Question:
Let $X(t)=\int_0^t W(s)dW(s)$ where $W(t)$ is the Brownian motion (Wiener process). What function $f(X(t))$ of $X(t)$ can be used with the Itô formula to get the explicit result $X(t)=\frac{1}{2} W(t)^2 – \frac{1}{2}t$?
Details (if context needed):
The textbook introduction of Itô's lemma is preceded/motivated by the Itô (stochastic) integral of the type $\int X(s)dW(s)$, where $X(t)$ is an Itô-integrable stochastic process. It is then shown, as a simple initial example, that $\int_0^tW(s)dW(s)=\frac{1}{2} W(t)^2 – \frac{1}{2}t$, by simply writing the integral as the limit of the sum of sub-intervals and rearranging as such:
$$
\int_0^tW(s)dW(s) \\
= \lim_{n \to \infty} \left ( \sum_{i=0}^{n-1} W \left ( \frac{t}{n}i \right ) \left ( W \left ( \frac{t}{n}i+1 \right ) – W \left ( \frac{t}{n}i \right )\right ) \right ) \\
= \frac{1}{2} W(t)^2 – \frac{1}{2} \underbrace{\lim_{n \to \infty} \left ( \sum_{i=0}^{n-1} \left ( W \left ( \frac{t}{n}i+1 \right ) – W \left ( \frac{t}{n}i \right )\right ) \right )^2}_{=<W(t)>=t} \\
= \frac{1}{2} W(t)^2 – \frac{1}{2}t \\
$$
where $<W(t)>$ is the quadratic variation. [explicit calculations are here chp19.2.2, p139; also cf. Bingham & Kiesel (2014). Risk Neutral Valuation. Chp5.5, Eq(5.1)]. Then the Itô formula is introduced as a generalization.
Can the above result be shown via the Itô formula, and not via the limiting sum as was done above? If so, how?
For example, for the geometric Brownian motion with SDE $dX(t)=X(t)(\mu dt + \sigma dW(t))$ the function $f(X(t))=\log(X(t))$ is used with the Itô formula. Some terms cancel nicely out so that all integrals become easy to calculate and then we can solve for $X(t)$. Is something like this possible for the integral above?
Best Answer
You're seeking an explicit formula for $$X(t)= \int_0^t W(s)dW(s),$$ which is of the form $\int_0^t g(W(s))dW(s)$ for $g(x)=x$. You want to calculate it by Ito's Formula, which in the case of Wiener Process boils down to $$f(W(t)) = f(0) + \int_0^t f'(W(s))dW(s) + \frac{1}{2}\int_0^t f''(W(s))ds. $$ The only term which has integral with respect to Wiener Process (that is, $dW(s)$ term) is the middle one, hence we need $f$ such that $X(t) = \int_0^t f'(W(s))dW(s)$ for any $t$, so the "simpliest" candidate is to try to find $f$ such that $f'(x) = x$ (then $f'(W(s)) = W(s)$). We see that $f(x) = \frac{x^2}{2}$ works, which gives $$ \frac{W^2(t)}{2} = 0 + \int_0^t W(s)dW(s) + \frac{1}{2}\int_0^t 1 ds = \int_0^t W(s)dW(s) + \frac{t}{2},$$ hence $$ X(t) = \int_0^t W(s)dW(s) = \frac{1}{2}\left( W^2(t) - t\right).$$