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Evolutionary Game Theory question

  1. Jan 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Quite a long intro to the question so I thought it easier to include it as an image:

    http://img96.imageshack.us/img96/7264/78941753.jpg [Broken]
    http://img686.imageshack.us/img686/7780/39557949.jpg [Broken]


    3. The attempt at a solution

    I can do Q2.3 and get the payoff matrix given when V=4 and C=6.

    For Q2.4a I get

    [tex]E_{H,x}=-x_{H}+4x_{D}+x_{B}[/tex]
    [tex]E_{D,x}=2x_{D}+x_{B}[/tex]
    [tex]E_{B,x}=-0.5x_{H}+3x_{D}+2x_{B}[/tex].

    For Q2.4b I normalize the payoff matrix to get

    [tex]\[ \left( \begin{array}{ccc}
    0 & 2 & -0.5 \\
    1 & 0 & -1 \\
    0.5 & 1 & 0 \end{array} \right)\][/tex]

    Now comes the problems.

    For an ESS we must have

    [tex]E_{H,x}=E_{D,x}=E_{B,x}[/tex] (*)

    By using the normalized matrix we can rewrite these as

    [tex]E_{H,x}=2x_{D}-0.5x_{B}[/tex]
    [tex]E_{D,x}=x_{H}-x_{B}[/tex]
    [tex]E_{B,x}=0.5x_{H}+x_{D}[/tex].

    Let x=(h,d,b) be our interior ESS, then by (*) we have

    2d - 0.5b = 0.5h + d and h - b = 0.5h + d .

    The first of these can be rearranged to give h=2d-b while the second can be rearranged to give h=2d+2b. Clearly these can only both be satisfied when b=0. But this contradicts the fact that x=(h,d,b) is an interior ESS. Hence there can be no interior ESS's.

    Now that seemed correct to me, but it doesn't tie-in with Q2.4c. This question claims that the only ESS is the pure strategy B. By considering the H-D subgame I get an ESS at (2/3,1/3,0).

    Assuming the question is written correctly, where am I going wrong?

    Thanks for any help!!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 4, 2010 #2
    Anyone? :-(
     
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