# Condition on density function such that related function is increasing

Consider a continuously differentiable and strictly positive probability density function $f(x)$ defined on some support $[\underline{x}, \overline{x}]$. The corresponding CDF $F(x)$ is given by $F(x) = \int_{\underline{x}}^{x} f(z)dz$, with $F(\underline{x}) = 0$ and $F(\overline{x}) = 1$.

Consider next two values $x_l$, $x_h$, with $\underline{x} \leq x_l < x_h \leq \hat{x} \leq \overline{x}$.

I want find to find a condition on the density function $f(x)$, preferably as general and simple as possible, such that

$g(x_l, x_h, \hat{x}) := \frac{f\left(x_l\right)+\int_{x_h}^{\hat{x}} (1-F(z)) f'\left(z-x_h+x_l\right) \, dz}{1-F\left(x_l\right)-\int_{x_h}^{\hat{x}} (1-F(z)) f\left(z-x_h+x_l\right) \, dz}$

is weakly increasing in $x_l$ for all $\underline{x} \leq x_l < x_h \leq \hat{x} \leq \overline{x}$.

(Notice that the nominator is given by the negative derivative of the denominator with respect to $x_l$. It can moreover be shown via integration by parts that both the nominator and denominator are strictly positive.)

Taking the partial derivative of $g(x_l,x_h,\hat{x})$ with respect to $x_l$, this amounts to showing that

$\left[f'\left(x_l\right)+\int_{x_h}^{\hat{x}} (1-F(z)) f''\left(z-x_h+x_l\right) \, dz\right]\left[1-F\left(x_l\right)-\int_{x_h}^{\hat{x}} (1-F(z)) f\left(z-x_h+x_l\right) \, dz\right] + \left[f\left(x_l\right)+\int_{x_h}^{\hat{x}} (1-F(z)) f'\left(z-x_h+x_l\right) \, dz\right]^2 \geq 0.$

Since the expressions in the second and third squared bracket are strictly positive, an easy sufficient condition is that the expression in the first squared bracket is weakly positive. Indeed, it can be shown (again via integration by parts of the integral in the first bracket) that this is the case whenever $f'(x)$ is weakly positive.

However, I am interested in weaker conditions like this. In particular, I have the conjecture -- but cannot prove it unfortunately -- that the above condition will be satisfied whenever $f(x)$ is weakly log-concave (meaning that $\log(f(x))$ is a weakly concave function in $x$).

Two properties of log-concave densitites might be helpful to show this (if it is even true). First, log-concavity of $f(x)$ implies that both $F(x)$ and $1-F(x)$ are log-concave as well (I am happy to outline a simple proof on request). Second, log-concavity of $1-F(x)$ (as implied by log-concavity of $f(x)$) is equivalent to the property that $\frac{f(x)}{1-F(x)}$ is weakly increasing in $x$.

The rationale for suspecting that log-concavity of $f(x)$ might be sufficient arose when considering the density $f(x) = e^{-x}$ (where $e$ is Euler's constant, with corresponding CDF $F(x) = 1-e^{-x}$ and support $[0, \infty)$). It is then easy to see that $f(x)$ is exactly on the boundary of being log-concave. Moreover, when plugging $f(\cdot)$ and $F(\cdot)$ into $g(x_l, x_h, \hat{x})$, it turns out that $g(x_l,x_h, \hat{x})$ is exactly constant, such that my conditon is also on the boundary of being satisfied.

So to sum up, my question is as follows: Can it somehow be proven that $g(x_l, x_h, \hat{x})$ is weakly increasing, provided that $f(x)$ is weakly log-concave? Alternatively, can any counterexample be provided?

Any help would be greatly appreciated. Thanks for reading this lenghty question!