Consider a sealed-bid second price auction with two bidders. Standard textbooks claim that bidding one's true valuation $v_i$ weakly dominates bidding lower than one's valuation $b_i<v_i$. But I am not sure of this.
If bidder 2 bids $b_2$, and bidder 1 bids $b_1<v_1$, the possible cases are
- $b_2>b_1$: Then, $u_1=0$ because bidder 1 loses.
- $b_2<b_1$: Then, $u_1=v_1-b_2$ because bidder 1 wins and pays the lower bid.
- $b_2=b_1$: Then, $u_1=0.5(v_1-b_2)>0$, assuming that in the case of a tie, a winner is randomly chosen.
Now, if bidder 2 bids $b_2$, and bidder 1 bids $v_1$, the possible cases are
- $b_2>b_1$: Then, $u_1=0$ because bidder 1 loses.
- $b_2<b_1$: Then, $u_1=v_1-b_2$ because bidder 1 wins and pays the lower bid.
- $b_2=b_1$: Then, $u_1=0$, assuming that in the case of a tie, a winner is randomly chosen. Since bidder 1 bids his valuation, even if he wins, he gets a payoff of $0$.
So, bidding one's true valuation gives the same payoff as bidding lower than the valuation in two cases, and $b_1=v_1$ gives a lower payoff in the case of a tie. I don't understand how do the textbooks claim this means $b_1=v_1$ weakly dominates $b_1<v_1$.
Could someone point out the flaw in my logic?