I am looking for a result on the ordering of distribution functions.
The probability density functions $f(x)$ and $g(x)$ bear the Monotone Likelihood Ratio Property (MLRP) if
$$ \frac{f(x)}{g(x)} $$
is increasing in $x$.
By First Order Stochastic Dominance (FOSD) of the probability distribution function $F(x)$ over $G(x)$, it follows that
$$ F(x) \leq G(x) $$
for all $x$, with strict inequality for some $x$.
It is well known that the MLRP implies First Order Stochastic Dominance (FOSD) but not vice versa.
I am interested in sufficient conditions on $f(x)$ and $g(x)$ that, together with $F(x)$ FOSD $G(x)$ imply the MLRP. It seems to me that assuming that both $f(x)$ and $g(x)$ are log-concave should be sufficient. It holds in the following example:
Consider $g(x) = f(x + a)$, where $a>0$. I.e., $g$ is a version of $f$ shifted to the left. FOSD obviously holds. In this case, MLRP also holds because
$$ \frac{\partial\frac{f(x)}{g(x)}}{\partial x} > 0 \Leftrightarrow \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} > 0\Leftrightarrow \frac{f'(x)}{f(x)} - \frac{f'(x+a)}{f(x+a)} >0.$$
The last condition holds by log-concavity, as $\frac{f'(x)}{f(x)}$ is decreasing in $x$.
I'm not so sure how to generalize from this to other cases where $f$ and $g$ are both log-concave but where $g$ might not be a simple left shift of $f$.
Has anyone got ideas how to show this or approach this problem?
A counterexample:
Let \begin{align*} f(x) &= \frac{2x^2 + 2x + 4}{5\sqrt \pi}\mathrm{e}^{-x^2}, \\ g(x) &= \frac{2x^2 + x + 4}{5\sqrt \pi}\mathrm{e}^{-x^2}. \end{align*} We have $f, g > 0$ for all $x\in (-\infty, \infty)$. We have, for all $x \in (-\infty, \infty)$, \begin{align*} (\ln f)'' &= - \frac{2x^4 + 4x^3 + 12x^2 + 10x + 5}{x^2+x+2} < 0,\\ (\ln g)'' &= - \frac{8x^4 + 8x^3 + 42x^2 + 20x + 17}{2x^2+x+4} < 0. \end{align*}
Let \begin{align*} F(x) &= \int_{-\infty}^x f(t) \mathrm{d} t = \frac12 \mathrm{erf}(x) + \frac12 - \frac{2x + 2}{10\sqrt \pi}\mathrm{e}^{-x^2}, \\ G(x) &= \int_{-\infty}^x g(t) \mathrm{d} t = \frac12 \mathrm{erf}(x) + \frac12 - \frac{2x + 1}{10\sqrt \pi}\mathrm{e}^{-x^2}. \end{align*} Clearly, $F(x) < G(x)$ for all $x \in (-\infty, \infty)$. Also, $\lim_{x\to \infty} F(x) = 1$, $\lim_{x\to \infty} G(x) = 1$, $\lim_{x\to -\infty} F(x) = 0$, and $\lim_{x\to -\infty} G(x) = 0$.
However, we have $$\frac{f(x)}{g(x)} = \frac{2x^2 + 2x + 4}{2x^2 + x + 4}$$ and $$\left(\frac{f(x)}{g(x)}\right)' = \frac{4 - 2x^2}{(2x^2 + x + 4)^2}.$$