 Dear students, let me present to you the concept of transformation in the case of a bivariate probability mass function using the MGF technique. And I will do it with the help of an example, let X1 and X2 be discrete random variables and let them have the joint PDF p of X1, X2 equal to mu 1 raised to X1 into mu 2 raised to X2 into e raised to minus mu 1 into e raised to minus mu 2 over X1 factorial into X2 factorial where X1 takes on the values 0, 1, 2, 3 and so on and X2 also takes on the same values and the PMF is equal to 0 elsewhere mu 1 and mu 2 being fixed positive real numbers. Now given this scenario, suppose that we define Y as X1 plus X2 and we are interested in determining the PMF of Y. Obviously, if we are dealing with a discrete variable situation, if X1 and X2 are discrete random variables then Y also will be a discrete random variable. Before we go to the MGF of Y, let me talk about the MGF of X1 and the MGF of X2. It is the product of two PMFs which are the PMFs of the Poisson distributions having mu 1 and mu 2 as their parameters respectively. Poisson distribution, would you not agree that for the random variable X1, we will say that p of X1 is equal to e raised to minus mu 1 into mu 1 raised to X1 over X1 factorial where X1 is equal to 0, 1, 2, 3 and so on and similarly for the random variable X2, if it is Poisson with parameter mu 2, its PMF is given by e raised to minus mu 2 into mu 2 raised to X2 over X2 factorial where X2 is equal to 0, 1, 2, 3 and so on. So if you multiply these two and construct this bivariate PMF, that means that we are dealing with that distribution, that bivariate distribution which has come about in that situation where we have two independent Poisson random variables. So by multiplying them, we get the joint PMF. By the basic definition of the MGF, the MGF of X1 is the expected value of e raised to TX1 and after solving and doing the algebra, we obtain e raised to mu 1 into e raised to T minus 1. In any power kender, we have mu 1 into e raised to T minus 1. Similarly for the MGF of the other Poisson random variable X2, we will have exactly the same kind of an expression e raised to mu 2 into e raised to T minus 1. Now let us try to find the MGF of Y. We have defined the new variable that Y is equal to X1 plus X2. If we want to remove this MGF, students, what will be the formula? It will be the expected value of e raised to TY and how do you define expected value in the case of a bivariate PMF? Well it is the double summation, summation over X1 and summation over X2 of e raised to T into X1 plus X2 because Y is equal to X1 plus X2. So e raised to T into X1 plus X2 and this whole thing multiplied by P of X1, X2. You need double summation to have this product over all possible values of X1 and X2. But students from basic algebra, we know that we can take some of the things out of the second summation sign. If the first summation sign is with respect to X1 and the second summation sign is with respect to X2, then all the terms of X1 will come out. We have the expression inside e raised to T into X1 plus X2. You can write e raised to TX1 plus TX2 and you can write e raised to TX1 into e raised to TX2. This is the E portion with which P of X1, X2 is written. As you know that is mu1 raised to X1 into mu2 raised to X2, e raised to minus mu1, e raised to minus mu2, this whole thing over X1 factorial, X2 factorial. They will come out of the second summation which is over X2. So what will be the expression then? The expression will be expected value of e raised to TX1, mu raised to X1, e raised to minus mu1 over X1 factorial. After this, we will be writing summation over X2 of e raised to TX2, mu raised to X2, e raised to minus mu2 over X2 factorial. Now, let's further solve this and you will find as you can now see on the screen that after doing a few steps, you obtain the product e raised to mu1 into e raised to T minus 1. This thing multiplied by e raised to mu2 into e raised to T minus 1. If we solve this product further, you will know that power will be added and e raised to T minus 1 common will come out and we are left with e raised to mu1 plus mu2 and this whole thing multiplied by e raised to T minus 1. But students, the product form. Now, a little while ago, I had discussed MTF of X1 and MTF of X2. Are not these two exactly the same? Yes, of course, e raised to mu1 into e raised to T minus 1 is the MTF of X1. Similarly, the other one is the MTF of X2. So the point is that here you have to see that after solving it, the final expression is that its form is not exactly the same form which is different from the first one. In power, mu1 is written in the beginning and that is the MTF of X1. In the second one, mu2 is written in power and that is the MTF of X2. Now, the final answer is that in power, mu1 is written in the beginning and mu2 is written as mu1 plus mu2. But the rest of its overall form is exactly the same as that of those two. So this means that the random variable Y also has a Poisson distribution. The only thing that the parameter of this particular Poisson distribution is neither mu1 nor mu2 but mu1 plus mu2. Then look again at the MTF, which is the uniqueness property of the MTF. The form of the MTF is recognized by that distribution. So the form of the MTF is exactly that of the Poisson distribution and therefore, students, we can write the PMF of Y as follows. P of the random variable Y at the point small y is equal to e raised to minus mu1 plus mu2 equal to mu1 plus mu2 whole raised to y over y factorial. Now, we will write y and of course, after this we have to put a comma and after that we have to write the values of y and students y will be equal to 0, 1, 2 and so on. Exactly the same for the Poisson distribution. If you want to know how come y is equal to 0, 1, 2 and so on, then go back to the point where y was defined. Y was equal to y is equal to x1 plus x2 or x1 key all possible values 0, 1, 2, 3, x and so on. x2 key all possible values 0, 1, 2, 3 and so on. Now, you start adding. What might be the minimum possible sum? 0 plus 0 is equal to 0. After that, additions. 0 plus 1 is 1, 1 plus 0 is also 1 and 0 plus 2 is 2, 1 plus 1 is 2 and 2 plus 0 is also 2. So, you see and you will have no doubt that y also assumes the values 0, 1, 2 and so on. So, this is a very, very interesting way of finding the PMF of a transformed variable y based on original variables x1 and x2. Whose MGFs are available are known and if later we find that the MGF of this y is also of the same form, we conclude that this new variable also has the same distribution.