 So, now let us go back to our example where we have this x 1, x 2 up to x n, where each of this x i's are normal with mu and sigma square, where I assume ok let us call it as theta, theta is unknown and sigma square is known. My unknown parameter is only theta here. What was our estimator? Sorry, what was our hypothesis testing here? Ok, let me now this one I want to see whether my hypothesis 0 is whether theta is equals to theta naught and h 1 is theta is not equals to theta naught ok. Let us consider these two scenarios. What was our rejection LRT? What is the value? What is the condition it gave us? What was the value of this LRT? x bar minus theta naught divided by sigma square by n and our rejection region was if lambda x is less than or equals to c, you reject all those samples ok. Now, was it this or there was an exponent? It was exponential value of this. I think this was exponential value of this right, can you check? And also there is a square. I mean of course, this the Gaussian has to play a role here right, this structure is coming from the Gaussian. This is the thing and now this is my lambda of x and I want this lambda of x to be less than r c and that is my rejection r is. Now, what does this translate to? If I have to plug in this. So, this will translate to x bar minus theta naught square greater than or equals to minus log c and sigma square by n log c minus maybe sigma square when I will write like this sigma square by n that is it right. Is there a square root here? Can you check? No, I think there is no square root there was n here and this was like a simply sigma square. So, this part was numerator was simply the exponent in the Gaussian term right that is simply sigma square and we did some manipulation and we got an n term here just cross verify this. Now, this is going to be minus log c and sigma square by n. So, there is also 2 here which I will write 2 here minus. So, what is the value of c? What is the range of c here? c 0, 0 to 0. We know that this has to be between 0 to 1 because lambda x has is a ratio which is always between 0, 1 and because log c is between 0, 1 we know that log of c is going to be a negative quantity and because of this whole quantity is a positive quantity. Now, how does this translate to? This translate to x bar minus theta naught I can write it as to be greater than or equals to minus 2 log of c into sigma square by n. Agree? Which I can further say now this is now x bar this is like x bar minus theta naught less than or equals to minus 2 log c sigma square by n less than or equals to square root of minus of minus 2 log c sigma square by n, right? Everybody agree with this computation? Now, what I will do is x bar minus theta naught I will get this sigma square by n from both sides. Now, this is going to be minus of 2 log c and minus of minus 2 log c and as always we know that this is going to be distributed what? So, I am going to write this is going to be minus 2 log c minus of minus 2 log c agree? Now, see that it is kind of giving me already some range here what I am doing is. So, what I am going this is going to give me if this condition holds if my x bar is such that x bar minus theta is going to lie in this interval I am going to accept that x bar to coming from parameter theta naught or no. This is basically rejection region, right? I have basically translated that rejection region into this interval here, ok. So, if my x bar so, if this x bar right now I do not know this x bar let me just put it in terms of x bar is now this is going to be theta naught plus minus 2 log c sigma square by n this one theta naught minus minus 2 log c, right? Ok, maybe I will just get rid of this sigma square by n here and this is this quantities are known and this is like. So, there is going to be a sigma square by n here sigma square by n here and another sigma square by n. What happened this one? So, I have taken theta naught this side, right? That is fine and then theta naught also this side. What is the issue? I am not ready. No, this is correct, right? This is correct. When I remove the modulus, this is correct, no? Ok, let us do this when, ok. Suppose like say when this is a positive quantity, ok you are saying this should be there and when it is a negative quantity, ok let me write it, ok let me write this, this quantity whenever this suppose this quantity happens to be positive the equality remains like this minus 2 log c sigma square by n and when there is a negation of this I have to take negation both sides this is like x bar minus theta naught and so, this is going to be less than or equals to minus of minus 2 log sigma square by like this and now because of this, ok let us write this. Now, if I have to get this x bar minus theta naught is going to be upper bounded by minus 2 log sigma square by n and then this is going to be minus. But see there is a issue here, right? Something is wrong. We know that 2 log c is going to be negative quantity minus 2 log c is going to be a positive quantity. So, this entire thing this entire thing is a positive quantity, but now this entire thing is a negative quantity. How can this happen? Ok fine let us do that. Now, what we have probability that x bar equals to this and this. Now, how you are going to write this? What we have written is actually and right like this and this. So, how can this happen like? So, this that is what I am saying this is a inside that is a positive quantity. It has to be greater than positive quantity, but less than a negative quantity. How can that happen? Ok we have to start with this maybe let us find out. So, this is the condition we have right and we know that ok this x bar minus theta naught we know this has to satisfy. How you are going to define in terms of the intervals? Now we have to work out. This is true. With this is this true? If this is x bar minus theta this happens to a positive quantity this either this or this has to happen. This is or this is not an AND condition ok. So, now probability that ok now how we will write it x bar minus theta naught ok. Now, let us say let us define this I know that my quantity minus 2 log c sigma square by n has to be here and minus of this and minus 2 log c sigma square by n has to be here. And what I am asking is this x bar minus theta naught should be greater than this and less than this ok. Whenever my x bar minus theta naught it is so happening whether it is falling in this region or this region what I am going to do? I am going to reject and I am going to accept whenever it is going to be in this region ok. So, let us focus on that acceptance region I think that is where the confusion came. We were not interest ok. Now this is going to give me the rejection region and if I have to get the acceptance region then this has to be x bar minus theta naught has to be less than or equals to minus 2 log c sigma square by n and minus of minus 2 log c sigma square by n. So, this is the rejection and this is the acceptance region this is clear. Now, I am going to call it as A this region all the x xers such that x bar minus theta naught. So, maybe now I can take that theta naught on the other side whenever this x bar is such that it is theta naught plus minus 2 log c sigma square by n and theta naught minus of minus 2 log c sigma square by square by n this is going I am going to accept that ok. Now, this is for a given theta naught right theta naught is fixed and for that parameter I am trying to define this acceptance region and I am going to call it as A of theta naught. So, whenever my sample is falling in this region that x bar minus theta I am going to accept it. Now the question is is it possible for us to translate this into the confidence interval ok. So, we will discuss that in the next class, but first thing we need to note here is that does it matter what particular theta 0 that we are looking here this could be any theta ok. So, for any theta I can look for such conditions and this is what the rest of the things remain the same and this is what we will use to come up with our confidence interval that should work for any given sample x ok. So, we will continue that in the next class.