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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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    What do you mean by $\frac{d}{dn}$?2011-10-08
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    @Srivastan: I tried to mean the derivative of the integral with respect to $n$. Is that not the correct notation?2011-10-08
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    It's acceptable to differentiate w.r.t. $n$, provided $n$ takes real values. But you constrain $n$ to be an integer, so differentiation doesn't make much sense.2011-10-08
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    @Srivatsan: Oh, I see what you mean. I guess I was thinking the rate of change between the terms in the series that would result (is there a word for that?) But I will take away the integer constraint.2011-10-08
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    There's [this](http://en.wikipedia.org/wiki/Finite_difference) idea of finite differences. For e.g., if $a$ is a sequence, then the "difference sequence" (I made up that term just now) $\Delta a$ is defined by $\Delta a(n)= a(n+1)-a(n)$. I am not sure if you are talking about something similar.2011-10-08
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    @Srivatsan: yes, that is what I meant. thank you! I have learned. But for this question, I will leave it as is (with n now as a real number) since i don't think the integer constraint is important to this question.2011-10-08
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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 $I_1(F)(a+1)(2a+1)$. Consider for example $a=1/2$.