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Let $X$ be a real random variable with log-concave density $f$. Assume that $E(X) =0$ and $E(X^2)=1$.

Show that there is a universal (independent of $X$) constant $c>0$ such that:

$$P(X\in[-1/2;0])\geq c.$$ I'm looking for the simplest possible proof of this result.

I think there should be various ways to prove this without too much effort, but for some reason I could not find an elementary way. Of course the log-concavity assumption is crucial otherwise a Bernoulli random variable gives a very easy counterexample.

EDIT : To explain the motivation behind the question, it can be shown (using somewhat evolved results) that for a log-concave random variable $Z$, then

$$Var(Z) \leq c_1Var(Z^2)^{1/2}$$ for a universal constant $c_1>0$. One possible way to prove this in a more elementary way is to set $Z=a+X$ with $X$ centered log-concave. Assuming by homogeneity that $Var(Z)=1$, then $X$ satisfies the assumptions above. Then without loss of generality assume that $a\geq0$. A quick computation shows that $$E(Z^2) = 1+a^2$$ Now if $X\in [0,-1/2]$ then $$Z^2 = a^2+X^2 +2aX \leq a^2+X^2 \leq a^2 + 1/4$$ Thus on the event $X\in [0,-1/2]$, $Z^2$ is at distance at least $3/4$ from its expectation. Thus we get that

$$Var(Z^2) \geq (3/4)^2 P(X\in [0,-1/2]).$$

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  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Commented May 17, 2023 at 13:33

1 Answer 1

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$\newcommand{\ep}{\varepsilon}\newcommand\R{\mathbb R}$Let $L$ denote the set of all real random variables (r.v.'s) $X$ with log-concave density such that $EX=0$ and $EX^2=1$.

Let \begin{equation*} c:=\inf\{P(X\in(-1/2,0))\colon X\in L\}. \tag{10}\label{10} \end{equation*} We have to show that $c>0$.

For some sequence $(X_n)$ in $L$ we have $P(X_n\in(-1/2,0))\to c$ (as $n\to\infty$). In view of the condition $EX^2=1$ for $X\in L$, the sequence of the distributions of the $X_n$'s is tight. So, passing to a subsequence, without loss of generality (wlog) assume that $X_n\to Y$ in distribution for some r.v. $Y$. Moreover, the condition $EX^2=1$ for $X\in L$ implies the uniform integrability of the $X_n$'s. Hence, \begin{equation*} EY=\lim_n EX_n=0 \tag{20}\label{20} \end{equation*} and \begin{equation*} E|Y|=\lim_n E|X_n|\le1 \tag{25}\label{25} \end{equation*} Furthermore, by the Portmanteau theorem, \begin{equation*} p_0:=P(Y\in(-1/2,0))\le c. \tag{30}\label{30} \end{equation*}

We will need the following lemma, which will be proved at the end of this answer.

Lemma 1: For all $X\in L$ we have $E|X|\ge1/2$.

It follows immediately from \eqref{25} and Lemma 1 that \begin{equation*} E|Y|\ge1/2>0. \tag{40}\label{40} \end{equation*}

So, by Theorem 2.7 and Proposition 3.6 and in view of \eqref{30}, $Y$ has a log-concave density. So, in view of Definition 2.4., the function $$\mathbb Z\ni n\mapsto p_n:=P(Y\in(n/2-1/2,n/2))$$ is a discrete log-concave function. So, if $p_0=0$, then either (i) $p_n=0$ for all integers $n\ge0$ or (ii) $p_n=0$ for all integers $n\le0$. So, if $p_0=0$, then either $P(Y>0)=0$ or $P(Y<0)=0$. But this conclusion contradicts the conjunction of conditions \eqref{20} and \eqref{40}. So, $p_0>0$. So, by \eqref{30}, $c>0$, as desired.


It remains to prove Lemma 1. To do that, take any $X\in L$. Let $m$ be the median of $X$, so that $P(X<m)=P(X>m)=1/2$. Let $Z^\pm$ be r.v.'s whose respective distributions are the conditional distributions of $\pm(X-m)$ given $\pm(X-m)>0$. Then $Z^\pm$ are positive r.v.'s with log-concave densities. So, by a well-known inequality (see e.g. inequality (0.3)), \begin{equation*} E(Z^\pm)^2\le2(EZ^\pm)^2. \tag{45}\label{45} \end{equation*} Since $EZ^\pm=2E(X-m)_\pm$ and $E(Z^\pm)^2=2E(X-m)_\pm^2$, where $u_\pm:=\max(0,\pm u)$, we can rewrite \eqref{45} as $4(E(X-m)_\pm)^2\ge E(X-m)_\pm^2$. Therefore, and because (i) $m$ is a minimizer of $E|X-a|$ in real $a$ and (ii) $EX=0$ is a minimizer of $E(X-a)^2$ in real $a$, we get \begin{equation*} 4(E|X|)^2\ge4(E|X-m|)^2\ge4(E(X-m)_+)^2+4(E(X-m)_-)^2 \\ \ge E(X-m)_+^2+E(X-m)_-^2=E(X-m)^2\ge EX^2=1. \end{equation*} This completes the proof of Lemma 1. $\quad\Box$

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  • $\begingroup$ first link goes to michigan tech single sign on $\endgroup$ Commented May 14, 2023 at 22:53
  • $\begingroup$ @mathworker21 : Thank you for your comment. The link is now fixed. $\endgroup$ Commented May 14, 2023 at 23:38

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