Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a continuous local martingale, and $\langle X \rangle$ be its quadratic variation process. The "standard" proof of Burkholder-Davis-Gundy inequalities found in books yields $(\mathsf{E} |X|^{p})^{1/p} \le O(p) \cdot (\mathsf{E} \langle X \rangle ^{p/2})^{1/p}$ for large $p$.

Can the growth rate be improved to, say, $O(p^{1/2})$? For example, if $\langle X \rangle$ is bounded, this estimate gives exponential tails for $|X|$, which is clearly suboptimal, since they should be Gaussian.

share|cite|improve this question
What is $\langle X \rangle$? – Bill Johnson May 31 '12 at 20:49
Quadratic variation. Updated the post to clarify this. – Alexander Shamov May 31 '12 at 20:52
The best constants are known, and you can't do better than p-1 for p > 2. This was proven by Davis I think, but I'm not sure if that applies specifically to continuous martingales. – George Lowther May 31 '12 at 21:56

2 Answers 2

up vote 4 down vote accepted

I know a version which exactly gives the constant $O(p^{1/2})$ for $p\ge 2$. It is contained in a lecture note by D. Khoshnevisan on SPDE.

share|cite|improve this answer
Do you mean Theorem 5.27 in On the bottom of p.17 he says, in his notation, something equivalent to $\left(\phi\left(t\right)\right)^{1/p}\le a_{p}^{1/2}\left(\mathsf{E}\left\langle N\right\rangle _{t}^{p/2}\right)^{1/p}$, where $\phi\left(t\right)=\mathsf{E}\sup_{\left[0,t\right]}\left|N\right|^{p}$ and $a_{p}=\frac{p\left(p-1\right)}{2}\left(\frac{p}{p-1}\right)^{p}$. This seems to give the $O(p)$ bound that I was talking about, not $O(p^{1/2})$. – Alexander Shamov Jan 23 '14 at 23:19
Please see p. 196 of the file: – epsilon Jan 24 '14 at 14:40

You are correct that for bounded $<X>_T$ the tails of $X_T$ should be Subgaussian. However, the Burkholder-Davis-Gundy inequality gives an upper bound for the $L^p$-norm of the running supremum $X_T^* = \sup_{t \le T} |X_T|$, of $X$ not just for $X_T$ itself.

I do not see a reason why $X_T^*$ should have Subgaussian tails, even if $<X>_T$ is bounded. In fact it cannot always have Subgaussian tails, otherwise the known optimal constant $p-1$ for $p \ge 2$ (see George Lowthers remark) would not be optimal.

share|cite|improve this answer
The question was about continuous martingales. A continuous martingale is a time-changed Brownian motion, and the maximum of a Brownian motion over a bounded interval is sub-Gaussian. – Alexander Shamov Nov 13 at 14:53
The time-change may be random and unbounded, thus your argument regarding Brownian Motion does not transfer to general continuous martingales. – Martin Keller-Ressel Nov 17 at 10:05
The time change is the quadratic variation. I thought you were talking about the case when it's bounded. – Alexander Shamov Nov 17 at 14:02
Yes, sorry, in the context of my answer your comment makes perfect sense, of course. – Martin Keller-Ressel Nov 17 at 15:29

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.