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I have two cdfs, both distributed over 0 to 1. Let's call them $F(x)$ and $G(x)$.

If I know that $\int_0^1 F(x) \,dx < \int_0^1 G(x) \,dx$

then, does it follow that

$ \left|\frac{d}{dn} \int_0^1 F(x)^n \,dx\right| > \left|\frac{d}{dn}\int_0^1 G(x)^n \,dx\right|$ where $n$ is a positive real number >1? (I used to say integer but corrected due to comments)

I would think this would be the case, because the exponent will have a greater effect on a smaller fraction, and $F$ has to have smaller fractions on average. But, generally whenever I say, "I would think this would be the case" I am wrong and not thinking of something.... so I pose the question to the awesome SE community...

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    @Sri: "difference sequence" - plain "difference" is fine. ;) If you must be precise, that's$a$"forward difference". A "backward difference" goes like $\nabla a(n)=a(n)-a(n-1)$.2011-10-08

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For every CDF $F$ on $[0,1]$ and nonnegative $s$, let $I_s(F)=\displaystyle\int\limits_0^1 F^s(1-F)$. You are asking whether $I_0(F)>I_0(G)$ implies that, for every positive real number $n$, $I_n(G)>I_n(G)$ (or maybe $n$ is an integer and/or maybe n>1).

My first reaction is to ask why should that be the case. My second reaction is to note that, if this was true then $I_0(F)=I_0(G)$ would imply $I_n(G)=I_n(G)$ for every $n$, a remark which furthers my skepticism. My third and final reaction is to look for a simple counterexample.

Let $F(x)=\sin(\pi x/2)^2$ and $G(x)=x^a$ for a given positive $a$. Then $I_0(F)=1/2$ and $I_0(G)=a/(a+1)$ hence every $a<1$ yields $F$ and $G$ such that $I_0(F)>I_0(G)$.

And $I_1(F)=1/8$ while $I_1(G)=a/((a+1)(2a+1))$ hence $I_1(F) for every $a$ such that $8a>(a+1)(2a+1)$. Consider for example $a=1/2$.