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Let $M_n$ be a vector-valued $F_n$-martingale ($M_n:\Omega \rightarrow R^p$). Is then $\lVert M_n \rVert $ also a martingale?

I have $E(M_{n+1} | F_n)= M_n$ and liked to say something about $E(\lVert M_{n+1} \rVert \mid F_n )= \lVert M_n \rVert$.

I tried to construct a counter-example, let $M_2=Y_1+Y_2$, then it would be enough to find RV $Y_1, Y_2$ such that $E(\lVert Y_1+Y_2 \rVert \mid F_1) =\lVert Y_1 \rVert $ and $E(Y_2 \mid F_1) \neq 0$ is possible at the same time.

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    @Gortaur If $||M_n||$ is a martingale concerning a different filtration than $\mathcal{F}^M_n$ or $\mathcal{F}^{\lVert M_n \rVert}_n$. Right now I conclude "probably not" because the counter-example seems like it could be extendible to other filtrations2011-10-19

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If $M$ is a martingale in the filtration $(F_n)$, $\|M\|$ is a sub-martingale in $(F_n)$. Assume that $\|M\|$ is a martingale in another filtration $(G_n)$, then $\mathrm E(\|M_n\|)$ is constant. Any submartingale with constant expectation is a martingale hence $\|M\|$ is a martingale in $(F_n)$.

Finally, it seems that a martingale $M$ is such that there exists a filtration in which $\|M\|$ is also a martingale if there exists a deterministic nonzero vector $\vec{u}$ such that $M_n=\lambda_n\cdot\vec{u}$ almost surely, for a nonnegative real valued martingale $(\lambda_n)$.

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    @JohannesL : my comment was referring to an earlier version of Didier Piau's answer. You only need convexity (not strict) to have that $\varphi(M_n)$ is a submartingale . However it will be a martingale iff all the $M_n$ take value (a.s.) on a subset where $\varphi$ is affine (this simply comes from the case of equality in Jensen's inequality). I hope this answers your comment.2011-10-26