Auctions

1

want to maximize surplus (sum of benefit for both seller and buyer)

first-price auction: difficult to analyze, will do so in ch 2/6

ascending-price auction strategy:

• increase price from 0 until all but one person drops out, then give that last person the current price
• assume that agents act to maximize their own benefit, then clearly there is no benefit to dropping out when price < WTP
• therefore agent $i$ will drop out at price $v_i$, so agent $1$ will gain $v_1-v_2$ and seller will gain $v_2$, thus achieving the maximum possible benefit of $v_1$

second-highest bid strategy:

• have everyone bid, offer second-highest price to highest bidder
• importantly, this encourages everyone to bid truthfully
• prove sketch: fix x to be the max of everyone else’s bids. if x < v_i, any bid >= x is optimal. and if v_i < x, any bid ⇐ x is optimal. therefore, v_i is always an optimal bid (and the only bid that is optimal in all scenarios)

maximizing consumer surplus rather than surplus is difficult, as if $v_i$ is constant, lottery is best: but if $v_1=1$ and all other $v_i=0$, second-highest bid is better. In general, no single mechanism works, whereas when maximizing surplus, second-highest bid is clearly optimal.

2

let $x$ and $p$ denote whether an agent receives a good and their payment, respectively

in order to be in equilibrium:

$x$ must be monotone as bid increases

intuitively, this makes sense: why would you bid higher only to get less chance of getting the item?

RSOP

b/c price given to each person doesn’t depend on what they bid, it’s clearly optimal for them to always tell the truth

1/4 lower bound: consider 2 bidders with $v_1>v_2$. With probability $\frac{1}{2}$ they are in different sets, and therefore seller gains revenue of $v_2$ with probability $\frac{1}{2}$, so expected revenue is $\frac{v_2}{2}$, which is $\frac{1}{4}$ of maximum revenue $2v_2$.

4.68 bound paper: $S_j$ = how many of the first $j$ bidders are in the sample $Z_j$ = fraction of first $j$ bidders in market to sample $Z$ = $\min Z_j$

• 1/15 proof considers scenarios where $Z\ge \frac{1}{3}$, and shows that this occurs with high probability

$E_{\alpha }$ ⇒ probability that the maximum ratio of $\frac{S_j}{j}$ does not exceed $\alpha$

$E_{\alpha }^T$ and $E_{\alpha }^{T\prime }$ are positively correlated for any $T$, $T\prime$