# First Price Auction/Analysis

## Analysis of First Price Auction

Let $G$ denoted the game under analysis, that is, the first 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 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 $v_i - b_i$.

For the other players the payoff is $0$.

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

$p_i \left({b}\right) = \begin{cases} v_i − b_i & : i = m \left({b}\right) \\ 0 & : \text{otherwise} \end{cases}$

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

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

$\displaystyle \max_{j \mathop \ne 1} v_j \le \max_{j \mathop \ne 1} b_j = b_1 \le v_1$

That is:

$(1): \quad b_1 \le v_1$ (that is, player $1$ does not suffer from the winner's curse)
$(2): \quad \displaystyle \max_{j \mathop \ne 1} v_j \le b_1$ (that is, player $1$'s bid was high enough to win)
$(3): \quad \displaystyle b_1 = \max_{j \mathop \ne i} b_j$ (that is, another player submitted the same bid as player $1$)

and so player $1$ wins the auction.

## Proof

### Sufficient Condition

Suppose that $b_1 > v_1$.

Then $1$'s payoff is negative.

It follows that his payoff increases to $0$ if he were to submit a reduced bid that is equal to $v_1$, whether this ends up with him winning the auction or not.

Thus if $b_1 > v_1$, $b$ is not a Nash equilibrium.

Suppose that $\displaystyle \max_{j \mathop \ne 1} v_j > b_1$.

Then player $j$ such that $v_j > b_1$ can win $F$ by submitting a bid in the open interval $\left({b_1 \,.\,.\, v_j}\right)$, say $v_j − \epsilon$.

Then his payoff increases from $0$ to $\epsilon$.

Thus if $\displaystyle \max_{j \mathop \ne 1} v_j > b_1$, $b$ is not a Nash equilibrium.

Suppose $\displaystyle b_1 > \max_{j \mathop \ne 1} b_j$.

Then player $1$ can increase his payoff by submitting a bid in the open interval $\displaystyle \left({\max_{j \mathop \ne 1} b_j \,.\,.\, b_1}\right)$, say $b_1 + \epsilon$.

Then his payoff increases from $v_1 − b_1$ to $v_1 − b_1 + \epsilon$.

Thus if $\displaystyle b_1 > \max_{j \mathop \ne 1} b_j$, $b$ 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 $(1)$ to $(3)$.

By the rules of a first price auction, it follows that player $1$ is the winner.

By $(1)$ his payoff is non-negative.

His payoff can increase only if he bids less.

But by $(3)$, at least one other player submitted the same bid as player $1$.

It follows by the rules of a first price auction that one of those players becomes the winner instead.

Thus player $1$'s payoff becomes $0$.

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

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

But then by $(2)$:

$\displaystyle \max_{j \mathop \ne 1} v_j < b_j$

and so his payoff becomes negative.

So $b$ is a Nash equilibrium.

$\blacksquare$