 Dear students, I would like to discuss with you the concept of independent random variables. If X1 and X2 are random variables of the continuous type where joint PDF, f of X1, X2 and marginal PDFs f1, X1 and f2, X2, then in accordance with the definition of the conditional PDF, we can write the joint PDF f of X1, X2 as follows, f of X1, X2 is equal to f1, X1 multiplied by f of X2 given X1. Do we not know the basic theorem of probability that probability of A intersection B is equal to probability of A multiplied by probability of B given A. So, now you can write what I just said and I will repeat it for you, f of X1, X2 is equal to f1, X1 into f of X2 given X1. Now suppose that we have an instance where f of X2 given X1 does not depend on X1. First, it does not depend on X1. Then what will we have? If our variables are continuous, then my dear students, let us consider the marginal PDF of X2. As you can see on the screen, we can write f2, X2 is equal to the integral of f of X1, X2 with respect to X1. So if I write this, then instead of f of X1, X2, I can substitute what I presented to you just about a minute ago. So what will my integral become? It will become like this. The integral from minus infinity to infinity of f1, X1 into f of X2 given X1 and the integral itself is with respect to X1. Now look carefully at this. This f of X2 given X1, this will come out of the integral sign. Now the question is, why would it come out of the integral sign? Look, I told you earlier that this is that situation where f of X2 given X1 does not depend upon X1. So if it does not depend on X1, then the expression will remain regardless of whatever X1 might be. It will be an expression containing X2 and X1 will not be involved. And so it will come out because the integral here, that is with respect to X1. So if it is given X1, then we could not have taken it out. But because it is independent of X1, that's why it is coming out. All right. So now what do we have? Our equation has been written as F2 X2 is equal to f of X2 given X1 multiplied by the integral from minus infinity to infinity of f1 X1 with respect to X1. Now look carefully at the integral students. Do you realize that this integral is equal to one? Yes, it is equal to one. Do we not know that the area under the curve for any PDF, the total area has to be equal to one? So when you are integrating it from minus infinity to infinity, then obviously that has to be equal to one. Now what do we have left? All we have now got is F of X2 given X1. What was the left hand side of our equation? F2 X2. So therefore our equation now is this. That F2 X2 is equal to F of X2 given X1. In other words, the unconditional PDF is the same as the conditional PDF. Now if this is the first equation, how can we rewrite it? What was our first equation? F of X1 X2, the joint PDF is equal to F of X1 X1 into F of X2 given X1. Now substitute F of X2 given X1 instead of F2 X2. So therefore we get a simple and neat attractive equation. F of X1 X2 is equal to F1 X1 into F2 X2. Yanni, the joint PDF is the product of the two marginal PDFs. This is what we have if X2 is independent of X1. So this is the case, all this that I have said is the case when we have the continuous variables and in a very similar manner, it can be shown that a similar equation holds in the case of discrete random variables. If X1 and X2 are discrete random variables having joint PMF, you know that in the case of discrete, we do not say PDF, we say PMF, probability mass function. So if X1 and X2 have the joint PMF, P of X1 X2 and the marginal PMFs P1 X1 and P2 X2, then if X1 and X2 are independent of each other, we will have P of X1 X2 equal to P1 X1 multiplied by P2 X2. This is the story of independent random variables.