Let $G \colon [0,1] \to [0,1]$ be a differentiable cumulative distribution function (monotonically non-decreasing function with $G(0) = 0$ and $G(1) = 1$). We say that $G$ is regular if $$ x - \frac{1- G(x)}{G'(x)}$$ is monotonically non-decreasing.


Let $F\_1,F\_2 \colon [0,1] \to [0,1]$ be regular twice differentiable cumulative distribution functions. 



> 
> Must $F = F\_1F\_2$ be regular?
> 
> 
> 



Is regularity closed under products?
The answer is no. In the definition of the regularity, we assume that $G'>0$ on $(0,1)$ and the monotonicity in question is over $(0,1)$. Suppose for a moment that $G$ is any cumulative distribution function (cdf) on $\mathbb{R}$ such that $G=0$ on $(-\infty,0)$, $G=1$ on $[1,\infty)$, $G$ is twice differentiable on $(0,1)$, and $ x - \dfrac{1- G(x)}{G'(x)}$ is nondecreasing in $x\in(0,1)$. Let 
$$g:=\frac1{1-G},$$
so that $G=1-\dfrac1g$,
assuming that $\frac10=\infty$ and $\frac1\infty=0$. 
The above conditions on $G$ can be translated into the following: 
$$(\*)\qquad\text{$g$ is nondecreasing on $\mathbb{R}$, $g=1$ on $(-\infty,0)$, $g=\infty$ on $[1,\infty)$, 
}$$ 
$$(\*\*)\qquad\text{and on $(0,1)$ the function $g$ is twice differentiable, with $g'>0$ and $g''\ge0$.}
$$ 


Let $g\_1(x):=2 + 5 (x - a) + \frac14 (x - a)^2$ and $g\_2(x):=\frac54 + \frac12 (x - a) + \frac14 (x - a)^2$, where $a:=\frac34$. Let then $G\_j:=1-\dfrac1{g\_j}$ for $j=1,2$, 
$$(!)\qquad G:=G\_1G\_2,$$ 
and $g:=\dfrac1{1-G}$. Then $g''(a+\delta)<-0.037<0$ for $\delta:=\frac1{200}$ and hence $ x - \dfrac{1- G(x)}{G'(x)}$ is decreasing in $x$ in a neighborhood of the point $a+\delta[\in(0,1)]$. 


However, $G\_1$ and $G\_2$ are not cdf's, and so, we need to work a bit more. 
Again for $j=1,2$, let $h\_j$ be the function on $(-\infty,1)$ such that 
$$
h\_j(x)=
\begin{cases}
g\_j(x)&\text{ if }a\le x<1,\\
1\vee[g\_j(a)+g\_j'(a)(x-a)]&\text{ if }x\le a.
\end{cases}
$$ 
Note that $h\_j$ is nondecreasing and convex, and $h\_j=1$ on $(-\infty,\frac14]$; that is, conditions $(\*)$--$(\*\*)$ are "almost" satisfied with $h\_j$ in place of $g$. Also, $h\_j=g\_j$ on $[a,1)$. 
It is easy (but a bit tedious) to construct a "smooth" enough approximation $f\_j\colon\mathbb{R}\to\mathbb{R}$ of $h\_j$ such that $f\_j=h\_j$ in a neighborhood of the point $a+\delta$, conditions $(\*)$--$(\*\*)$ hold for $f\_j$ in place of $g$, and $F\_j:=1-\dfrac1{f\_j}$ is a cdf twice differentiable on $\mathbb{R}$, with $F\_j=0$ on $(-\infty,0)$ and $F\_j=1$ on $[1,\infty)$. 
Then in a neighborhood of $a+\delta$ one has $f\_j=g\_j$ and hence $F:=F\_1F\_2=G$, where $G$ is as in $(!)$. So, $ x - \dfrac{1- F(x)}{F'(x)}$ is decreasing in $x$ in a neighborhood of the point $a+\delta[\in(0,1)]$.


**Addendum:** One way to to get the mentioned "smooth" approximations $f\_j$ of $h\_j$ is to use splines. Thus, one e.g. may choose 
$$
f\_1(x):=1 + 5\Big(\frac15\Big)p\_1\Big(\frac{x-11/20}{1/5}\Big) + \frac14\,\Big(\frac1{200}\Big)^2 p\_2\Big(\frac{x-3/4}{1/200}\Big)+ \frac1{10^6}\Big(\frac{x\_+}{(1 - x)\_+}\Big)^3
$$
and 
$$
f\_2(x):=1 + \frac12\,\Big(\frac15\Big)p\_1\Big(\frac{x-1/4}{1/5}\Big) + \frac14\,\Big(\frac1{200}\Big)^2 p\_2\Big(\frac{x-3/4}{1/200}\Big)+ \frac1{10^6}\Big(\frac{x\_+}{(1 - x)\_+}\Big)^3 
$$
for all real $x$, where $x\_+:=0\vee x$ and 
$$
p\_1(x):=x\_+^3+\frac{(x+1)\_+^3}2-\frac{(x+1)\_+^4}4+\frac{(x-1)\_+^3}2+\frac{(x-1)\_+^4}4=
\begin{cases}
 0 & \text{ if } x\leq -1, \\
 \frac14 (1+2x+2|x|^3-x^4) & \text{ if } -1<x\leq 1, \\
 x & \text{ if } x>1
\end{cases}
$$
and 
$$
p\_2(x):=3x\_+^3 - 3x\_+^4 + x\_+^5 - (x - 1)\_+^3 - 
 2(x - 1)\_+^4 - (x - 1)\_+^5=
\begin{cases}
 0 & \text{ if } x\leq 0, \\
 x^5-3 x^4+3 x^3 & \text{ if } 0<x\leq 1, \\
 x^2 & \text{ if } x>1
\end{cases}
$$
are twice continuously differentiable splines, with $p\_j'\ge0$ and $p\_j''\ge0$ for $j=1,2$, which approximate $x\_+$ and $x\_+^2$, respectively. 


Then, as desired, for each $j=1,2$ the function  

$F\_j:=1-\dfrac1{f\_j}$ is a cdf, which is regular in the sense defined in the question and twice continuously differentiable on $\mathbb{R}$, with $F\_j=0$ on $(-\infty,0)$ and $F\_j=1$ on $[1,\infty)$. 
On the other hand, for $F:=F\_1F\_2$, the derivative of $x - \dfrac{1- F(x)}{F'(x)}$ in $x$ at $x=a+\delta=\frac34+\frac1{200}$ is $\approx-0.016<0$, and so, $x - \dfrac{1- F(x)}{F'(x)}$ is decreasing near $x=a+\delta$.


