# Second Price Auction/Analysis

## Contents

## Analysis of Second Price Auction

Let $G$ denoted the game under analysis, that is, the second price auction under discussion.

Let $F$ be the object being bid for.

As in the formal definition of the sealed-bid auction, let the players be labelled in order of their valuations of $F$:

- $v_1 > v_2 > \cdots > v_n > 0$

In this context, a move by player $i$ is the bid that $i$ places on $F$.

Let $b$ be the profile of moves made by all players.

Let $m \left({b}\right)$ be defined as the lowest $j$ such that $\displaystyle b_j = \max_{k \mathop \in \left\{ {1, 2, \ldots, n}\right\} } b_k$.

That is, $m \left({b}\right)$ is the index of the player who has the lowest index of all players who place the highest bid on $F$.

Let player $i$ win the auction by move $b_i$.

By hypothesis, player $i$'s valuation for $F$ is $v_i$.

Hence $i$'s payoff is $v_i - \max_{j \mathop \ne 1} b_j$.

For the other players the payoff is $0$.

Thus the payoff function of player $i$ is defined as follows:

- $p_i \left({b}\right) = \begin{cases} v_i − \max_{j \mathop \ne i} b_j & : i = m \left({b}\right) \\ 0 & : \text{otherwise} \end{cases}$

$b$ is a Nash equilibrium if and only if:

- $\displaystyle \max_{j \mathop \ne i} v_j \le \max_{j \mathop \ne i} b_j = b_i \le v_1$

That is:

- $(1): \quad \displaystyle \max_{j \mathop \ne i} v_j \le b_i$ (that is, player $i$'s bid was high enough to win)

- $(2): \quad \displaystyle \max_{j \mathop \ne i} b_j \le v_i$ (that is, player $i$'s valuation was high enough to avoid suffering the winner's curse).

## Proof

### Sufficient Condition

Suppose that $\displaystyle \max_{j \mathop \ne i} v_j > b_i$.

Then player $j$ such that $v_j > b_i$ can win $F$ by submitting a bid in the open interval $\left({b_i \,.\,.\, v_j}\right)$, say $v_j − \epsilon$.

Then his payoff increases from $0$ to $v_j - b_i$.

Thus if $\displaystyle \max_{j \mathop \ne i} v_j > b_i$, $b$ is not a Nash equilibrium.

Suppose $\displaystyle \max_{j \mathop \ne i} b_j > v_i$.

Then player $i$ has a negative payoff.

$i$ can then increase that payoff to $0$ by submitting a losing bid.

Thus if $\displaystyle \max_{j \mathop \ne i} b_j > v_i$, $b$ is not a Nash equilibrium.

Thus it can be seen that if any of the conditions $(1)$ to $(3)$ do not hold, then $b$ is not a Nash equilibrium.

### Necessary Condition

Let $b$ be a profile of moves that satisfies $(1)$ and $(2)$.

Thus player $i$ is the winner.

By $(2)$, his payoff is non-negative.

By submitting another bid either:

- $i$ remains the winner with the same payoff

or:

- $i$ becomes a loser, with a payoff of $0$.

The payoff of any other player $j$ is $0$.

This can increase only if $j$ bids more and becomes the winner.

Then his payoff is $v_j - b_i$.

But by $(1)$, this payoff is negative.

So $b$ is a Nash equilibrium.

$\blacksquare$

## Also see

- Definition:Winner's Curse: when the valuation is less than the bid, and the payoff is thereby negative.

## Sources

- 1994: Martin J. Osborne and Ariel Rubinstein:
*A Course in Game Theory*... (previous) ... (next): $2.3$: Examples: Exercise $18.3$