It$\hat{o}$'s stochastic integral

Question

I am currently taking a course on Stochastic Analysis, and in the chapter entitled It$\hat{o}$'s Calculus, I am introduced to this integral: $$ \int^{t}_{0}Y_s\,dW_s$$ as a stochastic process, where $Y_s$ is a simple process. I have understood up to now what a Brownian Motion is, what a Martingale, Filtration, probability space, etc... is. But this, I am really stumped on. The notion of $dW_s$ is confusing me, and why this integral is so important in the grand scheme of things? What is the significance of this integral, and what exactly is $Y_s$? I feel that I can't move on with this course until I understand the very essentials of what Ito calculus is and why we are concerned with it. A answers that shed some light on this topic would be vastly appreciated.

Answer 1

You can think of the Itô integral much like the Riemann-Stieltjes integral: you are integrating with respect to a Wiener process. In general, you need $Y$ to be a locally square-integrable process which is adapted to the filtration generated by $W$. It seems you know what $W$ is, but for the sake of completeness: $W$ is usually a Brownian motion, but in general it can be a semimartingale. Itô integrals are important for giving a rigorous meaning to stochastic differential equations, which are extremely useful for modeling all kinds of things. For example, SDE are incredibly important in mathematical finance.

In defining the Itô integral, one generally starts by defining it for simple processes (much like defining the Lebesgue integral for simple functions) and then shows that you can approximate other processes by limits of these simple processes.

Edit: A simple process is like a simple function, it is of the form $$X_t = \sum_j x_j \chi_{E_j}(t),$$ where $E_j$ is a measurable set and $\chi_E$ denotes the characteristic function of the set $E$. For example, one might take $E_j = [t_{j-1},t_j)$. I generally think of a simple functions and simple processes as those with finite image.

Answer 2

Just to add some information to other answers.

$d W_s$ is change in Brownian process in time. (So it is indeed random variable). As any random process is random function of time (so there is something governing randomness and $\omega$, and $t$), it can be described as by its path (non-random function of time for each fixed $\omega$.). Suppose the general space (with filtration), is described by $\Omega$. So each $\omega$ defines path for $Y_s$ and $W_s$. In that case (i.e. fixing $\omega \in \Omega$), this integral will be fixed number (if integrable).

From this you can say that this integral depends on $\omega$, and for each $\omega$ it takes some value.

So this integral gives you random variable.

So this integral takes random process and gives you random variable.

Getting back to $dW_s$. It can be shortly written as $W(s+\Delta s)-W(s)$ in limit of $\Delta s$ going to zero.

So the general integral is of form like $\sum_{i} Y(s_i)(W(s_{i+1}-W(s_i))$ with partition length going to zero.

For each path (each fixed $\omega$) it should converge to limit regardless of partition.

And answering the question why it is important.

Suppose you buy something (for example stock, or gold, or oil, or something else) changing in value during time governed by random process $W_{s}$ (So describes your price of that). And suppose you buy $Y_{0}$ numbers of that. Additionally you may change this number during time, facing new prices. Then what you get (your gain) in small time $Y_{s}\times(W(s_{i+1})-W(s_i))$. After that you change the quantity $Y_{s}$. So this is gain in small time. So in general the integral will describe your gain during whole period $[0,T]$.

In general $Y_{s}$ will indeed be random process, but with some additional restrictions. (Not always you can change the quantities that fast).

And lastly why mostly one take $dW_s$. There is a famous martingle representation theorem, which roughly states that in some probability spaces any martingle can be represented by $M_{t}=M_{0}+\int _{0}^{t}Y_{s}dW_{s}$, with some good enough process(not necessarily simple) $Y_{s}$.

This may lead to some thoughts that this integral representation cover enough, to work with.