 students, I would now like to explain to you how to utilize the probability generating function in order to obtain the probabilities of a discrete distribution and also later on how to find the mean and variance of the distribution through the probability generating function. The relationship between the individual probabilities and the probability generating function is as given in equation one that you now see on the screen. It is expressed as follows. The probability of x is equal to k is the kth derivative of the probability generating function and after taking the derivative, we have to substitute in that z is equal to 0 and then this thing divided by k factorial. So let me give you an example to explain this point. Considering the probability generating function of a Bernoulli random variable, when for example, we are tossing a fair coin only once, you know that the probability generating function for this case is given by 1 by 2 plus z by 2. I want to find what is the probability of x is equal to 0. So according to the formula that I have just given you, I have to find the 0th derivative of the pgf the probability generating function and then I have to put z equal to 0 in that and then I have to divide that by 0 factorial. So students, what you have to do is to just simply put the value of the pgf, the expression of the pgf over there. So that is half plus z by 2 or 0 factorial saves you have divided and now you keep the value of z or z as 0. So what do you get? You get half plus 0 by 2 divided by 0 factorial and that is equal to half. So you know that the Bernoulli random variable that we are considering right now, the first probability is equal to half. So the formula is the same as the answer that is exactly the same. Similarly, the second value of x is equal to 1. If we want to get this probability through pgf, then according to our formula, we have to take the first derivative of the pgf and divide it by 1 factorial and then we have to put z equal to 0 in that expression. So in this case, what is the first derivative? Half plus z by 2, the first derivative, 0 plus 1 by 2, 1 by 2 divided by 1 factorial, that is 1 by 2. So once again, you have seen that the probability x equal to 1, the answer that you should have got is half. So this is a simple example to illustrate how you can generate probabilities from the probability generating function. After this, as I said earlier, I would also like to tell you how to find the mean and the variance through the probability generating function. Expected value of x, the mean of the random variable x, is given by g prime of 1 minus, as you can see on the screen. Before I do anything else, I have to explain to you what is meant by g prime 1 minus. And z instead of 1 minus, we are taking the limiting value of the prime z as z tends to 1 from the left hand side. So probability generating function, that is given by e raised to lambda z minus 1. Obviously, it will be by simple rules of calculus lambda e raised to lambda z minus 1. So the limiting value that z is tending to 1, that will be lambda e raised to lambda into 1 minus 1, that is lambda e raised to lambda into 0, that is lambda into 1 and that is lambda. You already know that the mean of the Poisson distribution is equal to lambda. So this is again a simple illustration of how the mean can be found through the pgf. As far as the variance is concerned, my dear students, the formula goes like this. The variance of x is equal to g double prime 1 minus plus g prime 1 minus minus g prime 1 minus whole square. G double prime means that we will take the second derivative. Now instead of z, we say that z is tending to 1 from the left hand side. So g double prime 1 minus comes out to be lambda square because the other part, that becomes equal to 1. Lambda square, your value is equal to g double prime 1 minus. Plus g prime 1 minus, lambda minus g prime 1 minus whole square, and we are left with the variance of x is equal to lambda. The mean and variance are equal.