Arguing uniqueness of Bayes-Nash equilibrium in an auction setting In an auction setting with interdependent values, let $\theta_i$ denote the type of player $i$ and $m_i$ that player's message (a bid, essentially). I have calculated the best response function as:
$$m_i^* = \theta_i + \gamma \sum_{j \ne i} (\theta_j-m_j)$$
Obviously, then, truth-telling by all parties ($m_k = \theta_k$ for all $k$) is a symmetric BNE, regardless of $\gamma$. I want to also argue that it is the unique BNE. If $\gamma = 1$, it seems like there would be assymetric BNE if there are only two players in which one player's positive misreport ($m_i - \theta_i$) would equal the other player's negative misreport ($\theta_j - m_j$). If $\gamma&lt;1$, it looks like the truth-telling BNE is unique, but how would I prove that?

Also, so far I have not used a common prior assumption. Do I need it here to argue the existence of the truth-telling BNE, or to argue it's uniqueness? If so, we can suppose $\theta_i$ is drawn i.i.d. uniformly from $[0,1]$.
Let $BR=(BR_1,BR_2,BR_3,...)$ denote the best response mapping. This gives a mapping given message vector $m$. (For any $BR_i$ only the messages of non $i$ players are used to derive the best response mapping.)

Let $m^* = (m_1^*,m_2^*,m_3^*,...)$ denote the symmetric equilibrium. As it is an equilibrium, for any $i$ you have $BR_i(m^*) = m_i^*$, so
$$
BR(m^*) = m^*.
$$
Suppose we have a non truth-telling equilibrium $m'$. As this is an equilibrium
$$
BR(m') = m'
$$
so
$$
\forall i\in N: m_i' = \theta_i + \gamma \sum_{j \ne i} (\theta_j-m_j').
$$
Rearranging this
$$
\forall i\in N: 0 = \theta_i - m_i' + \gamma \sum_{j \ne i} (\theta_j-m_j').
$$
Adding up all the equations yields
\begin{eqnarray*}
0 &amp; = &amp; \sum_i (\theta_i - m_i') + \gamma \sum_i \sum_{j \ne i} (\theta_j-m_j')
\\
0 &amp; = &amp; \sum_i (\theta_i - m_i') + \gamma \cdot (n-1) \cdot  \sum_i (\theta_i - m_i') \\
0 &amp; = &amp; (1 + \gamma \cdot (n-1)) \cdot \sum_i (\theta_i - m_i')  .
\end{eqnarray*}
There are two ways this can hold. Either $\sum_i (\theta_i - m_i') = 0$ or $\gamma = -\frac{1}{n-1}$. 

Case 1. Suppose $\sum_i (\theta_i - m_i') = 0$. Then from
\begin{eqnarray*}
m_i' &amp; = &amp; \theta_i + \gamma \sum_{j \ne i} (\theta_j-m_j') \\
m_i' - \theta_i &amp; = &amp; \gamma \sum_{j \ne i} (\theta_j-m_j') \\
(m_i' - \theta_i) - \gamma \cdot (m_i' - \theta_i)  &amp; = &amp;  \gamma \cdot (\theta_i - m_i') + \gamma \sum_{j \ne i} (\theta_j-m_j') \\
(1 - \gamma) \cdot (m_i' - \theta_i) &amp; = &amp; \gamma \sum_{j} (\theta_j-m_j') = 0.
\end{eqnarray*}
In the truth-telling equilibrium this is true for all $i \in N$. If $\gamma = 1$ non truth-telling equilibria are also possible for any number of players, e.g. $n-1$ players reporting their type $+1$ and one player reporting her type $-n+1$. If $\gamma &lt; 1$ and the equilibrium is non truth-telling then this is not possible.

Case 2.
Suppose $\gamma = -\frac{1}{n-1}$.   Let $x_i$ denote $\theta_i - m_i'$, so the equation system
$$
\forall i\in N: 0 = \theta_i - m_i' + \gamma \sum_{j \ne i} (\theta_j-m_j')
$$
becomes
$$
\forall i\in N: 0 = x_i + \gamma \sum_{j \ne i} x_j.
$$
For $\gamma = -\frac{1}{n-1}$ the matrix representation of this equation system is invertible so there is only a single solution, and $\forall i: x_i = 0$ is clearly a solution. This means that the only equilibrium is the truth-telling equilibrium.