Is it a weakly dominant strategy to bid your true value in a third-price auction with two identical goods and N uniform distributed bidders?

Question

I am interest in finding the optimal strategy for a single bidder (the BNE), when I have a third-price auction with N bidders and two identical goods. The bidders have iid valuations: U(0,1)

I have been told by my teacher that it is a weakly dominant strategy for the bidders to bid their valuations, but I cannot see why this would be?
(and I know from the literature that it is optimal to bid slightly above your true valuation if only one item is auctioned off, i.e. to bid $b_i=\frac{n-1}{n-2}v_i$)

The utility function for bidder i looks like this:
$ u_i(b_i, b_{-i}) = \left\{ \begin{array}{lr} v_i-\max_{(2)}\{b_{-i}\} & \text{if } b_i>\max_{(1)}\{b_{-i}\}>\max_{(2)}\{b_{-i}\} \\ v_i-\max_{(2)}\{b_{-i}\} & \text{if } \max_{(1)}\{b_{-i}\}>b_i>\max_{(2)}\{b_{-i}\} \\ 0 & \text{if } b_i<\max_{(2)}\{b_{-i}\} \end{array} \right. $
since in this auction, the highest and second highest bidder wins a good each, and both pay the third highest bid, because the bidders gain equal value from either good.

If anyone knows why it supposedly is a weakly dominant strategy to bid your true value in this type of auction, or where I might find a good explanation, I would really appreciate any help I can get!

Answer 1

Consider bidder $i$. Let $p = \max_{(2)}\{b_{-i}\}$.

By bidding $v_i$, she wins if $v_i > p$ and not if $v_i < p$ (and indifferent if $x_i = p$).

Suppose, however, she bids $z_i < v_i$.

(i) If $v_i > z_i > p$, she still wins the auction and still gets $v_i - p$.

(ii) If $p > v_i > z_i$, she still loses the auction and still gets $0$.

(iii) However, if $v_i > p > z_i$, she now loses instead of winning.

Thus, bidding less than $v_i$ can never improve $i$'s payoffs.

Can you finish the argument for the case $z_i > v_i$?