Random probability following a log concave distribution of order p In the article "Concentration of the information in data with
Log-concave distributions" of Bobkov and Madiman, it is written that if $X$ is a positive random variable following a log concave distribution of order $p$, then one has $V(X) \leq \frac{E(X)^2}{p}$.
A reference is given, but I don't understand how the result follows from the reference.
Also, it seems quite hard to prove, and the problem where those variables came from is said to be "easy" (it's the one dimension optimal matching problem), so I start to feel like I have misunderstood something.
Have you seen this inequality? Is it possible to give a relatively short proof?
 A: $\newcommand{\tla}{\tilde\lambda}\newcommand{\Ga}{\Gamma}$By Definition 4.1 in the paper by Bobkov and Madiman (BM), a positive random variable (r.v.) $\xi$ has a log-concave distribution of order $p\ge1$ if the pdf $f$ of $\xi$ is such that
\begin{equation*}
f(x) = x^{p-1}g(x) \tag{1}\label{1}
\end{equation*}
for $x > 0$, where the function $g$ is log-concave on $(0,\infty)$.
Corollary 3.2 in BM states that, if a positive r.v. $\eta$ has a log-concave pdf, then
\begin{equation*}
    \tla_p:=\frac{E\eta^p}{\Ga(p+1)}
\end{equation*}
is  log concave in $p\ge0$. It follows then that $\tla_{p+1}\tla_{p-1}\le\tla_p^2$ for $p\ge1$, that is,
\begin{equation*}
    E\eta^{p+1}\,E\eta^{p-1}\le\frac{p+1}p\,(E\eta^p)^2. \tag{2}\label{2} 
\end{equation*}
Suppose now that a positive r.v. $\xi$ indeed has a log-concave distribution of order $p\ge1$, so that \eqref{1} holds for some log-concave function $g$ and all $x > 0$. Let
\begin{equation*}
    h:=g/c,
\end{equation*}
where $c:=\int_0^\infty g$, so that $h$ is a log concave pdf on $(0,\infty)$. Let then $\eta$ be a r.v. with pdf $g$, so that \eqref{2} holds and
\begin{equation*}
    E\xi^k=\int_0^\infty x^k f(x)\,dx=\int_0^\infty x^{k+p-1}g(x)\,dx=c\,E\eta^{k+p-1} \tag{3}\label{3}
\end{equation*}
for all $k\in\{0,1,\dots\}$.
Using \eqref{3} with $k=0,1,2$, we rewrite \eqref{2} as
\begin{equation*}
    E\xi^2\le\frac{p+1}p\,(E\xi)^2,
\end{equation*}
which can be further rewritten as
$$Var\,\xi\le\frac1p\,(E\xi)^2,$$
as desired.
