# Anti-concentration inequalities: lower bound on realized second moment

Let $$X:(\Omega,\Sigma)\rightarrow (\mathbb{R}^d;\mathbb{B}(\mathbb{R}^d))$$ be a Borel-measurable random vector. What are some general classes of such random vector for which one can give a "lower concentration inequality" of the form: $$\mathbb{P}(\|X\|^2>\lambda) \geq \mbox{(insert non-trivial lower bound)}$$ where $$\lambda>0$$.

Suppose e.g. that $$X=X_1+\dots+X_n$$, where the $$X_i$$' are independent zero-mean random vectors with $$\|X_i\|\le1$$ for all $$i$$. Then the Hoeffding--Azuma inequality (see e.g. Wikipedia) yields $$P(\|X\|>u)\ge1- e^{-(E\|X\|-u)^2/(2n)}$$ for $$u\le E\|X\|$$.

A number of lower bounds on $$P(\|X\|\ge u)$$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $$p=2$$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $$u>0$$, where $$X$$ is just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $$X_i$$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $$u>0$$.

• Hmm. The Hoeffding-Azuma inequality you quote is not the one in the reference, and is actually wrong as the example $n=1$ and $u=E|X|$ shows. I believe there is a typo also in the reference you give (Bartlett's notes) - it should be an upper bound, not a lower bound, as the example t=0 (and the proof given there) show. Feb 19 '20 at 20:07
• @oferzeitouni : Thank you for your comment. This is now corrected. Feb 20 '20 at 0:30

Well, in your case notice that $$\|X\|^2>0$$ and so $$\mu\triangleq \mathbb{E}[\|X\|^2]>0$$. Thus, for any $$\lambda \in \left(0,\mu\right)$$ the Cantelli Inequality gives $$\Pr(X\ge\lambda) \ge 1 - \frac{\sigma^2}{\sigma^2 + \lambda^2},$$ where $$\sigma\triangleq \mathbb{E}\left[\left(\|X\|^2 - \mu\right)^2\right]$$.

• This is interesting, but it only holds for small $\lambda$...not sure I'm I'm asking for too much...
– BLBA
Feb 19 '20 at 17:18
• $0<\lambda<\mu$? Feb 19 '20 at 21:27