Before I ask my question I will remind some core concepts.

Consider an underlying filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\geq0},P)$ satisfying the usual conditions. Then, denote with $\mathbf{M}^2$ the space of $L^2$-martingales on the underlying filtered probability space.

Let $\mathcal{A}\subseteq \mathbf{M}^2$. The set of $\mathbf{M}^2$ martingale measures for $\mathcal{A}$, denoted by $\mathcal{M}^2(\mathcal{A})$, is the set of all probability measure $Q$ on $\mathcal{F}$ such that

  1. $Q\ll P$
  2. $Q=P$ on $\mathcal{F}_0$
  3. if $X\in\mathcal{A}$ then $X$ is a $(Q,\{\mathcal{F}_t\}_{t\geq0})$-$L^2$ martingale.

A measure $Q$ is an extremal point of $\mathcal{M}^2(\mathcal{A})$ if whenever $Q=\lambda R +(1-\lambda)S$, with $R,S\in\mathcal{M}^2(\mathcal{A})$, $R\neq S$, $0\leq\lambda\leq 1$, then either $\lambda=0$ or $\lambda=1$.

Now, my question is the following: if we fix a filtered probability space as above, is it possible to construct a martingale $M\in\mathbf{M}^2$ (or alternatively a finite set of martingales $(M^i)_{i=1}^n\subset\mathbf{M}^2$) for which $P$ is an extremal point of $\mathcal{M}^2(\mathcal{A})$, where $\mathcal{A}=\{M\}$ (or alternatively $\mathcal{A}=\{M^1,\ldots,M^n\}$)? One possibility could be to construct $M$ for which $P$ is the only martingale measure or for which there exists at most another martingale measure. Is it possible?

Any idea would be really welcome.

Thank you.

  • $\begingroup$ It might be helpful to look at the Radon-Nikodym derivatives. For instance, if $\mathcal F_t=\sigma(X_t)$, $\mathcal F_t\uparrow\mathcal F$, then $Q\in\mathcal M^2(\{X\})$ implies $t\mapsto Y_t:=\frac{dQ|_{\mathcal F_t}}{dP|_{\mathcal F_t}}$ is a martingale and so is $t\mapsto X_tY_t$. This seems like a fairly restrictive condition, and it wouldn't surprise me if it implied $Y_t\equiv1$, or equivalently $Q=P$. $\endgroup$
    – Jason
    Aug 24, 2017 at 16:27


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