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I've been studying martingale property under change of measure and I came up with a following observation.

I took a random example with a state space $\Omega = \{ -2,1,2\}$ and two equivalent probability measures $P = \{ 2/5, 2/5, 1/5\}$ and $Q = \{ 3/7, 2/7, 2/7\}$. A stochastic process $Z_{n} = \sum_{i=1}^{n} X_{i}$ is then a martingale with respect to both $P$ and $Q$ where $X_{i}: \Omega \rightarrow \mathbb{R}$ is a random variable.

Then it seems true that $Z_{n}$ remains a martingale for any convex mixture of probability measures $P$ and $Q$. Is this a general property? Is it true that whenever a stochastic process is a martingale for a number of probability measures, it remains so for any (countable?) convex mixture of these? What kind of (implicit) hypothesis I should be aware of?

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Beware your case is ot the general situation and a (general) change of measure does not preserve martingale property. For example look at Girsanov theorem for Brownian motion which can introduce a drift after a measure change which then loose its martinalge property. Regards – TheBridge Oct 11 '11 at 9:50
up vote 2 down vote accepted

The answer is that this holds in general, and the proof amounts to studying the following situation.

One is given a random variable $X$ defined on a measurable space $(\Omega,F)$, two probability measures $P$ and $Q$ on $(\Omega,F)$, and a sub-sigma-algebra $G\subseteq F$. One assumes that $X$ is integrable with respect to $P$ and $Q$ and that $\mathrm E_P(X\mid G)=\mathrm E_Q(X\mid G)$ almost surely.

Call $Y$ this random variable. This is equivalent to asking that $Y$ is $G$-measurable and that, for every bounded $G$-measurable random variable $Z$, one has both $\mathrm E_P(XZ)=\mathrm E_P(YZ)$ and $\mathrm E_Q(XZ)=\mathrm E_Q(YZ)$.

Now, consider the probability measure $R=tP+(1-t)Q$, for a given $t$ in $(0,1)$. Then $Y$ is still $G$-measurable, naturally, and $$ \mathrm E_R(XZ)=t\mathrm E_P(XZ)+(1-t)\mathrm E_Q(XZ)=t\mathrm E_P(YZ)+(1-t)\mathrm E_Q(YZ)=\mathrm E_R(YZ), $$ hence $Y=\mathrm E_R(X\mid G)$ as well.

In particular, if $(X_n)$ is an $(F_n)$-martingale under $P$ and $Q$ and if $R=tP+(1-t)Q$ with $t$ in $(0,1)$, then, for every $n$, $\mathrm E_P(X_{n+1}\mid F_n)=X_n=\mathrm E_Q(X_{n+1}\mid F_n)$ hence $\mathrm E_R(X_{n+1}\mid F_n)=X_n$ as well, which implies that $(X_n)$ is an $(F_n)$-martingale under $R$ as well.

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Thank you. This seems pretty neat. "This is equivalent..." part made me stop for a moment it but it seems to come simply from the definition of conditional expectation of which I have seen a slightly different formulation. Thanks again! – johnny Oct 11 '11 at 10:01
Yes, this uses the definition of E(X|G) as the unique (up to null sets) integrable random variable Y which is at the same time G-measurable and such that E(XZ)=E(YZ) for every bounded G-measurable random variable Z. // Something else: @TheBridge made an important comment, which I should have mentioned in my answer. – Did Oct 11 '11 at 10:09

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