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I really couldn't solve this problem. Can someone help me with this?

Consider the sealed-bid auction for a single item, where there are only two allowable bids. There are two risk-neutral bidders who have valuations that are private info and which are drawn from i.i.d random variables that are uniformly distributed over $[0,1]$. After observing their own valuation, each of the two bidders simultaneously and individually submit a bid selected from the two-element set $\{0,1/4\}$. The high bidder wins the item and pays the amount of her bid. However, instead of paying his price to the seller, he pays to the other bidder. So the losing bidder receives the other bidder's payment. When tied, each bidder takes the role of the winning bidder with probability $1/2$ and takes the role of the losing bidder with prob. $1/2$. Solve for Bayesian Nash Equilibrium.

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We are given that $V_1, V_2$ are i.i.d $\text{Unif}(0,1)$. We can look for a Bayesian Nash equilibrium where players will play a strategy of the following type: \begin{eqnarray*} b_i(v_i) = \begin{cases} \frac{1}{4} & \text{if } v_i > c_i \\ 0 & \text{if } v_i \leq c_i \end{cases} \end{eqnarray*}

i.e. they will bid $\frac{1}{4}$ if and only if their valuations are high enough.

We just need to find these cut-offs $c_1 \in [0,1]$ and $c_2 \in [0,1]$ for the two players.

Given that player $2$'s cutoff is $c_2$ and player $1$'s valuation is $v_1$,

  • player $1$'s expected payoff from bidding $0$ equals $\frac{1}{4}\Pr(V_2 > c_2)+v_1\left(\frac{1}{2}\Pr(V_2 \leq c_2)\right) = \frac{1}{4}(1-c_2) + \frac{1}{2}v_1c_2 = \frac{1}{4}-\frac{1}{4}c_2+\frac{1}{2}v_1c_2$
  • player $1$'s expected payoff from bidding $\frac{1}{4}$ equals $\frac{1}{4}\left(\frac{1}{2}\Pr(V_2 > c_2)\right)+\left(v_1-\frac{1}{4}\right)\left(\frac{1}{2}\Pr(V_2 > c_2)+\Pr(V_2 \leq c_2)\right) = \frac{1}{8}(1-c_2) + \left(v_1-\frac{1}{4}\right)\left(\frac{1}{2}(1+c_2)\right) = -\frac{1}{4}c_2+\frac{1}{2}v_1+\frac{1}{2}v_1c_2$

So, we get that player $1$ is indifferent between the two bids at $v_1=\frac{1}{2}$. Therefore, his best response choice of cutoff is $c_1=\frac{1}{2}$. By symmetry, $c_2=\frac{1}{2}$. Therefore, Bayesian Nash Equilibrium strategy of each player $i\in\{1,2\}$ is as follows: \begin{eqnarray*} b_i(v_i) = \begin{cases} \frac{1}{4} & \text{if } v_i > \frac{1}{2} \\ 0 & \text{if } v_i \leq \frac{1}{2} \end{cases} \end{eqnarray*}

Please note that at the cutoff valuation $\frac{1}{2}$, since the players are indifferent between the available actions, so they can choose any of the two actions at that valuation and therefore, there are a total of $4$ pure-strategy Bayesian Nash equilibrium profiles.

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