Non-uniqueness in the $L^1$ martingale representation

Question

Let $\xi \in L^1(P,\mathfrak F_T)$ on some probability space with measure $P$, supporting a Brownian motion, we consider the augmented filtration $\mathfrak F$ associated to $W$, and a time $T>0$. Then, there exists a predictable process $H$ such that $\int_0^T H^2_t ~ dt < \infty$ and \begin{equation*} \xi = E[\xi] + \int_0^T H_t ~ dW_t. \end{equation*} If $\xi \in L^2(P)$, then $H$ is unique (this follows from the Ito isometry). In general, the Ito isometry argument does not work. Do you know an interesting counterexample (i.e. two different martingale representations for $\xi \in L^1(P)$)?

Answer 1

Suppose $\xi$ is only integrable. Let $X=(X_t)_{t\in[0,T]}$ denote a continuous version of the martingale $\Bbb E[\xi|\mathfrak F_t]$, $0\le t\le T$. By localizing the usual $L^2$ martingale representation, there is a predictable process $K$ with $\int_0^T K_s^2\,ds<\infty$ a.s., such that $$ X_{t}=\Bbb E[\xi]+\int_0^t K_s\,dW_s,\qquad 0\le t\le T, \hbox{ a.s.} $$ The process $K$ is uniquely determined by the martingale $X$. On the other hand, using the representation you cite, we can define a local martingale $Y$ by $$ Y_t=\Bbb E[\xi]+\int_0^t H_s\,dW_s,\qquad 0\le t\le T. $$ The difference $X_t-Y_t$ is a local martingale on $[0,T]$ with terminal value $0$. If we knew that this implied $X_t=Y_t$ for all $t$, then we would have $H=K$ and the uniqueness you seek. But examples show that a local martingale need not be determined by its terminal value.

For example (modifying an example found in R. Williams' text on Math. Finance), take $T=1$ and define $Z_t=\int_0^t (1-s)^{-1/2}\,dW_s$ for $0\le t<1$. Then $Z$ is a local martingale on $[0,1)$, and one checks $\langle Z\rangle_t\uparrow+\infty$ as $t\uparrow 1$. It follows that the stopping time $\tau:=\inf\{t>1/2: Z_t=0\}$ satisfies $0<\tau<1$ a.s. The stopped process $Z_{t\wedge\tau}$ is then a local martingale on $[0,1]$ with terminal value $Z_1=0$, but also $Z_1=\int_0^1 J_s\,dW_s$, where $J_s=1_{\{s\le\tau\}}(1-s)^{-1/2}$. Notice that $J$ is non-zero!

More intriguing, R.M. Dudley showed in 1977 that any $\mathfrak F_T$ measurable r.v. $\xi$ admits a representation $\xi=c+\int_0^T H_s\,dW_s$ with $H$ predictable such that $\int_0^T H_s^2\,ds<\infty$ a.s., and the constant $c$ chosen arbitrarily by you ahead of time.