I am trying to show that $$\int\_0^1F\_n(x) dx \leq \int\_0^1F\_{n+1}(x) dx$$ when $$F\_n(x) = (1-(1-F\_{n-1}(x))^c)^c$$ and $F\_0(x) = x$ and $n$ and $c$ are integers, $n\geq 1$ and $c \geq 2$ 


Note that $F$ performs a "minimax type opertation". Starting with the uniform distribution, it produced the cdf of the distribution of the maximum of $c$ draws from the minimum of $c$ draws from the distribution before it. 


Here is a list of things I know that might be helpful:


1)
$$\int\_0^1 F\_0(x) dx = \int\_0^1 x dx < \int (1-(1-x)^c)^c dx= \int\_0^1 F\_1(x) dx$$


2)


The distribution $F\_n$ converges to Dirac mass at the unique solution to $ L= (1-(1-L)^c)^c$, and $\frac{1}{2}^c< L <\frac{1}{2}$ The integral of $F\_n$ converges to $1-L$


3)


The integrals of the sequence defined by $F$ are identical to those defined by
related functions. $$\int\_0^1 F\_n dx = \int\_0^1 G\_n dx = \int\_0^11-F\_n^{-1} dx= \int\_0^1 1-G\_n^{-1}dx = $$ $$\int\_0^1 1-(F\_n(x^c))^{\frac{1}{c}} dx= \int\_0^1 (1-F\_n(x^c))^c dx$$ where $G\_n = (1-(1-G\_{n-1}^{\frac{1}{c}})^{\frac{1}{c}} $ 


4) 


$$\int\_0^1 F\_{n+1}(x)\,dx = \int\_0^1F\_n(x)dF^{-1}\_1(x)dx$$
(and various other ways of rewriting $\int F\_i$ in terms of $\int F\_j$)


5) 


The sequence operator $F= (1-(1-x)^c)^c$ is really two iterations of the simpler sequence operator $(1-x)^c$


6)


The derivative of $F\_n$ wrt $x$ = 
$$f\_n = \displaystyle\prod\_{k=1}^{n-1}c^2\left(1-\left(1-F\_k\right)^c\right)^{c-1}(1-F\_k)^{c-1}$$ $$=c^2\left(1-\left(1-F\_{n-1}\right)^c\right)^{c-1}(1-F\_{n-1})^{c-1}f\_{n-1} $$


Does anyone have any ideas?



Proving a sequence of integrals increases (iterated minimax distributions)
Let $c>1$ be any fixed real exponent and let us denote, for any $x\in I:=[0,1]$, $f(x):=(1-x)^c$, and $g(x):=1-x^{\frac{1}{c}}$. So $f$ is a strictly decreasing homeomorphism of $I$ into itself, with inverse map $g$ and with the interior fixed point $0 < L< 1/2$.
Also, since all even-order iterated of $f$ are strictly increasing, for any $x\in I$ and for any $n\in\mathbb{N}$, we have
$$x\le L\quad\mathrm{iff}\quad f(x)\ge x\quad\mathrm{iff}\quad F\_1(x)\le x \quad\mathrm{iff}\quad F\_{n+1}(x)\le F\_n(x)\, .$$
The sets 
$$A:=\{(x,y)\in I^2\, : \, F\_{n+1}(x)\le y \le F\_n(x)\}$$
and
$$B:=\{(x,y)\in I^2\, : \, F\_{n+1}(x)\ge y \ge F\_n(x)\}$$
are therefore included in the squares $[0,L]^2$, respectively $[L,1]^2$, so we have 
$$-\int\_0^L(F\_{n+1}-F\_n)dx=|A|$$
and
$$\int\_L^1(F\_{n+1}-F\_n)dx=|B|\, .$$
Moreover, $(x,y)\in A$, that is $ F\_{n+1}(x)\le y \le F\_n(x)$, if and only if 
$$F \_ {n+1}(f(x))=f(F \_ {n+1}(x))\ge f(y) \ge f(F\_ n(x))=F\_ n(f(x))\, ,$$ that is, $(f\times f)(x,y):=(f(x), f(y))\in B$. So $A=(g\times g)(B)\, .$ The map $g\times g$ has Jacobian determinant $g'(x)g'(y)=\frac{1}{c^2}(xy)^{\frac{1}{c}-1}\, ,$ and since $x\ge L$ and $y\ge L$ in $B$
$$|A|=|(g\times g)(B)|=\int\_{B} g'(x)g'(y)dxdy\le \left( \frac{L^{\frac{1}{c}-1}}{c}\right)^2|B|\, . $$ 


Let's prove that the factor in front of $|B|$ is less than or equal to $1$. So we wish to show that
$$L^{\frac{1}{c}-1}\le c\, . $$
Recalling that $L=(1-L)^c$, that is, $c=\frac{\log L}{\log (1-L)}$, this reduces to 


$$\frac{(1-L)\log(1- L)} {L\log L}\le 1 $$


which is indeed true exactly for $L\le 1/2$. So we have
$$\int\_0^1 F\_{n+1}dx -\int\_0^1 F\_n dx = \int\_0^L(F\_{n+1}-F\_n)dx + \int\_L^1(F\_{n+1}-F\_n)dx=$$ $$= -|A|+|B|\ge 0\,, $$
as required.


