log-concavity of a function of a log-concave CDF

Question

Let $f(x)$ be a log-concave probability distribution with support on $\mathbf{R}$ and that is symmetric around zero (for example a zero-mean Gaussian distribution). Denote by $F$ its cumulative density function. We have

• $F(x)$ is log-concave
• $F(-x) = 1 - F(x)$ (by symmetry around $0$)

I am trying to prove (or disprove) the following statement for all $x, y \in \mathbf{R}$ such that $x+y \le 0$:

$$ \frac{F(x + y)}{F(-(x +y))} \stackrel{?}{\ge} \frac{F(x)}{F(-x)} \cdot \frac{F(y)}{F(-y)} $$

My feeling is that the above inequality is correct for $x+y \le 0$, and reversed for $x+y \ge 0$. Here is my (unsuccessful, so far) attempt at trying to show this:

1. Try to show something about the log-concavity of $$ \frac{1}{F(-x)} = \frac{1}{1 - F(x)}. $$ Letting $f'(x)$ be the derivative of the density function, I get $$ \left[ \log \left( \frac{1}{1 - F(x)} \right) \right]'' = f'(x) \frac{1}{1 - F(x)} - f(x)^2 \frac{1}{(1 - F(x))^2} < 0 $$ for $x \in [0, \infty)$ because $f'(x)$ is negative on that interval. This means the the function $1 / (1 - F(x))$ is log-concave on that interval.
2. Use the the fact that the product of two log-concave functions remains log-concave, hence $$ \frac{F(x)}{F(-x)} $$ is log-concave on $[0, \infty)$
3. Then... I don't really know. I was hoping to use log-concavity somehow. But I don't see how.

Does someone see how I could make progress in solving this question?

(Source of the question: I am trying to understand how a result developed in this paper (Eq. 2) can be generalized to non-logistic distributions.)

Answer 1

I realized that the inequality is incorrect. (Furthermore, the second derivative of $\log \left( \frac{1}{1 - F(X)} \right)$ derived above contains mistakes).

First of all, I should have probably looked at the log-concavity of $$ \frac{F(x)}{1 - F(x)} $$ directly. Looking at the second derivative of the $\log$, we find $$ \left[ \log \left( \frac{F(x)}{1 - F(x)} \right) \right]'' = \frac{f'(x)F(x)[1-F(x)] - f^2(x)[1 - 2F(x)]}{F^2(x) (1 - F(x))^2} $$ There is no reason to believe that this expression will always be negative, so a priori no way to claim that the function is log-concave.

Second, some simulations with $F(x) = \Phi(x)$ (the Gaussian CDF) show that the inequality does not hold.

Here is some Python code:

import matplotlib.pyplot as plt
import numpy as np
from scipy.stats import norm

x = 3
ys = np.linspace(-5, 5, num=200)

lhs = np.log(norm.cdf(x + ys) / (1 - norm.cdf(x + ys)))
rhs = ((np.log(norm.cdf(x) / (1 - norm.cdf(x))))
       + (np.log(norm.cdf(ys) / (1 - norm.cdf(ys)))))

plt.plot(ys, lhs, label="log(left-hand side)")
plt.plot(ys, rhs, label="log(right-hand side)")
plt.title("both sides of the ineq. as a fct of $y$ for $x = 3$")
plt.xlabel("value of $y$")
plt.legend(loc="upper left")

Note to self: when possible, always run simulations to develop some intuition about the problem.