Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
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
show 4 more comments

1 Answer

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 $(\lambda_n)$.

share|improve this answer
$\|\cdot\|$ 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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.