Before I will present an approach for calculating the mathematical expectation of accounting process, I would like a shorter recall the definition and main properties of the Laplace transform. This transform can be defined for any function F from R plus to R. And the definition is very simple, Laplace transform of F is equal to the integral, from zero to plus infinity exponent in the power minus SX, F of x, dx. Main properties of this object has a following. First of all, if F is a density function of some random [inaudible] xai. Then, Laplace transform of F is actually equal to the mathematical expectation of exponent in the power minus xai. Secondly, if you have two functions F1 and F2, then the Laplace transform of the convolution is equal to the product of Laplace transform of F1 and Laplace transform of F2. And here, it's very important that the convolution is understood in the sense of densities. If F1 and F2 are identities are then this, fact basically follows from the first property. But if they are not densities this fact is still true, and actually, for any two F functions, F1 to F2, this property holds. The third property of which will be needed in this sequel is the following. That capital F be a distribution function of some positive, on a surely positive random variable, that is F0 is equal to zero, and let P, B's a derivative of F, that is a density function of the corresponding distribution. Then, the following statement is true, Laplace transform of the function capital F at point S is equal to the Laplace transform of the density function P divided by S. To prove it, let us apply the integration by parts formula to the left hand side of this equality. Actually, this Laplace transform of a function F is nothing more than the integral of R plus. And here, I can write F of x, D exponent minus S of x divided by S was a minus. Then, let me integrate it by parts. So, what we have here is one part F of x and the second part exponent minus S of x divided by S. And we will get the full length expression. It's basically, minus F of x exponent in the power minus S of x divided by S from zero to infinity plus integral of R plus, P of x exponent in the power minus S of x, tx. And here, I should also divide the second integral by S. As for the first [inaudible] this expression, it is equal to zero because when I substitute here, infinity exponent gives me zero. And when, I put here zero, I should use the properties of F and zero is equal to zero. So basically, this is equal to zero. And what we have here is exactly as the expression in right hand side. So, we conclude that this property is fulfilled and it will play some role in our method for estimations and mathematical expectation of NT. Now, let me provide just a short example, how one can calculate the Laplace transform. The first example, let me calculate Laplace transform of the function X in the power n, where n is some natural number. This is a very simple exercise for the application integration by parts formula. So, what we have here is the integral of R plus, X in the power n, d exponent to power minus sx divided by s. Integration by parts yields that this integral is equal to n divided by s, integral over R plus, x in the power n minus 1, exponent in the power minus sx dx. So, we have the same integral here as the original but with power n minus 1 instead of F. If now continue in the same manner, we get finally n divided by s multiplied by n minus 1 divided by s and so on, multiplied by 2 divided by s. And here, we have the integral over R plus exponent in the power minus S of x dx. And this integral is equal to 1 divided by s. So what we get finally, is n factorial divided by S with the power n. Okay. So, we conclude the Laplace transform of the [inaudible] function XN is equal to the N factorial divided by S divided by N. Just remember, this result, we will use it many times and what follows. And second example, which is also important for further study is the Laplace transform of the exponential function, that our calculation shows that this is now seen more than one divided by S minus a. If a is a number smaller than S. I advise you to remember also this result. It will be also very useful in what follows. And now, let me show how we can apply all of this stuff to the renewal theory and how we can get the direct approach for calculating with mathematical expectation of NT.