# Assumptions and notation

Let $f$ be a twice-differentiable log-concave density function on $[0,1]$, and let $F$ be the corresponding distribution function. Define $x^D$ by: $$\frac{1-2F(x^D)}{f(x^D)}=x^D-\frac{1}{2}.$$

Note that the equation has exactly one solution because log-concavity of $f$ implies that $(1-F)/f$ is decreasing and $F/f$ is increasing so that $(1-2F)/f$ is a decreasing function.

# Conjecture I wish to prove/disprove:

I conjecture that $f(x^D)\geq 1$.

The diagram below plots the two functions on each side of the equation defining $x^D$. The diagram is drawn for the distribution function $$F(x)=\frac{e^{10x}-1}{e^{10}-1}$$ which has a log-concave (in fact log-linear) density. In this example, it can be seen that $$1-2F(x^D)>\frac{1-2F(x^D)}{f(x^D)}$$ so that $f(x^D)>1$.

http://i66.tinypic.com/2duwk92.jpg

# Progress so far:

## Conjecture is true if the median $m$ is $1/2$:

In this case $x^D=m=1/2$ and it must be that $f(x^D)=f(m)=f(1/2)\geq 1$.

Proof: Suppose that $m=1/2$ and that without loss of generality the mode $M$ is at least $1/2$. Then $$\frac{1}{2}=\int_0^mf(x)dx \leq \int_0^mf(m)dx= mf(m)=\frac{1}{2}f(m),$$ where the inequality follows because $f$ is log-concave and hence unimodal, and because $m\leq M$.

## Conjecture is true for:

Log-linear densities: $$F(x)=\frac{e^{ax}-1}{e^a-1},\qquad f(x)=\frac{ae^{ax}}{e^a-1}.$$

Power functions: $$F(x)=x^n,\qquad f(x)=nx^{n-1}.$$

Numerical computations indicate it is also true for any beta distribution with a log-concave density.

## An example where the result does not hold ($f$ not log-concave)

Below is the diagram for the distribution $$F(x)=1-(1-x)^{1/4},$$ which is concave (and so log-concave), but has a density that is not log-concave. Here $f(x^D)<1$. Notice that the $\frac{1-2F(x)}{f(x)}$ is not decreasing on $[0,1]$. This example shows that log-concavity (or concavity) of $F$ is not sufficient for the result.

http://i65.tinypic.com/30stx95.jpg

## Limiting result:

Finally from numerical computations it also appears that as a sequence of distributions ordered by first-order stochastic dominance converges to the degenerate distribution at $1$, the value of $f(x^D)$ converges from above to 2.

# Context

In case anyone is interested, here $x^D$ is the equilibrium indifferent consumer in a Hotelling model of product differentiation. Showing that $f(x^D)\geq 1$ would allow me to put an upper bound on the equilibrium prices and profits in a duopoly.

• There might be some missing assumptions here, because f(x) = exp(- (x-1)^2 + B) is a log concave density on [0,1] when integration constant exp(B) is set properly. With this density, (F/f)' = 1 - 2*(1-x)* F(x)/f(x) which is positive near 0 (since F(0)=0). – VictorZurkowski Aug 1 '16 at 4:52
• As stated in the second line under the equation defining $x^D$, log concavity of a density function $f$ implies $F/f$ is increasing so that $(F/f)'$ is positive for all $x$ in the support. Your example does not contradict that. – smcc Aug 1 '16 at 8:58
• My bad. I mis-read the question. Somehow I read "F/f is decreasing" – VictorZurkowski Aug 1 '16 at 16:52
• Did you mean $f/f^{'}$ monotonically increasing? – theoGR Aug 3 '16 at 1:44
• Log concavity of $f$ is equivalent to $f/f'$ being increasing, but log concavity of $f$ also implies log concavity of $F$ and $1-F$ (which are, respectively, equivalent to $F/f$ being increasing and $(1-F)/f$ being decreasing). – smcc Aug 3 '16 at 12:03