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I am stuck at some point in part b, I would be very happy if you could help. My calculations are below the question.

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a)

P1 plays N with probability p. P2 plays N with probability q. Mixed Nash: $((\frac{1}{2},N),((\frac{1}{2},N))$

b)

P1 plays N if $(1+x\epsilon_1)q+x\epsilon_1(1-q)>q0+(1-q)1$

$\epsilon_1>\frac{1-2q}{x}$

P2 plays N if $(x\epsilon_2-1)p+x\epsilon_2(1-p)>p0+(1-p)(-1)$

$\epsilon_2>\frac{2p-1}{x}$

The distribution is not necessarily uniform. pdf's are not specified explicitly, so I need to find a general solution.

$p=Pr(\epsilon_1>\frac{1-2q}{x})=1-Pr(\epsilon_1<\frac{1-2q}{x})=1-F_{\epsilon_1} (\epsilon_1<\frac{1-2q}{x})$ $q=Pr(\epsilon_2>\frac{2p-1}{x})=1-Pr(\epsilon_2<\frac{2p-1}{x})=1-F_{\epsilon_2} (\epsilon_2<\frac{2p-1}{x})$

I am stuck at this point.

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  • $\begingroup$ Why exactly are you stuck? And why exactly are you looking at $\epsilon_2>\frac{2p-1}{x}$? Are you in general familiar with mixed equilibria? $\endgroup$
    – Giskard
    May 18, 2021 at 16:25
  • $\begingroup$ Of course I am familiar with mixed equilibria. But this is not about mixed equilibria. I am trying to find the threshold strategies in this perturbed game. $\epsilon_2>\frac{2p-1}{x}\equiv\epsilon_2^*$ is the threshold strategy of player 2. $\endgroup$
    – bkthn
    May 18, 2021 at 16:37
  • $\begingroup$ Which players know $\epsilon_1$, $\epsilon_2$ before choosing their action? If player 1 does not know $\epsilon_1$ then it seems like her expectations in $(1+x\epsilon_1)q+x\epsilon_1(1-q)>q0+(1-q)1$ are not handled correctly. $\endgroup$
    – Giskard
    May 18, 2021 at 16:49
  • $\begingroup$ Player 1 knows $\epsilon_1$ and player 2 knows $\epsilon_2$. So, they know their own types but they don't know each others' types. $\endgroup$
    – bkthn
    May 18, 2021 at 17:20
  • $\begingroup$ I will state the threshold strategy for player 1 explicitly. $\sigma_1(N|\epsilon_1) =\begin{cases} 1 & \text{if $\epsilon_1\geq\epsilon_1^*$}\\ 0 & \text{if $\epsilon_1\leq\epsilon_1^*$} \end{cases} $ I am trying to find a pure strategy NE in which players play their threshold strategies. This is about "Purification Theorem" due to Harsanyi. $\endgroup$
    – bkthn
    May 18, 2021 at 17:32

1 Answer 1

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I have never done such purification exercises, but I would approach it like that. As you state, $p=Pr(\epsilon_1>\frac{1-2q}{x})=1-Pr(\epsilon_1<\frac{1-2q}{x})=1-F (\frac{1-2q}{x}),$ $q=Pr(\epsilon_2>\frac{2p-1}{x})=1-Pr(\epsilon_2<\frac{2p-1}{x})=1-G (\frac{2p-1}{x}),$

where $F$ and $G$ are the cdfs corresponding to densities $f=F'$ and $g=G'$. Because they are cdfs, they are weakly increasing. For simplicity, suppose they are strictly increasing so that their inverses clearly exist (i.e., I assume full support. If there were holes, you could still define appropriately some inverse for the flat parts).

Then, combine those two equations to $1+x F^{-1}(1-p) - 2 G (\frac{2p-1}{x})=0$ (and another similar formulation for q). The LHS of this formula decreasing in $p$. Note that for $p=0$, we have $1+xF^{-1}(1)-2G(\frac{-1}{x})=1+x-0>0$ on the LHS such that the equation never holds, $p$ must be larger than zero. For $p=1$, have on the LHS $1+xF^{-1}(0)-2G(1/x)=1-2G(1/x)$ which can be more or less than 0. That is, for some $x$ there is a pure strategy equilibrium with $p=1$ and for others there is a mixed strategy equilibrium with $p<1$.

Now consider the same equation that solves for the mixed strategy and plug in $p=\frac{1-x}{2}$ such that the LHS above is $1+xF^{-1}(\frac{1+x}{2})- 2 G(-1)=1+x F^{-1}(\frac{1+x}{2})- 0>0$ for small $x$, meaning we have to increase $p$ to fit the equilibrium equation LHS=0.

Next consider the same equation that solves for the mixed strategy and plug in $p=\frac{1+x}{2}$ such that the LHS above is $1+xF^{-1}(\frac{1-x}{2})- 2 G(1)=x F^{-1}(\frac{1+x}{2})-1<0$ for small $x$, meaning we have to decrease $p$ to fit the equilibrium equation LHS=0.

Thus, for small $x$, the equilibrium $p$ is such that $p\in [\frac{1-x}{2},\frac{1+x}{2}]$, leaving only candidate $p\to \frac{1}{2}$ as $x\to0$.

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    $\begingroup$ Perhaps it is a bit neater to replace $p=0$ and $p=(1-x)/2$ immediately with $p=1/2$. Making the same argument that it must be that $p>1/2$ for all x, but $p\to 1/2$ as $x\to 0$. $\endgroup$
    – Bayesian
    May 19, 2021 at 13:18
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    $\begingroup$ Thank you very much for your effort. I understand it perfectly. $\endgroup$
    – bkthn
    May 19, 2021 at 13:28
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    $\begingroup$ Glad it's helpful. Let me know if it is along the lines of your teacher's solution, and consider accepting it then that it doesn't remain unsolved. $\endgroup$
    – Bayesian
    May 19, 2021 at 14:30

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