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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.

Thanks.

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  • $\begingroup$ 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, 2013 at 19:31
  • $\begingroup$ yes, this is exactly what I am trying to show. $\endgroup$
    – jim
    Dec 15, 2013 at 20:09

1 Answer 1

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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.

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  • $\begingroup$ Are you missing come expectations in the right-hand side? $\endgroup$
    – SBF
    Apr 8, 2014 at 15:34
  • $\begingroup$ yeah, thanks. this was a while ago, i have to try and understand what i wrote. $\endgroup$
    – quasi
    Apr 8, 2014 at 19:22
  • $\begingroup$ Sure :) also, what does these square mean? That you take an expectation/integral of squared R-N derivative? $\endgroup$
    – SBF
    Apr 9, 2014 at 7:32
  • $\begingroup$ yeah. you apply $f(x)$ pointwise. $\endgroup$
    – quasi
    Apr 9, 2014 at 8:01

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