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Consider a filtered probability space $(\Omega, (\mathcal F_n), \mathcal F, \mathbb P)$, where $\Omega$ is the set of sequences with value in some $E \subseteq \mathbb R^d$, and $\mathcal F$ is the canonical filtration generated by the coordinate maps $(X_n: \Omega \to \mathbb R^d)$. We will assume that $X$ is a martingale.

Given a non-negative martingale $M$ with expectation $1$, there exists a probability measure $\mathbb Q$ on $(\Omega, \mathcal F)$ such that: $$M_n = \left.\frac {\mathrm d \mathbb Q}{\mathrm d \mathbb P}\right|_{\mathcal F_n}.$$ This is a consequence of Kolmogorov's extension theorem. Conversely, for any $\mathbb Q \ll \mathbb P $, the expression above defines a martingale in the original probability space.

Now, let's look at martingale transforms. All martingale transforms of $X$ are martingales, but apart from the very specific case of the binomial model, the converse is false.

First question: is there a characterization of the measures $\mathbb Q \ll \mathbb P$ for which $\left(\left.\frac {\mathrm d \mathbb Q}{\mathrm d \mathbb P}\right|_{\mathcal F_n}\right)$ is a martingale transform of $X$?

Another interesting property of these Radon Nikodym representations is that, given some non-negative martingale tranform $(X \cdot H)$ with expectation $1$, the associated probability measure $\mathbb Q$ satifies: $$\mathbb E^{\mathbb Q} (\log (X \cdot H)) \ge \mathbb E^{\mathbb Q} (\log (X \cdot K)),$$ for all processes $K$ for which $(X \cdot K)$ is non-negative with expectation $1$. The proof is easy, just write both martingale transforms as Radon Nikodym derivatives and apply Jensen's inequality.

Second question: suppose I take $\mathbb Q \ll \mathbb P $. Consider the process $H$ that maximizes $\mathbb E^{\mathbb Q} (\log (X \cdot H))$. Are there known bounds linking $\left(\left.\frac {\mathrm d \mathbb Q}{\mathrm d \mathbb P}\right|_{\mathcal F_n}\right)$ and $(X \cdot H)_n$?

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  • $\begingroup$ 1. Under mild conditions on $E$, the Kolmogorov extension theorem does grant the existence of a unique $\Bbb Q$ on $\mathcal F$ such that $d\Bbb Q|_{\mathcal F_n}/d\Bbb P|_{\mathcal F_n}=M_n$, $\Bbb P$-a.s., for each $n$. The probability measure $\Bbb Q$ need not be absolutely continuous with respect to $\Bbb P$ on $\mathcal F$, and indeed it may be that $\Bbb Q\perp\Bbb P$. 2. ``All martingale transforms of $X$ are martingales": You seem to be implicitly assuming that $X$ is a martingale. $\endgroup$ Commented Nov 10, 2017 at 16:26
  • $\begingroup$ Thank you for your comments! Indeed, my statement was inexact. I corrected the question and added the assumption that X is a martingale. $\endgroup$
    – Tartrate
    Commented Nov 10, 2017 at 21:00

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