Let $F:\mathbb{R} \rightarrow [0,1]$ be a cumulative distribution function on $\mathbb{R}$, and suppose that $F$ is a continously differentiable function, with derivative $f > 0$. I would like to prove that if $\log f$ is a strictly concave function, then, by defining for every $x > y $ \begin{equation} G(x,y)=\log(F(x)-F(y)), \end{equation} $G$ turns out to be a strictly concave function on $\{ (x,y) \in \mathbb{R}^2 | x > y \}$.

Does someone have any hint? Thank you very much.

NOTE 1. This result can be proved by means of calculus if we make the strong assumption that $F$ is twice continuously differentiable (see e.g. Nakamura and Hirai, On the Goodness of a Criterion for the Existence of MLE's Based on Interval-censored Data from Some Three-Parameter Distribution with a Shifted Origin, Lemma 3.3).

NOTE 2. The analogous result with "strict concavity" everywhere replaced by "concavity" has been proved without any differentiability assumption on $f$ by Pratt in Concavity of the Log Likelihood. Here the author uses the following wonderful theorem by Brascamp and Lieb, which is Theorem (1) in Brascamp and Lieb, Some inequalities for Gaussian measures and the long-range order of the one-dimensional plasma, in Arthurs (ed.), Functional Integration and Its Applications, which is also in Lieb, Inequalities, Selecta of Elliott H. Lieb. To state this theorem, let us recall that a function $G:\mathbb{R}^n \rightarrow [0,\infty)$ is said to be log-concave if for every $\lambda \in (0,1)$ and every $x, y \in \mathbb{R}^n$ we have \begin{equation} G^{\lambda}(x)G^{1-\lambda}(y) \leq G(\lambda x + (1- \lambda) y). \end{equation}

Theorem. Let $F:\mathbb{R}^{m+n} \rightarrow [0,\infty)$ be a log-concave function, with $F:(x,y) \mapsto F(x,y)$ for $x \in \mathbb{R}^m$ and $y \in \mathbb{R}^n$. Then, if we have \begin{equation} G(x) = \int_{\mathbb{R}^n} F(x,y) dy < \infty \quad \forall x \in \mathbb{R}^m, \end{equation} $G$ is a log-concave function.

This theorem is actually a simple corollary of the Prékopa-Leindler Inequality: see Prékopa, Logarithmic concave measures with applications to stochastic programming, Leindler, On a certain converse of Hölder's inequality II and in particular Prékopa On logarithmic concave measures, Theorem (6) and finally Brascamp and Lieb, On extensions of the Brunn-Minkowski and Prékopa-Leindler theorems, Corollary (3.5).


1 Answer 1


Assume first only that $f > 0$ and that $\log(f)$ is concave. Define the function \begin{equation} I(x,y,u) = \begin{cases} 1 &\quad\text{if} & x > u > y\\ 0 &\quad\text{otherwise.} \\ \end{cases} \end{equation} Since the set $\{(x,y,u) : I(x,y,u) = 1 \}$ is convex, $I$ is log-concave and so also the function $(x,y,u) \in \mathbb{R}^3 \mapsto h(x,y,u)f(u)$ is log-concave. By the Theorem quoted in the post we have then that \begin{equation} H(x,y)= \int_{\mathbb{R}} I(x,y,u)f(u) du \end{equation} is log-concave. Since for $x > y$, we have $H(x,y)=F(x) - F(y) > 0$, we get that the function $G(x,y)= \log(H(x,y))$, defined for $x > y$, is concave. This is Pratt's proof quoted in Note (2).

Now, let us consider the hypotheses in the post, that is $f > 0$ and $\log(f)$ strictly concave. We must show that $G$ turns out to be strictly concave. This follows by applying to our particular case the argument given in Brascamp and Lieb, Some inequalities for Gaussian measures and the long-range order of the one-dimensional plasma to prove the Theorem stated in the post.

We choose $x > y$ and $x'> y'$, with $(x,y) \neq (x',y')$, and we consider the log-concave function $Q: \mathbb{R}^2 \rightarrow [0,\infty)$ defined by \begin{equation} Q(t,u)=I(t x + (1-t)x',t y + (1-t)y',u)f(u), \end{equation} so that $H(t x + (1-t)x',t y + (1-t)y')=P(t)= \int_{\mathbb{R}} Q(t,u) du$.

Consider $t=1, t'=0$, fix $\lambda \in (0,1)$, and let $b \in \mathbb{R}$ such that $\sup_{u \in \mathbb{R}} Q(0,u)= e^{b} \sup_{u \in \mathbb{R}} Q(1,u)$. Define the log-concave function $\bar{Q}(t,u)=e^{bt}Q(t,u)$, and put \begin{equation} \bar{P}(t)=\int_{\mathbb{R}}\bar{Q}(t,u)du=e^{bt}P(t). \end{equation} Clearly, $P(t)$ is strictly log-concave if and only if $\bar{P}(t)$ is so.

Now, let us put, for any fixed $z \geq 0$: \begin{equation} C(z)= \{ (t,u) \in [0,1] \times \mathbb{R} : \bar{Q}(t,u) \geq z \}, \end{equation} and for any fixed $(t,z) \in [0,1] \times [0,\infty)$: \begin{equation} C(t,z)= \{ t \} \times \{u \in \mathbb{R}: \bar{Q}(t,u) \geq z \}. \end{equation} $C(z)$ is a convex set, and $C(t,z)$ is a segment.

Then let $T(z)$ be the trapezoid having as parallel sides the segments $C(0,z)$ and $C(1,z)$. $C(z)$ is the intersection of $T(0)$ with the set \begin{equation} B(z) = \{ (t,u) \in [0,1] \times \mathbb{R} : e^{bt} f(u) \geq z \}, \end{equation} which is convex since $(t,u) \mapsto e^{bt}f(u)$ is (strictly) log-concave. Now, let us note that since $\log(f)$ is strictly log-concave, in particular it is not constant on any interval, so that $C(0,z)$ and $C(1,z)$ shrink simultaneously to a single point when $z$ approaches $\bar{z}=\sup_{u \in \mathbb{R}}Q(0,u)$. We have then, by convexity of $C(z)$, that if $g(t,z)$ is the length of the segment $C(t,z)$, we have for any $z \geq 0$: \begin{equation} g(\lambda,z) \geq \lambda g(0,z) + (1- \lambda) g(1,z) \quad (I). \end{equation} From a well known result about distribution functions (see e.g. Rudin, Real and Complex Analysis, 3rd Edition, Section (18.5)) if $\mu$ is the one dimensional Lebesgue measure, we have \begin{equation} \bar{P}(t)=\int_{\mathbb{R}} \bar{Q}(t,u)du=\int_{0}^{\infty} \mu(\{u : \bar{Q}(t,u) > z \}) dz = \int_{0}^{\infty} \mu(\{u : \bar{Q}(t,u) \geq z \}) dz = \int_{0}^{\infty} g(t,z) dz, \end{equation} (where the third equality follows by continuity of $f$ or more in general because the nonincreasing functions $z \mapsto \mu(\{u : \bar{Q}(t,u) > z \})$ and $z \mapsto \mu(\{u : \bar{Q}(t,u) \geq z \})$ differ only on a countable set). We then get \begin{equation} \bar{P}(\lambda) \geq \lambda \bar{P}(0) + (1 - \lambda) \bar{P}(1) \quad (II). \end{equation} Now assume that (II) holds with equality. Set for any $t \in [0,1]$: \begin{equation} L(t,z)= \inf \{u \in \mathbb{R} : \bar{Q}(t,u) \geq z \}, \end{equation} \begin{equation} L(z)=\{(t,L(t,z)) \in \mathbb{R}^2 : t \in [0,1] \}, \end{equation} \begin{equation} U(t,z)= \sup \{u \in \mathbb{R}: \bar{Q}(t,u) \geq z \}, \end{equation} \begin{equation} U(z)=\{(t,U(t,z)) \in \mathbb{R}^2 : t \in [0,1] \}, \end{equation} and note that $L(0,z)$, $L(1,z)$, $U(0,z)$ and $U(1,z)$ are the vertices of $T(z)$ (in particular $L(0,0)=y$, $L(1,0)=y'$, $U(0,0)=x$ and $U(1,0)=x'$) and that $z \mapsto L(t,z)$ and $z \mapsto U(t,z)$ are continuous.

Since (II) holds with equality, (I) must hold with equality for a.e. $z \in [0,\bar{z}]$, and so actually for every $z \in [0,\bar{z}]$, given the contintuity of $z \mapsto g(t,z)$. This in turn implies that for every $z \in [0,\bar{z}]$ $L(z)$ and $U(z)$ must be segments, so we conlude that $C(z)$ must coincide with the trapezoid $T(z)$ for all $z \in [0,\bar{z}]$.

But then, since for $z$ approaching $\bar{z}$ from the left, $C(0,z)$ and $C(1,z)$ shrink to single points and since $(x,y) \neq (x',y')$, there exists some $z \in (0,\bar{z}]$ such that at least one between the segments $L(z)$ and $U(z)$ is not horizontal and not coinciding with a side of $T(0)$. Fix such a $z$ and let e.g. $L(z)$ be the segment with the stated property. Then we have \begin{equation} z=e^{bt}f(u) \end{equation} for every $(t,u) \in L(z)$, so that if $t=c+mu$ is the equation of the straight line containing $L(z)$, we get \begin{equation} f(u)= z \exp(-bc-bmu), \end{equation} for $u$ ranging between $-c/m$ and $(1-c)/m$. So $f$ would be not strictly log-concave, a contradiction.

We conclude that (II) holds strictly, and so \begin{equation} \bar{P}(\lambda) > \lambda \bar{P}(0) + (1 - \lambda) \bar{P}(1) \geq [\bar{P}(0)]^{\lambda} [\bar{P}(1)]^{1-\lambda} \end{equation} where the last inequality is the classical inequality between arithmetic mean and geometric mean. So we have \begin{equation} P(\lambda) > [P(0)]^{\lambda} [P(1)]^{1-\lambda}, \end{equation} that is, by taking the logarithms, \begin{equation} G(\lambda x + (1 - \lambda) x', \lambda y + (1- \lambda) y') > \lambda G(x,y) + (1 - \lambda) G(x',y'). \end{equation} QED


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