Anti-concentration inequalities: lower bound on realized second moment

Question

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 $\lambda>0$.

Answer 1

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 $u>0$.

Answer 2

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 $\sigma\triangleq \mathbb{E}\left[\left(\|X\|^2 - \mu\right)^2\right]$.