Anti-concentration of gaussian variable

Question

Let $X$ be $\mathcal{N}(\mu,\sigma^2)$ gaussian. Its expectation $\mu$ is positive. Can we derive a lower bound on $$\mathbb{P}(X\geq\epsilon)\geq g(\epsilon,\mu,\sigma) \text{ where } \epsilon\leq\mu$$ such that

1. $g$ is decreasing along with $\epsilon$ and
2. $g$ depends on $\mu$ and $\sigma$ as well?

Answer 1

$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\vpi\varphi$By shifting and rescaling, without loss of generality $\mu=0$ and $\sigma=1$. Then many lower bounds on $P(X\ge t)$ are available in the form of algebraic expressions in terms of $t$ and $\vpi(t)$, where $\vpi$ is the standard normal density. E.g., one has $$P(X\ge t)>p(t):=\frac{t\vpi(t)}{1+t^2}$$ for $t>0$; see e.g. the Laplace continued fraction (16) for $R(t):=P(X\ge t)/\vpi(t)$. Note that $p(t)$ is decreasing in $t\ge t_*:=\sqrt{\sqrt2-1}=0.643\ldots$.

For $t\in(0,t_*]$, one can use the following: $$P(X\ge t)=\frac12-\int_0^t du\,\vpi(u) =\frac12-\vpi(0)\int_0^t du\,\sum_{k=0}^\infty\frac{(-u^2/2)^k}{k!} \\ =\frac12-\vpi(0)\sum_{k=0}^\infty\frac{(-1)^k t^{2k+1}}{k!(2k+1)2^k}>l_{2m}(t) :=\frac12-\vpi(0)\sum_{k=0}^{2m}\frac{(-1)^k t^{2k+1}}{k!(2k+1)2^k}$$ for any $m=0,1,\dots$; actually, the inequality $P(X\ge t)>l_{2m}(t)$ holds for all real $t>0$. In particular, $l_0(t)=\frac12-\vpi(0)t$ and $l_2(t)=\frac12-\vpi(0)(t-t^3/6+t^5/40)$. Alternatively, for $t\in[0,t_*]$ one can use formula (7) from the linked paper.

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As an illustration, below are graphs

• $\{(t,P(X\ge t))\colon 0<t<2\}$ (black);
• $\{(t,\max[l_0(t),p(t)])\colon 0<t<2\}$ (green);
• $\{(t,p(t))\colon 0<t<2\}$ (dotted blue and then green);
• $\{(t,l_0(t))\colon 0<t<2\}$ (green and then dotted red);
• $\Big\{\Big(t,\dfrac{1-\sqrt{2e}\,t}2\Big)\colon 0<t<t_0\Big\}$ for some $t_0\in(1,2)$ (orange), regarding the lower bound $1-\sqrt{2e}\,t$ on $P(|X|\ge t)=2P(X\ge t)$ at the end of the other answer;
• $\Big\{\Big(t,\dfrac{\max[l_2(t),p(t)]}{P(X\ge t)}\Big)\colon 0<t<6\Big\}$ (magenta).

We see that the lower bound $\max[l_2(t),p(t)]$ on $P(X\ge t)$ is close to $P(X\ge t)$ for all real $t>0$ and very close to $P(X\ge t)$ for $t\in(0,1]$ and for large $t$.

Using mentioned results, we can get however close (even though somewhat more complicated) lower (and upper bounds) on $P(X\ge t)$.

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To get lower bounds on $P(X\ge t)$ for real $t<0$, simply write $P(X\ge t)=1-P(X\le t)=1-P(X\ge|t|)$ and then use upper bounds on $1-P(X\ge|t|)$ from the cited paper and/or the inequality $P(X\ge|t|)<l_{2m+1}(|t|)$ for $m=0,1,\dots$. This way, you e.g. get $$P(X\ge t)>1-\min\Big(\frac{\vpi(t)}{|t|},l_{2m+1}(|t|)\Big)$$ for real $t<0$.

Answer 2

There are various papers in the literature regarding anti-concentration in polynomials of Gaussian random variables. I'll point you to this, which discusses a few other papers as well. The result (discussed in that paper) below is due to Carbery and Wright.

There exists a universal positive constant $C$ such that the following holds. Let $f : \mathbb{R}^n \to \mathbb{R}$ be any degree $d$ polynomial, and let $\mu$ denote any log-concave measure on $\mathbb{R}^n$. For every $\epsilon > 0$: $$\Pr_{x\sim \mu}[|f(x)| \leq \epsilon \cdot \mathbb{E}_{x\sim\mu}[|f (x)|]] \leq C_d \epsilon^{1/d}.$$

Here, "log concave" means it admits a density with a concave logarithm, i.e. of the form $\exp(-V(x))$ for $V(x)$ convex. Gaussians are of this form. Note that this result does not indicate how large the constant $C_d$ is, though the linked paper has pointers to things that work out the constant in simpler cases. For example, the main result of that paper is

Let $f : \mathbb{R}^n \to\mathbb{R}$ be any non-negative degree two polynomial. For every $\epsilon > 0$: $$\Pr_{x\sim \mathcal{N}(0, I_n)}[f(x) \leq \epsilon \cdot \mathbb{E}_x[f(x)]] \leq (2e \epsilon)^{1/2}$$

One can apply this to $f(x) = x^2$ and rearrange algebraically to get that

$$\Pr_{x\sim \mathcal{N}(0, \sigma^2)}[|x| \leq \sqrt{\epsilon} \sigma] \leq (2e\epsilon)^{1/2},$$

or equivalently

$$\forall t > 0 : \Pr_{x\sim \mathcal{N}(0, \sigma^2)}[|x| > t\sigma] \geq 1 - \sqrt{2e}t.$$