How to show f/F is decreasing where f and F are pdf and cdf of normal distributions, respectively. I read in a few econ journals this results that they use without proof.
I was trying to prove it, but couldn't figure out how.
It is to show:
$f/F$ is decreasing where $f$ and $F$ are pdf and cdf of a normal distribution, respectively. The domain of normal distribution takes interval $[-\infty,+\infty]$.
 A: \begin{align}
f(x) & = \frac 1 {\sqrt{2\pi}} e^{-x^2/2}, \\[10pt]
F'(x) & = f(x).
\end{align}
Therefore
\begin{align}
\frac d {dx} \, \frac{f(x)}{F(x)} & = \frac{F(x) f'(x) - f(x) F'(x)}{(F(x))^2} & & \text{This is the quotient rule.} \\[10pt]
& = \frac{F(x) e^{-x^2/2}(-x) - (f(x))^2}{(F(x))^2} \\[10pt]
& = \frac{F(x) e^{-x^2/2}(-x) - (e^{-x^2/2})^2}{(F(x))^2} \\[10pt]
& = \frac{e^{-x^2/2}}{(F(x))^2} \cdot \underbrace{(-x - e^{-x^2/2})}_{\text{This is negative} \\  \quad \text{if } x\,\ge\,0.}
\end{align}
The first factor on the last line above is always positive. The second factor is negative not only when $x\ge0,$ but also for some negative values of $x.$ Therefore $f/F$ decreases on an interval from some negative number to $+\infty.$
A: The derivative of the pdf of the normal distribution function is a negative number for $x > \mu$, where $\mu$ is the mean of the distribution. Therefore, as $x \to \infty$, the pdf will decrease to zero, while the value of the cdf will increase to 1. Therefore, $f/F$ should be a strictly decreasing function, but it is not guaranteed to hold for all $x > \mu$.
