My problem is the following. Given $F$ and $G$ cumulative distribution functions, with densities $f,g$ (for example on $[0,1]$), what can I say on the monotonicity of $F(x)(1-G(x))$? More specifically: I would like to conclude that $F(1-G)$ should be increasing for low enough $x$ and decreasing for high enough $x$. I feel this should be true under quite general regularity conditions, but I could prove it only in the case of log concave densities (but it true is in many non log concave examples, as in the Pareto distribution). Maybe I am missing the obviouys, but are there more general regularity conditions that ensure the result?


All I could do is the following reasoning, proving that there is an interval where the product is increasing arbitrarily close to 0 if the cdfs are strictly increasing and the densities are continuous.


Indeed, if they are strictly increasing $F(0)(1-G(0))=0$ and $F(x)(1-G(x))>0$ if $x>0$, so by Lagrange theorem for any $x>0$ there is a point $\zeta \in (0,x)$ such that:
$D(F(1-G))(\zeta)=\frac{F(x)(1-G(x))}{x}>0$
hence by continuity there exists an interval around $\zeta$ in which $F(1-G)$ is strictly increasing.



When a product of cdf and tail distribution is increasing?
One has $(F(1-G))'=(1-G)f-Fg$. So, for $F(1-G)$ to be increasing (that is, nondecreasing) in a right neighborhood (r.n.) of $0$, it is necessary and sufficient that $(1-G)f-Fg\ge0$ in a r.n. of $0$. 


For the latter condition to hold, it is enough that 
\begin{equation}
\liminf\_{x\downarrow0}\frac{f(x)}{F(x)g(x)}>1, \tag{1} 
\end{equation}
since $(1-G)f-Fg=[(1-G)\frac f{Fg}-1]Fg$ and $G(0+)=0$. 


In turn, for (1) to hold, it is enough that one of the following conditions take place:


(i) $\liminf\_{x\downarrow0}\frac{f(x)}{g(x)}>0$;


(ii) $f$ is increasing in a r.n. of $0$ and $\limsup\_{x\to\infty}xg(x)<1$ for all $x$ in a r.n. of $0$; 


(iii) $f$ is increasing in a r.n. of $0$ and $xg(x)$ is monotonic in $x$ in a r.n. of $0$; 


(iv) $f$ is increasing in a r.n. of $0$ and $g$ is bounded in a r.n. of $0$. 


Indeed, (i) implies (1), since $F(0+)=0$. 


If (ii) holds, then for all $x$ in a r.n. of $0$ one has $F(x)\le\int\_0^x f(t)\,dt\le f(x)x$ and hence $\frac{f(x)}{F(x)g(x)}\ge\frac1{xg(x)}$, so that (1) again holds. 


Suppose now condition (iii) holds. Then $\lim\_{x\downarrow0}xg(x)=:c\in[0,\infty]$ exists. If $c>0$, then for any $d\in(0,c)$ and all $x$ in a r.n. of $0$ one has $g(x)\ge\frac dx$ and hence $G(x)=\int\_0^x g(t)\,dt=\infty$, a contradiction. So, $c=0$. Thus, (iii) implies (ii). 


Also, clearly (iv) implies (ii). 


Moreover, (iii) and (iv) can be considered pure regularity conditions. 



> 
> Added: Let us also show the following: Under the apparently quite general ``non-oscillation'' regularity condition that $c:=\lim\_{x\downarrow0}\frac{f(x)}{F(x)g(x)}$ exists in $[0,\infty]$ (which will follow if e.g. $\frac f{Fg}$ is monotonic in a r.n. of $0$), we must have $c=\infty$ and hence (1) will hold, and hence 
>  $F(1-G)$ will be increasing in a r.n. of $0$. 
> 
> 
> Indeed, suppose that $c<\infty$. Then 
>  \begin{equation}
>  \ln\frac{F(z)}{F(x)}=\int\_x^z\frac{f(t)}{F(t)}\,dt=(c+o(1))\int\_x^z g(t)\,dt
>  =(c+o(1))[G(z)-G(x)]\to0
> \end{equation}
>  as $0<x<z\to0$ and hence $\frac{F(z)}{F(x)}\to1$ as $0<x<z\to0$. However, since $F(0+)=0$, for each $z>0$ we can always find some $x\in(0,z)$ such that $F(x)\le zF(z)$, which shows that $\limsup\_{0<x<z\to0}\frac{F(z)}{F(x)}=\infty$. This contradiction completes the proof. 
> 
> 
> 




---


The monotonicity of $F(1-G)$ in a left neighborhood (l.n.) of $1$ can be considered quite similarly. Alternatively, it can be reduced to the monotonicity of $F(1-G)$ in a r.n. of $0$ by introducing distribution functions $F\_1$ and $G\_1$ defined by the conditions $F\_1(y)=1-G(1-y)$ and $G\_1(y)=1-F(1-y)$, so that for $x=1-y$ one has $F(x)(1-G(x))=F\_1(y)(1-G\_1(y))$ and $x\uparrow1\iff y\downarrow0$. 


