Do you have any idea about how we can prove, and under which conditions, that an equivalent martingale measure (EMM) in an incomplete market is unique? The assumptions we have made are:

1) that the stochastic process St of the asset is a semi martingale (continuous) and

2) that this EMM exists.

In other words, that the variance optimal measure is unique.


  • $\begingroup$ Can you clarify your question a little? A market being incomplete is equivalent to there being multiple equivalent martingale measures. Are you asking how to show that the variance optimal EMM is unique? The answer to that question is basically strict convexity, I can elaborate if that's what you're after. $\endgroup$ – quasi Dec 15 '13 at 19:31
  • $\begingroup$ yes, this is exactly what I am trying to show. $\endgroup$ – jim Dec 15 '13 at 20:09

Suppose that there are multiple martingale measures $Q_1$ and $Q_2$ that attain the minimal variance. Then the convex combination $Q_* := \frac{1}{2}Q_1 + \frac{1}{2}Q_2$ is also a martingale measure. Due to the strict convexity of $f(x) = x^2$, it can be shown that $$ E_P \left[\frac{dQ_*}{dP}^2 \right] < \frac{1}{2} E_P \left[ \frac{dQ_1}{dP}^2 \right] + \frac{1}{2} E_P \left[ \frac{dQ_2}{dP}^2 \right]. $$

To make this completely airtight, you need to use the notion of uniform convexity, i.e. $f \left(\frac{x+y}{2} \right) < \frac{1}{2} \left( f(x) + f(y) \right) -\epsilon | x - y|$, for some $\epsilon$. The details are tedious but not hard.

In any case, you have a contradiction, and the minimizer must be unique.

| improve this answer | |
  • $\begingroup$ Are you missing come expectations in the right-hand side? $\endgroup$ – Ilya Apr 8 '14 at 15:34
  • $\begingroup$ yeah, thanks. this was a while ago, i have to try and understand what i wrote. $\endgroup$ – quasi Apr 8 '14 at 19:22
  • $\begingroup$ Sure :) also, what does these square mean? That you take an expectation/integral of squared R-N derivative? $\endgroup$ – Ilya Apr 9 '14 at 7:32
  • $\begingroup$ yeah. you apply $f(x)$ pointwise. $\endgroup$ – quasi Apr 9 '14 at 8:01

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