Now, I would like to discuss the concept of Stochastic Integration. Actually we'll have a separate lecture about this topic, but at the moment I need to define integrals of the following type. Actually this is the simplest type of integrals, here X_t is the Stochastic process. We take integral as on some compact interval or some of these numbers can be infinite, so actually it doesn't matter. The definition of this integral is very close to what we have in the theory of Riemann Integrals in Calculus. Actually, this integral can be defined as a limit of the following sums. We'll take the interval a b, and then divide this integral to n small sub-intervals. So, point a is equal to t_0, point b to t_n and here we have points t_1, t_2, t_3 and so on. To take a limits: k from one to n, X_t_k - 1 multiplied by t_k - t_k - 1. And here's the limit, it's meant in the following sense: so, take this as a maximum, t_i - t i - 1 maximum over all i is tending to zero. Moreover, the limit here is taken as a sense of mean square. This means that this sum minus the integral squared and taking mathematical expectation should converge to zero as its maximum is tending to zero. So, this is very close to what we have in the theory of Riemann Integral in Calculus. Basically, you know that any continuous functions is integratable. Here we have something similar or more precisely, the following theory holds: So, let X_t be a stochastic process we'll find a second moment. So let me assume that mathematical expectation of the process X_t is a continuous function as a function of t. Let me also assume that the co-variance function is also continuous as a function of two parameters t and s. Therefore, if both of these conditions are fulfilled, the theorem states that the process X_t is integratable, that is integral X_t dt from a to b exists. Well, let me first comment on this conditions. Let me start with the second one. So, this is the first one, it's very clear, so you have one dimensional function and it's clear how to show that it's continuous or not. But as for the second, it can be a little bit difficult. So, actually K_t is a function of two variables and sometimes this analysis can be complicated. But, in this case, in the context of covariance functions, everything is not really difficult because the following statement holds: It turns out that the function K_t_s is continuous at any point t_0 s_0 if and only if this function is continuous at any point t_0, t_0. It is any point such that the first and second coordinates coincide. In this case it is very common to say that the function K_t_s is continuous on the diagonal. Well, this is a very interesting property of the covariance function. And of course not all functions depending on two variables has this property and this is a very special thing which is true only for covariance functions. And the origins of this statement is as the following: So, actually if the covariance function is continuous at some point t_0 t_0, then the process X_t is continuous at t_0 is a mean square sense, that is mathematical expectation of X_t minus X_t_0 converges to zero as t tends to t_0. On the other side, if the process X_t is continuous as the stochastic process in points t_0 and s_0, then, the function K at the point t_0 and s_0 is also continuous. So, we conclude the following: if the process K_t_s is continuous at any point t_0 t_0, then from the first statement we get that X_t is continuous in the mean square sense at any point t_0, and from the second part we conclude that the function K is continuous at any point t_0 s_0. So, this statement is correct. This means that we can check the second condition of the theorem only on the diagonals, that is, on all points such as the first and second coordinate coincide. Let me now discuss the main properties of the Stochastic integral, so from now we'll assume that the integral exist, and for most cases this is exactly so because almost all of our examples are such that both of these conditions are satisfied. The first property, let me consider the mathematical expectation of the Stochastic integral. It turns out that the signs of integral and mathematical expectation can be changed in the places, so it's an integral expectation of X_t dt. The origins of this statement is a well known Fubini theorem. Mathematical expectations, also integral and actually what we have here is a doubled integral, one of which is integral with respect to some probabilistic measure, and the second one is the Stochastic integral. So, if you will apply for Fubini's theorem carefully, you will get that this change is always possible. But there's some logic if we consider mathematical expectation of the integral X_t dt squared. We can also conclude that the signs of expectation integral can be changed to the following form: let me first represent this square as a double integral. It's double integral X_t X_s dt ds. And now we have exactly the same situation, we have some integral and expectation and definitely we can change the places and get this as a doubled integral, mathematical expectation of X_t X_s dt ds. Third property which a direct corollary from the first two is that if you consider the variance of this integral, of course you can represent variance as the mathematical expectation of the square of this integral minus mathematical expectation of the integral squared and if you combine this to formulas, you will immediately get that this is a doubled integral of the covariance function t_s dt ds. Here I didn't write any bounds of this integral, but I assumed that they were taken by some closed interval a b and actually since the function K is symmetric, we can also represent this integral as two integrals from a to b and from a to s K_t_s dt ds. Well, these are the basic properties of Stochastic integrals and as I said before it will be a separate lecture about this topic. But to continue our studies, to continue with the introduction of various properties, we already need the notion of this integral.
