Second-Price Auction: Truth-Telling Is Dominant

My value is $v = 60$ and the opponent bids truthfully.

If I overbid with $b = 80$ , compute my expected payoff, win probability, and expected payment, and compare with truthful bidding ( $b = 60$ ).

U(b; v) = \int_0^b \frac{v - v_o}{100}\, dv_o = \frac{vb - b^2/2}{100}

U(80; 60) = \frac{4800 - 3200}{100} = 16 < 18 = U(60; 60)

v = 60

b = 80

— overbidding loses 2.

Show proof

Expected payoff is $U(b; 60) = (60b - b^2/2)/100$ . At $b = 80$ , $U = (4800 - 3200)/100 = 16$ ; at truthful $b = 60$ , $U = (3600 - 1800)/100 = 18$ — overbidding loses $18 - 16 = 2$ .
The source of the loss: when $v_o \in (60, 80]$ you win but pay $v_o > 60 = v$ , earning negative surplus.
The win probability rises to $80/100 = 0.8$ , but expected payment rises with it to $80^2/200 = 32$ , and the net effect is negative.
Since $dU/db = (60-b)/100 < 0$ for $b > 60$ , raising the bid past the value only shrinks the payoff.
$\frac{dU}{db} = \frac{v - b}{100}, \qquad \left.\frac{dU}{db}\right|_{b=80} = -0.2$

In the same situation, what is the expected payoff of underbidding with $b = 40$ ?

Compare with truthful bidding and prove in general that no bid can beat the truth.

U(40; 60) = \frac{2400 - 800}{100} = 16 < 18 = U(60; 60)

v = 60

b = 40

— underbidding also loses 2.

Show proof

$U(40; 60) = (2400 - 800)/100 = 16$ — lower than truthful bidding by $2$ .
Whenever $v_o \in (40, 60]$ you lose, yet winning there would have earned the positive surplus $60 - v_o > 0$ .
The loss from underbidding is the opportunity cost of forfeited wins.
In general, $U(v; v) - U(b; v) = \frac{v^2}{200} - \frac{vb - b^2/2}{100} = \frac{(v-b)^2}{200} \ge 0$ — no bid beats truthful bidding. This is the computational proof that truth-telling is a weakly dominant strategy.
$U(v; v) - U(b; v) = \frac{(v-b)^2}{200} \ge 0$

Part of these learning paths