 Let us quickly and go through it is a proof. I just want to go through the proof because if that is the constant, then it will be always 40 of x equal to 20 of y. It is an if and not. So, how you have to interpret? Ok, let us say our claim, final claim is here what? Whether t is a minimum sufficient statistics or all. This is our final claim. Now, what we are checking is for any pair t of x y is this constant is this ratio is going to be constant if and only if t of x equals to t y. If this is the case, then we are going to talk about t is minimum sufficient statistics. So, here basically you are going to check some condition right. Yes, this check is and that checking condition itself has if and only if in it ok and if that ratio is going to constant that means. So, first of all that ratio is constant if t of x equals to t y and on the other hand if t x equals to t y that ratio should be constant. If that is happening, then we are going to say that. So, in a way you are right like this is kind of giving a sufficient conditions in terms of another condition which is necessary and I mean which has both forward and reverse direction. It itself is not like a complete characterization it is basically giving a condition sufficient condition under which t is a minimal sufficient statistic, but what is that condition? It has implication both forward and reverse direction ok. All of you see this point see I am not saying this I am not saying if and only if here I am only saying then that means if this condition holds then t is a minimal sufficient statistics that condition holds means basically I am giving a sufficient condition and that sufficient condition itself is characterized in terms of both forward and reverse condition ok. Now, let us understand this ok this is the first step you give me a statistic t what I did is I construct a partition. What I do is I look into all t such that t of x equals to t for some x right. What I did is this was my space like last time maybe let us take this at my 3 all these points here they mapped into let us say t 2 all this mapped into t 1 and let us call all these points here mapped into t 3. So, this is my image right. So, t 1 t 2 t 3 are my image of my space under statistic t let us call this as capital tau. And now partition we know that this already defines a partition right you take one point t here then I know then I have t equals to set of all points such that t of x goes to t. So, for example, you give me t 1 then that set is all the points which are mapping to that t 1. Now I am going to so, in this case I have 3 ranges right a t 1 a t 2 what I am going to do is I am going to select 3 point from each one of this what are these 3 points let us call them x t 1 x t 2 x t 3 ok. So, x t 1 is coming from here x 2 2 is coming from here and x t 3 is coming from here suppose if I take x of t 1 my claim is this is equals to t 1 all of you agree with this because every point here this x t 1 is coming from this space right every this point when I applied t capital this has to give me. So, these are let us say these are I am fixing these 3 points let us take them to be 3 points. Now what we will do is take any point x somewhere in this one of this I do not know what it is and I know that this point I can pair like with its let us call this some point of representative of each of this set. Suppose if x comes from x t 1 then I am going to associated with this x t 1 if this x come from this region I am going to associated with x t 2 like that. So, for each x I can associate this with some some t let us simply call it some t which is one of these points right you give me any x I am able to find what which which group it belongs to and find out the representative point if from there and I can associate like this can I do this right ok. Now, let us see. So, on this point I know that t of x equals to t of x t you agree. So, x is a point and I have paired it with its representative point from that set. So, I know that if x and x t both belong to the same set if that is the case t will be assigned the same value to both of them ok. Now, I want to see given a point x and theta what I will do is I am going to divide this guy by x of x t the representative point there theta divided by f of x t by theta. So, what now if I want to apply the factorization theorem here to conclude that this is a sufficient statistic what I want to do here is I want to write it in the form of g of t given t and h of theta like this ok. So, now if this t is a sufficient statistics I need to write it like this and now what I am going to assume I am going to assume that this guy is a constant because t of x equals to t x t. So, if that is the case can I say that this factor here is a constant in the sense it does not depend on theta. Can I take that as my h of x if it does not depend on theta it better depend on x quantity right we have and this quantity here now it depends through t and now I can take it as g of t given theta. Now, if t of x equals to t y what I have just argued is this ratio is a constant and I mean yeah this sorry I am use the fact that if this t of x equals to t y this is a constant under that condition and I have just demonstrated that if that is the case t is a sufficient statistic ok. Right now all I have done is I have established the fact that t is a sufficient statistic. Now, let us use the other fact what we will do is we will now assume that this is a constant ok. Now, I have partitioned in such a way that you give me a point x you you give me a point x and I came up with an another pair with it I mean I came with another pair another point to pair with this. Now, I know that these two pair will be such that t of x equals to t of x t ok. Now, under this this is this ratio is a constant if and only if right now I have point x and y for which this is same because of this I can now this is I am assuming that this is a constant that is what exactly I am doing this. So, I have written f of x theta here I need to show that this pdf factorizes into g and h function. So, to get that I have divided and multiply them by f of x t and now I am exploiting the fact that this ratio when I look into the pair x and x t that does not depend on theta that is when I use the fact and then it is ok. Now, what I want to do is all we said that when this under this condition I am able to demonstrate that if I have a t which is satisfying this I have shown that t is a sufficient statistic. Now, let us say how to show that sufficient statistics is going to be a minimal sufficient statistic now. So, now, let us take another sufficient statistic t prime and we know that if t prime is a sufficient statistics f of x theta I should be able to find a h prime into g prime t given theta right. Is this my factorization theorem correctly applied here? I have simply applied my factorization theorem here. Now, what I will do is assume now I am going to going to assume this holds let us say t prime equals to t prime on this sufficient statistics t prime and now I am going to apply my factorization theorem and see how does this behave x of theta divided by f of y of theta and I know that this is equals to under this h prime of x g prime of t prime of x given theta divided by h prime of y g prime of t prime of y given theta, but now this points x and y are such that t prime of x equals to t prime of y. So, this quantities this quantity are the same g prime of t of x and g prime of t prime of y because both t prime of x and t prime of y are the same. So, knock of it. Now, it is now this ratio is h prime of x h prime does not depend on theta and I have h prime of y which also does not depend on theta. Now, can I argue that this ratio does not depend on theta and we know that if this is the case, now we have just demonstrated that this ratio is a constant does not that depend. So, this should imply t of x equals to t of y on these two points. So, we just demonstrated that if they exist are sufficient a sufficient statistic. So, first of all we argued that if there is a statistics which satisfies this condition in the forward direction t is a sufficient statistic and I said that if there is another sufficient statistics for which this ratio holds I just argued that this ratio does not depend on theta. And now if this ratio does not depend on theta we have this condition t of x equals to t of y this implies now what we started with this and we argued that this implies t of x equals to t of y. So, if t prime of x equals to t prime of y we are arguing that t of x equals to t of y what does this mean? t is a minimal sufficient statistics ok. So, do not get lost in this that is what I did not discuss this yesterday, but go back and you just see that like how we are exploiting the sufficient sorry factorization theorems here in deriving all these properties. Let us quickly discuss this application now. Suppose I have some samples drawn from Gaussian distribution and where both mu and sigma square is unknown and what is one possible sufficient statistics for this? Sample mean and sample variance. Let us say I have two points x and y for which I have this sample mean and sample variance and another ok this should have been y here y and s y. And now let us see if I have this two I have this sufficient statistics which is computed on the two points x and y and let us compute this ratio. If I take this ratio we know how to write the joint pdf in the case of Gaussian distribution here. You can write this expression and you will see that the only way this guy you can make it independent of mu and sigma square by setting x bar equals to y bar and s x square equals to s y square. If they are not equal there is no way you can make it independent of mu and sigma square ok. So, this ratio becomes independent only when x bar equals to 1 that is your t of x the first part is sorry t 1 equals to y and the second part is also same on those two points. So, here t 1 is your x bar and the t 2 are sample variance ok. Now, one more quick example we will talk about uniform minimal statistic. So, this instead of Gaussian now you will be looking into uniform ok. Let us say I have a uniform distribution, but the parameter is theta and theta plus 1. Now, I have this uniform distribution, but now I am looking into uniform the interval length is 1, but it is starting at theta and ending at theta plus 1 ok. Now, I can write the joint pdf as it is going to be 1 if all the samples are between theta and theta plus 1. If any of the sample is outside this theta then the joint pdf is going to be 0 right that is obvious. Now, I can manipulate this if I have to take all the samples right and take max value of them do you think the max value is going to be outside theta plus 1 no it has to be below right. So, that is why this max value of x i is going to be less than or equals to theta plus 1. So, this max value has to be less than theta plus 1 and similarly if you take the minimum value of all the samples, can this be smaller than theta? No, it has to be greater than theta. So, we can write theta to be minimum these samples. Now, you can check that if you take the ratio of this joint pdf at point x and y they will be constant if and only if these two quantities are the same at points x and y. So, this will give me a hint about what could be the possible minimum sufficient status for sticks for this uniform distribution. It says that if you take t of x to equals to minimum of this maximum of that that is what is minimum of x i we called it as first order statistic and max value as that the nth order statistics right that is the last one then that becomes your minimal sufficient statistics and you can also check that instead of looking into the minimal maximal you can look into the difference between these two that is x n minus x 1 and their average of the maximum minimum value that is also minimal sufficient statistics. So, I am saying that this is also sufficient statistic this is also sufficient statistic. I am giving you two sufficient statistics for uniform distribution then minimum then what does it imply can minimal sufficient statistics be unique? No. No. So, there could be multiple sufficient statistics and there could be multiple minimal sufficient statistics ok. This last one slide we will cover about this statistics and this will conclude our discussion on sufficiency principle. So, we just said that all our focus so far have been constructing a statistics which constraints all the necessary information about the parameter of interest, but here ancillary statistic something which is not providing information about the parameter of my interest, but something compliment about that ok. So, our statistics S of x whose distribution does not depend on the parameter is called sufficient sorry ancillary sufficient statistic. So, if I just to give you ancillary sufficient statistics it does not provide maybe any information about your parameter theta, but when it is used in conjunction with other statistics maybe it will provide more information ok. Just to give you a quick example I said ok what does x n minus x 1 give you range, but if I just to give you a range does it tell you information about theta here? So, here let us say theta equals to 1 and then it becomes 2 I could take theta equals to 2. So, in one case the range is 1 to 2 and in other case range is 2 to 3. So, the first one is telling you just the range information though sample could be coming from here or here, but on the other hand what this is giving you x n what this part is giving you average. So, will this average if I am going to take the sample average of this sample it is always going to lie in this right and if I am going to take the samples of this interval they are going to lie somewhere here. So, to know if I give a sample whether it belongs here or here is it enough to give me the range information? No right I need to give you some mean value also where potentially it can lie. So, here this x n minus x 1 can I treat it as ancillary statistics about my parameter theta, but if I conject if I use it in conjunction with x n this average here it is providing me full information about my parameter theta and about the distribution. So, in fact, it becomes minimal sufficient statistics in this case. So, that is the difference between your ancillary sufficient statistics ok. So, there is an example I will leave you to look into this example because it has some more computation in this. So, with this we will conclude our discussion on the statistics. So, I hope you will all have understood what is the statistics, what is the sufficient statistics, what is a sufficient statistics, how to check for sufficiency and then how to check for minimal sufficient statistics and then this ancillary statistics.