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EDIT (for bounty):

Consider the differential equation

$G(p;x,\lambda)p \left[1-\lambda-x(1+\lambda)\right] + x(1+\lambda)p + (1-x)(1-\lambda) \int_{p}^{1} z G'(z;x,\lambda) dz - (1-\lambda) = 0$,

where $G(p;x,\lambda)$ is a c.d.f. with parameters $x \in (0,1)$ and $\lambda \in (0,1)$. The boundary condition is $G(1,x,\lambda) = 1$.

This equation has the solution

$G(p;x,\lambda) = \frac{x(1+\lambda) - (1-\lambda)p^{-\frac{x(1+\lambda)-(1-\lambda)}{2x \lambda}}}{x(1+\lambda)-(1-\lambda)}.$

Solving $G(p;x,\lambda) = 0$, one can find its lower bound $\underline{p} = \left(\frac{1-\lambda}{x(1+\lambda)}\right)^{\frac{2\lambda x}{(1+\lambda)-(1-\lambda)}}$.

Define $C_1(x) := \int_{\underline{p}}^{1} (1+\lambda) G(p;x,\lambda) - G(p;x,\lambda)^2 dp$

and $C_2(x) := \lambda + (1-\lambda) \int_{\underline{p}}^{1} G(p;x,\lambda)^2 dp$.

One can show that both $C_1(x)$ and $C_2(x)$ are increasing in $x$. Also, $G(p;x,\lambda)$ is increasing in $x$ (over the relevant range). Moreover, I know that

$x C_1(x) + (1-x)C_2(x) = \lambda$.

OBJECTIVE: Show that $C_2(x)-C_1(x)$ is decreasing in $x$ over the relevant parameter range. Bonus objective: Show that it is convex (both of these features can be shown numerically).

Possibly helpful hint: $C_2(x)-C_1(x)$ can also be expressed as

$\frac{1-\lambda}{x} \int_{\underline{p}}^{1} G(p;x,\lambda)^2 dp$.

Many thanks in advance for any suggestions that might lead to a proof!

EDIT: Actually, the problem doesn't look too hard, but I still cannot solve it. Any clever ideas what to do here?

Let $x \in (0,1), \lambda \in (0,1)$.

Let $p(x,\lambda) = \left(\frac{(1+\lambda) x}{1-\lambda}\right)^{-\frac{2 \lambda x}{x(1+\lambda)-(1-\lambda)}}$.

Show that

$\frac{(1-\lambda)(\lambda + x) - x(1+\lambda)^2 p(x,\lambda)}{(1-x)(1-x-\lambda)}$

is strictly decreasing in $x$ [for $x \in (0,1), \lambda \in (0,1)$].

Original post

For a couple of days I've been struggling with confirming that $\frac{\partial S(x)}{\partial x}$, with

$$S(x) = v \dfrac{(1+\lambda)^2 x \left(\dfrac{1-\lambda}{(1+\lambda) x}\right)^{\tfrac{2 \lambda x}{x(1+\lambda)-(1-\lambda)}}-(1-\lambda) (\lambda+x)}{(1-x) [x-(1-\lambda)]}\; \;,$$

is strictly negative for $\lambda \in (0,1)$, $x \in (0,1)$ [$v>0$ is irrelevant]. I can show this numerically, but a general proof would be desirable.

For what it's worth, the derivative of this function can be shown to have the same sign as $(1-\lambda)\left[1+(1-\lambda) \lambda-2 \lambda x-x^{2^{\vphantom{a}}}\right]+ \frac{\Large\left(\frac{(1+\lambda) x}{1-\lambda}\right)^{-\frac{2 \lambda x}{x(1+\lambda)-(1-\lambda)}} \left(\left(1-\lambda^2\right)^2-\left(1-\lambda^2\right)^2 x+(1+\lambda)^2 (-1+\lambda (-3+2 \lambda)) x^2+(1+\lambda)^2 (1+3 \lambda) x^3\right)}{\Large x(1+\lambda)-(1-\lambda)}$

$+\dfrac{2 (1-\lambda) \lambda (1+\lambda)^2 (1-x) x \left(\frac{(1+\lambda) x}{1-\lambda}\right)^{-\tfrac{2 \lambda x}{x(1+\lambda)-(1-\lambda)}} [x-(1-\lambda)] \text{Log}\left[\frac{(1+\lambda) x}{1-\lambda}\right]}{[x(1+\lambda)-(1-\lambda)]^2},$

provided that $x \neq 1-\lambda.$ The expression gets much simpler by using the upper bound $Log[z] \leq z-1$ (for the case that $x > 1-\lambda$), but then the sign of the simplified expression unfortunately becomes ambiguous.

A possibly simpler representation of $\frac{\partial S(x)}{\partial x}$ can be given by

$$(1-\lambda)\int_{\underline{p_G}(x)}^{v} 2 x G(p;x) \frac{\partial G(p;x)}{\partial x} - G(p;x)^2 dp,$$ where $$G(p;x) = \frac{x(1+\lambda) - (1-\lambda)\left(\frac{v}{p}\right)^{\tfrac{x(1+\lambda)-(1-\lambda)}{2x \lambda}}}{x(1+\lambda)-(1-\lambda)},\quad\underline{p_G}(x) = v \left(\frac{1-\lambda}{x(1+\lambda)}\right)^{\tfrac{2\lambda x}{x(1+\lambda)-(1-\lambda)}},$$ $$G(\underline{p_G}(x);x) = 0,\quad G(v) =1.$$ Moreover, I can show that $\frac{\partial G(p;x)}{\partial x} \geq 0.$

Any help would be welcome! I will happily include a thanking note to my (economics) research article for suggestions that lead to a solution.

EDIT: Two other things that might help:

$G(p;x)$ is actually the solution to the differential equation $G(p;x)p \left[1-\lambda-x(1+\lambda)\right] + x(1+\lambda)p + (1-x)(1-\lambda) \int_{p}^{v} z G'(z;x) dz - (1-\lambda)v = 0$,

and $S(x) = \left(1-\lambda \right)\frac{\int_{\underline{p_G}(x)}^{v} G(p;x)^2dp}{x}$.

share|cite|improve this question
I've tried to make the math expressions a bit larger to improve readability, but you're welcome to change it back if you want. – Zev Chonoles Jun 26 '12 at 14:51
Thank you, that's great. – Martin Jun 26 '12 at 14:56
@Martin Did you get a proof of this? – user31373 Jul 27 '12 at 19:23
Unfortunately, I had to give up for now... I can give a simpler representation of $S(x)$ just as a function of $\int G(p;x)$ (instead of $\int G(p;x)^2$) or alternatively as a pure function of $\underline{p_G}(x)$. If that is interesting to you, I can post them here. Other than that, I had no real progress. – Martin Jul 30 '12 at 9:17
What is the range of values for $\lambda$ ? All of $\mathbb R$ ? – Ewan Delanoy Oct 8 '12 at 13:08

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