Let me recall the concept of convolution. There are a couple of statements in the probability theory which actually reveal what does this term mean. All of these statements are about independent variables but let me take two independent variables x and y, and let me first formulate what does convolution mean in terms of distribution functions, and what follows, I will denote the distribution function of X by Fx, and distribution function of Y by Fy. And general theory says that the distribution function of the sum X plus Y, is equal to the integral of R Fx as a point X minus Y dF(y). This integral is actually still ts integral. And, this operation is denoted by star. Fx star Fy. And other theorem says that if X has a density Px and Y has a density Py. Then the sum of X and Y also has a density which we denote by Px plus y. The point X it is equal to the integral over R. Px_x-y Py(y)dy. This operation is also denoted by star. Actually, is Px star Py. We were speaking about probability theory. There is no problem that these two operations are denoted by the same sign star. Because, if y and x have densities then these two statements completely coincide. But, actually, we'll also defines this operations, not for probability distribution functions and probability densities but for any two functions. And in this case, we should specify which kind of convolution do we use. To avoid any misunderstanding, I will call this kind of convolution as a convolution in terms of distribution function. And, this type of convolution, convolution in terms of densities. But both types of convolutions are both be denoted by this star. Now, if speculating to our case, to the case of renewal processes as already mentioned, the process Sn is equal to the sum of many Xi, Xi_1 plus so on plus Xi_m. Therefore, we should somehow extend the notion of convolution. There's a case when we have not two but many random variables. All of them are independent so theory really works. And to work with this object, I would like to introduce a full notation. So, let me denote by FN Star is the convolution of N functions F in the sense of distribution functions. Let me shortly discuss the properties of this object actually, to play some role in the following. So, let me now formulate and show at least four properties of this object. So first property of this object is the following one, actually FN Star(x) is less or equal than F(x) in the power N provided that F at zero is equal to zero. Once more, this condition means that there's a random variable with distribution function F is all almost surely positive. How to show that there's an equality holes. This is nothing more than the following thing. So, if we take Xi_1 and so on Xi_m and independent identical distributed random variables with distribution function F, then the event that Xi_1 plus so on plus Xi_m is less or equals and X is included in the event that Xi_1 is less or equal an X and so on Xi_n is less or equals than X. This is just because if the sum of all Xi_n is less or equal than X, then all of Xi's should be also less or equals than X. This is a simple call away from the fact that Xi's almost surely positive. If you take probabilities of left and right-hand side of these string, we get the probability that Xi_1 plus Xi_n is less or equals than X is less or equal than the probability of this event. But this event has affect N intersection of N independent events. And therefore, this is a product K from 1 to N. Also probability that Xi_k is less or equals than X. What you have from the left-hand side is exactly FN Star(x) and what you have on right-hand side is exactly F(x) multiplied by itself n times. So, we conclude that, the statement is proven. Second property is that FN Star(x) is large or equals than F_n plus 1 Star(x). This is also a very simple property and it follows from the fact the event Xi_1 plus so on plus Xi_n is less or equals than X includes the event that Xi_1 plus so on plus Xi_n plus 1 is less or equals an X. This is once more because all sides are non-negative. And from here, we immediately conclude there's this unequality holds. Well, this is basically all that you should know about this operation. There are also a couple of properties like committed to return the assets activity of the convolution, I guess it is also quite trivial. Now, let me show how you can apply these properties in the context offering UL theorem. The following theorem holds. Let Sn be a renewal process. These Sn equals to Sn minus 1 plus Xi_n, Xi_1, Xi_2 and so on is a sequence of independent and integral distributed random variables with the distribution function F and we assume that these random variables are almost surely positive so this F is equals to zero at zero. Okay. The theorem has two parts. The first part would claim that the serious UT equals to the sum N from one to infinity Fn Star(t). This sum series converge. And second item stances that for renewal process Sn the mathematical expectation of its counting process N_t is equal to exactly N U(t). So both of the statements are quite important. So, the first statement. We have series converge. And the second that this series give us actually the mathematical expectation of N_t. Well, the first item can be proving by applying the properties which I discussed previously. This is a very technical result and I would like to skip it. For the second part, let me prove this statement. The proof is quite simple. We just show to apply the definition of the counting process. That mathematical expectation of t is a mathematical expectation of the maximum index of N such that S_n is less and equal N_t. But S_n itself is increasing process at least not decreasing. Therefore, this is exactly the same as to calculate the mathematical expectation of the amount of N such that S_n is less or equals than t. Well, this equal to the mathematical expectation of the sum N from one to infinity indicator that S_n is less or equals than T. Here you can put the sum outside the mathematical expectation and what you will have here, is a mathematical expectation of the indicator. This is nothing more than a probability. So it's sum from one to infinity probability is that S_n is less or equals N_t. But the distribution function of S_n is exactly FN Star. And so, what you have here is this sum FN Star(t). Okay, we have proven this fact. And actually it is a very nice result which gives us a precise formula for the mathematical expectation of N_t. There is only one very important issue related to this fact, is that it's almost impossible to calculate the sum directly. In fact, it should calculate not only the convolution of N distribution function but you also should find the limit of this series. And it turns out to be completely impossible task if you have a distribution of F in general form. And to solve this problem, to calculate mathematical expectation from F, we shall somehow apply this result but non direct way. In the next subsection, I would like to show you one approach. How we can reach this goal, how we can estimate mathematical expectation of N_t from F.