calculations see this term will be there. You will have actually double integral from say for example, if you firstly do with respect to y r then it will be from 0 to y s and then it will be from 0 to 1 all these terms d y r d y s. So, in the first one you make the transformation y s minus y r is equal to something then or basically you make it as t y r is equal to t times y s then that thing will come out you will get 1 minus t to the power something and here it will become t to the power something. Then this will become t will be from 0 to 1 then it becomes a beta function that can be evaluated. At the next stage you can evaluate the integral whatever term you are getting as a beta function in y s. So, I am not writing all the calculations here by using all these transformations it turns out that this term will be equal to r into s plus 1 divided by n plus 1 into n plus 2. Actually it is instructive to compare it with this term we had calculated expectation of x r square as r into r plus 1 by n plus 1 into n plus 2. So, here you can see this r has been replaced by s here. So, based on this if I calculate the covariance between y r and y s then this is turning out to be r into n minus s plus 1 divided by n plus 1 square into n plus 2. So now, we can write down the formula for the correlation coefficient between y r and y s let me call it rho r s. So, that will be equal to r into n minus s plus 1 divided by square root of r into n minus r plus 1 square root s into n minus s plus 1. So, that will be equal to this can be simplified as r into n minus s plus 1 divided by s into n minus r plus 1 whole to the power half. So, if r is less than s then this is the form that you are getting r is less than s. So, this is less than 1 and n minus s plus 1 will be less than n minus r plus 1. Another point which you may note down that this term will be positive. So, order statistics are always positively correlated which is understandable because they are in the same direction here. Another thing is that let us consider the minimum and the maximum. In particular what will be say if I consider correlation between say minimum and the maximum that is rho 1 n what it will be equal to. So, r is equal to 1 n minus n plus 1 that is 1 divided by n and n minus 1 plus 1. So, it is becoming 1 by n square to the power half. So, this is very interesting as the distance between the minimum and maximum increases. So, the correlation is actually inversely proportional to that. That means, it is simply inversely proportional to the length of this. So, actually this is going to 0 as n tends to infinity. This is a very interesting observation that is the correlation between the minimum and the maximum is inversely proportional to the sample size. So, you may also think it like this that if I have too many observations then the relationship between the minimum and the maximum will be of very mild nature. That means, they may become almost uncorrelated that means how they will vary, but if the sample size is small. So, suppose I say 4, 5 etcetera then it is still having some significance. Of course, for samples of size bigger than 20 or something like that this will have a very minor value because suppose I say n is equal to 20 then I get only 5 or 0.05 basically. Usually we have this like variance of a summation is equal to sum of the variances if the correlations are 0. That means, if the random variables are uncorrelated. Now certainly your random variables say which are now I am considering order statistics they are correlated. Therefore, if I consider say variance of say x 1 plus variance of x 2 etcetera then it will not be same as variance of x 1 plus variance of x 2. Now just as an example you can consider say variance of suppose n is equal to 2 then what is variance of x 1 that will be equal to let us consider this r by n plus 1. So, if I take n is equal to 2 then this is simply becoming 1 by 3 and what is variance of x 2 that will become s by n plus 1. So, that is equal to 2 by 3 because if I am taking 2. So, this will become 2 by 3 and what will be the covariance between x 1 and x n x 2. So, from the formula that we have derived just now here I am putting r is equal to 1 and 2 minus 2 plus 1 divided by n is equal to 2. So, this is 3 square into 4. So, this is becoming 1 by 36. So, if I look at say what is variance of say x 1 plus x 2 that is equal to variance of x 1 plus variance of x 2 plus twice covariance between x 1 x 2 that is equal to 1 by 3 plus 2 by 3 plus 1 by 18 that is equal to say 19 by 18 which is certainly in this case greater than variance of x 1 plus variance of x 2 here. This is actually equal to 1 here because the variance of x 1 is 1 by 3 and variance of x 2 is 2 by 3. Another thing which may look somewhat interesting see here you have derived the exact expressions for the I think I made some mistakes here this is actually not variance this is expectation variance will become equal to 1 and then this will be 2 minus 1 plus 1. So, that will become 2 here and this will become 2 by and 3 square into 4 that is 36. So, that is 1 by 18 here and this one will become corresponding to 2 then this is becoming 2 minus 2 plus 1 that is 2 by 36 that is equal to 1 by 18. So, I made a mistake here then this is becoming equal to 1 by 9 plus 1 by 18. So, that is equal to 1 by 6 which is obviously greater than 1 by 9 that is equal to variance of x 1 plus variance of x 2. So, this is a mistake here, but anyway now the fresh calculations show the exact values here. You also look at this the expectation of y r is directly dependent upon the positioning among the order statistics. So, if I increase r this is increasing. So, this is another interesting fact here that this term is strictly increasing function of r here. So, let me just mention this remarks here expectation of x r that is equal to r by n plus 1 is increasing in r. But if I consider the variance here that is not showing the same thing because it is r into n minus r plus 1 divided by some n plus 1 square into n plus 2. So, this term does not show the similar behavior actually it will be maximized for r equal to say n by 2 kind of thing whereas, if I take minimum or the maximum that like if I consider what is the variance of x 1 here that will give me here n divided by n plus 1 square into n plus 2. And similarly if I consider what is the variance of x n then this is again becoming equal to n divided by n plus 1 square into n plus 2. So, this is a symmetric function here this is same this function is becoming equal here ok. So, expectation increases, but the variability increases and then decreases here. So, that is the way it will come here. In the next part I will consider general order statistics right now I have considered the order statistics from the uniform distribution where we are able to obtain certain exact expressions for the moments of the order statistics. In fact, if you see I can in fact, write down even the third order moment fourth order moment and the third order and fourth order central moment. So, we can actually study the measures of skewness and kurtosis also, but if I am considering the general f here in the general f none of these calculations can be done unless I have the explicit form of the capital F function there. Even if that is available in many times for example, you consider simple normal distribution then capital F is nothing, but the capital phi function where capital phi is the cdf of the standard normal distribution. So, if I have capital phi to the power something and 1 minus capital phi to the power something and then small phi certainly this type of integrals are not easy to evaluate because I will be multiplying by x to the power k there. So, if I even if I make the transformation phi x is equal to something the distributional part will become in the form of uniform, but then this x will become phi inverse of something. So, I would not be able to evaluate that. In such cases it is useful to have certain tools which can give approximate expressions or the bounds for the moments of order statistics. So, in the next lecture I will be discussing the bounds on the expectations of order statistics.