# Compute Nash Equilibrium in median voter game

Consider a spatial model in which two candidates A and B compete for office. The policy space ranges from -1 to 1 and each candidate can take one of three positions, -1, 0, and 1 (so that they have only three pure strategies). Voters vote for the candidate closer to their ideal points, but the location of the median voter is not known to the candidates: the median voter is located at either -1 or 1 with probability s < 1/2 and is located at 0 with probability 1 - 2s. Assume that candidate A is slightly more advantaged than B: A prevails if A and B are equidistant from the median voter but B wins if he is closer to the median voter. Winning yields a payoff of 1 and losing 0. The candidates simultaneously choose a position. Answer the following questions.

1. Draw a 3-by-3 payoff matrix and show if there are any pure-strategy Nash equilibria.

My attempt: $\begin{array}{r|ccc} A\backslash B & -1 & 0 & 1\\ \hline -1 & 1, 0 & s, 1-2s+s & 1-2s+s, s\\ 0 & 1-2s+s, s & 1, 0 & 1-2s+s, s\\ 1 & 1-2s+s, s & s, 1-2s+s & 1, 0 \end{array}$

This simplifies to $\begin{array}{r|ccc} A\backslash B & -1 & 0 & 1\\ \hline -1 & 1, 0 & s, 1-s & 1-s, s\\ 0 & 1-s, s & 1, 0 & 1-s, s\\ 1 & 1-s, s & s, 1-s & 1, 0 \end{array}$

• Computing PSNE:
• If A chooses -1 -- if s>1/2, B chooses 1 -- if s=1/2, B chooses either 1 or 0 -- if s<1/2, B chooses 0
• If A chooses 0 -- B chooses either -1 or 1
• If A chooses 1 -- if s>1/2, B chooses -1 -- if s=1/2, B chooses either -1 or 0 -- if s<1/2, B chooses 0
• If B chooses -1 -- A chooses -1
• If B chooses 0 -- A chooses 0
• If B chooses 1 -- A chooses 1

There's no PSNE.

Or can I say the following? Since the BR for either player should result in a strategy that helps them win the median voter, it can't happen that both players win the median voter. Therefore, there exists no PSNE.

1. Characterize a MSNE if it exists. Hint: let A's mixed strategy to choose -1, 0, and 1 be p, q, 1-p-q, respectively, and B's mixed strategy be x, y, and 1-x-y, respectively.

For B to be indifferent, EU(-1)=EU(0)=EU(1). Thus, sq+s(1-p-q)=(1-s)p+(1-s)(1-p-q)=sp+sq

Solution: p=s/(2-s), q=(3s-2)/(s-2)

For A to be indifferent, EU(-1)=EU(0)=EU(1). Thus, x+sy+(1-s)(1-x-y)=(1-s)x+y+(1-s)(1-x-y)=(1-s)x+sy+(1-x-y)

Solutions: s=o, y=0, or x=(s-1)/(s-2), y =s/(2-s)

I don't know how to translate these into MSNE. Can you also let me know if I have got everything right so far?

• "Winning yields a payoff of 1 and losing 0." Note that this is actually meaningless: positive affine transformations result in equivalent utility functions, so the payoffs for only two outcomes does not affect what is a NE (as long as the ordering of those results is preserved). Aug 14 '18 at 18:24

The calculations look correct.

For the PSNE, I would just say since there is no mutual BRs in pure strategy, there is no PSNE, and leave it at that.

In the MSNE, A would play $-1$ with probability $\frac{1-s}{2-s}$, $0$ with $\frac{s}{2-s}$ and $1$ with $\frac{1-s}{2-s}$; B would play $-1$ with $\frac{s}{2-s}$, $0$ with $\frac{2-3s}{2-s}$ and $1$ with $\frac{s}{2-s}$.

Corrections

s cannot be greater than 1/2.

If A chooses 0 -- if s>0, B chooses either -1 or 1 -- but if s=0, B chooses either -1 or 0 or 1. 0 is only weakly dominated.

If B chooses -1 -- if s>0, A chooses -1 -- but if s=0, A chooses either -1 or 0 or 1. 0 and 1 are only weakly dominated.

If B chooses 0 -- A chooses 0

If B chooses 1 -- if s>0, A chooses 1 -- but if s=0, A chooses either -1 or 0 or 1. -1 and 0 are only weakly dominated.

PSNE

if s=0, (0, -1) and (0, 0) and (0, 1) are pure strategy Nash equilibria. Otherwise, there are none.

Mixed Strategy

B is indifferent when:

A plays $-1$ or $1$ with probability $-\frac{s}{s-2}=\frac{s}{2-s}$ (from 0 to 1/3), and $0$ with probability $\frac{3s-2}{s-2}$ (from 1 to 1/3)

A is indifferent when:

B plays $-1$ or $1$ with probability $\frac{s-1}{s-2}$ (from 1/2 to 1/3) and $0$ with probability $-\frac{s}{s-2}$ (from 0 to 1/3)