<p>I have never done such purification exercises, but I would approach it like that. As you state,
<span class="math-container">$p=Pr(\epsilon_1&gt;\frac{1-2q}{x})=1-Pr(\epsilon_1&lt;\frac{1-2q}{x})=1-F (\frac{1-2q}{x}),$</span>
<span class="math-container">$q=Pr(\epsilon_2&gt;\frac{2p-1}{x})=1-Pr(\epsilon_2&lt;\frac{2p-1}{x})=1-G (\frac{2p-1}{x}),$</span></p>
<p>where <span class="math-container">$F$</span> and <span class="math-container">$G$</span> are the cdfs corresponding to densities <span class="math-container">$f=F'$</span> and <span class="math-container">$g=G'$</span>. 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).</p>
<p>Then, combine those two equations to
<span class="math-container">$1+x F^{-1}(1-p) - 2 G (\frac{2p-1}{x})=0$</span> (and another similar formulation for q). The LHS of this formula decreasing in <span class="math-container">$p$</span>. Note that for <span class="math-container">$p=0$</span>, we have <span class="math-container">$1+xF^{-1}(1)-2G(\frac{-1}{x})=1+x-0&gt;0$</span> on the LHS such that the equation never holds, <span class="math-container">$p$</span> must be larger than zero. For <span class="math-container">$p=1$</span>, have on the LHS <span class="math-container">$1+xF^{-1}(0)-2G(1/x)=1-2G(1/x)$</span> which can be more or less than 0. That is, for some <span class="math-container">$x$</span> there is a pure strategy equilibrium with <span class="math-container">$p=1$</span> and for others there is a mixed strategy equilibrium with <span class="math-container">$p&lt;1$</span>.</p>
<p>Now consider the same equation that solves for the mixed strategy and plug in <span class="math-container">$p=\frac{1-x}{2}$</span> such that the LHS above is <span class="math-container">$1+xF^{-1}(\frac{1+x}{2})- 2 G(-1)=1+x F^{-1}(\frac{1+x}{2})- 0&gt;0$</span> for small <span class="math-container">$x$</span>, meaning we have to increase <span class="math-container">$p$</span> to fit the equilibrium equation LHS=0.</p>
<p>Next consider the same equation that solves for the mixed strategy and plug in <span class="math-container">$p=\frac{1+x}{2}$</span> such that the LHS above is <span class="math-container">$1+xF^{-1}(\frac{1-x}{2})- 2 G(1)=x F^{-1}(\frac{1+x}{2})-1&lt;0$</span> for small <span class="math-container">$x$</span>, meaning we have to decrease <span class="math-container">$p$</span> to fit the equilibrium equation LHS=0.</p>
<p>Thus, for small <span class="math-container">$x$</span>, the equilibrium <span class="math-container">$p$</span> is such that <span class="math-container">$p\in [\frac{1-x}{2},\frac{1+x}{2}]$</span>, leaving only candidate <span class="math-container">$p\to \frac{1}{2}$</span> as <span class="math-container">$x\to0$</span>.</p>

CC BY-SA 4.0
2021-05-19T13:18:29.040Z
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 $x\to 0$.
Bayesian
CC BY-SA 4.0
2021-05-19T13:28:45.977Z
Thank you very much for your effort. I understand it perfectly.
bkthn
CC BY-SA 4.0
2021-05-19T14:30:06.287Z
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.
Bayesian
CC BY-SA 4.0
2021-05-19T12:42:42.657Z
2021-05-19T13:08:37.227Z
2021-05-19T13:08:37.227Z
Bayesian
<p>I am stuck at some point in part b, I would be very happy if you could help. My calculations are below the question.</p>
<p><a href="https://i.sstatic.net/AOtrF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AOtrF.png" alt="enter image description here" /></a></p>
<p>a)</p>
<p>P1 plays N with probability p. P2 plays N with probability q.
Mixed Nash: <span class="math-container">$((\frac{1}{2},N),((\frac{1}{2},N))$</span></p>
<p>b)</p>
<p>P1 plays N if <span class="math-container">$(1+x\epsilon_1)q+x\epsilon_1(1-q)&gt;q0+(1-q)1$</span></p>
<p><span class="math-container">$\epsilon_1&gt;\frac{1-2q}{x}$</span></p>
<p>P2 plays N if <span class="math-container">$(x\epsilon_2-1)p+x\epsilon_2(1-p)&gt;p0+(1-p)(-1)$</span></p>
<p><span class="math-container">$\epsilon_2&gt;\frac{2p-1}{x}$</span></p>
<p>The distribution is not necessarily uniform. pdf's are not specified explicitly, so I need to find a general solution.</p>
<p><span class="math-container">$p=Pr(\epsilon_1&gt;\frac{1-2q}{x})=1-Pr(\epsilon_1&lt;\frac{1-2q}{x})=1-F_{\epsilon_1} (\epsilon_1&lt;\frac{1-2q}{x})$</span>
<span class="math-container">$q=Pr(\epsilon_2&gt;\frac{2p-1}{x})=1-Pr(\epsilon_2&lt;\frac{2p-1}{x})=1-F_{\epsilon_2} (\epsilon_2&lt;\frac{2p-1}{x})$</span></p>
<p>I am stuck at this point.</p>

CC BY-SA 4.0
2021-05-18T16:25:49.757Z
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?
Giskard
CC BY-SA 4.0
2021-05-18T16:37:23.860Z
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.
bkthn
CC BY-SA 4.0
2021-05-18T16:49:33.097Z
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.
Giskard
CC BY-SA 4.0
2021-05-18T17:20:33.487Z
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.
bkthn
CC BY-SA 4.0
2021-05-18T17:32:59.013Z
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.
bkthn
<game-theory>
A question related to purification theorem
bkthn
# A question related to purification theorem

(Asked by: username_0 on 2021-05-18)

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

[![enter image description here](https://i.sstatic.net/AOtrF.png)](https://i.sstatic.net/AOtrF.png)

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.

**Tags:** game-theory

> **Comments:**
> **username_1**: 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?  
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: 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.  
> **username_0**: 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.  

---

## 1 answers

### Answer by username_2 on 2021-05-19. Score: 6 | Accepted answer

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 $x\to0$.

> **Comments:**
> **username_2**: 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 $x\to 0$.  
> **username_0**: Thank you very much for your effort. I understand it perfectly.  
> **username_2**: 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.  

---

