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LectureFeb15_MATH4321_12S

Course: MATH 4321, Spring 2012
School: HKUST
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of Reduction a Game in Extensive Form to Strategic Form. Pure strategy. A pure strategy is a players complete plan for playing the game. It should cover every contingency. A pure strategy for a Player is a rule that tells him exactly what move to make in each of his information sets. It should specify a particular edge leading out from each information set. Example: Player I has 3 information sets. The set of...

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of Reduction a Game in Extensive Form to Strategic Form. Pure strategy. A pure strategy is a players complete plan for playing the game. It should cover every contingency. A pure strategy for a Player is a rule that tells him exactly what move to make in each of his information sets. It should specify a particular edge leading out from each information set. Example: Player I has 3 information sets. The set of edges coming out from the information sets are{A,B}, {E,F}, {C,D}. Player II has two information sets. The set of edges coming out from the information sets are {a,b}, {c,d}. Set of Pure Strategies for I: {A,B}x{C,D}x{E.F} ={ACE, ACF, ADE, ADF, BCE, BCF, BDE, BDF} Set of Pure Strategies for II: {a,b}x{c,d}={ac,ad,bc,bd} Remark: Even for a simple game, there may be a large number of pure strategies. Example: (Two move Tic-Tac-Toe) Player I and II take turn to make a move on the Tic-Tac-Toe board. The game stops after each player has made one move. How many pure strategies are there for Player I and for Player II? Example: (Two move chess) There are 20 possible moves for Player I and 20 possible moves for Player II. How many pure strategies are there for Player II? Reduced pure strategy: Specification of choices at all information sets except those that are eliminated by the previous moves. Remark: It is not too easy to find the set of reduced pure strategies. Also we need to use full pure strategies for another important concept. Example: Set of Reduced Pure Strategies for I:{AE, AF, BC, BD} Set of Reduced Pure Strategies for II: {ac,ad,bc,bd} Now that we have set of pure strategies for each player, we need to find the payoffs to put the game in strategic form. Random payoffs. The actual outcome of the game for given pure strategies of the players depends on the chance moves selected, and is therefore a random quantity. We represent random payoffs by their average values. Example: Suppose Player I uses BCF and Player II uses ac. Then, the payoff will be (-5, 1) with probability 0.7 and (4,5) with probability 0.3. Then, the expected payoff is 0.7(-5,1) + 0.3(4,5) = (-2.3, 2.2) Remark: In representing the random payoffs by their averages, we are making a rather subtle assumption. We are saying that receiving $5 outright is equivalent to receiving $10 with probability 0.5. The proper setting for this concept is Utility Theory developed by von Neumann and Morgenstern. Game in strategic form: Set of players: {1,,n} A set of pure strategies for player i, Xi i=1,,n. Payoff function for the ith player. ui : X1 xx Xn R Payoff function tabulated in bimatrix form: ac ad ACE ACF ADE ADF BCE BCF BDE BDF bc bd Set of reduced strategies for I:{AE, AF, BC, BD} Set of reduced strategies for II:{ac, ad, bc, bd} Straegic Form: ac AE AF ad bc bd (10, 0) (0, 20) (10, 0) (0, 20) (0,30) (40, 0) (0,30) (40, 0) (2.6,3.7) (2.6,3.7) (2.3, 2.2) (0.5, 2.3) (2.6,3.7) (2.6,3.7) BC (2.3, 2.2) BD (0.5, 2.3) Example: Strategic Form: r1r2 R1R2 (3,3) r1d2 (1,4) d1r2 (0,3) d1d2 (0,3) R1D2 (2,2) (2,2) (0,3) (0,3) D1R2 (1,1) (1,1) (1,1) (1,1) D1D2 (1,1) (1,1) (1,1) (1,1) Definition. A vector of pure strategy choices (x1, x2, . . . , xn) with xi Xi for i = 1, . . . , n is said to be a pure strategic equilibrium, or PSE for short, if for all i = 1, 2, . . . , n, and for all x Xi, u i (x1, . . . , xi-1, xi, xi+1, . . . , xn) u i (x1, . . . , xi-1, x, xi+1, . . . , xn). (1) Equation (1) says that if the players other than player i use their indicated strategies, then the best player i can do is to use xi. Such a pure strategy choice of player i is called a best response to the strategy choices of the other players. The notion of strategic equilibrium may be stated: a particular selection of strategy choices of the players forms a PSE if each player is using a response best to the strategy choices of the other players. face.jpg Finding All PSEs. (2 Person) Put an asterisk after each of Player Is payoffs that is a maximum of its column. Put an asterisk after each of Player IIs payoffs that is a maximum of its row. Then any entry of the matrix at which both Is and IIs payoffs have asterisks is a PSE, and conversely. Example:Prisoner's Dilemma Confess Confess (9* , 9* ) Silent * (10, 0 ) Silent (0* , 10) ( 1, 1) Remark:Many games have no PSE's. Example: * * (2, 2 ) (3 , 3) * * (3 , 3) (4, 4 ) Games of perfect information always have at least one PSE that may be found by the method of backward induction. Method of Backward Induction: Starting from any terminal vertex and trace back to the vertex leading to it. The player at this vertex will discard those edges with lower payoff. Then, treat this vertex as a terminal vertex and repeat the process. Then, we get a path from the root to a terminal vertex Theorem: The path obtained by the method of backward induction defines a PSE. SREN KIERKEGAARD (1813-1855) Life can only be understood backwards; Life but it must be lived forwards but Example: Zermelos Theorem Zermelo Theorem (Zermelo (1912)): In chess either white can force a win, or balck can force a win, or both can force at least a draw. In fact, the PSE obtained by the method of backward induction satisfies stronger properties so that it is called a perfect pure strategy equilibrium. Definition: A subgame of a game presented in extensive form is obtained by taking a vertex in the Kuhn tree and all the edges and paths originated from this vertex. Definition: A PSE of a game in extensive form is called a Perfect Pure Strategy Equilibrium (PPSE) if it is a PSE for all subgames. Theorem: The path obtained by the method of backward induction defines a PPSE. Remark: Subgame perfect implies that when making choices, a player look forward and assumes that the choice that will subsequently be made by himself and by others will be rational. Threats which would be irrational to carry through are ruled out. It is precisely this kind of forwardlooking rationality that is most suited to economic applications. Example: Incredible Threats Example: Strategic Form Monopolist no entry fight (0,2) no fight (0,2) entry (-1,-1) (1,1) Entrant Example: Strategic Form U u (1,1) d (1,1) D (-1,-1) (2,0) Incredible Threats Incredible ? PSE allow players to make noncredible threats provided they never have to carry them out. Decisions on the equilibrium path are driven in part by what the players expect will happen off the equilibrium path. Example: Strategic Form a AC AD B b (20,7) (20,7) (5,-20) (3,6) (4,8) (5,-20) PPSE: (BD, b) Remark: PPSE is not so perfect. Example: Centipede Game I II (1,1) (0,3) I II I I (2,2) (1,4) (3,3) II (99,99) (98,101) Player I will go down in the 1st move. (100,100) Application of Backward Reasoning Application Sophisticated Voting Sincere Voting: Vote for most preferred outcomes. Sophisticated Voting: Individual actors vote against their preferences in the earlier votes if they can anticipate the outcome of later votes. Example: Assume there are three alternatives- x, y, and z- and three voters- Player 1, 2, and 3with the following preferences: Player 1: x y z Player 2: y z x Player 3: z x y The three alternatives are voted in pairwise comparisons: first x versus y, and then the winner of that vote versus z. At the last round of voting, sincere voting is the dominated strategy. Then, we can figure out the Voting Tree. Therefore, Player 1 has a sophisticated voting strategy in round 1 by voting for y. The outcome is then y which Player 1 prefers over z. Player 2 will vote for y in round 1 and 2. Player 3 does not has the option of sophisticated voting.
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HKUST - MATH - 4321
Part II. Two-Person Zero-Sum GamesExample: Odd or EvenPlayers I and II simultaneously call out one of the numbersone or two. Player Is name is Odd; he wins if the sum of thenumbers if odd. Player IIs name is Even; she wins if the sumof the numbers is
HKUST - MATH - 4321
Remark: In the above analysis, we used the basicassumption of Common Knowledge.A fact is common knowledge if everyone knows it, everyoneknows that everyone knows it, everyone knows thateveryone knows that everyone knows it,., and so on adinfinitum.,
HKUST - MATH - 4321
Equilibrium Principle: BR to each otherMaximin Principle: Safety FirstFor Player I: Find p so that MinqpTAq islargest. p is called the Safety Strategy orOptimal Strategy.MinqpTAq is the lower value.For Player II: Find q so that MaxppTAq issmallest.
HKUST - MATH - 4321
Recall that Equilibrium Principle: BR to each other Maximin Principle: Safety firstFor Player I: Find p so that MinqpTAq is largest. p iscalled the Safety Strategy or Optimal Strategy.MinqpTAq is the lower value.For Player II: Find q so that MaxppTA
HKUST - MATH - 4321
MATH 4321Game TheorySpring 2012Assignment 1Exercise I.1.5: 1, 2, 4Exercise I.2.6: 1(a), 1(b), 2, 3, 4Exercise I.3.5:1, 2, 3, 5Exercise I.4.5: 2, 3, 6, 8.Mission : Exercise I.1.5: 6(b), 8.
HKUST - MATH - 4321
MATH 4321Game TheorySpring 2012Assignment 21. Problem II.5.9.1, II.5.9.4.2. The Hidden pearl: There are two dark boxes. Player I hides a pearlin one of them. Then Player II, not knowing which box contains the pearl,peeks into one of them. If the pe
HKUST - MATH - 4321
MATH 310Game TheorySpring 2012Assignment 31. Reduce the Kuhn Tree of Exercise 2 in Assignment 2 to strategic formand then find all PSEs.2. Find the PPSE of the Votes by Veto game in Assignment 2.3. Reduce the following Kuhn tree to strategic form.
HKUST - MATH - 4321
MATH 310Game TheorySpring 2012Assignment 41. Problems II.1.5.1, II.1.5.2, II.1.5.32. Problems II.2.6.2, II.2.6.4, II.2.6.5, II.2.6.6, II.2.6.7, II.2.6.8, II.2.6.103. Prove that for an mxn matrix game, any two saddle points have thesame value.Missi
HKUST - MATH - 4321
MATH 4321Game TheorySpring 2012Assignment 51. Problems II.3.7.3, II.4.7.2, II.5.9.9, II5.9.10(d)2. Solve the following matrix game using the method of linearprogramming.1 31 53. Given a matrix game suppose p1, p2 are optimal strategies for Player
HKUST - MATH - 4321
HKUST - MATH - 102
Math 100 - Introduction to Multivariable CalculusFINAL EXAMINATIONFall Semester, 1999Time Allowed: 2.5 Hours.Total Marks: 100Student Name:Student Number:1. (10 marks) Locate all relative maxima, relative minima and saddlepoints of the functionf (
HKUST - MATH - 102
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HKUST - MATH - 102
Problem 1 (12 points)HKUSTYour Score:MATH 1021(a) If f (x, y ) = (x3 + y 2 ) 3 , nd fx (0, 0) and fy (0, 0).Final ExaminationMultivariable Calculus(b) Let z = f (x, y ), where x = g (t) and y = h(t).(i) Show thatAnswer ALL 8 questionsddtTime
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HKUSTMATH 102Final ExaminationMultivariable CalculusAnswer ALL 8 questionsTime allowed 3 hoursProblem 11(a) If f (x, y ) = (x3 + y 2 ) 3 , nd fx (0, 0) and fy (0, 0).(b) Let z = f (x, y ), where x = g (t) and y = h(t).(i) Show thatddtzx= 2
HKUST - MATH - 102
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HKUST - MATH - 102
HKUST - MATH - 102
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HKUST - MATH - 102
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HKUST - MATH - 102
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HKUST - MATH - 102
HKUSTMATH 101Midterm ExaminationProblem 1(a) If e is any unit vector and a an arbitrary vector show thata = (a e)e + e (a e).Multivariable CalculusThis shows that a can be resolved into a component parallel to and one perpendicular to an14 October
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Math102 Midterm-2005Problem 1(a) (a b) c = (a c)b (b c)a,i.e. = a c, = b c and = 0.The resulting vector of (a b) c is a linear combination of the vectors a and b, hence it lieson the plane containing the vectors a and b.(b)(i) r = 3.(ii) 0(iii)
HKUST - MATH - 102
HKUSTMATH 102Midterm ExaminationMultivariable and Vector Calculus21 Dec 2005Answer ALL 8 questionsTime allowed 180 minutesDirections This is a closed book examination. No talking or whispering are allowed. Work mustbe shown to receive points. An a
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HKUST - MATH - 102
HKUSTMATH 102Midterm ExaminationMultivariable and Vector Calculus6 Nov 2006Answer ALL 5 questionsTime allowed 120 minutesDirections This is a closed book examination. No talking or whispering are allowed. Work mustbe shown to receive points. An an
HKUST - MATH - 102
HKUSTMATH 102Second Midterm ExaminationMultivariable and Vector Calculus15 Dec 2006Problem 1(a) x + 2y z = 10(b) x2 + (y 1)2 + z 2 = 1which is the equation of a sphere with center (0, 1, 0) and radius 1.(c) The distance between two planes isd =
HKUST - MATH - 102
HKUSTMATH 102Second Midterm ExaminationMultivariable and Vector Calculus15 Dec 2006Answer ALL 8 questionsTime allowed 180 minutesProblem 1(a) Find an equation of the plane through (1, 4, 3) and perpendicular to the linex = t + 2,y = 2t 3,z = t.
HKUST - MATH - 102
HKUSTMATH 102Third Midterm ExaminationMultivariable and Vector Calculus12 April 2007Answer all ve questionsTime allowed 120 minutesDirections This is a closed book examination. No talking or whispering are allowed. Workmust be shown to receive poi
HKUST - MATH - 102
Problem 1 (20 points)HKUSTYour Score:MATH 102Identify the following surfacesMidterm One Examination(a) r u = 0.Multivariable and Vector Calculus(b) (r a) (r b) = k .30 Oct 2007(c)r (r u)u = k . [Hint: What are the vectors (r u)u and r (r u)u?]
HKUST - MATH - 102
HKUSTMATH 102Midterm One ExaminationMultivariable and Vector CalculusProblem 4(a) Find the parametric equation of the curve of intersection C between the plane z = 2y + 3and the surface z = x2 + y 2 . Find also the equation of the projection curve o
HKUST - MATH - 102
HKUSTMATH 102Third Midterm ExaminationMultivariable and Vector Calculus12 April 2007Answer all ve questionsTime allowed 120 minutesDirections This is a closed book examination. No talking or whispering are allowed. Workmust be shown to receive poi
American University in Cairo - EENG - 320
Signals and Systems (EENG 320)Assignment (5)1. Use either the Fourier transform analysis and synthesis equations or theFourier transform properties to:a. Compute the Fourier transform of each of the following signals:i. [e t cos(o t )]u (t ), > 0ii.
American University in Cairo - EENG - 320
American University in Cairo - EENG - 320
Signals and Systems (EENG 320)Assignment (4)1- Determine the Fourier series representation for the following signals:i. x(t ) periodic with period 2 andx(t ) = e tfor - 1 < t < 1ii. Each x(t ) illustrated in figures (1)-(3).x(t)1-5-3-4-1153
American University in Cairo - EENG - 320
Signals and Systems (EENG 320)Assignment (3)1) Compute the convolution y[ n] = x[ n] h[ n] of the following signals :x[n] = nu[n]y[n] = nu[n], 2) Compute the convolution y (t ) = x (t ) h(t ) of the following signals, sketch the result:x (t ) = e t
American University in Cairo - EENG - 320
Signals and Systems (EENG 320)Assignment (2)1) Consider a continuous time signal :x (t ) = (t + 2) (t 2)If a signal :2ty (t ) = x( )da. Is y(t) an energy signal or power signal?b. Calculate its energy and power2) Consider the discrete time signa
American University in Cairo - EENG - 320
Signals and Systems (EENG 320)Assignment (1)fig(1)fig(2)1) A continuous-time signal x(t) is shown in fig(1). Sketch and label carefully each of thefollowing signals: The original signal x(t)a. x(t-1)b. x(2-t)c. x(2t+1)d. x(4-t/2)e. [x(t) + x(-t)
American University in Cairo - EENG - 320
Signals and Systems (EENG 320)Assignment (1)fig(1)fig(2)1) A continuous-time signal x(t) is shown in fig(1). Sketch and label carefully each of thefollowing signals:a. x(t-1)b. x(2-t)c. x(2t+1)d. x(4-t/2)e. [x(t) + x(-t)] u(t)f. x(t) [ (t+2/3)
ADFA - MATH - 101
CHAPTER 22.1DYNAMIC MODELS AND DYNAMIC RESPONSERevisiting Examples 2.4 and 2.5 will be helpful.(a)R (s) = A;Y ( s) =y(t ) =(b)KA - t / te;tR(s) =A;sKAt s +1yss = 0Y(s) =KAs (s + 1)y(t) = KA (1 e t / ) ;(c)R(s) =yss = KAAKA; Y(
IIT Kanpur - STATISTICS - SI406
DATA MINING ASSIGNMENTGROUP MEMBERS:1. LOHIT GUPTA2. SUNIL KUSHWAHA3. NATARAJANIn this problem, we are going to analyze three kernel functions Uniform,Gaussian and Epanechnikov.Cluster Analysis:The analysis is started off by clustering the depende
IIT Kanpur - STATISTICS - SI406
SI515 (Autumn 2010) Assessment of Solutions of Assignment 1Common Mistakes/Flaws:C1: Quantitative Feature Values are NOT normalized.C2: How the training sample was selected randomly from the given data is NOTSpecified/justified.C3: Your own code and/
IIT Kanpur - STATISTICS - SI406
Department of Mathematics, IIT BombaySI 515 (Data Mining): Autumn 2010Assignment Sheet-IIEvery question is a Test Assignment for individual groups and carries max. of 10 marks.Q1. Make clusters of Fishers iris data using (i) suitable 0-1 Integer progr
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Chapter 2The Backprop AlgorithmCopyright 1995 by Donald R. Tveter, drt@christianliving.net. Commercial UseProhibited2.1 Evaluating a NetworkFigure 2.1 shows a simple back-propagation network that computes the exclusive-or(xor) of two inputs, x and y
IIT Kanpur - STATISTICS - SI406
Biostatistics 695 HW # 3Jian KangOctober 3, 20072.7 In the United States, the estimated annual probability that a woman over the ageof 35 dies of lung cancer equals 0.001304 for current smokers and 0.000121 fornonsmokers (M. Pagano and K. Gauvreau, P
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Mathematical Geology, Vol. 14, No. 6, 1982U se of the Bray-Curtis Similarity Measure inCluster Analysis of Foraminiferal Data jMichael G. Michie 2Transformation of data effectively limits the distortion by outlying values on the BrayCurtis similarity
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Stat1301B Probability& Statistics IChapter II 2.1RandomVariablesDistributionsandFall 2011-2012ProbabilityRandom VariablesDefinitionA random variable X : is a numerical valued function defined on a samplespace. In other words, a number X ( ) ,
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