Deriving the definition of stochastic integrals with respect to Ito processes from first principles

Question

When I first encountered the definition of integrals with respect to Ito processes (Shreve's Stochastic Calculus for Finance Vol II), I didn't think twice. However, I wanted to see if the definition could be derived.

In the rest of this post $\bar{f}$ is such that $\bar{f}'=f$ and $t_{j}^{f}$ is such that $t_{j}\leq t_{j}^{f}\leq t_{j+1}$ according to the mean-value theorem, for some process $f$.

Consider the process $$X(t)=X(0)+\int_{0}^{t}\Delta(u)\;dW(u)+\int_{0}^{t}\Theta(u)\;du$$ where the processes $\Theta_{t}(\omega)$ and $\Delta_{t}(\omega)$ are adapted to the Brownian motion $W_{t}(\omega)$. Then if $\Gamma_{t}(\omega)$ is adapted to $X_{t}(\omega)$, we define $$\int_{0}^{t}\Gamma(u)\;dX(u)=\int_{0}^{t}\Gamma(u)\Delta(u)\;dW(u)+\int_{0}^{t}\Gamma(u)\Theta(u)\;du,$$ which is of course what we would expect by writing (formally)

$$\int_{0}^{t}\Gamma(u)(\Delta(u)\;dW(u)+\Theta(u)\;du)=\int_{0}^{t}\Gamma(u)\Delta(u)\;dW(u)+\int_{0}^{t}\Gamma(u)\Theta(u)\;du.$$

It seems this should be derivable (rather easily) from the original definition of the Ito stochastic integral. However, when I tried to do this, I failed:

$$\begin{align*} \int_{0}^{t}\Gamma(u)\;dX(u)&=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\left(\int_{t_{j}}^{t_{j+1}}\Delta(u)\;dW(u)+\int_{t_{j}}^{t_{j+1}}\Theta(u)\;du\right)\\ &=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\cdot\int_{t_{j}}^{t_{j+1}}\Delta(u)\;dW(u)+\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\cdot\int_{t_{j}}^{t_{j+1}}\Theta(u)\;du\\ &=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\cdot\int_{t_{j}}^{t_{j+1}}\Delta(u)\;dW(u)+\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})(\bar{\Theta}(t_{j+1})-\bar{\Theta}(t_{j}))\\ &=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\cdot\int_{t_{j}}^{t_{j+1}}\Delta(u)\;dW(u)+\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\Theta(t_{j}^{\Theta})(t_{j+1}-t_{j})\\ &=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\cdot\int_{t_{j}}^{t_{j+1}}\Delta(u)\;dW(u)+\int_{0}^{t}\Gamma(u)\Theta(u)\;du. \end{align*}$$ where $t_{j}\leq t_{j}^{\Theta}\leq t_{j+1}$ and $\bar{\Theta}'=\Theta.$

At this point it would appear nothing can be done with the first sum since we don't have a mean-value theorem for Ito stochastic integrals (as quantified by Ito's lemma); i.e. there does not in general exist a $t_{j}^{\Delta}\in[t_{j},t_{j+1}]$ such that $$\int_{t_{j}}^{t_{j+1}}\Delta(u)\;dW(u)=\Delta(t_{j}^{\Delta})(W(t_{j+1}-W(t_{j})).$$

And in any event it doesn't really matter, since even if we had this result, it would not be known a priori whether the resulting sum converges (or whether it converges to the correct stochastic integral) due to the sensitivity of the limiting sums with respect to the sampling point used.

To get a more concrete sense of the difficulty, consider the special case $\Delta(u)=f(W(u))$.

Then $$\int_{t_{j}}^{t_{j+1}}\Delta(u)\;dW(u)=\bar{f}(W(t_{j+1}))-\bar{f}(W(t_{j}))-\frac{1}{2}\int_{t_{j}}^{t_{j+1}}f'(W(u))\;du,$$

and the first sum becomes $$ \begin{array}{l} \lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\left(\bar{f}(W(t_{j+1}))-\bar{f}(W(t_{j}))-\frac{1}{2}\int_{t_{j}}^{t_{j+1}}f'(W(u))\;du\right)\\ \;\;\;\;=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\left(\bar{f}(W(t_{j+1}))-\bar{f}(W(t_{j}))\right)-\lim_{n\to\infty}\frac{1}{2}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\left(\int_{t_{j}}^{t_{j+1}}f'(W(u))\;du\right)\\ \;\;\;\;=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})f(W(t_{j}^{f}))(W(t_{j+1})-W(t_{j}))-\lim_{n\to\infty}\frac{1}{2}\sum_{j\in\Pi_{n}}\Gamma(t_{j})f'(W(t_{j}^{f'}))(t_{j+1}-t_{j})\\ \;\;\;\;=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\Delta(t_{j}^{f})(W(t_{j+1})-W(t_{j}))-\frac{1}{2}\int_{0}^{t}\Gamma(u)f'(W(u))\;du.\end{array} $$

Putting this together yields

$$\int_{0}^{t}\Gamma(u)\;dX(u)=\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\Delta(t_{j}^{f})(W(t_{j+1})-W(t_{j}))-\frac{1}{2}\int_{0}^{t}\Gamma(u)f'(W(u))\;du+\int_{0}^{t}\Gamma(u)\Theta(u)\;du,$$ and it's hard to see how this ends up being equal to the original definition.

One would need to somehow show that the mean-value sampling $\Delta(t_{j}^{f})$ in the first sum results in

$$\lim_{n\to\infty}\sum_{j\in\Pi_{n}}\Gamma(t_{j})\Delta(t_{j}^{f})(W(t_{j+1})-W(t_{j}))=\int_{0}^{t}\Gamma(u)\Delta(u)\;dW(u)+\frac{1}{2}\int_{0}^{t}\Gamma(u)f'(W(u))\;du.$$

And in any event, this is just a special case of the process $\Delta_{t}(\omega)$.

I have no doubt the issue can be resolved analytically; however, while the difficulties remain unresolved, it tends to make (in my mind) the definition somewhat artificial.

Answer 1

Let me first start with your very last equation. There, you just stumbled into an application of the Wong-Zakai decomposition: http://projecteuclid.org/euclid.aoms/1177699916

It essentially says that if a sequence of processes $( M_n )$ converges weakly to a martingale $M$, then the sequence of processes, $ t \mapsto \int_0^t \Gamma(s)dM_n(s) $ converges weakly to the process $ t \mapsto \int_0^t \Gamma(s)\circ dM_n(s)$, where $\circ$ is the Stratonovich integral (which is Ito integral plus a quadratic variation correction). In your case the approximating process con be made to be $M_n(t) = \Delta(t_j^f) W(t)$, $t \in [(t_j, t_{j+1})$, where $( t_j ) $ is a uniform partition on $n$ disjoint intervals of $[0,t]$ and $t_j^f$ is as you defined (which dependens on $n$ ).

Having said that the construction of the stochastic integral from first principles starts by noticing that no path wise approach can be applied and hence studying the convergence in the $L^2$ sense. This is standerd and it can be found for exempla in the Karatzas & Shreve stochastic integration and Brownian Motion book, or a different approach can be found in the Stochastic Integration book by Philip Protter.

Answer 2

Ito integrals differ in their complexity depending on which function you'd like to integrate. Let's talk about the simplest but still powerful version - the one mentioned by Innombrabre, for integrands from the $L^2$ space.

In general, you can think of the procedure as follows. You have two complete metric spaces $(X,d_X)$ and $(Y,d_Y)$. Let $A$ be a subset of "simple" elements of $X$ for which we can easily define a function $f$. If $f$ is Lipschitz, then every Cauchy sequence $x_0,x_1,\dots$ on $X$ generates a Cauchy sequence $y_0,y_1,\dots$ on $Y$ where $y_i = f(x_i)$. As a result, if you define $f$ on $A$ then you can for free continuously extend it to $\bar A$. Moreover, it will still stay Lipschitz.

In your case $X$ is the path space, and $Y$ is the space of random variables, $A$ are simple (piece-wise constant) paths and $f$ is the Ito integral.