# Nash equilibrium in second price sealed-bid auction

I'm trying to understand the basics of game theory and the topic of auctions has arisen. I understand the basic concepts of auctions but I'm struggling with second price sealed-bid auctions.

I understand that a weakly dominant strategy in a second price sealed-bid auction is to always bid the amount how much the item is worth to you. However I'm having trouble with some example questions that I've been given.

We have $$n$$ bidders, $$n \ge 2$$. There is only one object in the auction. Player $$i, i = 1, . . . , n,$$ evaluates the object by giving it a valuation $$v_i$$ , where:

$$v_1 > v_2 > v_3 > . . . > v_n > 0$$

Each player i submits a sealed bid $$b_i \ , i = 1, . . . , n$$. So, we can describe a bidding profile of all players as $$(b_1, b_2, b_3, . . . , b_n)$$.

Now what I what I want to understand is are both the listed bidding profiles below nash equilibrium?

A) bidding profile $$(v_1, 0, 0, . . . , 0)$$

B) bidding profile $$(v_2, v_1, 0, . . . , 0)$$

Surely both are a nash equilibrium because every bidder will bid there valuation for the item thus no one has any incentive to change there bid? E.G a weakly dominant strategy

I just wanted some clarification, any help on the matter would be greatly appreciated.

The two bidding profiles (A) and (B) are Nash equilibria, but the reasoning has a couple more steps than "every bidder will bid their valuation for the item." To show a bidding profile is a Nash equilibrium, you must show, for each player, that no alternate bid would yield a higher payoff (holding the bids of all other players static).

I will adopt the convention that bidding ties break in favor of the earliest-indexed player among those with the same bid. I'll also make explicit an assumption that bids must be nonnegative.

For (A), bidding profile $$(v_1, 0, \ldots, 0)$$:

• Player 1, who wins the auction, gets a payoff equal to his valuation $$v_1$$ minus the second price, which is $$0$$. Because ties break in favor of the earlier player, any bid by player 1 will lead to the exact same outcome: player 1 wins, pays $$0$$, and gets the good that he values at $$v_1$$, giving a payoff of $$v_1$$. So there is no move player 1 could make to secure a better outcome.
• Player $$j \ge 2$$ bids $$0$$ and loses to player 1, giving a payoff of $$0$$ (not because he bid $$0$$; regardless of bid, losers in an action gain nothing and pay nothing, receiving a payoff of $$0$$). This outcome is unchanged unless player $$j$$ raises his bid to $$b_j > v_1$$. In this case, player $$j$$ wins the auction and receives a good he values at $$v_j$$. Because the second price is $$v_1$$, this is what he pays, yielding a payoff of $$u_j(v_1, 0, \ldots, 0, b_j, 0, \ldots, 0) = v_j - v_1 < 0$$ because $$v_1 > v_j$$. Thus $$j$$ has no move that could improve his payoff (raise it above $$0$$).

Thus since no player can unilaterally secure a higher payoff by changing his bid, $$(v_1, 0, \ldots, 0)$$ is a Nash equilibrium.

Similarly, for (B), bidding profile $$(v_2, v_1, 0, \ldots, 0)$$:

• Player 1 bids $$v_2 < v_1$$ and thus loses the auction, getting a payoff of $$0$$. This outcome is unchanged unless he raises his bid to $$b_1 \ge v_1$$, since $$v_1$$ is the highest bid by any other player. If $$b_1 = v_1$$, then the tie breaks to player 1, who wins and pays the second price $$v_1$$, securing a payoff of $$v_1 - v_1 = 0$$; no improvement (in spite of winning the auction). If $$b_1 > v_1$$, then player 1 wins and gets a payoff of $$v_1 - b_1 < 0$$. Thus there is no move player 1 could make to secure a better outcome.
• Player 2 bids $$v_1$$ and wins, paying the second price $$v_2$$ and securing a payoff of $$v_2 - v_2 = 0$$. Any other price $$b_2 > v_2$$ (the highest bid by any other player) would still win and pay the same second price, not changing the outcome. A price $$b_2 \le v_2$$ would cause player 1 to win instead of player 2, and player 2 would then still get a payoff of $$0$$; no improvement. Note that in this last case player 1 would be very happy: he would win and pay the second price $$b_2 \le v_2 < v_1$$, securing a positive payoff. But since this is a move by player 2 that doesn't improve player 2's payoff, it is irrelevant to the discussion of Nash equilibria. There is no move player 2 could make to secure a better outcome.
• The argument that players $$j \ge 3$$ cannot secure a better outcome is exactly the same as the second bullet point under (A) above.

Thus since no player can unilaterally secure a higher payoff by changing his bid, $$(v_2, v_1, 0, \ldots, 0)$$ is also a Nash equilibrium.

The latter equilibrium has an interesting implication: in a second-price auction, in spite of the existence of the weakly dominant strategy of everyone bidding their own valuation, there still exist ("inefficient") equilibria in which the winner is not player 1.

In scenario $$A$$, only the highest bidder bids their true value and everyone else bids $$0$$. None of the other players can increase their payoff by changing their strategy alone, as no one values the item above $$v_1$$. This makes it a weakly dominant strategy.

In scenario $$B$$, the bidder valuing the item at $$v_1$$ bids $$v_2$$ and the bidder valuing the item at $$v_2$$ bids $$v_1$$, with $$v_1 > v_2$$. In this scenario, Player 2 is actually bidding $$v_1 - v_2$$ above their true value, making it beneficial for this player alone to switch to bidding lower than $$v_2$$ to obtain $$0$$ value instead of his current negative value. Thus this is not a weakly dominant strategy.

• The question is about second price auction and thus Player 2 actually has 0 utility as he/she pays v$_{2}$ for the object and not v$_{1}$ Commented Oct 29, 2019 at 14:31