 Let me present to you a theorem pertaining to the moment generating function for a joint distribution. The statement of the theorem is as follows. Suppose that the joint Mgf capital M of t1 t2 exists for the random variables x1 and x2. Then x1 and x2 are independent if and only if M of t1 t2 is equal to M of t10 multiplied by M of 0 t2. In other words, the joint Mgf is identically equal to the product of the marginal PDFs. So in order to prove this one, we note that in the statement of the theorem, it says if and only if. In this case, first we will prove that x1 and x2 are independent, then that equation holds or after that we will assume that the equation holds and then we will show that x1 and x2 are independent. So let us begin. First part, suppose that x1 and x2 are independent. Now what is the definition of the joint Mgf? By definition, capital M of t1 t2 is the expected value of e raised to t1 x1 plus t2 x2. It is the expected value of e raised to t1 x1 and this thing multiplied by e raised to t2 x2. Now since we are assuming that x1 and x2 are independent, then we can apply that basic theorem which says that in such a case the expected value of ux1 into vx2 is equal to the expected value of ux1 multiplied by the expected value of vx2, ux1 means some function of x1, vx2 means some function of x2. If we apply this here, then ux1 is e raised to t1 x1 and vx2 is e raised to t2 x2 and since we have assumed that x1 and x2 are independent, therefore what we will write is that that expression is equal to the expected value of e raised to t1 x1 into the expected value of e raised to t2 x2. Now let us take another step in a small source. I can write this first one as you can see on the screen as the expected value of e raised to t1 x1 plus 0 x2. That one I can write as expected value of 0 x1 plus t2 x2. Why did I do this? So that I can write this first expression as capital M of t1 comma 0 and I can write the second expression as capital M of 0 comma t2. This means that if you 0 to t2, then the first one becomes M of t1 comma 0 and if you 0 to t1, then the second one becomes M of 0 comma t2. So therefore students, the independence of x1 and x2 implies that the Mgf of the joint distribution factorizes into the product of the Mgf of the two marginal distributions. So therefore what we have to prove from both the sides is that the first part is completed. Now let us begin the second part. In the second part, we will come in reverse direction. So what are we going to assume now? We are going to assume that the Mgf of the joint distribution of x1 and x2 is given by the following equation. Capital M of t1 t2 is equal to capital M of t1 comma 0 multiplied by capital M of 0 comma t2. Now let us look at the definition of the Mgf. Now x1 has a unique Mgf. Mgf is the Mgf of the joint distribution of x1 and x2 and x2 is the Mgf of the joint distribution of x1 and x2. So let us look at the definition of the Mgf of the joint distribution of x1 and x2. Now x1 has a unique Mgf. Mgf is unique, isn't it? So in the continuous case, which we are showing at this time, we can write capital M of t1 comma 0, i.e. marginal Mgf of x1 that is equal to the integral from minus infinity to infinity of e raised to t1 x1 into f1 x1 and the integration is being carried out with respect to x1. The expected value of e raised to t1 x1 is the Mgf of the marginal distribution of x1. We can write capital M of 0 comma t2 is equal to the integral from minus infinity to infinity of e raised to t2 x2 into f2 x2 and the integration is being carried out with respect to x2. Now if we multiply these two, look carefully at the students, it is actually very simple, you multiply these two. Now the way a double integral can come out of it, if it is out then it can also go inside. That is why these two integrals we have written as a product job. The double integral minus infinity to infinity again minus infinity to infinity e raised to t1 x1 into e raised to t2 x2 into f1 x1 into f2 x2 dx1 dx2 and after that add the e k powers. Now the next thing to note is that this product this is the product of the marginal Mgfs. But we have assumed that this second part of the proof that the product of marginal Mgfs is equal to capital M of t1 t2. So then we can write that capital M of t1 t2 is equal to the double integral of e raised to t1 x1 plus t2 x2 f1 x1 f2 x2 dx1 dx2. So we have got this now. Now after this note that by definition M of t1 t2 is the Mgf of the joint distribution of x1 and x2. So therefore by definition capital M of t1 t2 is the expected value of e raised to t1 x1 plus t2 x2 in other words the double integral of e raised to t1 x1 plus t2 x2 into f of x1 x2 joint dx1 dx2. The product of those two is equal to M of t1 t2. So students equation one or equation two Do you not see that the left hand side of both of them is the same M of t1 t2? Those two right hand sides have to be equal and because of the uniqueness of the Mgf we can see that we can now say that f1 x1 into f2 x2 that product which is contained in equation one is equal to f of x1 x2 that joint PDF that is contained in equation two. So the moment you are able to write this that f of x1 x2 is equal to f1 x1 into f2 x2 for all values of x1 and x2 then it means that x1 and x2 are independent in this way we have been able to prove this theorem from both sides and this is it.