I run through many iterations of numerical simulations and I am pretty sure that this property of a normal distribution is true, but I cannot find a way to prove it. I am wondering if it is possible to formally prove it?
$\frac{\phi(u-h)-\phi(u)}{\Phi(u)-\Phi(u-h)} > \frac{\phi(l-h)-\phi(l)}{\Phi(l)-\Phi(l-h)}$
where $h>0$, $l<0<u$, $|l|>|u|, $ $\phi$ is the pdf, and $\Phi$ is the cdf of a normal distribution $N(0,1)$, respectively.
Consider the limit case $h=\infty$, where you have $f(x)=-\frac{\phi(x)}{\Phi(x)}$.
For the standard normal distribution you have $$f(x)=-\frac{\sqrt{\frac 2 \pi} e^{-x^2/2}}{\mathrm{erfc}(-\frac{x}{\sqrt 2})}$$ Here's the plot of the function:
It's increasing function, so you get $f(u)>f(l)$, then your inequality holds.
I think that you can prove a general case of finite $h$ using the fundamental theorem of calculus.
It's easy to show that your inequality holds for $h\le 2u$ because in this case $\phi(u-h)- \phi(u)\ge 0$ and $\phi(l-h)- \phi(l)< 0$.
