paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0010309 | We observe that MATH . That is MATH where MATH and MATH . Observe that MATH, where MATH is the NAME matrix MATH . Its determinant is given by the formula MATH. Therefore MATH, where MATH. Therefore, MATH is an invertible matrix, and one has MATH where MATH is a matrix with entries in MATH. Therefore MATH for MATH. NAME... |
math/0010309 | The proof uses REF and an analysis of the denominators of the relevent characteristic classes as discussed in Subsection REF By assumption, there is a rational cohomology class MATH such that MATH. Therefore there is a positive integer MATH such that MATH is an integral cohomology class. If MATH is a MATH-cycle on MATH... |
math/0010309 | Let MATH be a projection in MATH. Then MATH is also a projection in MATH, and one has MATH . Each term in the above equation is non-negative. By the Theorem B and by hypothesis, it follows that MATH belongs to the finite set MATH . Since MATH is an ICC group, therefore the enveloping NAME algebra of MATH is a factor, c... |
math/0010309 | To prove this, we will define an auxilliary algebra, which is a formal deformation quantization of MATH. Since MATH, there is a MATH-valued REF-cocycle MATH on MATH such that MATH. Define MATH as being the associative algebra of formal power series over the ring MATH of power series over MATH, where the multiplication ... |
math/0010310 | Let MATH and MATH be the northern and southern hemispheres of MATH, that is, two disks in MATH such that MATH. Assume that MATH and MATH. Then MATH is MATH. We can extend any MATH to a diffeomorphism of the whole sphere by setting it to be the identity on MATH. Let MATH be the homomorphism defined in this way. This wil... |
math/0010310 | That the induced map is epic follows from NAME 's theorem CITE that that the genus two mapping class group is generated by five NAME twists, all of which happen to be MATH equivariant. See REF . This is the point where the analogous theorem fails for higher genus surfaces. That the induced map is one-to-one is more dif... |
math/0010310 | Onto is easy: Each of the five NAME twists shown in REF is sent to a half NAME twist around a curve separating two puncture points from the rest. Two examples are shown in REF . The definition of a half NAME twist is as illustrated in REF . These half NAME twists are the standard generators of the mapping class group o... |
math/0010321 | We have to prove REF for MATH. For MATH it is just MATH, for MATH it is MATH . But we have from REF : MATH and MATH for any MATH-morphism MATH. The general case is analogous. The statement of Lemma is true for any MATH-morphism MATH, not only for the NAME 's one. |
math/0010321 | We need to prove REF for MATH defined as above. The l.h.s.of REF has a form MATH where MATH is the set of admissible graphs with MATH vertices of the first type, MATH vertices of the second type, and MATH edges, and MATH. The numbers MATH are linear-quadratic expressions in the weights MATH,defined in REF, and the NAME... |
math/0010321 | It is straightforward. |
math/0010321 | REF follows from REF , because MATH is a quasi-isomorphism by REF follow from the definitions. |
math/0010321 | CASE: In MATH we have: MATH . CASE: By definition, MATH. We have: MATH . |
math/0010321 | It is straightforward. |
math/0010321 | The numbers MATH, MATH in REF do not depend on the NAME algebra MATH. Therefore, one can suppose that MATH is semisimple. We proved in REF that if MATH is semisimple, that for MATH and MATH one has: MATH where MATH and MATH are both induced by the map MATH. Next, consider the isomorphism MATH, defined as MATH . It is p... |
math/0010323 | Let MATH be a root of unity of order MATH. Let MATH be the subset of MATH of those MATH such that MATH. By CITE , the ideal in MATH of functions vanishing on MATH is MATH. Saying that MATH is regular is the same as saying that MATH. As MATH, this is the same as saying that MATH, which, using the isomorphism MATH, is eq... |
math/0010323 | CASE: Let MATH be a system of basic invariants. Let, as in the proof of the previous lemma, MATH be the subset of those MATH such that MATH does not divide MATH, and MATH the ideal generated by MATH. If MATH is monic in MATH, then MATH, so MATH is regular (by REF ). Now assume that MATH is regular, that is, that MATH. ... |
math/0010323 | If MATH is regular, then by REF we can find MATH such that the monomial MATH appears in MATH; this monomial must be a valuation monomial. Now assume that the valuation of the discriminant is MATH. Choose a system of basic invariants MATH. By weighted homogeneity, a valuation monomial in MATH can only involve those MATH... |
math/0010323 | As MATH is obtained from MATH by removing complex codimension MATH subvarieties, MATH is surjective. In terms of paths, REF is the following: let MATH be a path in MATH from MATH to a divisor MATH; then in the MATH-homotopy class of MATH relatively to MATH, there exists a path avoiding MATH. This follows from standard ... |
math/0010323 | The space MATH is trivial bundle of fiber MATH over MATH, so MATH can be lifted to a generator MATH of the monodromy around MATH in MATH (any lifting of a defining path suits). Let MATH be the set of irreducible components of MATH, let MATH the set of irreducible components of MATH which are not in MATH. The point REF ... |
math/0010323 | We describe an inductive construction procedure for such generating sets. We keep the notations used in the previous subsection. Let MATH be such that MATH is MATH-dominant. If MATH, then MATH is the head monomial of MATH, MATH and REF gives generating sets with MATH generators-of-the-monodromy. Now assume MATH. Let MA... |
math/0010323 | Note that MATH central implies MATH and that both being pure and being regular are invariant by conjugation properties. Hence the statement is independent from the choice of the starting leg MATH (only the cyclic order matters). This will allow us, later in the proof, to choose MATH at our convenience. Let us check tha... |
math/0010323 | It is enough to prove the proposition when MATH is irreductible: if MATH is reductible, it is a direct product of irreductible groups; except in a few degenerate and straightforward cases, MATH decomposes as a corresponding direct product; then MATH and MATH also decompose, and the reduction to the irreducible case fol... |
nlin/0010040 | Consider the following commutative diagram: and let us set MATH, MATH, MATH, MATH, MATH, MATH, where MATH. Then the above diagram is infinitely continued to the left, while by setting MATH and passing to the inverse limit, we obtain REF with MATH, MATH, and MATH. |
nlin/0010040 | Let MATH and MATH. Then due to the definition of the NAME - NAME bracket one has MATH . The first summand vanishes, since MATH. On the other hand, MATH and consequently is a one-form on MATH. Hence, the second summand vanishes as well. |
nlin/0010040 | The family of coverings MATH is a deformation of MATH. Since we work with deformations which leave the equation structure unchanged, then, by REF , their infinitesimal parts are elements of MATH. Let MATH be such an element. Now, by REF , the deformation we are dealing with is to be trivial as a deformation of MATH end... |
nlin/0010041 | It suffices to take the trivial bundle MATH for the bundles MATH and MATH and an arbitrary vector field MATH for the operator MATH. Flatness is a consequence of REF. |
nlin/0010045 | The NAME transform in REF establishes the correspondence between the terms in the asymptotic expansion of MATH and the terms of the small MATH expansion of MATH. This correspondence is MATH . Our plan is to modify the integrand in the definition of MATH, getting rid of the factor MATH, expand the integral in inverse po... |
nlin/0010045 | To show that the statement is true we need to prove that the function MATH is a nowhere decreasing function of MATH. On the line MATH we have MATH . Thus MATH and its derivative is, after simplification, MATH. |
nlin/0010058 | Any hydrodynamic conservation law MATH may be expanded, using REF , yielding MATH . The integrability condition for this is MATH . By differentiating this with respect to MATH the unity element, one finds that MATH satisfies the same equation, and hence is also conserved. These conserved densities are all homogeneous a... |
nlin/0010059 | The proof is obtained finding both the linear MATH and MATH satisfying REF . So doing, we introduce in REF the general form for MATH and MATH, namely MATH, MATH. The problem consits then in the evaluation of the coefficients of MATH and MATH by solving the algebraic system obtained identifying to zero the coefficients ... |
nlin/0010059 | The cofactors of MATH are MATH, MATH, MATH respectively. REF use only these three algebraic solutions. For the other statements, it is easy to check, that under suitable assumptions, MATH is an algebraic solution of the LVREF, that is, that MATH verifies REF where MATH, the cofactor of MATH takes the values MATH for st... |
nlin/0010059 | The first integral can be written as MATH with MATH. The cofactors of MATH are MATH, MATH, MATH respectively. Now it is easy to check that for REF - REF MATH and MATH are the algebraic solutions of REF , for REF MATH and MATH are the algebraic solutions of REF , for REF MATH and MATH are the algebraic solutions of REF ... |
quant-ph/0010031 | Note that we can write MATH where MATH. MATH is in MATH by definition. Since MATH contains MATH and hence MATH, it must also contain the identity matrix MATH. Hence, noting that MATH generates a one-dimensional NAME algebra isomorphic to MATH we have indeed MATH. |
quant-ph/0010031 | Using the relation MATH which follows from the definition of MATH, it can be verified by direct computation that the skew-Hermitian matrices MATH and MATH satisfy the equations MATH and a bit of algebra therefore shows that the MATH elements MATH and MATH for MATH generate the entire NAME algebra MATH. |
quant-ph/0010031 | If MATH then MATH where MATH. This leads to MATH . Since MATH by hypothesis it follows that MATH. Repeating this procedure MATH times shows that all the generators MATH and MATH for MATH are in MATH. Similarly, we can show that MATH and MATH in MATH implies that MATH contains all the generators MATH and MATH, MATH. |
quant-ph/0010031 | We evaluate MATH . Using MATH and MATH we obtain MATH, where MATH . Repeating the previous steps for MATH and MATH leads to MATH, where MATH . After MATH iterations, we have MATH where MATH . Since by REF this means MATH and noting that MATH we have MATH . The conclusion now follows from REF . |
quant-ph/0010031 | Let MATH. In this case the element MATH is in MATH and its sum and difference with MATH give rise to MATH which, along with their commutator MATH are in MATH. Starting with MATH and MATH, we have MATH . Since by REF for MATH, we have MATH. Similarly, starting with MATH and MATH, we can also prove MATH. This implies MAT... |
quant-ph/0010034 | We begin by deriving a more usable expression for MATH. MATH where we have used the fact that MATH is periodic of period MATH. Since MATH is one-to-one when restricted to its period MATH, all the kets MATH are mutually orthogonal. Hence, MATH . If MATH, then since MATH is a MATH-th root of unity, we have MATH . On the ... |
quant-ph/0010034 | We begin by noting that MATH where we have made use of the inequalities MATH . It immediately follows that MATH . As a result, we can legitimately use the inequality MATH to simplify the expression for MATH. Thus, MATH . The remaining case, MATH is left to the reader. |
quant-ph/0010034 | Since MATH we know that MATH which can be rewritten as MATH . But, since MATH, it follows that MATH . Finally, since MATH and hence MATH, the above theorem can be applied. Thus, MATH is a convergent of the continued fraction expansion of MATH. |
quant-ph/0010034 | From the above theorem, we know that MATH where MATH is a monotone decreasing sequence of positive reals converging to zero. Thus, MATH . |
quant-ph/0010040 | CASE: The probability of error MATH of finding the hidden state MATH is given by MATH where MATH where MATH is the function that rounds to the nearest integer. Hence, MATH . Thus, MATH CASE: The computational cost of the NAME transform MATH is MATH single qubit operations. The transformations MATH and MATH each carry a... |
quant-ph/0010101 | By REF , we need to compute MATH. For MATH, MATH and so MATH is the sign representation of MATH. Then MATH is one-dimensional and carries the trivial representation if MATH is even, or the sign representation if MATH is odd. Now for MATH even, we can easily compute a non-zero function MATH in MATH. For MATH is the top ... |
quant-ph/0010101 | Let MATH. Then MATH where MATH is the character of MATH. If MATH acts trivially on MATH, then MATH. If MATH acts non-trivially, we claim MATH. To show this, it suffices to show that MATH has at least two distinct eigenvalues on MATH; this is because MATH is the sum of the MATH eigenvalues of MATH. Now the set MATH of M... |
quant-ph/0010101 | We have isomorphisms MATH since the representation of MATH on MATH factors through MATH. Thus the algebra MATH identifies with MATH. Now MATH is the algebra of MATH-invariant polynomial functions on traceless MATH diagonal matrices, and so is a polynomial algebra on the functions MATH and MATH. These invariants corresp... |
quant-ph/0010111 | We say the protocol has failed if many of the quantum states measured and published by NAME do not match what NAME sent or what NAME sent. If there are only very few cases with discrepancies the protocol is considered a success. If the protocol does not fail then the protocol is concealing to NAME. NAME has sent some s... |
quant-ph/0010111 | The bit commitment used for forcing measurments need only be shortly binding. It need only be binding until the honest measurement is performed unreversibly. Hence the collusion MATH, which would be able to violate the binding condition but not the concealing condition of the above bit commitment, cannot cheat after th... |
quant-ph/0010111 | By REF we can have oblivious transfer from NAME to NAME and from NAME to NAME. As NAME cannot be in conflict with NAME or NAME (or a cheater can be identified) we can realize GCOT from NAME to any other player by the protocols of CITE with the security of the oblivious transfer channel of REF . The above protocol for G... |
quant-ph/0010111 | If one looks at CITE one can see that the above theorem is proven there for all cases but one. The case left open is that two collusions MATH, MATH, covering MATH for a player MATH, are in conflict such that no player from MATH can use the oblivious transfer to any player of MATH (and vice versa). In this case we can p... |
quant-ph/0010112 | If the players of two possible collusions MATH covering MATH cannot cooperate then it is not clear for MATH which collusion is cheating. To continue with the protocol all messages between players who are complaing about each other have to be exchanged over the broadcast channel or over secure channels via MATH. Obvious... |
quant-ph/0010112 | We show that it is impossible to realize a bit commitment for a party NAME MATH and NAME MATH. This is simple as we are almost in the two party scenario: Assume the collusion MATH can by no means measure the bit NAME committed to, then the collusion MATH can, by keeping every action at the quantum level cheat analogous... |
quant-ph/0010112 | Assume there exists a MATH with MATH for MATH. We would like to implement a bit commitment from a player from MATH to the player MATH. To prevent the players from MATH to jointly change the committed bit it must be possible for the players of MATH to measure the committed bit. Only the assumption that MATH does not col... |
quant-ph/0010112 | One can define a bit commitment protocol where NAME has to perform this measurement and send the information measured to NAME. As NAME can still not distinguish between the commitments zero and one the attack of NAME, Lo/NAME applies and there exists a unitary transform changing the bit. |
quant-ph/0010112 | If the honest but curious party remains independent of the two parties NAME and NAME bit commitment can be implemented by classical multiparty protocols. Now assume the honest but curious third party joins NAME or NAME after the commit phase is completed. The honest but curious third party will follow the protocol, but... |
quant-ph/0010112 | Let MATH denote two sets of possibly colluding players and let MATH be an adversary structure containing the complements of MATH and MATH. We show that it is impossible to implement an oblivious transfer from NAME MATH to NAME MATH which is MATH-partially robust and MATH-secure after termination. Assume such an oblivio... |
quant-ph/0010112 | We will let NAME commit to a bit string MATH. Commit via Secret Sharing REF CASE: NAME sends NAME a random string MATH. CASE: NAME shares the string MATH using a secret sharing scheme with access structure MATH. This protocol shares NAME secret MATH with the access structure MATH. If the receiver NAME is honest this pr... |
quant-ph/0010112 | The partial robustness is the same as claimed by REF so we need to prove only the improved security. We have to see first that MATH is an adversary structure. The set MATH is an adversary structure and it contains all proper subsets of MATH. Hence the set MATH is an adversary structure, too. To obtain the higher securi... |
quant-ph/0010112 | The security of a bit commitment scheme is only relevant if the sender is honest. So we can assume throughout the proof that the secret is properly shared. Because no two collusions of MATH cover MATH every set of the access structure MATH contains at least two honest players. So even if all players of a collusion MATH... |
quant-ph/0010112 | We are proving our claim using the above protocol. Whenever the receiver of the bit commitment complains about the sender then we need not be able to implement a bit commitment hence in the following we analyse only conflicts between the sender of the bit commitment and players other than the receiver. We consider two ... |
cs/0011004 | Let MATH and MATH be two possible collusions covering MATH, then oblivious transfer cannot be implemented MATH-robustly between players of MATH and players of MATH. Between any two players NAME MATH and NAME MATH the oblivious transfer channel does not work, but it is not obvious for the player MATH who is refusing to ... |
cs/0011004 | We will not state a full proof here as it can be found in CITE. But we will restate the copying procedure as it is an important subprotocol of all of the following protocols. Suppose NAME is committed to NAME to a bit MATH and wants two instances of this commitment. Then NAME ceates MATH pairs of global bit commitments... |
cs/0011004 | Each player generates a GBCX and opens the commitment to NAME. In case of a conflict the player opens his GBCX publicly. Then NAME creates a GBCX such that the parity bit is the bit she wanted to create a DBC for. Only NAME knows how to open all commitments as she created one herself. |
cs/0011004 | We will essentially restate the GCOT protocol of CITE and see that with one-out-of-two UOT instead of one-out-of-two OT any conflict results in the identification of a cheater. MATH CASE: All participants together choose one decodable MATH linear code MATH with MATH and MATH for positive constants MATH, efficiently dec... |
cs/0011004 | First we note that we do not need a broadcast channel as generating a GBC and unveiling it can be viewed as broadcasting. We now sketch the phases of a multi party protocol following CITE. To implement oblivious circuit evaluation to realize arbitrary functions we have to show the existence of an AND and a NOT function... |
cs/0011004 | Every player sends MATH times anonymously a random number to NAME. NAME sends her message MATH to every player together with MATH for all random numbers MATH . NAME received. Then every pair of players compares the message they received. Either they are all the same and the protocol was successful or two different mess... |
cs/0011004 | To send a message anonymously one has to encode the message with an error correcting code to cope with the erasures of the AOT. For an anonymous broadcast NAME sends her message MATH anonymously to a player MATH. This player broadcasts the message. If he broadcasts something wrong NAME is in conflict with this player, ... |
cs/0011004 | For an anonymous broadcast with later identification NAME authenticates her message MATH with MATH random numbers which she sends anonymously to the players. Each player receives one random number. Then she anonymously broadcasts the thus authenticated message according to REF . No other player is later able to imperso... |
cs/0011004 | We let all players create GBCX according to the protocol of CITE, but anonymously, using AOT and anonymous broadcast. Then after some time no new conflicts occur for MATH anonymous GBCX of each player (MATH is a security parameter which is polynomial in MATH). If a player NAME was unable to create a GBCX we will split ... |
cs/0011004 | NAME creates a GBCX following REF and NAME publishes positions of two substrings of the strings NAME sent to him. One substring where he knows all the bits and one substring where he knows nothing. The substrings must have approximately the same length. NAME publishes the bits of one of the substrings. Then NAME either... |
cs/0011004 | NAME sends, as a commitment, MATH bit strings of length MATH (MATH, MATH are security parameters which are polynomial in MATH) with parity MATH. Then the knowledge all other players have about MATH is negligible in MATH. Because the probability that a bit is received by at least one player is MATH and the probability t... |
cs/0011004 | NAME creates a NAME and NAME publishes positions of two substrings of the strings NAME sent over the oblivious broadcast. One substring where he knows all the bits and one substring where he knows nothing. The substrings must have approximately the same length. NAME publishes the bits of one of the substrings. Then NAM... |
cs/0011009 | We use induction on MATH; in the base case MATH, there is one (empty) maximal independent set, and for any MATH, MATH. Otherwise, we divide into cases according to the degrees of the vertices in MATH, as follows: CASE: If MATH contains a vertex MATH of degree three or more, then each maximal independent set MATH either... |
cs/0011009 | We use a recursive backtracking search, following the case analysis of REF : if there is a high-degree vertex, we try including it or not including it; if there is a degree-one vertex, we try including it or its neighbor; if there is a degree-zero vertex, we include it; and if all vertices form chains of degree-two ver... |
cs/0011009 | Clearly this is true of the initial values of MATH. Then for any MATH and any independent set MATH, we can color MATH by using a coloring of MATH and another color for each vertex in MATH, so MATH, and each step of our algorithm preserves the invariant. |
cs/0011009 | If we have any MATH-coloring of MATH, then the partition formed by separating the largest MATH color classes from the smallest color class satisfies the inequality MATH, so clearly this also is true when MATH is the partition maximizing MATH. If MATH were not maximal, due to the existence of another independent set MAT... |
cs/0011009 | Clearly, the initialization phase of the algorithm causes this to be true when MATH. Otherwise, let MATH be as in REF . By induction on MATH, MATH at the time we visit MATH. Then MATH, and MATH, so the inner loop for MATH will visit MATH and set MATH to MATH. |
cs/0011009 | MATH is itself a maximal MATH-chromatic subset of MATH, so REF shows that the algorithm correctly computes MATH. Clearly, the space is bounded by MATH. It remains to analyze the algorithm's time complexity. First, we consider the time spent initializing MATH. Since we perform a REF-coloring algorithm on each subset of ... |
cs/0011015 | See REF. |
cs/0011015 | The correctness of NAME follows from REF . Below, we analyze the running time. We initialize a maximum heap CITE in MATH time to store the edges of MATH according to their weights. Let MATH be the running time of NAME excluding this initialization. Let MATH be the set of the heaviest edges in MATH. Then REF takes MATH ... |
cs/0011015 | Suppose, for the sake of contradiction, that every minimum weight cover allows some bad node. Then we can obtain a contradiction by constructing another minimum weight cover with no bad node. Let MATH be a minimum weight cover of MATH with MATH as a bad node, that is, MATH. Recall that MATH is a minimum weight cover of... |
cs/0011015 | Consider any edge MATH of MATH. If MATH is not in MATH, then MATH. Assume that MATH is in MATH. Note that its weight in MATH is MATH. Since MATH is a cover, MATH. Thus, MATH. It follows that MATH is a cover of MATH. To show that MATH is a minimum weight one, we observe that MATH . By REF , MATH is minimum. |
cs/0011015 | The correctness of Compute-Min-Cover-MATH follows from REF . For the time complexity, the analysis is similar to that of REF . |
cs/0011015 | The running time of Recover-Max-Matching-MATH is dominated by the construction of MATH. Since MATH has at most MATH nodes and at most MATH edges, MATH can be constructed in MATH time using NAME algorithm CITE. It remains to show that MATH is a maximum weight matching of MATH. First, we argue that MATH has a perfect mat... |
cs/0011015 | REF follows from REF is proved as follows. Since MATH is a maximum weight matching of MATH, MATH. By REF , MATH if and only if MATH. We prove the latter as follows. Given a minimum weight cover MATH of MATH, we can obtain a cover MATH of MATH as follows. For any node MATH of MATH, MATH if MATH and MATH; otherwise, MATH... |
cs/0011015 | As MATH, let MATH be a shortest even-length alternating path for MATH where MATH. Based on MATH, we can construct an even-length alternating path MATH for MATH starting from MATH as follows. If MATH is not matched by MATH, MATH is simply a path of zero length. From now on, we assume that MATH is matched by MATH. As MAT... |
cs/0011015 | The two statements are proved as follows. CASE: Let MATH be the largest integer such that MATH. By REF , MATH for all MATH, and MATH otherwise. Note that if MATH, MATH must be matched by MATH. Thus, MATH. Below, we prove the following two equalities: CASE: MATH. CASE: MATH. Then, by REF , MATH and MATH. Thus, MATH and ... |
cs/0011015 | Follows from REF . |
cs/0011018 | This theorem follows from REF . |
cs/0011018 | This lemma follows from REF and basics of linear programming CITE. |
cs/0011018 | CASE: By direct verification, MATH and MATH. Then, since MATH, by REF MATH and MATH. CASE: Note that MATH by REF . Then, by REF , MATH is an optimal mixed strategy of the online player. Thus, this statement follows from REF . As pointed out in REF , MATH. By direct evaluation and REF , MATH. Then, this statement follow... |
cs/0011018 | Straightforward. |
cs/0011018 | REF follows from the fact that MATH if MATH and MATH otherwise. REF follows from REF . For REF , we let MATH be the online player's MATH-th pure strategy; and let MATH be the the adversary's MATH-th pure strategy. As in REF, the payoff matrix MATH is defined by MATH. The statement then follows from the facts that MATH ... |
cs/0011018 | We use MATH to emphasize the dimension MATH of MATH. Let MATH be the submatrix of MATH obtained by deleting row MATH and column MATH. To expand MATH along the first row of MATH, observe that MATH. Furthermore, the first column of MATH equals MATH times that of MATH, while the other columns of MATH equal the correspondi... |
cs/0011018 | Let MATH, and MATH. By REF , MATH exists. Below we prove MATH and MATH. Then, REF follows from REF and the fact that MATH, MATH, and MATH. REF follows from REF and the fact that every component of MATH and MATH is greater than MATH. REF follows from REF To prove MATH and MATH, observe that MATH can be obtained by swapp... |
cs/0011018 | Let MATH. By REF , MATH. By algebra, MATH is a decreasing function of MATH. Thus, MATH is a function of MATH whose minimum occurs at one end of the domain MATH. The lemma follows from this concavity. |
cs/0011023 | Since MATH knows MATH's bidding algorithm, MATH can perform at least as well as MATH by employing the same algorithm. Then this lemma follows from the fact that our auction is a zero-sum game. |
cs/0011023 | Let MATH be MATH's optimal bids for the MATH objects, respectively. MATH's probability of winning the MATH-th object is MATH. Since MATH's bids are within MATH, it is not to MATH's advantage to bid over MATH. Hence MATH and MATH. MATH's optimal bids maximize MATH as follows: MATH . |
cs/0011023 | Note that MATH. To show that MATH is a joint probability density function, we need only verify that the integral of MATH over MATH is MATH. Let MATH . Consider the case MATH. Then MATH . Hence if MATH, MATH which equals MATH . By symmetry, if MATH, MATH . It can be verified that MATH . Thus, MATH . |
cs/0011023 | Note that MATH, whether MATH is even or odd. CASE: This lemma is correct since if a random variable MATH is drawn from the uniform probability distribution on MATH, then MATH has the same probability distribution. CASE: The proof of REF shows that the marginal probability distribution of each MATH is MATH for MATH. It ... |
cs/0011023 | REF gives an upper bound for MATH. REF give an upper bound for MATH, which in turn gives a matching lower bound for MATH because MATH. |
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