Big Red Bits

More about hats and auctions

October 29th, 2009 renatoppl No comments

In my last post about hats, I told I’ll soon post another version with some more problems, which I ended up not doing and would talk a bit more about those kind of problems. I ended up not doing, but here are a few nice problems:

Those ${n}$ people are again a room, each with a hat which is either black or white (picked with probability ${\frac{1}{2}}$ at random) and they can see the color of the other people’s hats but they can’t see their own color. They write in a piece of paper either “BLACK” or “WHITE”. The whole team wins if all of them get their colors right. The whole team loses, if at least one writes the wrong color. Before entering the room and getting the hats, they can strategyze. What is a strategy that makes them win with ${\frac{1}{2}}$ probability?

If they all choose their colors at random, the probability of winning is very small: ${\frac{1}{2^n}}$ . So we should try to correlate them somehow. The solution is again related with error correcting codes. We can think of the hats as a string of bits. How to correct one bit if it is lost? The simple engineering solution is to add a parity check. We append to the string ${x_0, x_1, \hdots, x_n}$ a bit ${y = \sum_i x_i \mod 2}$ . So, if bit ${i}$ is lost, we know it is ${x_i = (y + \sum_{j \neq i} x_j) \mod 2}$ . We can use this idea to solve the puzzle above: if hats are places with ${\frac{1}{2}}$ probability, the parity check will be ${0}$ with probability ${\frac{1}{2}}$ and ${1}$ with probability ${\frac{1}{2}}$ . They can decide before hand that everyone will use ${y = 0}$ and with probability ${\frac{1}{2}}$ they are right and everyone gets his hat color right. Now, let’s extend this problem in some ways:

The same problem, but there are ${k}$ hat colors, they are choosen independently with probability ${\frac{1}{k}}$ and they win if everyone gets his color right. Find a strategy that wins with probability ${\frac{1}{k}}$ .

There are again ${k}$ hat colors, they are choosen independently with probability ${\frac{1}{k}}$ and they win if at least a fraction ${f}$ ( ${0 < f < 1}$ ) of the people guesses the right color. Find a strategy that wins with probability ${\frac{1}{fk}}$ .

Again to the problem where we just have BLACK and WHITE colors, they are chosen with probability ${\frac{1}{2}}$ and everyone needs to find the right color to win, can you prove that ${\frac{1}{2}}$ is the best one can do? And what about the two other problems above?

The first two use variations of the parity check idea in the solution. For the second case, given any strategy of the players, for each string ${x \in \{0,1\}^n}$ they have probability ${p_x}$ . Therefore the total probability of winning is ${\frac{1}{2^n}\sum_{x \in \{0,1\}^n} p_x}$ . Let ${x' = (1-x_1, x_2, \hdots, x_n)}$ , i.e., the same input but with the bit ${1}$ flipped. Notice that the answer of player ${1}$ is the same (or at least has the same probabilities) in both ${x}$ and ${x'}$ , since he can’t distinguish between ${x}$ and ${x'}$ . Therefore, ${p_{x} + p_{x'} \leq 1}$ . So,

$\displaystyle 2 \frac{1}{2^n}\sum_{x \in \{0,1\}^n} p_x = \frac{1}{2^n}\sum_{x \in \{0,1\}^n} p_x + \frac{1}{2^n}\sum_{x \in \{0,1\}^n} p_x' \leq 1$

. This way, no strategy can have more than ${\frac{1}{2}}$ probability of winning.

Another variation of it:

Suppose now we have two colors BLACK and WHITE and the hats are drawn from one distribution ${D}$ , i.e., we have a probability distribution over ${x \in \{0,1\}^n}$ and we draw the colors from that distribution. Notice that now the hats are not uncorrelated. How to win again with probability ${\frac{1}{2}}$ (to win, everyone needs the right answer).

I like a lot those hat problems. A friend of mine just pointed out to me that there is a very nice paper by Bobby Kleinberg generalizing several aspects of hat problems, for example, when players have limited visibility of other players hats.

hats1

I began being interested by this sort of problem after reading the Derandomization of Auctions paper. Hat guessing games are not just a good model for error correcting codes, but they are also a good model for truthful auctions. Consider an auction with a set ${N}$ single parameter agents, i.e., an auction where each player gives one bid ${b_i}$ indicating how much he is willing to pay to win. We have a set of constraints: ${\mathcal{X} \subseteq 2^N}$ of all feasible allocations. Based on the bids ${(b_i)_{i \in N}}$ we choose an allocation ${S \in \mathcal{X}}$ and we charge payments to the bidders. An example of a problem like this is the Digital Goods Auction, where ${\mathcal{X} = 2^N}$ .

In this blog post, I discussed the concept of truthful auction. If an auction is randomized, an universal truthful auction is an auction that is truthful even if all the random bits in the mechanism are revealed to the bidders. Consider the Digital Goods Auction. We can characterize universal truthful digital goods auction as bid-independent auctions. A bid-independent auction is given by function ${f_i(b_{-i})}$ , which associated for each ${b_{-i}}$ a random variable ${f_i(b_{-i})}$ . In that auction, we offer the service to player ${i}$ at price ${f_i(b_{-i})}$ . If ${b_i \geq f_i(b_{-i})}$ we allocate to ${i}$ and charge him ${f_i(b_{-i})}$ . Otherwise, we don’t allocate and we charge nothing.

It is not hard to see that all universal truthful mechanisms are like that: if ${x_i(b_i)}$ is the probability that player ${i}$ gets the item bidding ${b_i}$ let ${U}$ be an uniform random variable on ${[0,1]}$ and define ${f_i(b_{-i}) = x_i^{-1}(U)}$ . Notice that here ${x_i(.) = x_i(., b_{-i})}$ , but we are inverting with respect to ${b_i}$ . It is a simple exercise to prove that.

With this characterization, universal truthful auctions suddenly look very much like hat guessing games: we need to design a function that looks at everyone else’s bid but not on our own and in some sense, “guesses” what we probably have and with that calculated the price we offer. It would be great to be able to design a function that returns ${f(b_{-i}) = b_i}$ . That is unfortunately impossible. But how to approximate ${b_i}$ nicely? Some papers, like the Derandomization of Auctions and Competitiveness via Consensus use this idea.

Categories: puzzles, theory Tags: coding theory, mechanism design, profit maximization, puzzles, theory

Hats, codes and puzzles

October 3rd, 2009 renatoppl No comments

When I was a child someone told me the following problem:

A king promised to marry his daughter to the most intelligent man. Three princes came to claim her hand and he tryed the following logic experiment with them: The princes are gathered into a room and seated in a line, one behind the other, and are shown 2 black hats and 3 white hats. They are blindfolded, and 1 hat is placed on each of their heads, with the remaining hats hidden in a different room. The first one to deduce his hat color will marry the princess. If some prince claims his hat color incorrectly he dies.

The prince who is seated behind removes his blindfold, sees the two hats in front of him and says nothing. Then the prince in the middle removes his blindfold after that and he can see the hat of the prince in front of him. He also says nothing. Noticing the other princes said nothing, the prince seated in the first whole, without even removing his blindfold, gives the correct answer? The question is: what is the color he said?

This is a simple logical puzzle: we just write all the possibilities and start ruling them out given that the other princes didn’t answer and in the end we can find the color of his hat. I remember that this puzzle surprised me a lot as a kid. A found it extremly cool by then, what made me want to read books about logic problems. After that, I had a lot of fun reading the books by Raymond Smullyan. I usually would read the problems, think something like: there can’t ba a solution to that. Then go to school with the problem in mind and spend the day thinking about that. Here is a problem I liked a lot:

There is one prisoner and there are two doors: each has one guardian. One door leads to an exit and one door leads to death. The prisioner can choose one door to open. One guardian speaks only the truth and one guardian always lies. But you don’t know which door is which, which guardian is which and who guards each door. You are allowed to choose one guardian and make him one Yes/No question, and then you need to choose a door. What is the right question to ask?

hats1

But my goal is not to talk about logic puzzles, but about Hat problems. There are a lot of variations of the problems above: in all of them a person is allowed to see the other hats but not his own hat and we need to “guess” which is the color of our hat. If we think carefully, we will see that this is a very general kind of problem in computer science: (i) the whole goal of learning theory is to predict one thing from a lot of other things you observe; (ii) in error correcting code, we should guess one bit from all the others, or from some subset of the others; (iii) in universal truthful mechanisms, we need to make a price offer to one player that just depends on all other players bids. I’ll come back to this example in a later post, since it is what made me interested in those kinds of problems, but for now, let’s look at one puzzle I was told about by David Malec at EC’09:

There are black and white hats and ${3}$ people: for each of them we choose one color independently at random with probability ${\frac{1}{2}}$ . Now, they can look at each others hats but not at their own hat. Then they need to write in a piece of paper either “PASS” or one color. If all pass or if someone has a wrong color, the whole team loses (this is a team game) and if at lest one person gets the color right and no one gets wrong, the whole team wins. Create a strategy for the team to win with ${\frac{3}{4}}$ probability.

To win with ${\frac{1}{2}}$ probability is easy: one person will always write “BLACK” and the others “PASS”. A better strategy is the following: if one person sees two hats of equal color, he writes the opposite color, otherwise, he passes. It is easy to see the team wins except in the case where all hats are the same color, what happens with ${\frac{1}{4}}$ probability. We would like to extend this to a more general setting:

There are black and white hats and ${2^k - 1}$ people: for each of them we choose one color independently at random with probability ${\frac{1}{2}}$ . Now, they can look at each others hats but not at their own hat. Then they need to write in a piece of paper either “PASS” or one color. If all pass or if someone has a wrong color, the whole team loses (this is a team game) and if at lest one person gets the color right and no one gets wrong, the whole team wins. Create a strategy for the team to win with ${1-\frac{1}{2^k}}$ probability.

It is a tricky question on how to extend the above solution in that case. A detailed solution can be found here. The idea is quite ingenious, so I’ll sketch here. It envolves Error Correcting Code, in that case, the Hamming Code. Let ${F = \{0,1\}}$ with sum and product modulo ${2}$ . Let ${w_1, \hdots, 2^{2^k-1}}$ be the non-zero vector of ${F^k}$ and the following linear map:

$\displaystyle \begin{aligned} \phi: F^{2^k-1} \rightarrow F^k \\ (a_1,\hdots, a_{2^k-1}) \mapsto \sum_i a_i w_i \end{aligned}$

Let ${H}$ be the kernel of that application. Then, it is not hard to see that ${H, H+e_1, \hdots, H+e_{2^k-1}}$ is a partition of ${F^{2^k-1}}$ and also that because of that fact, for each ${x \in F^{2^k-1}}$ either ${x \in H}$ or exists a unique ${i}$ s.t. ${x + e_i \in H}$ . This gives an algorithm for just one player to guess his correct color. Let ${x \in F^{2^k-1}}$ be the color vector of the hats. Player ${i}$ sees this vector as:

$\displaystyle (x_1, \hdots, x_{i-1}, ?, x_{i+1}, \hdots, x_n)$

which can be ${(x_1, \hdots, x_{i-1}, 0, x_{i+1}, \hdots, x_n)}$ or ${(x_1, \hdots, x_{i-1}, 1, x_{i+1}, \hdots, x_n)}$ . The strategy is: if either one of those vector is in ${H}$ , write the color corresponding to the other vector. If both are out of ${H}$ , pass. The team wins iff ${x \notin H}$ , what happens with ${1 - \frac{1}{2^k}}$ probability. Is is an easy and fun exercise to prove those facts. Or you can refer to the solution I just wrote.

hats2

Now, we can complicate it a bit more: we can add other colors and other distributions. But I wanted to move to a different problem: the paper Derandomization of Auctions showed me an impressive thing: we can use coding theory to derandomize algorithms. To illustrate their ideas, they propose the following problem:

Color guessing problem: There are ${n}$ people wearing hats of ${k}$ different colors. If each person can see everyone else’s hats but not his or her own. Each person needs to guess the color of his own hat. We want a deterministic guessing algorithm that ${1/k}$ fraction of each color class is guessed correctly.

The problem is very easy if we have a source of random bits. Each person guesses some color at random. It seems very complicated to do that without random bits. Surprisingly, we will solve that using a flow computation:

Let ${c = (c_1, \hdots, c_n)}$ be an array of colors ${c_{-i}}$ the array with color ${i}$ removed. Consider the following flow network: nodes ${s}$ and ${t}$ (source and sink), nodes ${v_{c_{-i}}}$ for each ${c_{-i}}$ . There are ${n \cdot k^{n-1}}$ such nodes. Consider also nodes in the form ${u_{\gamma, c})}$ where ${\gamma}$ is a color ( ${1, \hdots, k}$ ) and ${c}$ is a color vector. There are ${k^{n+1}}$ such nodes.

hats_net

We have edges from ${s}$ to ${v_{c_{-i}}}$ for all nodes of that kind. And we have edges from ${u_{\gamma, c})}$ to ${t}$ . Now, if ${c = (\gamma, c_{-i})}$ , i.e., if ${c_{-i}}$ completed in the ${i}$ -th coordinate with ${\gamma}$ generates ${c}$ , then add an edge from ${v_{c_{-i}}}$ to ${u_{\gamma, c})}$ .

Consider the following flow: add ${1}$ unit of flow from ${s}$ to ${v_{c_{-i}}}$ and from ${v_{c_{-i}}}$ split that flow in pieces of size ${1/k}$ and send each to ${u_{\gamma, c}}$ for ${c = (\gamma, c_{-i})}$ . Now, each node ${u_{\gamma, c_{-i}}}$ receives ${\frac{n_\gamma(c)}{\gamma}}$ flow, where ${n_{\gamma}(c)}$ is the number of occurencies of ${\gamma}$ in ${c}$ . Send all that flow to ${t}$ .

We can think of that flow as the guessing procedure. When we see ${c_{-i}}$ we choose the guess independently at random and this way, each ${c}$ receives in expectation ${\frac{n_\gamma(c)}{\gamma}}$ guesses ${\gamma}$ . Notice that an integral flow in that graph represents a deterministic guessing procedure: so all we need is an integral flow so that the flow from ${u_{\gamma, c}}$ to ${t}$ is ${\lfloor n_\gamma (c) / k \rfloor }$ . The flow received is from nodes of the type: ${v_{c_{-i}}}$ and that means that bidder ${i}$ in ${c}$ , looking at the other hats will correctly choose ${c_i}$ , ${\lfloor n_\gamma (c) / k \rfloor }$ times.

Now, define the capacities this way: for all edges from ${s}$ to ${v_{c_{-i}}}$ and from ${v_{c_{-i}}}$ to ${u_{\gamma, c}}$ have capacity ${1}$ and from ${u_{\gamma, c}}$ to ${t}$ capacity ${\lfloor n_\gamma (c) / k \rfloor }$ . There is an integral flow that saturates all edges from ${u}$ to ${t}$ , because of the fractional flow showed. So, the solution gives us a deterministic decision procedure.

In the next blog post, I’ll try to show the result in the Derandomization of Auctions that relates that to competitive auctions.

Categories: puzzles Tags: coding theory, game theory, mechanism design, puzzles

Competitive Auctions

September 17th, 2009 renatoppl No comments

This week I will present the Theory Discussion Group about Competitive Auctions. It is mainly a serie of results in papers from Jason Hartline, Andrew Goldberg, Anna Karlin, Amos Fiat, … The first paper is Competitive Auctions and Digital Goods and the second is Competitive Generalized Auctions. My objective is to begin with a short introduction about Mechanism Design, the concept of truthfulness and the characterization of Truthful Mechanisms for Single Parameter Agents. Then we describe the Random Sampling Auction for Digital Goods and in the end we discuss open questions. I thought writting a blog post was a good way of organizing my ideas to the talk.

1. Mechanism Design and Truthfulness

A mechanism is an algorithm augmented with economic incentives. They are usually applied in the following context: there is an algorithmic problem and the input is distributed among several agents that have some interest in the final outcome and therefore they may try manipulate the algorithm. Today we restrict our attention to a specific class of mechanisms called single parameter agents. In that setting, there is a set ${N}$ consisting of ${n}$ agents and a service. Each agent ${i \in N}$ has a value ${v_i}$ for receiving the service and ${0}$ otherwise. We can think of ${v_i}$ as the maximum player ${i}$ is willing to pay for that service. We call an environment ${\mathcal{X} \subseteq 2^N}$ the subsets of the bidders that can be simultaneously served. For example:

Single item auction: ${\mathcal{X} =\{S; \vert S \vert \leq 1\}}$
Multi item auction: ${\mathcal{X} =\{S; \vert S \vert \leq k\}}$
Digital goods auction: ${\mathcal{X} =2^N}$
Matroid auctions: ${\mathcal{X}}$ is a matroid on ${N}$
Path auctions: ${N}$ is the set of edges in a graph and ${\mathcal{X}}$ is the set of ${s-t}$ -paths in the graph
Knapsack auctions: there is a size ${s_i}$ for each ${i \in N}$ and ${S \in \mathcal{X}}$ iff ${\sum_{i \in S} s_i \leq C}$ for a fixed ${C}$

Most mechanism design problems focus in maximizing (or approximating) the social welfare, i.e., finding ${S \in \mathcal{X}}$ maximizing ${\sum_{i \in S} v_i}$ . Our focus here will be maximizing the revenue of the auctioneer. Before we start searching for such a mechanism, we should first see which properties it is supposed to have, and maybe even first that that, define what we mean by a mechanism. In the first moment, the agents report their valuations ${v_i}$ (which can be their true valuations or lies), then the mechanism decides on an allocation ${S \subseteq N}$ (in a possibly randomized way) and charges a payment ${P_i}$ for each allocated agents. The profit of the auctioneer is ${\sum_{i \in S} P_i}$ and the utility of a bidder is:

$\displaystyle u_i = \left\{ \begin{aligned} v_i - P_i &, i \in S \\ 0 &, i \notin S \end{aligned} \right.$

The agents will report valuations so to maximize their final utility. We could either consider a general mechanism e calculate the profit/social welfare in the game induced by this mechanism or we could design an algorithm that gives incentives for the bidders to report their true valuation. The revelation principle says there is no loss of generality to consider only mechanisms of the second type. The intuition is: the mechanisms of the first type can be simulated by mechanisms of the second type. So, we restrict our attention to mechanisms of the second type, which we call truthful mechanisms. This definnition is clear for deterministic mechanisms but not so clear for randomized mechanisms. There are two such definitions:

Universal Truthful mechanisms: distribution over deterministic truthful mechanisms, i.e., some coins are tossed and based on those coins, we choose a deterministic mechanism and run it. Even if the players knew the random coins, the mechanism would still be truthful.
Truthful in Expectation mechanisms: Let ${u_i(b_i)}$ be the utility of agent ${i}$ if he bids ${b_i}$ . Since it is a randomized mechanism, then it is random variable. Truthful in expectation means that ${\mathop{\mathbb E}[u_i(v_i)] \geq \mathop{\mathbb E}[u_i(b_i)], \forall b_i}$ .

Clearly all Universal Truthful mechanisms are Truthful in Expectation but the converse is not true. Now, before we proceed, we will redefine a mechanism in a more formal way so that it will be easier to reason about:

Definition 1 A mechanism ${\mathcal{M}}$ is a function that associated for each ${v \in {\mathbb R}^N}$ a distribution over elements of ${\mathcal{X}}$ .

Theorem 2 Let ${x_i(v) = \sum_{i \in S \in \mathcal{X}} Pr_{\mathcal{M}(v)}[S]}$ be the probability that ${i}$ is allocated by the mechanism given ${v}$ is reported. The mechanism is truthful iff ${x_i(v)}$ is monotone and each allocated bidder is charged payment:

$\displaystyle P_i = v_i - \frac{1}{x_i(v_i, v_{-i})} \int_0^{v_i} x_i(w, v_{-i}) dw$

This is a classical theorem by Myerson about the characterization of truthful auctions. It is not hard to see that the auction define above is truthful. We just need to check that ${x_i(v_i, v_{-i}) (v_i - P(v_i, v_{-i})) \geq x_i(v'_i, v_{-i}) (v_i - P(v'_i, v_{-i}))}$ for all ${v'_i}$ . The opposite is trickier but is also not hard to see.

Note that this characterization implies the following characterization of deterministic truthful auctions, i.e., auctions that map each ${v \in {\mathbb R}^N}$ to a set ${S \in \mathcal{X}}$ , i.e., the probability distribution is concentrated in one set.

Theorem 3 A mechanism is a truthful deterministic auction iff there is a functions ${f_i(b_{-i})}$ such that for each we allocate to bidder ${i}$ iff ${b_i \geq f_i(b_{-i})}$ and in case it is allocated, we charge payment ${f_i(b_{-i})}$ .

It is actually easy to generate this function. Given a mechanism, ${b_i \mapsto x_i(b_i, b_{-i})}$ is a monotone and is a ${\{0,1\}}$ -function. Let ${f_i(b_{-i})}$ the point where it transitions from ${0}$ to ${1}$ . Now, we can give a similar characterization for Universal Truthful Mechanism:

Theorem 4 A mechanism is a universal truthful randomized auction if there are functions ${f_i(r,b_{-i})}$ such that for each we allocate to bidder ${i}$ iff ${b_i \geq f_i(r,b_{-i})}$ and in case it is allocated, we charge payment ${f_i(r,b_{-i})}$ , where ${r}$ are random bits.

2. Profit benchmarks

Let’s consider a Digital Goods auction, where ${\mathcal{X} = 2^N}$ . Two natural goals for profit extraction would be ${\mathcal{T}(v) = \sum_i v_i}$ and ${\mathcal{F}(v) = \max_i i v_i}$ where we can think of ${v_1 \geq v_2 \geq \hdots \geq v_n}$ , the first is the best profit you can extract charging different prices and the second is the best profit you can hope to extract by charging a fixed price. Unfortunately it is impossible to design a mechanism that even ${O(1)}$ -approximates both benchmarks on every input. The intuition is that ${v_1}$ can be much larger then the rest, so there is no way of setting ${f_1(b_{-1})}$ in a proper way. Under the assumption that the first value is not much larger than the second, we can do a good profit approximation, though. This motivates us to find an universal truthful mechanism that approximates the following profit benchmark:

$\displaystyle \mathcal{F}^{(2)}(v) = \max_{i \geq 2} i v_i$

which is the highest single-price profit we can get selling to at least ${2}$ agents. We will show a truthful mechanism that ${4}$ -approximates this benchmark.

3. Profit Extractors

Profit extractor are building blocks of many mechanisms. The goal of a profit extractor is, given a constant target profit ${C}$ , extract that profit from a set of agents if that is possible. In this first moment, let’s see ${C}$ as an exogenous constant. Consider the following mechanism called CostShare ${_C (v)}$ : find the largest ${k}$ s.t. ${k \cdot v_k \geq C}$ . Then allocate to

Lemma 5 CostShare ${_C}$ is a truthful profit-extractor that can extract profit ${C}$ whenever ${\mathcal{F}(v) = \max_i i v_i \geq C}$ .

Proof: It is clear that it can extract profit at most ${C}$ if ${\mathcal{F}(v) \geq C}$ . We just need to prove it is a truthful mechanism and this can be done by checking the characterization of truthful mechanisms. Suppose that under CostShare ${_C (v)}$ exacly ${k}$ bidders are getting the item, then let’s look at a bidder ${i}$ . If bidder ${i}$ is not getting the item, then his value is smaller than ${C/k}$ , otherwise we could incluse all bidders up to ${i}$ and sell for a price ${C/k_1}$ for some ${k_1 > k}$ . It is easy to see that bidder ${i}$ will get the item just if he changes his value ${v_i}$ to some value greater or equal than ${C/k}$ .

On the other hand, it ${i}$ is currently getting the item under ${v}$ , then increasing his value won’t make it change. It is also clear that for any value ${v_i \geq C/k}$ , he will still get the item. For ${v_i < C/k}$ he doesn’t get it. Suppose it got, then at least ${k+1}$ people get the item, because the price they sell it to ${i}$ must be less than ${v_i < C/k}$ . Thefore, increasing ${v_i}$ back to its original value, we could still sell it to ${k+1}$ players, what is a contradiction, since we assumed we were selling to ${k}$ players.

We checked monotonicity and we also need to check the payments, but it is straightforward to check they satisfy the second condition, since ${x_i(v_i) = 1}$ for ${v_i \geq C/k}$ and zero instead. $\Box$

4. Random Sampling Auctions

Now, using that profit extractor as a building block, the main idea is to estimate ${C}$ smaller than ${\mathcal{F}(v)}$ for one subset of the agents and extract that profit from them using a profit extractor. First we partition ${N}$ is two sets ${N'}$ and ${N''}$ tossing a coin for each agent to decide in which set we will place it, then we calculate ${\mathcal{F}' = \mathcal{F}(v_{N'})}$ and ${\mathcal{F}'' = \mathcal{F}(v_{N''})}$ . Now, we run CostShare ${_{\mathcal{F}'} (v'')}$ and CostShare ${_{\mathcal{F}''} (v')}$ . This is called Random Cost Sharing Auction.

Theorem 6 The Random Cost Sharing Auction is a truthful auction whose revenue ${4}$ -approximates the benchmark ${\mathcal{F}^{(2)}(v)}$ .

Proof: Let ${\mathcal{R}}$ be a random variable associated with the revenue of the Sampling Auction mechanism. It is clear that ${\mathcal{R} = \min \{ \mathcal{F}', \mathcal{F}'' \}}$ . Let’s write ${\mathcal{F}^{(2)}(v) = kp}$ meaning that we sell ${k}$ items at price ${p}$ . Let ${k = k' + k''}$ where ${k'}$ and ${k''}$ are the items among those ${k}$ items that went to ${N'}$ and ${N''}$ respectively. Then, clearly ${\mathcal{F}' \geq p k'}$ and ${\mathcal{F}'' \geq p k''}$ , what gives us:

$\displaystyle \frac{\mathcal{R}}{\mathcal{F}^{(2)}} = \frac{\min\{\mathcal{F}', \mathcal{F}''\}}{\mathcal{F}^{(2)}} \geq \frac{\min\{k'p, k''p\}}{kp} = \frac{\min\{k', k''\}}{k}$

and from there, it is a straighforward probability exercise:

$\displaystyle \frac{\mathop{\mathbb E}[{\mathcal{R}}]}{\mathcal{F}^{(2)}} = \mathop{\mathbb E}\left[{ \frac{\min\{k', k''\}}{k} }\right] = \frac{1}{k} \sum_{i=1}^{k-1} \min\{ i, k-i \} \begin{pmatrix} k \\ i \end{pmatrix} 2^{-k}$

since:

$\displaystyle \frac{k}{2} = \sum_{i = 0}^k i \begin{pmatrix} k \\ i \end{pmatrix} 2^{-k} \leq \frac{k}{4} + 2 \sum_{i = 0}^{k/2} i \begin{pmatrix} k \\ i \end{pmatrix} 2^{-k}$

and therefore:

$\displaystyle \frac{\mathop{\mathbb E}[{\mathcal{R}}]}{\mathcal{F}^{(2)}} \geq \frac{1}{4}$ $\Box$

This similar approximations can be extended to more general environments with very little change. For example, for multi-unit auctions, where ${\mathcal{X} = \{ S; \vert S \vert \leq k \}}$ we use the benchmark ${\mathcal{F}^{(2,k)} = \max_{2 \leq i \leq k} i v_i}$ and we can be ${O(1)}$ -competitive against it, by random-sampling, evaluating ${\mathcal{F}^{(1,k)} = \max_{\leq i \leq k} i v_i}$ on both sets and running a profit extractor on both. The profit extractor is a simple generalization of the previous one.