# Hanson-Wright inequality (quadratic form concentration inequality) for bounded random vectors

Is there a concentration inequality for quadratic forms of bounded random vectors $$X \in [-1, 1]^n$$ with zero mean and given covariance matrix $$\Sigma \in \mathbb{R}^{n \times n}$$ but otherwise unknown distribution, i.e. a bound on the tail probability $$\Pr(|X^T \Sigma^{-1} X - n| \ge t) \le \ldots$$ Since for sub-Gaussian random vectors there is the Hanson-Wright inequality $$\Pr(|X^T A X - \operatorname{E}[X^T A X]| > t) \le \ldots$$ for some matrix $$A \in \mathbb{R}^{n \times n}$$ and $$E[X^T \Sigma^{-1} X] = n$$, it seems like such a bound should be within reach for the stronger restriction of bounded random vectors, even if the variance of the quadratic form is not available.

Note that this is specifically a question regarding the concentration about the mean $$n$$, since else we have the straightforward bound $$\Pr(X^T \Sigma^{-1} X \ge t) \le n/t$$ from Markov's inequality.

Edit: Thanks to @felipeh's answer for pointing out that there need to be additional restrictions on the function to give a meaningful bound. It would be reasonable to ask for the $$X_i$$ to be some or all of continuous, unimodal, symmetric as helpful.

Let $$X$$ be the random vector that is identically $$0$$ with probability $$1/2$$, and with probability $$1/2$$ is sampled uniformly from the boolean cube $$\{-1,1\}^n$$. The covariance is $$\Sigma=\frac{1}{2} Id$$, and $$X^T\Sigma^{-1} X$$ is either equal to $$0$$ or $$2n$$, each with probability $$1/2$$.
• (I take it you meant the boolean cube $\{-1, 1\}^n$.) That is a good example, there need to be additional quantifiers on the distribution. I will edit the question, apologies for that. – student Mar 18 '19 at 16:30