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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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You have two different $n$'s running around in your question. (Maybe $M_n : \Omega \to R^p$ would be better). At any rate, let $M_n$ be real-valued. Since all norms are convex, then $\|M_n\|$ is a submartingale. Let $M_n$ be a simple random walk and consider what happens when $M_n = 0$. –  cardinal Oct 19 '11 at 9:42
    
$Y_i$ Bernoulli, $P(Y_i=1)=1/2=P(Y_i=-1)$. $n=2$ $M_2=0=Y_1+Y_2$ means that the signs of $Y_i$ turned out differently. $0=E(||M_2|| \mid F_1) \stackrel{?}{=} ||Y_1||$ and $Y_1\neq0$ –  Johannes L Oct 19 '11 at 9:55
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More explicitly, let $\{\xi_i\}$ be iid $\mathrm{Ber}(1/2)$. Take $M_n = \sum_{i=1}^n \xi_i$ and $M_0 = 0$. Now, $M_n$ is a martingale, but, $\mathbb E (\|M_1\| \mid \|M_0\|) = 1$ almost surely. –  cardinal Oct 19 '11 at 9:59
    
You wrote $E(||M_1|| \mid F_0)=1$ so this counter-example shows that ||M_n|| is not a martingale concerning the filtration $F^{(1)}_n=\sigma\{||M_0||, \dotsc, ||M_n||\}$.. This could still mean that $||M_{n}||$ is a $F_n$-martingale where $F_n$ is another filtration than $\sigma(\{M_0, \dotsc, M_n\})$ (the natural filtration of $M_n$) and than $F^{(1)}_n$ –  Johannes L Oct 19 '11 at 10:08
    
I got a little sloppy. It still holds under the original filtration $\{\mathcal F_n\}$ as you should be able to see/verify. –  cardinal Oct 19 '11 at 10:40

1 Answer 1

up vote 6 down vote accepted

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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$\|\cdot\|$ is not strictly convex. I believe the condition on $M_n$ to have $\|M_n\|$ martingale should be that there is a half-line (from the origin) that supports all the $M_n$. –  pgassiat Oct 19 '11 at 14:45
    
@pgassiat, thanks, I modified my post. –  Did Oct 19 '11 at 15:02
    
@pgassiat But the proposition that $\varphi(M_n)$ is a submartingale just demands $\varphi$ convex and for each $n$, $\varphi(M_n) \in \L^1$.. I want to apply a Inequality by Doob which - I looked at a reference - also holds for submartingales, so it would be good to know if the "strict convexity" is necessary –  Johannes L Oct 25 '11 at 14:13
    
@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. –  pgassiat Oct 26 '11 at 12:50

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