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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.

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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 $x\uparrow1\iff y\downarrow0$.

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  • $\begingroup$ Thanks, this is a good point. I expect the fact to be true more generally though: the only example that I can imagine in which fails is when the derivative of $F(1-G)$ oscillates a lot, so I am expecting that the most general condition should involve bounding some derivatives. $\endgroup$
    – Matteo
    Dec 3, 2017 at 10:05
  • $\begingroup$ It is unclear in what terms you want the condition(s) to be stated. Anyhow, I have now given a number of sufficient conditions of various degrees of generality. The condition given previously is now condition (i). Conditions (iii) and (iv) can be considered pure regularity conditions, if that is what you wanted. $\endgroup$ Dec 3, 2017 at 16:53
  • $\begingroup$ I have added another, apparently quite general ``non-ossilation'' regularity condition. $\endgroup$ Dec 3, 2017 at 20:47

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