Is the norm of a martingale a martingale? 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.

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 $(\lambda_n)$.
