7
$\begingroup$

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!

$\endgroup$
6
  • $\begingroup$ 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, 2017 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, 2017 at 15:41
  • 1
    $\begingroup$ @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, 2017 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 $i>0$. $\endgroup$ Sep 23, 2017 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, 2017 at 7:52

1 Answer 1

5
+50
$\begingroup$

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.)

$\endgroup$
1
  • $\begingroup$ 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, 2017 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.