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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.

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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
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.

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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.

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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 '15 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 '15 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 '15 at 14:02
Yes, sorry, in the context of my answer your comment makes perfect sense, of course. – Martin Keller-Ressel Nov 17 '15 at 15:29

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