A question related to purification theorem

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

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.

• 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? May 18, 2021 at 16:25
• 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. May 18, 2021 at 16:37
• 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. May 18, 2021 at 16:49
• 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. May 18, 2021 at 17:20
• 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. May 18, 2021 at 17:32

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 $$x\to0$$.

• 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 $x\to 0$. May 19, 2021 at 13:18
• Thank you very much for your effort. I understand it perfectly. May 19, 2021 at 13:28
• 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. May 19, 2021 at 14:30