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I am working with a log-concave real random variable, that has a density $f(x) = \exp(-\varphi(x))$ with $\varphi$ convex. Assuming that $X$ is centered and has unit variance ($\mathbb{E}X=0$, $\mathbb{E}X^2=1$), I want to prove that $X$ satisfies a small deviation inequality like $$\mathbb{P}(|X| \leq t) \geq c t$$ for a numerical constant $c$ and all $t\in (0,1)$ (emphasize on $t\leq 1$). I know that log-concave distributions have sub exponential tails but this allows to prove large deviation bounds, typically $\mathbb{P}(|X| > t)\leq 2\exp(-ct)$.

My progress so far:

  • the density satisfies $f(0)\geq 1/8$ (see Lovasz and Vempala, lemma 5.5) so I expect that some regularity of the density $f$ (or the potential $\varphi$) should allow to prove that $f$ is uniformly lower-bounded on the interval $[-1, 1]$.Essentially, if $f$ decays too fast around $0$, it would cause the variance to be too small, and we set the variance to $1$.

  • I know how to prove the result if $\varphi$ (hence $f$) is even, because in this case we can use the lower bound for the concentration function of Bobkov and Chistyakov, corollary 2.2. Is it possible to use a symmetrization argument to reduce the problem to the symmetric case ?

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We have $f=e^g$, $g$ is concave, $\int f=1$, $\int x f(x)\,dx=0$, and $\int x^2 f(x)\,dx=1$. As you noted, then $f(0)\ge 1/8$ and hence $$g(0)\ge-a,\tag{0}\label{0}$$ where $a:=\ln8$.

We have to show that then $$\int_{-t}^t f\overset{\text{(?)}}\ge ct \tag{1}\label{1}$$ for some real $c>0$ and all $t\in[0,1]$.

Suppose for a moment that $g(1/2)\le-3$. Then, by the concavity of $g$ and \eqref{0}, $$g(x)\le h(x):=-3-(x-1/2)\frac{3-a}{1/2}$$ for $x>1/2$ and hence $$\int_{1/2}^\infty x^2 f(x)\,dx\le\int_{1/2}^\infty x^2 e^{h(x)}\,dx<0.04.$$ Similarly, the assumption $g(-1/2)\le-3$ would imply $\int_{-\infty}^{-1/2} x^2 f(x)\,dx<0.04$. So, if $g(1/2)\le-3$ and $g(-1/2)\le-3$, then $$1=\int x^2 f(x)\,dx \\ =\int_{-\infty}^{-1/2} x^2 f(x)\,dx+\int_{-1/2}^{1/2} x^2 f(x)\,dx +\int_{1/2}^\infty x^2 f(x)\,dx \\ <0.04+1/4+0.04<1,$$ a contradiction. So, $g(1/2)\ge-3$ or $g(-1/2)\ge-3$.

By symmetry, without loss of generality $g(1/2)\ge-3$. So, by the concavity of $g$ and \eqref{0}, we have $g\ge\min(-3,-a)=-3$ on $[0,1/2]$. So, for $t\in[0,1/2]$, $$\int_{-t}^t f\ge\int_0^t e^g\ge e^{-3}t.$$ So, for $t\in[1/2,1]$, $$\int_{-t}^t f\ge\int_{-1/2}^{1/2} f\ge e^{-3}/2\ge( e^{-3}/2)t.$$

Thus, \eqref{1} does hold for $c:= e^{-3}/2>0$ and all $t\in[0,1]$. $\quad\Box$

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  • $\begingroup$ Thank you so much, your concise answer is tremendously helping ! $\endgroup$
    – Hugo Ch
    Commented Nov 12, 2023 at 22:10

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