# Compactness of the set of densities of equivalent martingale measures

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.

• @ Aldanor : Hi you say that Neyman-Pearson's lemma gives compactness of the set $Z_{\mathcal{P}^*}$. Could you give a reference for this statement and also give the topology with respect to which this set is compact. Best Regards – The Bridge Jul 2 '12 at 9:54
• @The Bridge: The theorem I was talking about is Theorem 3.1 on page 43 from this PhD thesis and also a more generalized version in Theorem 3.3 on page 46. If I understand correctly, the compactness is required in order to be able to apply Tonelli's theorem (Remark 3.4 on page 48). As stated in the text, the set is required to be compact "as a subset of $L^1$". – Aldanor Jul 2 '12 at 15:16
• @Aldanor: you may get more answers on the SE site which is devoted to Quantitative Finance quant.stackexchange.com – Andrey Rekalo Jul 3 '12 at 10:26
• I think that $Z_{\mathcal{P^\ast}}$ is never compact except in the case where $P^\ast$ is a singleton. – George Lowther Jul 3 '12 at 22:25
• ...I worked through all the cases and it is true that you can always construct an absolutely continuous but not equivalent martingale measure whenever $\mathcal{P}^\ast$ contains more than one element, so it is not compact. I'll post the proof when I have time, but it is a bit involved. – George Lowther Jul 9 '12 at 13:27

The set $Z_{\mathcal{P^\ast}}$ is never compact except in the case where it is a singleton (or empty). This is for the general case with $S=(S^1,S^2,\ldots,S^d)$ being an $\mathbb{R}^d$-valued semimartingale.

Let $\mathbb{Q}$ be an equivalent measure under which $S$ is a sigma-martingale, so $Z_\mathbb{Q}\in\mathcal{P^\ast}$. If there existed any measure $\mathbb{Q}^\prime$ absolutely continuous, but not equivalent, to $\mathbb{P}$ with respect to which $S$ is a sigma-martingale then we can argue as in weakstar's answer. The measures $\mathbb{Q}\_n=\frac1n\mathbb{Q}+(1-\frac1n)\mathbb{Q}^\prime$ are in $\mathcal{P}^\ast$, so $Z_{\mathbb{Q}\_n}=\frac1nZ_\mathbb{Q}+(1-\frac1n)Z_{\mathbb{Q}^\prime}$ are in $Z_{\mathcal{P}^\ast}$. However, $Z_{\mathbb{Q}^\prime}=\lim_{n\to\infty}Z_{\mathbb{Q}\_n}$ in $L^1$. Hence, $Z_{\mathcal{P^\ast}}$ is not closed in $L^1$ and cannot be compact. In general, you can show that if $\mathcal{P^\ast}$ contains more than one element, then there does exists an absolutely continuous measure which is not equivalent to $\mathbb{P}$ under which $S$ is a sigma-martingale, so $Z_{\mathcal{P^\ast}}$ is not closed in $L^1$. The proof of this which I give here is rather long, because it is broken down into 4 separate cases and a tricky lemma. Maybe there is a more efficient proof, but I can't think of one now.

Suppose that $\mathbb{Q},\mathbb{Q}^\prime$ are distinct elements of $\mathcal{P^\ast}$. Replacing the original measure $\mathbb{P}$ by $\mathbb{Q}^\prime$ if necessary, without loss of generality we can suppose that $\mathbb{Q}^\prime=\mathbb{P}$. Furthermore, replacing $\mathbb{P}$ by $(\mathbb{P}+\mathbb{Q})/2$, we can suppose that $d\mathbb{Q}/d\mathbb{P}$ is uniformly bounded by 2. Also, replacing $S$ by $\int\varphi\,dS$ for some strictly positive predictable process $\varphi$, we can suppose that $S$ is an $\mathcal{H}^1$ martingale rather than just a sigma-martingale (see Proposition 2.6 of The Fundamental Theorem Of Asset Pricing For Unbounded Stochastic Processes). That is, $\sup_t\lVert S_t\rVert$ is $\mathbb{P}$-integrable.

Define the $\mathbb{P}$-martingale $$U_t=\frac{d\mathbb{Q}}{d\mathbb{P}}\Bigg\vert\_{\mathcal{F}_t}=\mathbb{E}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}\;\Bigg\vert\mathcal{F}\_t\right].$$ This is a uniformly bounded martingale with $\mathbb{E}[U_t]=1$ such that $US$ is a martingale. We cannot have $U_t=1$ almost surely for each $t$, as $\mathbb{P},\mathbb{Q}$ are assumed to be distinct.

Assume that the filtration is right-continuous and complete (you can always pass to the right-continuous version with no problems, and completeness is generally assumed for concepts such as stochastic integration to make sense). In that case, every martingale has a cadlag modification. So, take a cadlag modification for $U$. Then, $U$ is a strictly positive and non-constant martingale with $S,US$ both martingales. By integration by parts, this implies that the quadratic covariation $[U,S]$ is a local martingale. Suppose that we can find a nonnegative martingale $M$ which is not identically zero such that $MS$ is a martingale and $\mathbb{P}(M_T=0) > 0$. Then, we can define an absolutely continuous but non-equivalent martingale measure $\mathbb{P}^\ast$ by $$\frac{d\mathbb{P}^\ast}{d\mathbb{P}}=M_T/\mathbb{E}[M_T].$$ This would imply that $Z_{\mathcal{P}^\ast}$ is not closed in $L^1$ by the argument above. Let's consider the following four exhaustive (and increasingly exhausting) cases.

1. $U_0$ is not almost-surely equal to 1. Then $\mathcal{F}_0$ would contain a set $A$ with probability neither 0 or 1, so $M$ can be taken to be the (almost-surely constant) martingale $$M_t=1_A/\mathbb{P}(A).$$

2. $U_0=1$ and $\Delta U_t\ge 0$ for all $t$. That is, $U$ only has positive jumps. As $U$ is assumed to be not almost-surely constant, there will exist $K\in(0,1)$ such that $\mathbb{P}(\inf_tU_t < K) > 0$. Let $\tau$ be the first time at which $U_\tau\le K$, and $\tau=\infty$ if this never happens. As $U$ has nonnegative jumps, $U_\tau=K$ whenever $\tau < \infty$. Then, the martingale $M$ can be defined by $$M_t=U_{t\wedge\tau}-K.$$ Note that this hits zero whenever $\tau < \infty$, which happens with positive probability.

3. There exists a predictable stopping time $\tau > 0$ at which $\mathbb{P}(\Delta U_\tau\not=0) > 0$. Then, as $U,S,US$ are martingales, \begin{align} \mathbb{E}[\Delta U_\tau \Delta S_\tau\vert\mathcal{F}\_{\tau-}]&=\mathbb{E}[\Delta(US)\_\tau-U_{\tau-}\Delta S_\tau - S_{\tau-}\Delta U_\tau\vert\mathcal{F}\_{\tau-}]\cr &=0. \end{align} As $U$ is bounded by $2$, $\Delta U_\tau \ge -2$. Also, as $\Delta U_\tau$ is not identically zero and $\mathbb{E}[\Delta U_\tau\vert\mathcal{F}\_{\tau-}]=0$, there is an $\epsilon > 0$ such that $\Delta U_\tau\ge\epsilon$ with positive probability. Applying the lemma below to the $\mathbb{R}^{d+1}$-valued $\mathcal{F}\_\tau$-measurable random variable $(1,\Delta S_\tau)$ and nonnegative random variable $1+\Delta U_\tau/2$, there exists a uniformly bounded $\mathcal{F}\_{\tau-}$-measurable random variable $Y\ge1+\epsilon/2$, a nonnegative $\mathcal{F}\_\tau$-measurable variable $X\le Y$ such that $\mathbb{P}(X=Y) > 0$ and, \begin{align} &\mathbb{E}[X\Delta S_\tau\vert\mathcal{F}\_{\tau-}]=\mathbb{E}[(1+\Delta U_\tau/2)\Delta S_\tau\vert\mathcal{F}\_{\tau-}]=0\cr &\mathbb{E}[X\vert\mathcal{F}\_{\tau-}]=\mathbb{E}[1+\Delta U_\tau/2\vert\mathcal{F}\_{\tau-}]=1. \end{align} The martingale $M$ is then given by $$M_t = 1 - 1_{\{ t\ge\tau\}}(X-1)/(Y-1).$$ Note that $M_T=0$ whenever $X=Y$, which has positive probability, and that the quadratic covariation $$[M,S]\_t=1_{\{t\ge\tau\}}(X-1)\Delta S_\tau/(Y-1)$$ is a martingale so, by integration by parts, $MS$ is a martingale.

4. $U$ is quasi-left-continuous (i.e., $\Delta U_\tau=0$ a.s. for each predictable stopping time $\tau$) and, with positive probability, $\Delta U_t < 0$ for some $t\in(0,T]$. Then, there exists $\epsilon > 0$ so that, with positive probability, $\Delta U_t < -\epsilon$ for some $t\in(0,T]$. Let $\tau$ be the first time at which $\Delta U_\tau < -\epsilon$ and $\tau=\infty$ if this never happens. By the lemma below, there exists an $\mathcal{F}\_{\tau-}$-measurable $Y\ge\epsilon$ and a nonnegative $\mathcal{F}\_\tau$-measurable random variable $X\le Y$ such that $\mathbb{P}(X=Y) > 0$ and $$\mathbb{E}[X\Delta S_\tau\vert\mathcal{F}\_{\tau-}]=\mathbb{E}[-\Delta U_\tau\Delta S_\tau\vert\mathcal{F}\_{\tau-}].$$ Now, let $V$ be the step process $V_t=1_{\{t\ge\tau\}}(X+\Delta U_\tau)$. The quadratic covariation $$[V,S]\_t=1_{\{t\ge\tau\}}(X+\Delta U_\tau)\Delta S_\tau$$ is a martingale. As $\mathbb{P}(\sigma=\tau < \infty)=0$ for each predictable $\sigma$, $\tau$ is totally inadmissible. Therefore, it has a compensator $V^p$, which is a continuous finite variation adapted process starting from zero such that $V-V^p$ is a martingale. Consider the martingale $$N_t=U_{t\wedge\tau}+V^p_t-V_t.$$ Then, $[N,S]=[U,S]^\tau-[V,S]$ is a local martingale. Also, $\Delta N_t\ge-\epsilon$ for $t < \tau$ and $\Delta N_\tau=-X\ge-Y$ (and equality holds with positive probability). As $Y$ is $\mathcal{F}\_{\tau-}$-measurable, there is a predictable process $\xi$ with $\xi_\tau=Y$. Replacing $\xi$ by $\xi\vee\epsilon$, we can suppose that $\xi\ge\epsilon$. Consider the martingale $$\tilde N=\int\xi^{-1}\,dN.$$ Then $\Delta\tilde N_t=\xi_t^{-1}\Delta N_t\ge-1$ and $\Delta\tilde N_\tau=-1$ with positive probability. Let $M$ be the solution to the SDE $$M_t = 1+\int_0^tM_{s-}\,d\tilde N_s.$$ This is the Doléans exponential of $\tilde N$ and is given explicitly by $$M_t=\exp\left(\tilde N_t-\frac12[\tilde N]\_t\right)\prod_{s\le t}e^{-\Delta\tilde N_s+\frac12(\Delta\tilde N_s)^2}\left(1+\Delta\tilde N_s\right).$$ As $\Delta\tilde N\ge-1$, this is nonnegative. Also, $M_T=M_\tau=0$ whenever $\Delta\tilde N_\tau=-1$, which occurs with positive probability. As $dM_t=M_{t-}\,d\tilde N_t$ and $d[M,S]\_t=M_{t-}\,d[\tilde N,S]\_t$, $M$ and $[M,S]$ are local martingales so, by integration by parts, $MS$ is a local martingale. By localization (replacing $M$ by $M^\sigma$ for a suitable stopping time $\sigma$ with $\mathbb{P}(\sigma\ge\tau) > 0$), we can assume that $M$ and $MS$ are both proper martingales. So, $M$ satisfies the required properties.

That's it! There does exist an absolutely continuous but not equivalent martingale measure. It just remains to state and prove the following lemma which was used in cases 3 and 4.

Lemma: Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G}$ be a sub-σ-algebra of $\mathcal{F}$, $S$ be an integrable $\mathbb{R}^d$-valued random variable and $Z$ be a uniformly bounded nonnegative random variable with $\mathbb{P}(Z\ge K) > 0$ for some $K > 0$. Then, there exists a nonnegative random variable $X$ and a uniformly bounded $\mathcal{G}$-measurable random variable $Y\ge K$ such that $X\le Y$, $\mathbb{P}(X=Y) > 0$ and $$\mathbb{E}[XS\vert\mathcal{G}]=\mathbb{E}[ZS\vert\mathcal{G}].$$

To prove the lemma, I'll make use of the relatively simple fact that if $\mu$ is a probability measure on $\mathbb{R}^d$ and $S\subseteq\mathbb{R}^d$ is Borel with $\mu(S)=1$, then $\int x\,d\mu(x)$ lies in the convex hull of $S$ (note: this only works in finite dimensions. In general Banach spaces you are only guaranteed that it lies in the closure of the convex hull). Let $V$ be an affine subspace with minimal dimension such that $\mu(V)=1$. Then, $y\equiv\int x\,d\mu(x)\in V$. Suppose that it did not lie in ${\rm conv}(S)$. Then, by the Hahn-Banach theorem/separating hyperplane theorem, there exists a nontrivial affine map $L\colon V\to\mathbb{R}$ such that $L(x)\ge0$ for $x\in S$ and $L(y)=0$. However, $L(y)=\int\_V L(x)\,d\mu(x)$. For this to be zero, $L(x)$ must be zero $\mu$-almost everywhere, so $\mu(W)=0$ where $W=L^{-1}(0)$ contradicting the minimality of $V$. So $y\in{\rm conv}(S)$.

Next, if $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space and $U\colon\Omega\times(0,1)\to\mathbb{R}^d$ is jointly measurable such that $U(\cdot,x)$ is $\mathbb{P}$-integrable for each $x$, then we can define a random variable $V(\omega)=\int_0^1U(\omega,x)\,dx$. By the paragraph above, $V(\omega)$ lies in the convex hull of $\{U(\omega,x)\colon x\in(0,1)\}$ for each $\omega$. The convex combinations can be done in a measurable fashion; there are sequences $p_n,x_n$ of random variables such that $p_n$ are nonnegative, eventually zero with $\sum_np_n=1$, $x_n\in(0,1)$ and $$V(\omega)=\sum_np_n(\omega)U(\omega,x_n(\omega))$$ for $\mathbb{P}$-almost every $\omega$. That this can be done in a measurable way is a consequence of the measurable section theorem (broken link) (that's the planetmath link I wrote, see also Donald L. Cohn, Measure theory. Birkhäuser, 1980).

Now, moving onto the proof of the lemma. Let $Y$ be the essential infimum of all $\mathcal{G}$-random variables bounded below by $Z\vee K$. By definition, $K\le Y$ and $Z\le Y$. If $\mathbb{P}(Z=Y) > 0$ then we can set $X=Z$ and we are done. Otherwise, for each $x\in(0,1)$, define the random variable $$U_x=\mathbb{E}\left[1_{\{Z\ge xY\}}YS\vert\mathcal{G}\right].$$ This can be chosen to be jointly measurable (e.g., take it to be finite variation in $x$) and, $$\int_0^1U_x\,dx=\mathbb{E}\left[\int_0^11_{\{Z\ge xY\}}YS\,dx\big\vert\mathcal{G}\right]=\mathbb{E}[ZS\vert\mathcal{G}].$$ By the discussion above, this can also be written in the form $\sum_np_nU_{x_n}$ for $\mathcal{G}$-measurable random variables $p_n\ge0$, $x_n\in(0,1)$ where $\sum_np_n=1$ and $p_n$ is eventually zero. Set $$X=\sum_np_n1_{\{Z\ge x_nY\}}Y.$$ This satisfies the required identity. Letting $x_\ast(\omega)$ be the maximum of $\{x_n(\omega)\colon p_n(\omega) > 0\}$, then $(x_\ast Y)\vee K < Y$ with positive probability so, by the definition of $Y$ as an essential infimum, $Z > x_\ast Y$ with positive probability. However, in this case, $X=Y$.

• Wow, George, thanks for input, didn't quite expect such a detailed answer! The proof looks valid in general, I'll just have to go through it again in detail to verify. Concerning the choice of measures $\mathbb{P}, \mathbb{Q}, \mathbb{Q'}$ - since we are basically looking for an arbitrary abs. continuous but not equivalent measure, the choice of $\mathbb{P}$ doesn't affect anything, is that what you meant by "without the loss of generality"? Basically, if $Z_{\mathcal{P}^\ast}$ is non-compact for some $\mathbb{P}$, it's non-compact for all $\tilde{\mathbb{P}}\sim\mathbb{P}$, right? – Aldanor Jul 10 '12 at 17:05
• I'm still a little confused about the third paragraph. You say we replace $\mathbb{P}$ with $\mathbb{Q}'$, then replace it with $\tfrac{1}{2}(\mathbb{P}+\mathbb{Q})$? Also, $d\mathbb{Q}/d\mathbb{P}$ is bounded uniformly with respect to what? Finally, what is the set $A$ you use in part (1)? – Aldanor Jul 10 '12 at 18:01
• To answer your questions: (i) Switching $\mathbb{P}$ to any equivalent measure makes no difference because we are only trying to show that there is an absolutely continuous but non-equivalent martingale measure. The set of such measures only depend on $\mathbb{P}$ up to equivalence. (ii) the measure $\mathbb{P}^\prime=(\mathbb{P}+\mathbb{Q})/2$ is an equivalent martingale measure satisfying $d\mathbb{Q}/d\mathbb{P}^\prime\le 2$. So, replacing $\mathbb{P}$ by $\mathbb{P}^\prime$ if necessary, we can assume that $d\mathbb{Q}/d\mathbb{P}\le2$. By bounded uniformly, I mean bounded in the uniform – George Lowther Jul 11 '12 at 0:00
• ($L^\infty$) norm. (iii) $A\in\mathcal{F}\_0$ is any set with probability in (0,1). The point is that $\mathcal{F}\_0$ has such sets in this case. Actually, you can always take $A=\lbrace U_0\ge1\rbrace$. – George Lowther Jul 11 '12 at 0:03
• you can contact me at my first name dot last name at blueyonder dot co dot uk. – George Lowther Jul 11 '12 at 0:04

It seems to me that in the statement of the Neyman-Pearson Lemma, equivalence isn't assumed, just absolute continuity. I think in general, it is these sets of absolutely equivalent local martingale measures which are closed.

If you think about it, $\mathbf{Z}_P$ will almost never be $L^1$-compact, the reason being that it lacks closedness. Suppose that you had $Z^e$ which comes from an equivalent local martingale measure and $Z^a$, which comes from an absolutely continuous but not equivalent local martingale measure. Then $Z^e > 0$. Therefore, consider the convex combinations $\frac{1}{n}Z^e + \frac{n-1}{n}Z^a$. These are equivalent local martingales measures since $Z^e$ is postive, but they converge to $Z^a$ (in $L^1$), which is not equivalent. So, as soon as you have any nonequivalent local martingale measure, you lose closedness.

• @weakstar: Thank you for the answer. Indeed, you are right in that the Neyman-Pearson lemma generally only requires absolute continuity of the measures contained in the hypotheses with respect to a reference probability measure. However, in certain applications (like the problem of hedging in incomplete markets, see Theorem 4.9 on page 55 here: princeton.edu/~brudloff/Diss_Rudloff.pdf), the compound null hypothesis is comprised of the equivalent martingale measures and is tested against a simple alternative. – Aldanor Jul 3 '12 at 5:49
• @weakstar: In the thesis I referred to above, the author spends considerable time on investigating the case when $Z_{\mathcal{P}^\ast}$ is compact, in which case it is possible to apply some convenient theorems. This kind of made me wondering - can it be compact at all given that $\mathcal{P}^\ast$ is a family of equivalent martingale measures? – Aldanor Jul 3 '12 at 5:51

I guess this answer comes way too late. In case you are still trying to see if the set of densities of absolutely continuous martigale measures may be compact in the incomplete case ... the answer is "almost never".

In " Representing Martingale measures when asset prices are continuous and bounded ", F. Delbaen roughly speaking proves that as soon as your filtration is well-behaved, the aforementioned set of densities is not weakly compact in the complete case. This rules out norm compactness of course.

Regards