# Small deviations of real log-concave random variable

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 ?

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 $$t\in[0,1]$$. $$\quad\Box$$