# How to show that bids are in ex post equilibrium in a Vickrey auction?

There is one indivisible item. We have 2 bidders, 1 and 2, with valuations:

$$v_1 \in [0,2]\\ v_2\in [0,1]$$

They present their bids simultaneously, $b_1\in[0,2]$ and $b_2\in[0,2]$. The winner pays the bid of the loser, so this is a Vickrey auction. Their bidding strategies are given by:

$$b_1(v_1)=\begin{cases}v_1 & \text{ if } & v_1\leq 1\\ 1 & \text{ if } & v_1>1\\ \end{cases} \quad b_2(v_2)=\begin{cases} v_2&\text{ if } &v_2\leq 1\\ 2&\text{ if }&v_2>1 \end{cases}$$

Show that $(b_1, b_2)$ is an ex-post equilibrium. Why is the resulting allocation (in)efficient?

To show that these strategies form a (Nash) equilibrium you must show that each agent would not like to deviate from their strategy taking as given the other agent's strategy.

Agent $1$

If agent $2$ is playing $$b_2(v_2)=\begin{cases} v_2&\text{ if } &v_2\leq 1\\ 2&\text{ if }&v_2>1 \end{cases}$$ agent $1$ has to consider three cases:

$[1]$ If $v_2\le 1$ and $v_1 \lt v_2$, then $b_2=v_2$ and following the strategy agent $1$ loses the auction getting a payoff of $0$. The only relevant deviation is to bid $b_1>v_2$ which would lead to a payoff of $v_1-v_2\lt 0$.

$[2]$ If $v_2\le 1$ and $v_1 = v_2$, then $b_2=v_2$ and following the strategy there is a tie. What happens in this case is not specified, but in any case it does not matter since if agent $1$ gets the item or not the payoff will be $v_1-v_2=0$.

$[3]$ If $v_2\le 1$ and $v_1 \gt v_2$, then $b_2=v_2$ and following the strategy agent $1$ wins the auction getting a payoff of $v_1-v_2>0$. Bidding any other $b_1>v_2$ would lead to exactly the same payoff, hence does not justify a deviation. Finally, bidding $b_1\le v_2$ would lead the agent to lose the auction getting a payoff $0$.

$[*]$ Notice that $v_2>1$ is ruled out by assumption.

Agent $2$

If agent $1$ is playing $$b_1(v_1)=\begin{cases} v_1&\text{ if } &v_1\leq 1\\ 1&\text{ if }&v_1>1 \end{cases}$$ agent $2$ has to consider the same three cases:

$[1]$ If $v_2\le 1$ and $v_1 \lt v_2$, then $b_1=v_1$ and following the strategy agent $2$ wins the auction getting a payoff of $v_2-v_1>0$. Bidding any other $b_2>v_1$ would lead to exactly the same payoff, hence does not justify a deviation. Finally, bidding $b_2\le v_1$ would lead the agent to lose the auction getting a payoff $0$.

$[2]$ If $v_2\le 1$ and $v_1 = v_2$, then $b_1=v_1$ and following the strategy there is a tie. Again, what happens in this case is not specified, but in any case if agent $2$ gets the item or not the payoff will be $v_1-v_2=0$.

$[3]$ If $v_2\le 1$ and $v_1 \gt v_2$, then $b_1>b_2$ and following the strategy agent $2$ loses the auction getting a payoff of $0$. The only relevant deviation is to bid $b_2>\min\{v_1,1\}$ which would lead to a payoff of $v_2-\min\{v_1,1\}\le 0$.

Efficiency

Notice that the agent that values the item the most always wins the auction so the resulting allocation is efficient.

• The OP asks why the strategies form an ex-post equilibrium, but you've shown why they form a Nash equilibrium. Apr 23, 2018 at 5:56
• As far as I understand it they are the same thing once you take the types as given, as I have done.
– mzp
Apr 23, 2018 at 9:50