Are any concentration inequality available for $d$-dimensional martingale. It is easy to find such inequality using the inequalities for single dimension, but that will contain the dimension $d$ in the inequality.
Thanks.
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Bounds (not depending explicitly on the dimension) on the moments of the norm of martingales in arbitrary 2-smooth Banach spaces (which of course include all finite-dimensional Euclidean spaces) can be found in [1]; see also further references there.
As usual, under appropriate conditions, such bounds on the moments imply the corresponding bounds on the tails of the distributions of such martingales; cf. e.g. the way Bernstein's exponential bound is derived from bounds on moments. Alternatively (and sometimes more efficiently), instead of deriving bounds on the exponential moments, one can use, more directly, an inequality of the form $$P(X\ge x)\le\inf_{p>p_0}E|X|^p/x^p $$ for $x>0$ and $p_0\ge0$.
answered Nov 30, 2015 at 16:22
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$\begingroup$ @losif: Thanks. I'll check. Actually, I want a Azuma kind of inequality. $\endgroup$– SoshaCommented Dec 1, 2015 at 5:13
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2$\begingroup$ I have expanded the remark about deriving bounds on tails from bounds on moments. Also, what is sometimes referred to as the Azuma inequality is in fact due entirely to Hoeffding (1963, Journal of the American Statistical Association) -- see Theorem 2 and the last paragraph of Section 2 therein. $\endgroup$ Commented Dec 1, 2015 at 20:07
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$\begingroup$ Thank you for this. Is that a typo - does one replace $p_0$ in the infimum by just $0$ (which I believe then simply follows from Markov's inequality)? $\endgroup$– Uzu LimCommented Jun 28, 2021 at 16:27
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1$\begingroup$ @FinnLim : Oftentimes, the bounds on moments $E|X|^p$ are only available for large enough $p$. That is why I had this provision, $p>p_0$ $\endgroup$ Commented Jun 28, 2021 at 17:00