Question

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. My question is: Do there exist absolute constants $C,c>0$ such that for every $\epsilon>0$ and positive real numbers $a_1,\dots,a_n>0$, we have the following bound

$\quad \Pr(\sum_{i=1}^n a_i X_i^2 \le \epsilon \sum_{i=1}^n a_i)\le C\epsilon^c\quad ?$

Background: Results of Carbery and Wright (2001) give powerful anti-concentration inequalities for polynomials of Gaussian variables. If we replace the term $\epsilon \sum_{i=1}^n a_i$ by $\epsilon\sqrt{\sum_{i=1}^n a_i^2}$ the bound follows directly from Carbery-Wright with $c=1/2$. Here, however, we need a stronger bound in terms of the $\ell_1$-norm rather than $\ell_2$-norm of the polynomial.

The reason why I believe the claim is still true is that all the variables $X_i$ appear squared so that there are no cancellations. Also, for $n=1$ the claim follows from simple Gaussian anti-concentration bounds with $c=1/2.$

Note: A simple application of Paley-Zygmund gives the bound $\Pr(\sum a_i X_i^2 \ge \epsilon\sum a_i )\ge (1-\epsilon)^2/3.$ This, unfortunately does not imply the statement above.

Answer 1

We can show that $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$ so that the inequality holds with $c=1/2$ and $C=\sqrt{e}$.

For $\epsilon\ge1$ the right hand side is greater than 1, so the inequality is trivial. I'll prove the case with $\epsilon < 1$ now. Without loss of generality, we can suppose that $\sum_ia_i=1$ (just to simplify the expressions a bit). Then, for any $\lambda\ge0$, $$ \begin{align} \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\right)&\le\mathbb{E}\left[e^{\lambda\left(\epsilon-\sum_ia_iX_i^2\right)}\right]\cr &=e^{\lambda\epsilon}\prod_i\mathbb{E}\left[e^{-\lambda a_iX_i^2}\right]\cr &=e^{\lambda\epsilon}\prod_i\left(1+2\lambda a_i\right)^{-1/2}\cr &\le e^{\lambda\epsilon}\left(1+2\lambda\right)^{-1/2}. \end{align} $$ Take $\lambda=(\epsilon^{-1}-1)/2$ to obtain $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\right)\le e^{(1-\epsilon)/2}\sqrt{\epsilon}. $$

Answer 2

I also would strongly suspect it holds with $c = 1/2$. Here's an argument which, if I haven't made a mistake, gives the upper bound $O(\epsilon^{1/2} \ln^{1/2}(1/\epsilon))$.

Assume without loss of generality that $a_1 \geq a_2 \geq \cdots \geq a_n$ and that $a_1 + \cdots + a_n = 1$. Divide into two cases:

Case 1: $a_1 \geq \frac{1}{100\log(1/\epsilon)}$. In this case it's sufficient to show that we already have $\Pr[a_1 X_1^2 \leq \epsilon] \leq O(\epsilon^{1/2} \ln^{1/2}(1/\epsilon))$. But this holds as you said because of the anticoncentration of a single Gaussian.

Case 2: $a_1 \leq \frac{1}{100\log(1/\epsilon)}$. In this case, note that $\sigma^2 := \sum_i a_i^2 \leq a_1 \sum_i a_i \leq \frac{1}{100\ln(1/\epsilon)}$. Now the random variables $X_i^2$ are nice enough that one should be able to apply a Chernoff Bound to them. (I think I could provide a source for this if necessary.) Since $Y := \sum a_i X_i^2$ has mean $1$ and standard deviation $\sqrt{2}\sigma$ (I think, maybe the constant $\sqrt{2}$ is wrong), we should have a statement like $\Pr[|Y - 1| \geq t \cdot \sqrt{2}\sigma] \leq \exp(-t^2/c)$, where $c$ is a quite modestly small universal constant. So $\Pr[Y \leq \epsilon] \leq \Pr[|Y-1| \geq 1/2] \leq \exp(-\frac{1}{8c\sigma^2}) \leq \exp(-\frac{12.5}{c} \ln(1/\epsilon)) = \epsilon^{12.5/c}$, which is smaller than $\epsilon^{1/2}$ if $c$ is not too large, and even if $c$ is too large, one can make $100$ larger.