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Week 61 COS 444 Internet Auctions: Theory and Practice Spring 2009 Ken Steiglitz

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Presentation on theme: "Week 61 COS 444 Internet Auctions: Theory and Practice Spring 2009 Ken Steiglitz"— Presentation transcript:

1 week 61 COS 444 Internet Auctions: Theory and Practice Spring 2009 Ken Steiglitz ken@cs.princeton.edu

2 week 62 Assignment 3: discussion 3.1 Techniques for average-of-losing-bids auction 3.2 Best response 3.3 How can we encourage (subsidize?) early entry?

3 week 63 Conditional expectation Intuitively clear, very useful in auction theory If x is a random variable with cdf H, and A is an event, define the conditional expectation of x given A: Not defined when prob{ A } = 0.

4 week 64 Two quick examples of conditional expectation 1)Suppose x is uniformly distributed on [0,1]. What is the expected value of x given that it is less than a constant c ≤ 1? 2)Given two independent draws x 1 and x 2, what is the expected value of x 1, given that x 1 ≤ x 2 ?

5 week 65 Interpretation of FP equil. Now take a look once more at the equilibrium bidding function for a first-price IPV auction: I claim this has the form of a conditional expectation. What is the event A?

6 week 66 Interpretation of FP equil. Claim: event A = you win! = {Y 1,(n-1) ≤ v }, where Y 1,(n-1) = highest of (n-1) independent draws To check this Let y = Y 1,(n-1), the “next highest value”. The cdf of y = Y 1,(n-1) is F(v) n-1 so the integral in the expected value of y given that you win is

7 week 67 Interpretation of FP equil. Therefore, That is, in equilibrium, bid the expected next-highest value conditioned on your winning. …Intuition?

8 week 68 Stronger revenue equivalence Let P sp (v) be the expected payment in equilibrium of a bidder in a SP auction (and similarly for FP). So SP and FP are revenue equivalent for each v !

9 week 69 Graphical interpretation Once again, by parts:

10 week 610 Bidder preference revelation Theorem: Theorem: Suppose there exists a symmetric Bayesian equilibrium in an IPV auction, and assume high bidder wins. Then this equilibrium bidding function is monotonically nondecreasing. *Thanks to Dilip Abreu for describing this elegant proof.

11 week 611 Bidder preference revelation Proof:Proof: Bid as if your value is z when it’s actually v. Let w(z) = prob. of winning as fctn. of z p(z) = exp. payment as fctn. of z For convenience, let w=w(v), w΄=w(v΄), p=p(v), p΄=p(v΄), for any v, v΄.

12 week 612 Bidder preference revelation From the definition of equilibrium, the expected surplus satisfies: v·w – p ≥ v·w΄ – p΄ v΄·w΄ – p΄ ≥ v΄·w – p for every v,v΄. Add: (v – v΄ )·(w – w΄) ≥ 0. So v > v΄ → w ≥ w΄ → b(v) ≥ b(v΄). □

13 week 613 Example of an IPV auction with no symmetric Bayesian equil.: third-price (see Krishna 02, p. 34)

14 week 614 Riley & Samuelson 1981: Optimal Auctions Optimal Auctions Elegant, landmark paper, constructs the benchmark theory for optimal IPV auctions with reserves Paradoxically, gets more powerful results more easily by generalizing

15 week 615

16 week 616

17 week 617 Riley & Samuelson’s class A rs 1.One seller, one indivisible object 2.Reserve b 0 (open reserve, starting bid) 3.n bidders, with valuations v i i=1,…,n 4.Values iid according to cdf F, which is strictly increasing, differentiable, with support [0,1] ( so f > 0 ) 5.There is a symmetric equilibrium bidding function b(v) which is strictly increasing (we know by preference revelation it must be nondecreasing) 6.Highest acceptable bid wins 7.Rules are anonymous

18 week 618 Abstracting away… Bid as if value = z, and denote expected payment of bidder by P(z). Then the expected surplus is For an equilibrium, this must be max at z=v, so differentiate and set to 0:

19 week 619 We need a boundary condition… Denote by v * the value at which it becomes profitable to bid positively, called the entry value: Now integrate d.e. from v * to our value v 1 :

20 week 620 Once more, integrate by parts… And use the boundary condition: A truly remarkable result! Why?

21 week 621 Once more, integrate by parts… And use the boundary condition: A truly remarkable result! Why? Where is the auction form? FP? SP? Third-price? All-pay? …

22 week 622 Revenue Equivalence Theorem Theorem: Theorem: In equilibrium the expected revenue in an (optimal) Riley & Samuelson auction depends only on the entry value v * and not on the form of the auction. □

23 week 623 Marginal revenue, or virtual valuation Let’s put some work into this expected revenue: And integrate by parts (of course, what else?)…

24 week 624 Marginal revenue, or virtual valuation

25 week 625 Interpretation of marginal revenue Because F(v) n is the cdf of the highest, winning value, we can interpret this as saying: The expected revenue of an (optimal) Riley & Samuelson auction is the expected marginal revenue of the winner.

26 week 626 Hazard rate Let the failure time of a device be distributed with pdf f(t) and pdf F(t). Define the “survival function” = R(t) = prob. of no failure before time t = prob. of survival till time t. Since F(t) = prob. of failure before t, R(t) = 1 – F(t).

27 week 627 Hazard rate The conditional prob. of failure in the interval (t, t+Δt ], given survival up to time t, is The “Hazard Rate” is the limit of this divided by Δt as Δt → 0 :

28 week 628 Hazard rate Thus, the marginal revenue, which is key to finding the expected revenue in a Riley- Samuelson auction, is 1/HR is the “Inverse Hazard Rate”

29 week 629 Why call it “marginal revenue”? Consider a monopolist seller who makes a take- it-or-leave-it offer to a single seller at a price p. The buyer has value distribution F, so the prob. of her accepting the offer is 1–F(p). Think of this as the buyer’s demand curve. She buys, on the average, quantity q = 1–F(p) at price p. Or, what is the same thing, the seller offers price p(q) = F -1 (1– q) to sell quantity q. after Krishna 02, BR 89

30 week 630 Why call it “marginal revenue”? The revenue function of the seller is therefore q·p(q) = q F -1 (1-q), the revenue derived from selling quantity q. The derivative of this wrt q is by definition the marginal revenue of the monopolist : F -1 (1-q) = p, so this is □


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