 Dear students, I would like to discuss with you how we can determine the expected value of the product of two continuous random variables and I will do it with the help of an example. Suppose that x1 and x2 are two random variables with the joint pdf small f of x1, x2 equal to 4x1x2 such that x1 goes from 0 to 1 and x2 also goes from 0 to 1. So we would like to compute the expected value of x1 into x2 and we would like to see whether this expectation is equal to the product of the expectation of x1 and the expectation of x2. If it is equal to 4 we can probably say, probably or certainly say that x1 and x2 are independent random variables or if it is not equal then it means that they are not independent. So let us see what is going to happen. First of all, e of x1. Now first of all I would like to give you a rule. The thing is that you know that if joint pdf is there then if you remove its marginal distribution of x1 then of course e of x1 will be given by the integral of x1 into that marginal density and that is it. But if you want then you can present it in a different way which you can now see on the screen. e of x1 is equal to the integral from minus infinity to infinity of x1 into f of the random variable x1 at the point x1 with respect to x1. So what do we get, we can say that the expected value of x1 is equal to the integral from minus infinity to infinity x1 and we can put a bracket so that we are not confused. The actual formula by which we obtained that marginal density and what is that? It is the integral from minus infinity to infinity of f of x1 x2 with respect to x2. You know that if you want to remove f of x1 then you integrate with respect to x2. So that bracket comes in and after that you have dx1. Next step may you can remove the bracket or the internal integrals outside of x1 you can take that inside. So when you take that inside students what is the final version of the formula that you have for the expected value of x1? It is the double integral from minus infinity to infinity of x1 into f of x1 x2 dx2 dx1. Yeah, it doesn't matter. Double integral minus infinity to infinity in both cases of x1 into f of x1 x2 dx1 dx2. So I have done all this with you because I have said that you should note that if this version of the formula, if you keep this version in your mind then you don't have to do those two steps that first you remove the marginal then you can do it straight from the joint PDF. This was for e of x1, you can see that e of x2 is equal to the double integral each one of them going from minus infinity to infinity x2 into f of x1 x2 dx2 dx1. Now that you are clear about all this, I think there is no problem. You can apply it to this particular problem. Now here x1 is going from 0 to 1 and x2 is also going from 0 to 1 therefore obviously we will not write minus infinity to infinity but we will write 0 to 1. As you can see on the screen, e of x1 is equal to integral 0 to 1, the second integral again 0 to 1 x1 into 4 x1 x2 dx2 dx1, which is the same formula that I have put in front of you. Now you can solve it step by step and what is the final answer? It is 2 by 3. In a very similar fashion, you can find the expected value of x2 and students, this one also is equal to 2 by 3. Now we come to our product which I am interested in, the expected value of x1 into x2. So what will be the formula for this one? In general, it would be the integral from minus infinity to infinity then the second integral again from minus infinity to infinity x1 x2 into f of x1 x2 dx2 dx1. Now of course we can write 0 to 1 and then x1 into x2 into 4 x1 x2 dx2 dx1 and what do you get? The answer is 4 by 9. Up they came the equation here that is being fulfilled. The product of e of x1 and e of x2 2 by 3 multiplied by 2 by 3 is exactly equal to 4 by 9. The answer that we obtained just now for e of x1 into x2, therefore we may conclude that in this particular problem the random variables x1 and x2 are independent.