Uniqueness of the Variance-Optimal Martingale Measure

Summary

The document gives a convexity argument for uniqueness of a variance-optimal equivalent martingale measure. It considers two distinct candidate measures that both minimize the second moment of their density relative to the original probability measure. Their average remains a martingale measure because the martingale-measure set is convex.

Strict convexity of the squared-density objective then implies that the average has a strictly smaller objective value if the two densities differ, contradicting the assumption that both were minimizers. The answer notes that a fully rigorous proof needs an appropriate uniform-convexity argument. The proposed reasoning is concise and does not spell out all admissibility, integrability, or existence conditions. It addresses uniqueness of a minimizer of the variance criterion, rather than uniqueness of equivalent martingale measures generally; incomplete markets can still admit many such measures.

Key ideas

• The set of martingale measures is convex, so averaging two candidates preserves the martingale property.
• The squared-density objective is strictly convex in the density.
• Two distinct minimizers would have an average with a lower objective value, yielding a contradiction.
• The argument requires technical conditions and applies to the variance-optimal minimizer, not all martingale measures.

Tags

# unique equivalent martingale measure in incomplete markets


# unique equivalent martingale measure in incomplete markets












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.

## Answer by quasi (score 4)

https://quant.stackexchange.com/a/9736

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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