# Uniqueness in martingale representation theorem

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?

• Doesn't $Y = 1$ work? – Steve Huntsman Jan 12 '15 at 15:37
• @SteveHuntsman: If $Y=1$ then $\int_0^1 Y_t\,dW_t = W_1$, not 0. – Nate Eldredge Jan 12 '15 at 17:55
• Smack my head... – Steve Huntsman Jan 12 '15 at 23:34

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 $\int_0^1 Y_s\,dW_s = Z_\tau = 0$.

To show non uniqueness it is sufficient to exhibit a non trivial process $\phi \in L^2_{loc}[0,T]$ such that $\int_{0}^{T} \phi \, dW =0$. This is because, in this case, the representation for $\xi$ is given by

$\xi = \int_{0}^{T} Y dW = \int_{0}^{T} Y dW + 0 = \int_{0}^{T} Y dW + \int_{0}^{T} \phi \, dW = \int_{0}^{T} (Y+\phi) \, dW$

and as $\phi$ is non trivial, we have $Y + \phi$ to be different to $Y$.

Our only restriction is that $\phi$ be (locally) square integrable - let us pick the example that Nate mentioned, as it is easy to understand.

Consider any real number $a$ in the open set $]0,T[$ and set

$\phi (t) = \mathbb{1}_{s \in [0,a[} + \frac{1}{(T-s)}\mathbb{1}_{s \in [a,\tau]}$

for $t \in [0,T]$, where $\tau$ is the stopping time

$\tau = \inf \lbrace t \geq a : \int_{a}^{t} \frac{1}{(T-z)} \, dW_z = -W_a \rbrace$.

Now, two important observations:

• The event $\lbrace \tau < T \rbrace$ occurs $\mathbb{P}$-as, where $\mathbb{P}$ is the probability measure governed by the (standard) Weiner process $W$.
• We have $\phi \in L^2_{loc}[0,T]$.

See that

$\int_{0}^{T} \phi \, dW = \int_{0}^{a} \phi \, dW + \int_{a}^{T} \phi \, dW = W_a - W_a = 0,$

therefore the element $Y+\phi$ is also a valid representation for $\xi$.