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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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  • $\begingroup$ What is $\langle X \rangle$? $\endgroup$
    – Bill Johnson
    Commented May 31, 2012 at 20:49
  • $\begingroup$ Quadratic variation. Updated the post to clarify this. $\endgroup$
    – Alexander Shamov
    Commented May 31, 2012 at 20:52
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    $\begingroup$ 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. $\endgroup$
    – George Lowther
    Commented May 31, 2012 at 21:56

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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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  • $\begingroup$ Do you mean Theorem 5.27 in math.utah.edu/~davar/ps-pdf-files/SPDE.pdf? 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 $O(p^{1/2})$. $\endgroup$
    – Alexander Shamov
    Commented Jan 23, 2014 at 23:19
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    $\begingroup$ Please see p. 196 of the file: stt.msu.edu/CBMS2013/D_Khoshnevisan_Lecture.pdf $\endgroup$
    – epsilon
    Commented Jan 24, 2014 at 14:40
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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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    $\begingroup$ 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. $\endgroup$
    – Alexander Shamov
    Commented Nov 13, 2015 at 14:53
  • $\begingroup$ The time-change may be random and unbounded, thus your argument regarding Brownian Motion does not transfer to general continuous martingales. $\endgroup$
    – MKR
    Commented Nov 17, 2015 at 10:05
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    $\begingroup$ The time change is the quadratic variation. I thought you were talking about the case when it's bounded. $\endgroup$
    – Alexander Shamov
    Commented Nov 17, 2015 at 14:02
  • $\begingroup$ Yes, sorry, in the context of my answer your comment makes perfect sense, of course. $\endgroup$
    – MKR
    Commented Nov 17, 2015 at 15:29
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This comment was too long.

For martingales of the form $X_{t} = \int_{0}^{t}f(s,\omega)dB_{s}$ where $B_{t}$ is the standard Brownian motion, $f(s, \omega)$ is non-anticipating function such that $\int_{0}^{\infty}f^{2}(s, \omega)ds <\infty$ a.s. (I guess they are called $L^{2}$ integrable continuous martingales) one can obtain the bound of order $\sqrt{p}$ for large $p$. This is due to Burgess Davis, (see Section 3) of the reference. The result essentially follows from the Brownian case with arbitrary $L^p$ integrable stopping times.

However, for discrete time martingales, this is not true, and the best possible bound is of order $p$ for large $p$, for example, see paper of G. Wang, Remark 2. On the other hand, there are very special ``conditionally symmetric'' discrete martingales for which one can obtain the bound of order $\sqrt{p}$. And it looks like continuous $L^{2}$ integrable martingales in this sense are similar to conditionally symmetric ones.

Wang, Gang, Sharp inequalities for the conditional square function of a martingale, Ann. Probab. 19, No.4, 1679-1688 (1991). ZBL0744.60046.

Davis, Burgess, On the $L^p$ norms of stochastic integrals and other martingales, Duke Math. J. 43, 697-704 (1976). ZBL0349.60061.

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