# Optimal auction for risk-averse seller

Consider an auction of a single unit of indivisible good. There are $n$ buyers whose values of the object is drawn independently from the uniform distribution on $[0,1]$. The buyers have interim linear utility:

$$u_i(k(\hat{\theta}), t(\hat{\theta}));\theta_i) = \theta_i k_i(\hat{\theta}) + t_i(\hat{\theta})),$$

where $\hat{\theta}$ is the profile of bids, $k \in \mathbb{R}^n$ is the allocation ($k_i(\hat{\theta})$ is $1$ is player $i$ gets object and $0$ otherwise), and $t \in \mathbb{R}^n$ is the transfer/payment profile.

Assumption/Requirement We only consider auction mechanisms where in the resulting Bayesian Nash equilibrium, the object goes to the bidder with highest value, i.e.

$$k_i(\theta) = 1 \Leftrightarrow \theta_i = \max_j \theta_j.$$

Quoting the revelation principle, I have replaced $\hat{\theta}$ by the true type profile $\theta$ above.

Buyer $i$'s interim expected utility as a function of his type $\theta_i$ is

$$U_i(\theta_i) = \theta_i \bar{k}_i (\theta_i) + \bar{t}_i (\theta_i),$$

where $\bar{k}_i(\theta_i)$ is the probability he gets the good and $\bar{t}_i(\theta_i)$ is expected transfer.

From the seller's perspective, the expected payment from a buyer of type $\theta_i$ is

$$- \bar{t}(\theta_i) = - [ \int_0 ^{\theta_i} \bar{k}_i (\theta_i ') d \theta_i ' + U_i(0) - \bar{k}_i(\theta_i) \theta_i].$$

The requirement above means $\bar{k}_i(\theta_i) = \theta_i^n$. So when the seller is risk-neutral, the maximum expected revenue for the seller is achieved by any mechanism that makes $U_i(0) = 0$, i.e. a buyer of type $0$ has interim expected utility $0$.

Question: What if the seller is risk-averse, for example with utility function $w(t) = \sqrt{t}$? What characterizes the seller's optimal auction in this case?

According to my calculations, the first price auction gives the seller expected utility $$\sqrt{n(n-1)} \frac{1}{n + \frac{1}{2}},$$

while the second price auctions gives

$$n(n-1) \frac{1}{n + \frac{1}{2}} \frac{1}{n - \frac{1}{2}}.$$

So revenue equivalence no longer holds.

• Cross-posted from MO. – Michael Jul 13 '13 at 4:20
• What do you mean "Cross-posted from MO"? It's already on MO. – Joel Reyes Noche Jul 14 '13 at 1:24

Because all auction rules with $U_i(0)$ yield the same expected revenue, the optimal auction will be one that returns that revenue to the seller with certainty.

Another way to say this is that each buyer shares the seller's uncertainty about all the other buyers, so each risk-neutral buyer should fully insure the risk-averse seller against that shared uncertainty.

In the case of two bidders, you can accomplish this by having bidder 1 pay

$${1\over 2} \left({\hat{\theta_1}^2-\hat{\theta_2}^2 }\right) + {1\over 6}$$

and symmetrically, of course, for bidder 2.

For the general case, Eso and Futo give a large class of auction rules providing optimal insurance in a paper in Economics Letters (1999).

To expand on my response to your first comment: Suppose we adopt the above bidding rule. Then I make the following claims:

A) In Nash equilibrium, both bidders bid their true valuations.

B) Each player has $U(0)=0$, so the seller's revenue is the same as for any other such auction.

C) The seller's revenue is certain.

To check A), suppose my true valuation is $x$ and I bid $B$. Suppose also that the other player is known to be bidding his true valuation $y$. Then my expected gain is

$$Prob(B>y) x-(1/2)B^2+(1/2)E(y^2)-1/6$$

Now note that $Prob(B>y)=B$, so this becomes $Bx-(1/2)B^2+\hbox{constant}$. Therefore if the other guy bids his true valuation, I bid mine.

To check B, note that $(1/2)E(y^2)=1/6$ so my expected gain with $B=x$ is $x^2/2$, whereas we know that with any auction rule, my expected gain is $x^2/2+U(0)$. Thus $U(0)=0$.

As for C), note that the sum of the two bids is (deterministically) 1/6, the same in expectation as for any other auction rule with $U(0)=0$, but with no risk to the seller.

If I've done the arithmetic right, the solution in the case of $n$ bidders is that you (as one of the bidders) pay the amount

$${k-1\over k}B^k-\sum_{i=1}^{k-1}{B_i^k\over k(k-1)}+{1\over k(k+1)}$$

where $B$ is your bid and the $B_i$ are the other people's bids. This has three key properties:

1) It induces you to bid honestly in equilibrium, because you're choosing $B$ to maximize $$Prob(B>B_1,\ldots,B_k)x-{k-1\over k}B^k+\hbox{constant}$$ and, if everyone else is bidding honestly, that first probability is just $B^{k-1}$. Therefore you optimize by setting $B=k$.

2) The expected value of your payment is the same as in an ordinary second bid auction.

3) The sum of your and everyone else's payments is deterministically equal to a constant (namely the same expected revenue that the seller would have gotten from a second-price --- or first-price or third-price, etc. --- auction).

• Thank you. How does one get that formula? So there are bids $(\hat{\theta}_1, \hat{\theta}_2)$ such that the bidder not getting the object also pays? This clearly violates the bidder's participation constraint, no? Or am I missing something? – Michael Jul 13 '13 at 22:17
• Yes, both bidders pay, but we still have $U(0)=0$ (in expectation), so bidders are just as willing to participate in this auction as in any other with $U(0)=0$. I am about to add a few details to the answer to flesh this out. – Steven Landsburg Jul 13 '13 at 22:47
• PS: I've added details to the answer as promised, but it occurs to me that maybe you're missing a much simpler point, namely that both bidders "pay", but these payments can be either positive or negative, and in expectation they are equal to what they'd pay in an ordinary second price auction. – Steven Landsburg Jul 13 '13 at 22:59
• Thanks. How does one come up with something like that? It looks like the seller charges an entry-fee $\frac{1}{n} \times$ expected revenue and devises a bidding scheme where the buyers pay each other: a buyer of type $\theta_i$ pays what he would pay in the second price auction but also receives payments that offsets his entry fee in expectation. Is there a systematic way to go about this? – Michael Jul 15 '13 at 11:08