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My research question in a dynamic model of political competition boils down to the following conjecture. I am confident that it holds (all simulations work), but I have not been able to prove it yet. Let $c,\eta,\mu,i$ be parameters such that $0<c$, $1<\eta \leq 2$, and $0<i<\mu$. Define $r\in (i,\mu)$ and $l\in (-\mu,i)$ such that $$2(\mu-r)=c\eta (r-i)^{\eta-1}$$ and $$2(l+\mu)=c\eta (i-l)^{\eta-1}.$$ It is easy to see that $r$ and $l$ are uniquely defined within the given ranges. Then, I would like to prove that $$r^2+c(r-i)^{\eta} \leq l^2+c(i-l)^{\eta}.$$ In this problem, $i$ is the status-quo policy, $c$ and $\eta$ determine the cost of moving policies, $\mu$ is the degree of polarization of political parties, and $r$ ($l$) is the policy that a right-wing (left-wing) candidate would choose if elected. The inequality that I want to prove then simply says that the utility that the median voter will derive if $r$ is elected is at least as much as the one he/she will derive if $l$ is elected. We can assume that $r$ is the incumbent. If $\eta=2$, the result is straightforward. Together with my coauthors, I have already spent some time (in vain) trying to prove the claim. I have the impression that there has to be some easy argument, but in all proof strategies I have tried expressions turn extremely long and cumbersome immediately. By simulations, I also know that if $\eta>2$ or $i<0$ the result does not hold, so $1<\eta\leq2$ and $0<i<\mu$ should be used in the proof in some step. Any help or hint would be greatly appreciated. Thank you for your time!

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  • $\begingroup$ Hi Glorfindel. That's great! Would you mind telling me how to get the solution? Thank you very much in advance. I need the solution to prove the existence of incumbency advantage in a dynamic model of political competition. $\endgroup$ – Roger Sep 20 '17 at 8:31
  • $\begingroup$ @Roger sorry, that was a sarcastic comment and I'll remove it. The question is much better now. $\endgroup$ – Glorfindel Sep 21 '17 at 15:41
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    $\begingroup$ @Glorfindel, no worries! At first I thought it was a good idea of depriving the question of any context, but it was obviously not a good one ;-) Context helps. For one thing, it gives a purpose. $\endgroup$ – Roger Sep 21 '17 at 15:49
  • $\begingroup$ I think the condition $\mu>i>0$ should be weakened to just $\mu>|i|$, as there is nothing in the problem statement that hinges on whether $i>0$. $\endgroup$ – Bjørn Kjos-Hanssen Sep 23 '17 at 8:43
  • $\begingroup$ Hi Bjorn! Thanks for your input! As you say, the problem is indeed very "symmetric": if the desired inequality holds for all $\mu>i>0$, however, the reverse inequality should then hold for all $-\mu<i<0$, I think. The interpretation within the model of election is simple: if the median voter (who is located at 0), prefers the right-wing candidate (who has a preferred policy $\mu>0$) when the status-quo policy is $i>0$, then he/she should prefer the left-wing candidate (who has a preferred policy $-\mu<0$) when the status-quo policy is $-i<0$. This follows from the eqs. determining $r$ and $l$. $\endgroup$ – Roger Sep 25 '17 at 7:52
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The following seems to work:

Change variables to simplify: $x := r-i$, $y := i-\ell$, $p := \eta - 1$, $a := \mu - i$, $b := \mu + i$. Your two equations defining $r$ and $\ell$ can then be solved for $a$ and $b$ and the result used to express the desired inequality in terms of $x$, $y$, $c$, and $p$ (eliminating $a$ and $b$). Then the constraints reduce to $0 < c$ and $0 < p \leq 1$ and $0<x<y$, and the inequality to be proved becomes $$ f(x,y) := \left[\left(\frac{x+y}{2}+\frac{c}{4}(p+1)(x^p-y^p)\right)^2+cy^{p+1}\right]-\left[\left(\frac{x+y}{2}+\frac{c}{4}(p+1)(y^p-x^p)\right)^2+cx^{p+1}\right] \geq 0. $$ You've already handled the $p=1$ case, so assume $p<1$. Clearly, $f(x,x)=0$. By differentiating with respect to the second slot of $f$, you can verify that the partial derivative $f_2(x,x) \geq 0$ and that the second-order partial derivative $f_{22}(x,y) \geq 0$ for every $y \geq x$. Thus, $f_2(x,y) \geq 0$ for every $y \geq x$, and consequently $f(x,y) \geq 0$ for every $y \geq x$.

(Details of the calculations available on request.)

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  • $\begingroup$ Thank you very much! I just tried on my own following your indications and it seems to be correct. The key of your proof is on your change of variables from $r$ and $l$ to $x$ and $y$. I had calculated the first and second derivative of $f$ before, but with your change of variables you realized that there is no need to consider arbitrary $x$ and $y$, but it is enough to focus on the case where $x<y$ (which as you say is given by the restrictions). That was very smart! $\endgroup$ – Roger Oct 11 '17 at 15:15

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