Consider an incomplete market $(\Omega,\mathcal F,\mathbb P)$ driven by a semimartingale $S=(S_t)_{t\in[0,T]}$. Under the no free lunch under vanishing risk (NFLVR) assumption, the set $\mathcal P^\ast$ of equivalent martingale measures under which $S$ is a sigma-martingale is non-empty: $$ \mathcal P^\ast\neq\emptyset,\quad\mathbb P^\ast\sim\mathbb P,\enspace \forall \mathbb P^\ast \in \mathcal P^\ast. $$

Denote densities $$ Z_{\mathbb{P^*}}=\frac{d\mathbb P^\ast}{d\mathbb P},\quad \text{for }\mathbb P^\ast\in \mathcal P^\ast, $$

and the set of densities $$Z_{\mathcal{P^\ast}}=\{Z_{\mathbb{P^\ast}}\ :\ \mathbb P^\ast\in \mathcal P^\ast\}$$

Now, a certain theorem (a generalized version of the Neyman-Pearson lemma for incomplete markets, see Theorem 4.9 p.55 in this thesis) requires that $Z_{\mathcal P^\ast}$ is compact in $L^1(\Omega,\mathcal F,\mathbb P)$. I would like to see how restricting this assumption is by trying to provide an incomplete market example (i.e., a model where $\mathcal P^\ast$ is not a singleton) where $Z_{\mathcal P^\ast}$ is compact. However, I haven't been able to do so.

I first tried the discrete-time one-period trinomial model, in which the set of martingale measures and the set of associated densities represent convex polyhedrons. However, due to the measure equivalence requirement, the polyhedron is open (at the extreme points, one or more nodes will have zero probability, hence the extreme points are excluded along with the entire boundary). Therefore, it is not compact?

Next thing, I tried the continuous-time jump-diffusion model. The set of equivalent martingale measures can be parametrized with two real numbers, and the Radon-Nikodym derivative can be written down explicitly, however I have no idea how to approach the problem of verifying the compactness property.

Could someone provide an example of an incomplete market (where the set of equivalent martingale measures is not a singleton) where the set of densities is indeed compact?

Alternatively, can it be proved that $Z_{\mathcal P^\ast}$ is compact if and only if $\mathcal P^\ast$ is a singleton?

**Update**
As pointed below by @weakstar, if there exists a martingale measure $\mathbb Q$ that is absolutely continuous with respect to $\mathbb P$ but not equivalent to $\mathbb P$, a convex combination of $\mathbb Q$ and an equivalent martingale measure $\mathbb P^\ast\in\mathcal P^\ast$ would also be an equivalent martingale measure. In this case, it is possible to construct a sequence of densities $Z_{\mathbb Q^\alpha}$ of equivalent martingale measures converging to density $Z_{\mathbb Q}$ of a non-equivalent martingale measure which violates the closedness. This example naturally eliminates models like the multinomial model.

nevercompact except in the case where $P^\ast$ is a singleton. $\endgroup$ – George Lowther Jul 3 '12 at 22:253more comments