Dudley's martingale representation theorem states that if $W=\{W_t,\mathcal{F}_t;0\le t<+\infty\}$ is a standard one-dimensional Brownian motion, $0<T<+\infty$ and $\xi$ is $\mathcal{F}_T$-measurable, then there exists a progressively measurable process $Y=\{Y_t,\mathcal{F}_t;0\le t\le T\}$ satisfying
$$\int_0^TY_t^2\mathrm{d}t<+\infty\text{ a.s.}$$
such that
$$\xi=\int_0^TY_t\mathrm{d}W_t\text{ a.s.}$$
But this theorem does NOT have uniqueness.
We need to find a progressively measureable process $Y$ such that
$$0<\int_0^1Y_t^2\mathrm{d}t<+\infty\text{ a.s.,}\quad\text{but }\int_0^1Y_t\mathrm{d}W_t=0\text{ a.s.}$$
Could anyone give such an example?
Best Answer
You can find a counterexample in J. Michael Steele's Stochastic Calculus and Financial Applications on page 196. Here's an essentially equivalent construction.
Let $r(t)$ be any positive continuous function on $[0,1)$ with $\int_0^1 r(s)^2\,ds = +\infty$ (for example, $r(t) = 1/(1-t)$). Set $s(t) = \int_0^t r(s)^2\,ds$ so that $s$ is continuous on $[0,1)$, strictly increasing, and $s(1-) = +\infty$. If we let $Z_t = \int_0^t r(s)\,dW_s$ then $Z_t$ is a time-changed Brownian motion; specifically, $\{Z_{s^{-1}(u)}, 0 \le u < +\infty\}$ is a Brownian motion with respect to the filtration $\{\mathcal{F}_{s^{-1}(u)}, 0 \le u < +\infty\}$. (Observe that $Z_{s^{-1}(u)}$ is continuous with independent Gaussian increments having the correct variances.) Let $$\tau = \inf \left\{t \in \left[\frac{1}{2}, 1\right] : Z_t = 0\right\}.$$ Since Brownian motion is recurrent, almost surely $Z_{s^{-1}(u)}$ hits 0 for some $u \ge s^{-1}(1/2)$, thus $Z_t$ hits zero for some $t \ge 1/2$. So $\tau < 1$ almost surely. Finally set $$Y(t,\omega) = \begin{cases} r(t), & 0 \le t \le \tau(\omega) \\ 0, & \tau(\omega) < t \le 1. \end{cases}$$ Note $\int_0^1 Y_t^2\,dt = s(\tau)$. Since $\frac{1}{2} \le \tau < 1$ almost surely, by construction, we have $0 < s(1/2) < s(\tau) < \infty$ almost surely, hence $0 < \int_0^1 Y_t^2\,dt < \infty$ almost surely. And $\int_0^t Y_s\,dW_s = Z_{t \wedge \tau}$ so $\int_0^1 Y_s\,dW_s = Z_\tau = 0$.