Proving a sequence of integrals increases (iterated minimax distributions)

Question

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?

Answer 1

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.

Answer 2

Use Jensen's inequality and induction on $n$.

Let $\phi_c(z) = (1- (1 - z)^c)^c$ , so that $\phi_c(F_i) = F_{i+1}$. As you noted, $\phi = f \circ f$, where $f(z) = (1-z)^c$. $f$ is convex for $0 \leq z \leq 1$, since $\frac{d^2f}{dz^2} = c^2(1-z)^{c-2} \geq 0$. By chain rule, $\phi$ is convex when $0 \leq z \leq 1 - \frac{1}{(c+1)^{(1/c)}}$, so the rest of this argument doesn't work.

Now apply Jensen's inequality to get $\phi(\int F_i) \leq \int F_{i+1}$ provided $0 \leq F_i(x) \leq 1$, which holds for all $0 \leq x \leq 1$. For $z > \frac{3 - \sqrt(5)}{2}$, $\phi_c(z) > z$ for every $c \geq 2$. So since $\int F_0 = \frac{1}{2}$, we have $\int F_0 \leq \phi(\int F_0)$, and the result follows by induction.

Jensen's inequality could still be used on $f$, but that doesn't seem to get any closer to a solution.