 Dear students, let me discuss with you the concept of the support of a discrete random vector. And to do that, we begin with the concept of the support of a discrete random variable. So in general, what is the support of a random variable? It is the set of values that the random variable can take. It is also called domain. In your univariate situation, if your continuous f is equal to something, and then you write, for example, 0 less than x less than infinity, then 0 less than x less than infinity is the support of that bdf. And what does that mean? It means that the x variable that can assume any value between 0 and infinity, but it cannot be negative. Because if it was negative, then what did we write there? Minus infinity less than x less than infinity. So you are well aware of this concept. Domain is also called support for Love's BC. Now after this, if we concentrate on the discrete random variable, then we must note that it is the set, I mean that the support is the set of all the realizations of the variable x that have a strictly positive probability of being observed. I mean that if there is zero probability, then that is an impossible event. So we don't need to write that. If we think of it that way, then this is the rigorous definition of it. I repeat that I am still talking about the univariate situation. So for a discrete random variable, the support of that probability mass function for that random variable is the set of all the realizations of that variable x that have a strictly positive probability of being observed. Now when all of this is clear, then of course we can extend this idea to the case of a discrete random vector. So let me do it with the help of an example and a simple but interesting example and the case when we have two variables x1 and x2. Suppose that a coin is tossed three times and our interest is in the ordered pair, number of heads on the first two tosses, comma, number of heads on all three tosses. So how do we proceed? Subsequently we will construct the sample space which we may denote by capital C of our experiment. We are tossing a coin three times. So what are the various possibilities? Tail, tail, tail, tail, tail, head, tail, head, tail, tail, head, head, head, tail, head, head, tail, and last but not the least head, head, and head. Now when we have this sample space, this is the set of all possible outcomes. Apart from this, nothing is possible if you are tossing a coin three times. Now after this, we are able to define our variables x1 and x2. So let x1 denote the number of heads on the first two tosses and x2 denote the number of heads on all three tosses. So now let us look at it very carefully. It is in front of you now on the screen and you can see that x1 for the outcome ttt, tail, tail, tail is equal to zero. Obviously because there is no head in the first two tosses and x2 for the outcome, tail, tail, tail is also equal to zero because there is no head in all three tosses. Similarly, you can see that every single one of them is in front of you. I will take up one more. Let us look at just the second outcome that is tail, tail, head. In this, x1 for this outcome is equal to zero because there is no head in the first two tosses. But x2 for this particular outcome is equal to one because tail, tail, head, there is one head. So therefore now we can write the ordered pairs. Our random vector. It is a discrete random vector because only these numbers which you just saw, zero, one, etc. This is possible. So it is not continuous. We are counting the number of heads. So count the discrete variable. So what are the probabilities for the various values of the ordered pairs? The probability of x1, x2 being equal to zero, zero is one by it. So the eight that we just saw, one of them is zero, zero. Similarly, the probability of x1, x2 being equal to zero, one is one by it. But the probability of x1, x2 being equal to one, one is two by eight. So when we take out the probability, it will be two by eight. Similarly, for one, two, two, three, one by eight. When you add them, students, you get eight by eight and that is equal to one. And that fulfills the basic property of bivariate probability mass function that the sum has to be one. And the first property was that all those probabilities will be between zero and one. So obviously, one by eight or two by eight, they are all between zero and one. What was happening was that what is the support of this random vector? Well, students, if you express this situation in tabular form, it is quite simple, easy to understand what is the support for the random variable x1 and what is the support for the random variable x2 so that we can get and talk about the combined support as well. This table in front of you, this is giving you everything exactly. Now, whatever I had told you, the probability of the ordered pair zero, zero is one by eight. So you see in this table, the first cell in which x1 is zero and x2 is zero is at the same place. So that is one by eight. You will fill it yourself just using a little bit of common sense. You can write the various values of x1, which are obviously in the ordered pair, in the ordered pairs, they are those numbers which are prior to the comma sign. And the x2, you will write it yourself, which in the ordered pair version, they are the numbers which are after the comma. Once you have written those, then you can fill the table. Now, look at the table for example, where x1 is zero and x2 is two. We have written zero in the cell there because there is no probability. It's not possible to have x1 equal to zero and x2 equal to two. But look at the physical situation of it as well. If you have zero head in the first two tosses, then is it possible to have two heads in three tosses? All right, I'm sure you can work this out and you can see that in this way, a number of probabilities are zero. What is the support? The numbers that you have in the first column or in the top row, that is the support. Zero, one, two for x1 and zero, one, two, three for x2.