 Dear students, I would like to present to you the concept of transformation in the case of a bivariate probability density function using the Jacobian of transformation technique and I would like to do it with the help of an example. Suppose that the random vector x1 x2 has the PDF f of x1 x2 is equal to 1 where x1 lies between 0 and 1 and x2 also lies between 0 and 1 and the PDF is equal to 0 elsewhere. The support of this PDF, my dear students, is then the set capital S consisting of all ordered pairs that comprise the region of the two dimensional floor I can call it the x1 x2 floor that part of the floor which is defined by x1 going from 0 to 1 and x2 also going from 0 to 1. And after this, note that the PDF that is the PDF of a uniform distribution because basically, it is like a box, it is like a box the volume under which or within which is equal to 1 which is exactly what you need in case of any probability distribution. Now that we have this joint PDF suppose that I am interested in the PDF of the random vector y1 comma y2 where y1 is equal to x1 plus x2 and y2 is equal to x1 minus x2. So first of all, please note that when I said that y1 is equal to x1 plus x2, we can write it formally as follows y1 is equal to u1 of x1 x2 and that is equal to x1 plus x2. Similarly, y2 is equal to u2 of x1 x2 and that is equal to x1 minus x2. Any u1 or u2 are the notations for the functions that we are having. The important point to note is that this transformation is one to one and the next thing is that let us try to find the inverse function. So how do we proceed? Look at y1 is equal to x1 plus x2 and y2 is equal to x1 minus x2. You will get y1 plus y2 is equal to 2 times x1 and that x2 will cancel out with minus x2. So if y1 plus y2 is equal to 2x1 then of course x1 is equal to y1 plus y2. This whole thing divided by 2. Similarly, if you subtract the second equation from the first equation, then you will get y1 minus y2 is equal to 2x2. So what do you get? x2 is equal to y1 minus y2. This whole thing divided by 2. Students, if we want to write this in a formal way, then what will we write? We will write x1 is equal to w1 of y1 y2 and that is equal to y1 plus y2 over 2. Also, x2 is equal to w2 of y1 y2 and that is equal to y1 minus y2 over 2. The purpose is that the original transformation that you applied, we were using u1 and u2 to represent that those functions or here w1 and w2, these are representing the inverse functions. The set t is that portion of the y1 y2 plane which occurs under this particular mapping that I am presenting to you. Now, the next thing is the Jacobian of transformation. Continuously, you know that the Jacobian of transformation technique that is a very well known and widely used technique and if the conditions are being fulfilled, then often it is the easiest and best technique to use. So, if the univariate case is there, then it is very very simple, then it is simply dx by dy and then we take the absolute value of it and that is called the Jacobian of transformation. Now, at this point in time, we have two variables and therefore this process of taking the derivative, this will happen four times. Now, j is equal to the determinant of curl x1 by curl y1 and then just go rho wise curl x1 by curl y2 and then go to the second row curl x2 by curl y1 and the last one curl x2 by curl y2 partial derivatives of x1 and x2 with respect to y1 and y2. So, now you take this derivative. Curl x1 by curl y1 is equal to half. Curl x1 by curl y2 is also half. Curl x2 by curl y1 is also equal to half. But curl x2 by curl y2 is equal to minus 1 by 2. Because it is just simply taking the derivative, if you look at your inverse functions, x1 is equal to y1 plus y2 over 2. It should be obvious to you that it is going to be equal to half. And now, once you have that, the leading diagonal minus the other diagonal. So, half into minus half which is equal to minus 1 by 4 minus the other diagonal half into half that is minus 1 by 4. So, minus 1 by 4 minus 1 by 4 is minus 2 by 4. And in other words, it is equal to minus half. Now, students, this is the Jacobian of transformation. It's not correct. We have to take the modulus of this thing and that is what is called the Jacobian of transformation. So, modulus of j is the modulus of minus half and that is equal to half. That is what we are going to use in a few moments from now when we write the pdf, the joint pdf of y1 and y2. But before we do that, there is another very, very important step and that is to determine the boundaries of t. So, each one of them is equally important and together they define the set capital T or in other words that region of the y1 by 2 plane over which your joint pdf of y1 by 2 is going to be defined. These four inequalities are as follows, minus y1 less than y2, y2 less than 2 minus y1, y2 is less than y1 and y1 minus 2 is less than y2. If you take one inequality and leave the other one, then take the other one and leave the first one, similarly for the other one. For example, if we concentrate on the first inequality, what are we saying? We are saying that 0 is less than y1 plus y2 over 2. Now, if we take 2 to that side, then it will multiply with 0 and become 0. And what will we have? We will have 0 less than y1 plus y2. Now, if we take y1 to that side, then obviously it will become minus y1 and what will we have? We will have minus y1 less than y2. You will get all of them. That region of the y1 by 2 plane over which the joint pdf of y1 and y2 is going to be defined, that is given by these four inequalities. None of these four inequalities should be violated. And as far as we are concerned, how we are going to write it, how we may write it in a simple way. Well, we can just write it as follows. The joint pdf of y1 by 2 is given by f of y1 by 2, f of the random variables y1 by 2 at the point small y1 small y2 is equal to small f of the random variables x1 x2 at the point y1 plus y2 over 2 comma y1 minus y2 over 2. And this whole thing multiplied by the Jacobian of transformation and this whole thing defined for those ordered pairs y1 y2 that belong to capital T where capital T is itself according to those four inequalities just that I mentioned a short while ago. And the pdf is equal to zero elsewhere. Basically, point it not so high that our old pdf is what we have to write there f of x1 x2 at the point small x1 small x2. x1 is equal to y1 plus y2 over 2. That is also in that same manner. But students, do you not remember that it was a very, very simple function? What was it? f of the random variables x1 x2 at the point small x1 small x2 is equal to 1 for x1 0 to 1 and x2 0 to 1. So now here, you have to write 1 and multiply it by the Jacobian of transformation which a short while ago we found was equal to half. So it is as simple as this. 1 into half for y1 y2 belonging to capital T and 0 elsewhere. So 1 into half means half and thus we obtain the final version of the joint pdf of y1 y2. f of y1 y2 at the point small y1 small y2 is equal to half for all ordered pairs y1 y2 belonging to capital T and equal to 0 elsewhere.