 So, any questions from the last lecture what all we discussed about this factorization theorem, how to use this factorization theorem to come up with some statistics which are sufficient. And then if you know already some function is a belongs to exponential family all we need to do is write it in its in this form and there all the ti functions will already give me the sufficient statistics for that ok. We also discussed that sufficient statistics need not be unique there could be multiple ways of getting sufficient statistics and if you give me one sufficient statistics I may there are multiple sufficient statistics by using some invertible function on it ok. Then we started talking about minimal sufficient statistics. So, let us repeat some of the part here what we have already done. So, I am slowing myself by repeating things. If you want to further slow you should interact or ask something related to this right. I mean I am trying to incorporating the bringing the slowness by repeating things, but if if you also want to do from your side you should act do not simply sit ok. So, by definition we said that we are going to call a statistic T or rather a sufficient statistic T to be minimal sufficient statistic. If you have any other sufficient statistics T prime and the T primes take same value on two samples x and y then it implies that my T also takes the same value on those two samples x and y. That means, if two samples are indistinguishable under sufficient statistics T then they are also indistinguishable under the sufficient statistic or minimal sufficient statistic T ok. Now, let us quickly understand this. Suppose let us say what our written in text let us understand that through a diagram. Let us hypothetically just make a two dimensional case x 1 and x 2 ok. Let us say this is my region for some reason and hypothetically assume that this region gets partitioned into three parts under my hypothesis T. What I mean by that? Take any point here let us say x 1 comma x 2 this is going to map if you apply T on that that your function T on that it is going to give a value small t and that value small t is the same in this region. You understand that point all the points in this region map to the small t and similarly all the points in this also map to one t and like this I have made it into three partition ok and let us call this region as a 1, this region as a 2 and this region as a 3. So, all the points in the region a 1 they have the same value ok. When you apply T on that they will map to the same point. Now let us take another T prime sorry let us take another statistics T prime this is another statistics. So, let us call this is T and let us call let us do another partition under this this is under T prime not necessary that this T prime will also result in the same partition. It may result in some other partition let us call that as simply like this ok. Under this let us call this a 1 prime and let us call this a 2 prime and let us call this a 3 prime ok. In fact, it may result in more than 3 partitions it may have more than 3 or may be less than 3 whatever I am just for representation purpose I am just adding it as 3. Now, suppose T prime is a sufficient statistic let us assume that and let us call the point to which all this maps here as call let us call T 1 prime let us call this as T 2 prime and let us call all the value it maps to all the points in this region maps to as T 3 prime and similarly here that is called T 1 and this is T 2 and let us call this T 3. Now, if from this region I have all the points gets either map to T 1 prime T 2 prime or T 3 prime ok. Now, arbitrarily let us pick 2 2 prime and let us take some point let us let us know I know let us some point here let us some point let us call this point some x again that consist of x 1 and x 2 let us say this T 1 prime x maps to T 2 prime ok. In fact, all the points in this region are going to map to T 2 prime. Now, this region let us call this whatever this where this T 2 2 belongs let us call this A 2 prime. I know that all these points here in this region they are all getting map to the same value of T 2 point. Now, if I look into this portion here will they also under T will they also map to the same point not necessarily not map to the same point same as T 2 prime, but all their values will be same if T happens to be minimal sufficient statistic. So, all these points are taking same values let us say any point you take x and y here they are going to take the same value here, but if T is a sufficient statistic by definition they should also be taking a same value right. So, all these points that are here they are also taking the same value under T even though I have partitioned incorrectly it is not necessary. So, then there should be region here in under T which incorporates all these points right some region here which incorporates all these points that are map to the same value under T prime and that is what we are saying. So, this region here which has become a partition this is going to be a subset of another partition under T maybe ideally I should have drawn this like this it would have been if I have to draw this first figure with respect to this this should have been this maybe this like this and not even this with this with this and this. So, how many partition it got now? So, this has 1 2 3 1 2 3 and here I wanted to have only yeah maybe I even draw only this much now it has only 2 partitions right in this actually all these points are also here and that is now incorporated in a bigger partition it is just a representation. So, if this happens to be sorry here it I call him B here right in this set I am calling them as B, B1, B2 and B3. So, this B2 prime here is going to be part of A2 A2 here just by this definition that if T is a sufficient statistic every point that are mapping to the same they should also be mapping to the same value under my T ok. So, so obviously the number of partition under minimal sufficient statistics is going to be large or small compared to another sufficient statistics it is going to be smaller right right. If this is some sufficient statistics we saw it is going to partition my space into 3, but if T is sufficient statistics the number of partition it is going to have is going to smaller than this. So, in this way among all the statistics which statistics is going to partition your space into small number of subsets minimal sufficient statistic ok. So, this example we also discussed last time suppose we have a samples which are drawn from Gaussian distribution with parameter mu and sigma square with unknown mean mu and known parameter sigma square. So, my only thing that is unknown to me is mu. Now, I know that sample mean is going to be a sufficient statistic. Now, if I am going to take another sufficient status another statistic will actually come consist of 2 components sample mean as well as sample variance. Now, actually this is also the sufficient statistics because every information all the information about my parameter mu is contained in this and in fact, I can ignore x s square and s bar I can just retain which already you know is a sufficient statistics for mu. But I know that T 1 is a function of T 2 right. If you give me T 2 by just a drop in the second component I can recover T 1 right. Now, even though both this statistics or sufficient statistics contain same amount of information about my unknown parameter mu T 1 is a offers me a better reduction ok. So, because of that it compresses information better that should also give a kind of intuition that why the number of partitions is less under minimal sufficient statistics compared to a just an ordinary sufficient statistics ok. So, now how to find minimal sufficient statistics how to test a sufficient statistics is a minimal sufficient statistic. So, first of all we started statistics is any function that is fine. Check whether something is sufficient statistic that was not obvious, but we come up with a method what was our method to check whether a statistic is sufficient statistic yeah. So, one thing was characterization theorem which is give a both sufficient necessary condition for something to be sufficient statistic and if you have to just do simply verify whether something is a sufficient statistics we had some other thing also right that ratio test we had. See whether the ratio of unconditional distribution and of your samples and unconditional distribution of your statistics is independent of theta. So, now we have minimal sufficient statistics something more than sufficient statistics right how to test that. So, for that we have this result which again gives some complete characterization because it is giving both sufficient and necessary conditions. So, what is says is suppose ok let us say f of x by theta is your population density. Now, if there exists a statistic T such that you take any pair any pair x and y two samples random samples and if it so happens that the ratio of the pdfs at those two points x and y happens to be constant when I say constant here independent of x sorry independent of theta here the parameter ok there are only two three things right that are x y and theta here when I say it is a constant it is independent of theta this is going to be constant if and only if on if those two points x and y are such that my statistics gives the same value on those two points ok. If this is the case then my T is going to be a minimal sufficient statistics ok first before we will just briefly discuss his proof anybody has any difficulty in understanding this statement or are interpreting this statement what does this mean what he is saying is ok you give me a statistic and what I am going to do is I take two points two random points and check the ratio of these two point check the pdf at these two points if this is independent of theta provided on these two points T of x equals to T y only way then it should happen if this is going to be independent of theta on points x and y where T of x is not equals to T of y then you are failing this statement ok if that is the case then T is going to be a minimal sufficient statistic