# Permutation Betting Markets: Singleton Betting with Extra Information

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5 0.6 u u 2 u 3 u u 2 u 3 0. l l 2 l 3 l l 2 l 3 G 0. Figure : definition of G place a vertex in the upper part of G I and for every position, we place a vertex in the lower part of G I. Let U G denote the set of vertices in the upper part and L G denote the set of vertices in the lower part. We denote the ith vertex of the upper part by u i and the jth vertex of the lower part by l j. Finally, for every triple (b i, x i, y i) I, we put an edge between u xi U G and l yi L G with weight b i. Note that it is possible to have multiple edges between two nodes in G I. Given a simple edge-weighted bipartite graph G(U G, L G, E), let w G be the weight of the edge between vertices u i U and l j L. Given a multigraph G(V, E), let G (V, E ) with edge weights w be a simple edge-weighted graph with the same set of vertices, i.e, V = V such that wij, the weight of the edge between vertices i, j V in G, is equal to the number of edges between vertices i and j in G (as shown in Figure ). Note that w G is equal to the number of edges between vertices i and j in G. Each bet has a corresponding edge in G I. Therefore, when the auctioneer accepts a bet, we can also say that the auctioneer accepts the corresponding edge. For example, consider the singleton betting market depicted in Figure. There are nine bets corresponding to edges of the bipartite graph. As a result, the set of bets, I contains triples: (0.6,, ),(0.,, ),(0.3,,),(0.02,,2),..., (0., 3,3),(, 3,3) In this example, if the auctioneer accepts all bets, he/she gets \$3.32 before the outcome. If candidates, 2 and 3 stand in positions, 2 and 3 respectively, the auctioneers should pay = \$6 to traders. Accepted graph. If the auctioneer accepts a subset of bets, there is a subgraph in G which is formed by the edges corresponding to the accepted bets. We call this graph the accepted graph. We say an auctioneer will win with respect to an accepted graph H, if accepting the bets corresponding to the edges of H gives a positive revenue to the auctioneer in every possible outcome. Graph theoretic preliminaries. In every graph G, let the value of the maximum weighted matching be M G. Vector α = (α, α 2,..., α V (G) ) is called a weighted vertex cover of graph G if for every edge e = (i, j) we have w G e α i +α j. The minimum weighted vertex cover is a weighted vertex cover which minimizes the sum i V (G) αi. 4.2 The existence problem In this section, we give a necessary and sufficient condition for the existence problem which can be checked in linear time. 3 G 2 Theorem 4. Given a set of bets I for the singleton betting problem, there exists a set of risk-free bets for the existence problem if and only if there exists a vertex i V (G I) and a set of edges A such that (i) edges in A are adjacent to vertex i, and each pair of edges in A has only one endpoint in common which is i; and (ii) the total weight of edges in A exceeds one, or equivalently e A wg I e >. Proof. Proof of sufficiency: Suppose that there is a vertex i with the desired properties. Without loss of generality, suppose i is in the upper part of G I. Assume that there exists a set A of edges which are adjacent to i and the sum of their weights is greater than. If the auctioneer accepts the bets corresponding to edges in A and rejects the other ones, the amount of money which is earned by him/her is greater than \$. On the other hand, the auctioneer must pay at most \$ in the worst case, because all the accepted bets have the candidate i in common and their positions (the third element in the triples of bets) are distinct, so in each outcome the candidate i stands at only one position, and we lose at most \$. Proof of necessity : Suppose there is no vertex i with the desired property of the theorem and there is a subgraph H of G I such that the auctioneer will win if he accepts the bets corresponding to edges of H. First, we find the subgraph Ĥ of G I such that the auctioneer will win if he accepts the corresponding bets of Ĥ and we have wĥ, for all i, j. This means that there exists at most one edge between every pair of vertices u i UĤ and l j LĤ in Ĥ. Finally, we prove that if a subgraph like Ĥ exists, we will reach a contradiction. In order to prove the existence of Ĥ, we need the following Lemma. Lemma 5. Let G be a weighted simple bipartite graph with integer weights. If the value of the maximum weighted matching in G is M G, then there exists a vertex i with the following property: If we decrease the weights of all edges adjacent to i by unit, the value of the maximum weighted matching in the remaining graph will be M G. Proof. The dual of the maximum weighted matching problem in a bipartite graph G is the following problem: assign values α i and β j to the vertices of G (α i to vertex u i U and β j to vertex l j L) such that for every edge e = (u i, l j), we have that α i + β j w, and we also want to minimize the objective function u i U αi + l j L βj. Based on the weak duality theorem, we know that the minimum feasible value of u i U αi + l j L βj is greater than or equal to M G in G. Consider the optimal dual solution α i and β j for u i U and l i L. Note that the weights of the edges in G are integer, therefore, there exists an optimal solution in which all values of α i and β j are integers. This is true, since the dual of the weighted matching is totally unimodular, and its integrality gap is [22]. We also know that M G, so at least one α i or one β j is greater than 0. Without loss of generality, suppose α k > 0, and because α k is an integer number, we conclude that α k. Now, we can decrease the weights of the edges adjacent to u k by unit and let G be the remaining graph. It is clear that (β j = β j for all l j L, α i = α i for all u i U, u i u k and α k = α k ) is a feasible solution for the dual problem

6 in graph G with value u i U G α i + l j L G β j = M G. Therefore, the value of every weighted matching in G is not greater than M G. On the other hand, consider the maximum weighted matching in G with value M G. It is clear that the value of this matching in G is M G. So the value of the maximum weighted matching in G is M G. Now we return to the proof of Theorem 4. Consider a graph H and assume that the auctioneer accepts the bets corresponding to the edges of H. (Note that we assume that the auctioneer will win by accepting these bets). Let the value of the maximum weighted matching in H be M H. It is clear that for some permutations, the auctioneer must pay M H to the traders. On the other hand, the auctioneer gets e H wh e amount of money from traders at first. Since the auctioneer is seeking a risk-free subset, we should have: M H < e H w H e () If H has an edge with weight greater than, there are at least two edges in H between the endpoints of that edge with weight greater than one. We repeat the following procedure iteratively, until there is no edge with weight greater than in H. - We know that there exists a vertex i in H with the desired property of Lemma 5. For every vertex j, remove one of the edges between vertices i and j in H. Let Ḣ be the remaining graph. According to Lemma 5, if we decrease the weights of edges adjacent to i in H by unit, the value of the maximum weighted matching in H will decrease by exactly unit. Therefore, we have MḢ = M H. We assume that there is no vertex with the desired property of Theorem 4. Therefore, the sum of the weights of the removed edges from H is not greater than, and we have e Ḣ wḣe e H wh e. Using Equation we conclude: MḢ = M H < e H w H e e Ḣ wḣe (2) This proves that if the auctioneer accepts the bets corresponding to the edges of Ḣ, he wins. Therefore, we can replace H by Ḣ and repeat this procedure iteratively until we reach a graph H with no edge weight greater than one. This proves that there exists a graph H such that the auctioneer wins with respect to H, and there is at most one edge between any pair of vertices in H. Now, we construct a network flow F using the graph H as follows.. Add two vertices s and t to H. Let s be the source of our network flow and t be its sink. 2. Put an edge between s and each vertex u i in the upper part of H with capacity of. 3. Put an edge between each vertex l j in the lower part of H and t with capacity of. 4. Let the capacity of each edge from vertex u i in the upper part to vertex l j in the lower part be equal to w H u i,l j. H U L t s 0 0 Figure 2: Constructing the network flow F using graph H We know that there is no vertex with the desired property of Theorem 4. Therefore, for every vertex i, the sum of the weights of the edges adjacent to i in H is not greater than. So, we have a flow with value e H wh e in the network flow F. Construct a new network flow F by rounding up the capacities of edges in F. It is clear that the value of the maximum flow in F is not less than the value of the maximum flow in F, because we do not decrease the capacities. On the other hand, we can see that the value of the maximum flow in F is equal to the value of the maximum weighted matching in H (M H ). Knowing these facts, we can conclude: M H = max flow in F max flow in F e H w H e (3) which is a contradiction (see Equation.) Verifying the necessary and sufficient condition of Theorem 4 for all vertices in graph G I can be done in running time O( I + m + n) where n and m are the number of candidates and positions respectively. As a result, Theorem 4 gives a linear-time algorithm for the existence problem. Now we are ready to solve the generalization in which traders can order more than one share of security. In the existence problem, we only need to compute the sum of weights of edges incident to a specific vertex u. Note that If we submit C copies of a bet, these are C parallel edges in G I with 2 common vertices. Therefore we should consider only one of them in our calculations. Therefore this generalization is not computationally harder. 4.3 The revenue maximization problem In this section, we propose a polynomial-time algorithm for finding a subset of bets with the maximum guaranteed revenue to the auctioneer. The algorithm is based on a linear programming (LP) formulation. We first characterize an LP whose optimal integer solution is equal to our optimal solution in the singleton betting problem. We relax the linear program to a fractional linear program, and then prove that we can change any optimal fractional solution of our LP to an integer solution with the same objective function in polynomial-time. Note that in the revenue maximization problem, we have a weighted bipartite graph G with multiple edges each of which corresponds to one bet, and we want

7 0.6 u 0. l u 2 u 3 u u 2 u 3 l 2 l 3 l l 2 l 3 H 0. H Figure 3: A solution with revenue 0.2 to find a subgraph H of G that maximizes: we H M H (4) e H In other words, if the auctioneer accepts the bets of the edges of graph H, he earns e H wh e amount of money from traders, and he should pay M H in the worst case outcome. For example, consider the singleton betting market shown in Figure. If the auctioneer accepts bets (, 2, 2), (, 2,3), (, 3, 2) and (, 3,3), the accepted graph is H that is shown in Figure 3. Note that e H wh e is equal to 2.2, and the maximum weighted matching in graph H, M H is 2. So the revenue is = 0.2 in this case. Now we characterize the structure of the maximum revenue solution. Let the subgraph H G be the maximum revenue solution. Consider the minimum weighted vertex cover of H. Assume that the value α i is assigned to the vertex u i U H, and β j is assigned to the vertex l j L H in this weighted vertex cover. Lemma 6. In the optimal accepted graph H, w H is equal to min(w G, α i + β j) for every pair of vertices i and j. Proof. We know that H is a subgraph of G, thus w H is not greater than w G. On the other hand, the values α, α 2,..., α V (H) and β, β 2,..., β V (H) form a weighted vertex cover of graph H, so we have that w H α i + β j. For the sake of contradiction, assume w H is not equal to min(w G, α i + β j) for some i and j, thus w H < w G is integer, for any pair and w H < α i + β j. The value w H of vertices i and j. Therefore, by definition, α i and β j are also integers, for every i and j. Thus, w H + α i + β j. We can add one of the edges between i and j in G H to H. Note that such an edge exists because we assume that w H < w G. It implies that the value of w H increases by one. But, it is clear that the minimum weighted vertex cover of H is still a weighted vertex cover. Using the fact that the value of the maximum weighted matching is equal to the value of the minimum weighted vertex cover, we conclude that the value of e H wh e M H will increase by adding one of edges between vertices i and j to the optimal accepted graph H. This contradicts the optimality of H. Lemma 6 implies that using values of α i and β j, we can determine the value w H by setting it to min(w G, α i +β j). Now, we can write an integer linear program whose optimal solution determines our optimal solution for the singleton betting problem. In this ILP, we want to find the values of α i and β j. We also want to choose edges which should be added to the optimal subgraph H. For ease of notation, let See the definition of G in the preliminaries. w,t G be the weight of the t th edge between vertices i and j in G. Without loss of generality assume that the edges between vertices i and j are sorted in decreasing order with respect to their weight such that we have w G,t w G,t+. The following program is the ILP which helps us in computing the minimum weighted vertex cover: max ( w G,ty,t x i x j) (5) w G t= y,t = Y i U, j L Y x i + x j i U, j L x i, x j 0 i U, j L y,t {0, } i U, j L, t k ij The ILP variables x i, x j, y,t and Y are defined as follows: x i is the value of α i in the minimum weighted vertex cover of H. x j is the value of β j in the minimum weighted vertex cover of H. Y is the value of w H which is equal to number of edges between vertices i and j in H. y,t is a number which is equal to 0 or, and indicates whether the t th edge between i and j in G belongs to H. By strong duality, the value of the maximum weighted matching is equal to the value of the minimum weighted vertex cover. Since the ILP variables x and x correspond to the vectors of the minimum weighted vertex cover (α and β), the value x i + x j in the objective function of the ILP is equal to M H. As a result, the optimal solution of the integer linear program 5 characterizes the maximum revenue of the singleton betting problem. In order to solve the integer linear program 5, we relax the integer constraints y,t {0, } to linear fractional constraints 0 y,t. As a result, we get the following linear programming relaxation: max ( w G,ty,t x i x j) (6) w G t= y,t = Y i U, j L Y x i + x j i U, j L x i, x j 0 i U, j L 0 y,t i U, j L, t k ij We can solve the linear program 6. Now, the question is how to round the solution of 6 and construct an integer solution for program 5. The following Lemma 7 shows that the integrality gap of this linear program is and any solution of this LP can be rounded to an integer solution in polynomial time without changing the value of the objective function. Here, we prove this fact by showing that LP 6 is totally unimodular. Lemma 7. The integrality gap of LP 6 is and an optimal integer solution of ILP 5 can be found in polynomial-time by solving the LP relaxation 6.

8 Proof. Here, we prove this lemma by showing that the LP is totally unimodular. There are four types of variables in LP 6, i.e. y,t, Y, x i, x j. Let v be a vector that contains all types of variables. We can write the constraints of LP 6 as a matrix inequality Av b where A and b are defined as follows: A is a matrix whose number of rows and columns are equal to the number of constraints and variables in the LP respectively, and entries of A correspond to coefficients in the linear constraints of this LP. The vector b contains the right hand side values of the constraints. There are some equality constraints in LP 6. We can use some slack variables, and replace these equalities with some inequalities, so the constariants can be written in the inequality form Av b. By the way, these slack variables do not disturb the totally unimodularity property of this inequality system. It is well known that in order to prove that the integrality gap of LP 6 is, it suffices to show that A is totally unimodular [8, 24]. This fact also implies a polynomial-time algorithm for rounding any fractional solution of LP 6 to an optimal integer solution to ILP 5. For contradiction, assume A is not totally unimodular. In that case, A should have a square submatrix with determinant not equal to 0, or. Suppose K is the smallest square submatrix of A with this property. It is not hard to see that each row or column of K has at least two non-zero entries. Since, if there is a row (or column) with only one non-zero entry a (a is either or ), we can say that the absolute value of determinant of K is equal to the absolute value of determinant of K where K is the matrix that is obtained from K by removing the row and column of entry a. So the determinant of K is also not equal to 0, or. This contradicts with the assumption that K is the smallest square submatrix with determinant not equal to 0, or. According to the fact that each row of K has at least two non-zero entries, we conclude that rows corresponding to constraints like y,t or y,t 0 are not selected as rows of K. Therefore, without loss of generality, we can delete these rows from A, and assume that K is a submatrix of the remaining matrix. Since each column of K also has at least two non-zero entries, we can say that the columns corresponding to variables y,t are not selected as columns of K, because each of these columns in the remaining matrix contains exactly one non-zero entry. Similarly we can remove these columns, and assume that K is a submatrix of the remaining matrix. Now consider a row corresponding to a constraint of the form y,t = Y. Since the columns of variables of the form y,t are removed later, the row corresponding to this constraint has only one non-zero entry. Again we can remove these rows from our matrix. In the remaining matrix, the columns of variables of form Y has only one non-zero entry, therefore we remove these columns too. The remaining matrix has only rows of constraints of form Y x i + x j and columns of variables of form x i or x j. Note that each row of this matrix has exactly two non-zero variable with the same sign. Partition the columns into two sets B = {x, x 2,..., x n} and C = {x, x 2,..., x n}. One of those two non-zero entries belongs to a column in set B, and the other one belongs to a column in set C. According to [2], the determinant of any square submatrix of this matrix, including K is equal to 0, or which is again a contradiction. Therefore we conclude A is totally unimodular, and thus, the integrality gap of the corresponding integer linear program is equal to. Using ILP 5 and the result of Lemma 7, it follows that the revenue maximization problem for singleton betting is polynomial-time solvable. We conclude this section by the following theorem. Theorem 8. The revenue maximization problem for the auctioneer in singleton betting can be solved in polynomial time. Note that linear program 6 is a small polynomial-size linear program that can be solved very efficiently in practice as well. This is different from the exponential-size linear programming formulation of Chen et.al [3] for divisible variant of the subset betting. Now consider the case that traders can order more than one share of security. The only change we should make is that the constraints of the form y,t {0, } should be replaces with y,t {0,,..., C} in ILP5 where C is the number of copies of a bet that the corresponding trader orders. It is clear that the integrality gap of LP6 remains in this case. 4.4 The revenue maximization problem with extra information In this section, we study the revenue maximization problem for singleton betting when we are given a set of pieces of extra information about the outcome of the betting market. Each piece of the extra information is a forbidden pair (x,y) which means that candidate x never ends up in position y. The auctioneer may gain this type of information from various sources, or can predict such forbidden pairs with such a high confidence that he/she does not bear any risk by assuming these forbidden pairs. Let F be the set of these pairs. Given a set of forbidden pairs, a possible outcome we can illustrated by a perfect matching among candidates and positions in which no forbidden pair appears. In that case, the auctioneer can use this information in his/her favor and in order to solve the revenue maximization problem, he should take into account such extra information. In this section, we show that the LP-based revenue maximization algorithm from the previous section can be extended to solve the revenue maximization problem with extra information. Before stating the algorithm, we discuss an example. Consider the singleton betting market in Figure. Suppose we know that candidates and 2 do not stand at position in any possible outcome (see Figure 4). In other words, the edges (, ) and (2, ) are forbidden pairs. Using this extra information, we can say that the candidate 3 necessarily stands at position. Therefor, the auctioneer gains the maximum revenue by accepting all bets except bet (0.02,,2). The auctioneer gets \$3.3 before the outcome, and will pay at most \$ to the traders after the outcome. First, we show how to calculate the minimum revenue over all possible outcomes with respect to a given accepted graph H. Then we propose a linear programming method to find, an accepted graph which maximizes this minimum revenue over the possible outcomes. Note that a possible outcome can be shown by a perfect matching M among candidates and positions that does not use the forbidden pairs. The sum of weights of edges that are in both M and H (See definition of G in Subsection 4.) is the value that we should pay to the traders in this outcome. Therefore, in order to find the minimum revenue over all possible outcomes, we should find the maximum weighted perfect matching in H

9 u u 2 u 3 u u 2 u l 0.3 l 2 l 3 l l 2 l 3 H 0. Figure 4: Best solution with extra information without using forbidden edges. We can solve this by finding the integer solution of the following LP: H max ( w H x ) (7) n x = i U j= n x = j L i= x = 0 (i, j) F x 0 i U, j L We consider the dual of the above program: min ( α i + β j) (8) α i + β j w H α i + β j + δ w H () F (i, j) F In the above LP, α i is the variable corresponding to the constraint of candidate i U, β j is the variable corresponding to the constraint of position j L, and δ is the variable corresponding to the constraint of the forbidden edge (i, j). Note that all variables including δ can get arbitrarily large positive or negative values. Variables δ does not contribute in the cost function, so by setting δ = +, we can eliminate the constraints of the form α i+β j+δ w H. Therefore, we can rewrite the dual program as follows: min ( α i + β j) (9) α i + β j w H (i, j) F This program finds the values α i and β j such that for each non-forbidden edge (i, j), we have α i+β j w H. Since LP 7 is the similar to the LP of the maximum weighted matching problem, the integrality gap of LP 7 is [24]. Therefore the best fractional solution of the dual program 9 is equal to the maximum integer solution of LP 7. Now we propose an algorithm to find an accepted graph maximizing this minimum revenue over the possible outcomes. With the same argument as Lemma 6, we can use values of α i and β j of dual program 9 to determine the value w H by setting it to min(w G, α i + β j). Now, we can write an integer linear program whose optimal solution determines our optimal solution for the singleton betting problem with extra information. The ILP is very similar to ILP 5. Again we want to find the values of α i and β j and we also want to choose edges which should be added to the optimal subgraph H. The following program is the ILP for computing 2 the minimum weighted vertex cover in this setting: max ( w G,ty,t x i x j) (0) w G t= y,t = Y Y x i + x j i U, j L (i, j) F y,t {0, } i U, j L, t k ij Theorem 9. The revenue maximization problem for the auctioneer in singleton betting with extra information can be solved in polynomial time. Proof. Similar to the proof of Lemma 7, we can show that the constraint matrix of ILP 0 is totally unimodular. Therefore if we relax the integer constraints y,t {0, } to linear fractional constraints 0 y,t, we can solve this linear programming relaxation and round it to optimal integer solution of ILP 0. The optimal solution of ILP 0, corresponds to the maximum revenue of singleton betting problem with extra information. 5. THE PROBABILISTIC SETTING In this section, we study the betting problem in the probabilistic setting. We first define the problem formally. Assume that the auctioneer has a probability distribution q over the possible outcome permutations, i.e., σ is a permutation q(σ) =. Given a probability distribution q, a desired revenue x, and a desired probability 0 p, we consider the following two problems: Definition 0. In the max-expected subset betting problem, given a probability distribution q and a set I of subset bets, our goal is to find a subset S of bets I such that accepting bets in S maximizes the expected revenue of the auctioneer. Definition. In the max-probability singleton betting problem, given a probability distribution q over the possible outcomes, a desired revenue x, a desired probability 0 p, and a set I of simple bets, our goal is to accept a subset S of bets in I in order to have revenue x with probability at least p, and refuse to return a subset if there does not exist such a subset. Here, we observe that max-expected subset betting problem can be solved easily, but the max-probability singleton betting problem is #P-complete. In order to solve the max-expected subset betting problem, for any subset S I, let E(S) be the expected revenue when the auctioneer accepts the bets in set S. For every bet i I, let E(i) be the revenue when we accept only bet i. Based on the linearity of the expectation, E(S) = E( i Si) = i S E(i). Thus, in order to maximize E(S), we should add a bet i I into S iff E(i) > 0. Therefore, it suffices to compute E(i) for each bet i. Let p i be the probability that the security of bet i happens. Given the probability distribution q, we can estimate p i by sampling the probability distribution q. Thus, we can estimate E(i) = b i p i. Note that in the probabilistic model, the input can have exponential size, but we can assume that the probabilities are given in a oracle model. Next, we prove the hardness of the max-probability singleton betting problem.

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