Imperfect Information

Similar documents
Lecture 18: The Laplace Transform (See Sections and 14.7 in Boas)

Solution in semi infinite diffusion couples (error function analysis)

PHYS 705: Classical Mechanics. Canonical Transformation

Advanced time-series analysis (University of Lund, Economic History Department)

Mechanics Physics 151

( ) () we define the interaction representation by the unitary transformation () = ()

TSS = SST + SSE An orthogonal partition of the total SS

Mechanics Physics 151

Density Matrix Description of NMR BCMB/CHEM 8190

EP2200 Queuing theory and teletraffic systems. 3rd lecture Markov chains Birth-death process - Poisson process. Viktoria Fodor KTH EES

Graduate Macroeconomics 2 Problem set 5. - Solutions

In the complete model, these slopes are ANALYSIS OF VARIANCE FOR THE COMPLETE TWO-WAY MODEL. (! i+1 -! i ) + [(!") i+1,q - [(!

FI 3103 Quantum Physics

. The geometric multiplicity is dim[ker( λi. number of linearly independent eigenvectors associated with this eigenvalue.

Density Matrix Description of NMR BCMB/CHEM 8190

How about the more general "linear" scalar functions of scalars (i.e., a 1st degree polynomial of the following form with a constant term )?

Epistemic Game Theory: Online Appendix

Endogeneity. Is the term given to the situation when one or more of the regressors in the model are correlated with the error term such that

UNIVERSITAT AUTÒNOMA DE BARCELONA MARCH 2017 EXAMINATION

. The geometric multiplicity is dim[ker( λi. A )], i.e. the number of linearly independent eigenvectors associated with this eigenvalue.

Lecture VI Regression

Let s treat the problem of the response of a system to an applied external force. Again,

Lecture 6: Learning for Control (Generalised Linear Regression)

Econ107 Applied Econometrics Topic 5: Specification: Choosing Independent Variables (Studenmund, Chapter 6)

Dynamic Team Decision Theory. EECS 558 Project Shrutivandana Sharma and David Shuman December 10, 2005

[ ] 2. [ ]3 + (Δx i + Δx i 1 ) / 2. Δx i-1 Δx i Δx i+1. TPG4160 Reservoir Simulation 2018 Lecture note 3. page 1 of 5

V.Abramov - FURTHER ANALYSIS OF CONFIDENCE INTERVALS FOR LARGE CLIENT/SERVER COMPUTER NETWORKS

Scattering at an Interface: Oblique Incidence

CS434a/541a: Pattern Recognition Prof. Olga Veksler. Lecture 4

Linear Response Theory: The connection between QFT and experiments

CS286.2 Lecture 14: Quantum de Finetti Theorems II

Political Economy of Institutions and Development: Problem Set 2 Due Date: Thursday, March 15, 2019.

THEORETICAL AUTOCORRELATIONS. ) if often denoted by γ. Note that

NPTEL Project. Econometric Modelling. Module23: Granger Causality Test. Lecture35: Granger Causality Test. Vinod Gupta School of Management

Variants of Pegasos. December 11, 2009

Advanced Machine Learning & Perception

Mechanics Physics 151

CH.3. COMPATIBILITY EQUATIONS. Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Ordinary Differential Equations in Neuroscience with Matlab examples. Aim 1- Gain understanding of how to set up and solve ODE s

1 Review of Zero-Sum Games

Notes on the stability of dynamic systems and the use of Eigen Values.

A New Generalized Gronwall-Bellman Type Inequality

Advanced Macroeconomics II: Exchange economy

( ) [ ] MAP Decision Rule

Online Appendix for. Strategic safety stocks in supply chains with evolving forecasts

Department of Economics University of Toronto

Pattern Classification (III) & Pattern Verification

Keywords: Hedonic regressions; hedonic indexes; consumer price indexes; superlative indexes.

Math 128b Project. Jude Yuen

FTCS Solution to the Heat Equation

1 Constant Real Rate C 1

(,,, ) (,,, ). In addition, there are three other consumers, -2, -1, and 0. Consumer -2 has the utility function

Computing Relevance, Similarity: The Vector Space Model

Chapter Lagrangian Interpolation

ECON 8105 FALL 2017 ANSWERS TO MIDTERM EXAMINATION

Uniform Topology on Types and Strategic Convergence

Chapter 6: AC Circuits

CHAPTER 10: LINEAR DISCRIMINATION

New M-Estimator Objective Function. in Simultaneous Equations Model. (A Comparative Study)

January Examinations 2012

Foundations of State Estimation Part II

On One Analytic Method of. Constructing Program Controls

Lecture 11 SVM cont

8. Queueing systems lect08.ppt S Introduction to Teletraffic Theory - Fall

F-Tests and Analysis of Variance (ANOVA) in the Simple Linear Regression Model. 1. Introduction

GENERATING CERTAIN QUINTIC IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS. Youngwoo Ahn and Kitae Kim

Example: MOSFET Amplifier Distortion

Li An-Ping. Beijing , P.R.China

Normal Random Variable and its discriminant functions

Online Supplement for Dynamic Multi-Technology. Production-Inventory Problem with Emissions Trading

CHAPTER 5: MULTIVARIATE METHODS

Fall 2010 Graduate Course on Dynamic Learning

National Exams December 2015 NOTES: 04-BS-13, Biology. 3 hours duration

, t 1. Transitions - this one was easy, but in general the hardest part is choosing the which variables are state and control variables

J i-1 i. J i i+1. Numerical integration of the diffusion equation (I) Finite difference method. Spatial Discretization. Internal nodes.

Robustness Experiments with Two Variance Components

II. Light is a Ray (Geometrical Optics)

A Reinforcement Procedure Leading to Correlated Equilibrium

Cubic Bezier Homotopy Function for Solving Exponential Equations

Volatility Interpolation

Dynamic Regressions with Variables Observed at Different Frequencies

HEAT CONDUCTION PROBLEM IN A TWO-LAYERED HOLLOW CYLINDER BY USING THE GREEN S FUNCTION METHOD

Reactive Methods to Solve the Berth AllocationProblem with Stochastic Arrival and Handling Times

The Finite Element Method for the Analysis of Non-Linear and Dynamic Systems

Laser Interferometer Space Antenna (LISA)

グラフィカルモデルによる推論 確率伝搬法 (2) Kenji Fukumizu The Institute of Statistical Mathematics 計算推論科学概論 II (2010 年度, 後期 )

Tjalling C. Koopmans Research Institute

Chapters 2 Kinematics. Position, Distance, Displacement

2. SPATIALLY LAGGED DEPENDENT VARIABLES

John Geweke a and Gianni Amisano b a Departments of Economics and Statistics, University of Iowa, USA b European Central Bank, Frankfurt, Germany

NATIONAL UNIVERSITY OF SINGAPORE PC5202 ADVANCED STATISTICAL MECHANICS. (Semester II: AY ) Time Allowed: 2 Hours

Bundling with Customer Self-Selection: A Simple Approach to Bundling Low Marginal Cost Goods On-Line Appendix

A Cell Decomposition Approach to Online Evasive Path Planning and the Video Game Ms. Pac-Man

Lecture 2 M/G/1 queues. M/G/1-queue

II The Z Transform. Topics to be covered. 1. Introduction. 2. The Z transform. 3. Z transforms of elementary functions

Testing a new idea to solve the P = NP problem with mathematical induction

LOCATION CHOICE OF FIRMS UNDER STACKELBERG INFORMATION ASYMMETRY. Serhij Melnikov 1,2

Appendix H: Rarefaction and extrapolation of Hill numbers for incidence data

THE PREDICTION OF COMPETITIVE ENVIRONMENT IN BUSINESS

Chapter 7: Solving Trig Equations

Transcription:

Imerfec Informaon Comlee Informaon - all layers know: Se of layers Se of sraeges for each layer Oucomes as a funcon of he sraeges Payoffs for each oucome (.e. uly funcon for each layer Incomlee Informaon - any of he four eces of nformaon above s unknown o a layer; also called asymmerc nformaon Prvae Informaon - ycal examle of ncomlee nformaon where layer knows somehng abou hs own acons ha oher layers don' observe; could be he oher way hough (e.g.ecals [docor mechanc] has more nformaon abou your ayoffs han you do Harsany - aer n Managemen Scence (967 on how o conver game wh ncomlee nformaon o game of merfec bu comlee nformaon by nserng random move by naure Incomlee n Payoffs - Harsany frs argued ha all yes of ncomlee nformaon can be convered o ncomlee nformaon on ayoffs Players - no sure f layer s n he game... equvalen o knowng he s n he game bu no sure abou hs sraegy sace (e.g. could have four oons or jus one effecvely no n he game because he has no decsons Player Sraeges Sraeges - no knowng se of sraeges could be ransformed by addng sraeges ha wll never be layed because of - ayoff; now all oons have same number of sraeges bu we don' know whch ayoffs are correc Sraeges Payoffs Oucomes - no knowng oucomes s equvalen o no knowng he uly funcons (ayoffs... echncally here s a dfference o knowng oucomes and no uly funcons f here are several sraeges (from dfferen nformaon ses ha have he same oucome (wll yeld he same ayoffs Infne Regress - n order o solve ncomlee nformaon roblem we have o worry abou belefs abou belefs abou belefs ec. Soluon - Harsany solved roblem by defnng yes of layers where layers dffer n ayoffs (uly funcon and/or belefs Smlfcaon - Harsany also argued ha all belefs could be caured by a sngle robably dsrbuon over he yes of he oher layers (could be correlaed by usng a jon dsrbuon Examle - 3 yes of "layer " and yes of "layer ": and of 7 3 3 Each ye of "layer " has belefs abou he robably ha an oonen wll be of a secfc ye: = (0.5 0.5 0.5 0.5 = (0. 0. 0. 0.7 = (0.5 0.5 0 0 In or Ou? or or or 3 - - -

Fudge - n order o be racable hs mehod requres a fne number of yes Sandard - ycally we only allow layers yes o dffer on reference (uly funcon; heory allows references and/or belefs o vary bu he full mlcaons of hs haven' been suded Problem? - may have assumed away he roblem of ncomlee nformaon raher han solved Formally - Players - n layers Sraeges - each ye of layer has he same sraegy sace S (oal of n Se of Tyes - T s se of dfferen yes of layer (oal of n ses of yes Examle - from revous age T = { 3 } T = { 3 } Examle - yes of layer yes of layer and 3 yes of layer 3: T a b T c d T x y z = { } = { } 3 = { } Secfc Player - s he j h ye of layer (hs was denoed of 7 j on revous age Oher Player Tyes - T = T T T T + Tn... he Caresan roduc of he ses of all layer yes exce for layer T = ( a c( a d( b c( b Examle - { } Oher Players - 3 d T s a secfc ye of each layer (exce layer Examle - one of he elemens of T 3 above Belefs - = Pr[ ] = subjecve belef of j h ye of layer abou he robably ha hs secfc combnaon of oonens wll occur = se of all for layer Examle - 3x = (0.5 0.5 0.5 0.5 3 y = (0. 0. 0.3 0.3 3z = (0.5 0.5 0 0 So ye x of layer 3 hnks 's equally lkely o see any combnaon of yes of layers and ; ye y of layer 3 hnks he'll face ( a c (.e. ye a of layer and ye c of layer abou a ffh of he me; ec. = { } 3 3x 3 y 3z Preferences - u ( s ; uly for layer deends on he acual yes of each layer ( and he sraeges ha each layer uses (s Γ S S T T u u n n n n Sraeges Tyes Belefs Prefs Bayesan Game of Incomlee Informaon - ( (assume hs s common knowledge Examle of Belefs - n oker here are 5!/(7!5! ossble hands; ror o dealng all hands have equal robably; afer seeng your cards (and ohers avalable he robables assgned o dfferen yes of oonens (.e. hands hey have change... for examle f you have aces he robably ha an oonen has a hree or four of a knd wh aces or any oher hand ha requres more han aces wll be zero Professonal Poker - assume layer has queens and layer has jacks; on TV hey say n hs scenaro layer has a 90% chance of wnnng and layer only has a 0% chanceuggesng ha 's a bad dea for layer o say n he game... bu ha analyss s based on comlee nformaon; he layers have ncomlee nformaon so he relevan robables are each layer's (subjecve robably of wnnng gven hs

own cards (and hose he's seen; ha s each layer s basng hs decson on hs belef abou hs oonen's hand (usually a lo more robably on oor hands han good hands; f boh layers are sll n he game ha means hey boh hnk hey have greaer han 50% chance of wnnng Conssency - Pr[ ] = Pr[ ] = where Pr[ ] = Pr[ ] Pr[ ] Subjecve Objecve Examle - 3 layers each wh yes has oal of 8 combnaons (so each ye of each layer faces combnaons of ossble oonens: 3 3 3 3 3 3 3 3 Pr[ 3] No Bg Deal - Almond showed nconssen belefs can be ransformed no conssen belefs; we usually jus assume belefs are conssen because of hs Max Execed Uly - max u ( s ( ; Pr[ ] s Suose belef = Pr[ ] s no conssen ( u ( s above s ncorrec Suose belef ˆ s conssen Always Conssen - le N ( T = # of vecors T = roduc of number of yes of all layers exce ; (examle above has N ( T 3 = Assume each s equally lkely so ˆ = Pr[ ] = / N( T Based on he defnon of conssency belef Defne new uly uˆ ( s N( T u ( s = Noe ha ˆ uˆ ( s = u ( s Pr[ 3] Pr[ 3] = Pr[ 3] + Pr[ 3 ] + Pr[ 3 ] + Pr[ 3 ] = 3 ˆ s conssen Tha means execed uly wh conssen and nconssen belefs s he same Cach - have o know he "werd" (nconssen robables o ncororae no references (new uly funcon 3 of 7

Equlbrum - wo deas on how o fnd Players Aroach - rea each ye of each layer as a searae layer (e.g. layers each wh yes becomes a layer game; "layers aroach" s Len's erm; Slusky called he ex os aroach (meanng layers choose her sraeges afer knowng her ye Game Srucure - no all layers wll face each oher (e.g. doesn' lay agans ; hs means he reacon funcon only deends on some of he oonens (n he examle would be oonens nsead of 3 Ineracon - (anoher Poker asde a layer of one ye may wan o affec ayoffs when he's anoher ye; Examles: Beng - be wh a bad hand so oonen knows you be wh a bad hand and hen doesn' know when you have a good hand Bluffng - f you bluff and he oonen folds and asks o see your cards here are wo oons: - Tell oonen ha o see he hand he needs o call (mach be... n oher words he has o ay for he nformaon - Show hm your hand so he knows you're bluffng; ha way f you be laer wh a good hand he mgh hnk you're bluffng and be agans you Problem - hese are use reeaed game reasonng (we're sll focusng on sngle-sho games σ ( s an equlbrum sraegy for (ye j of layer f and T T u (σ ( ; Pr[ ] u ( s ; Pr[ ] s S T Englsh - akng he oher layers' sraeges ( as gven layer 's bes rely s σ ( (.e. he execed ayoff s greaer han or equal o he execed ayoff of layng any oher sraegy s n hs sraegy sace S ; n order for hs o be an equlbrum all he layers mus be layng bes reles o her oonens' bes reles so hs condon holds for all yes of all layers ( and T Noe: alernave sraegy doesn' need o be ndexed by he layer ye ( j because we assumed ha all yes of he same layer have he same sraegy sace (o of. Sraeges Aroach - frs move n game s naure choosng layers yes; solve an merfec bu comlee nformaon game; "sraeges aroach" s Len's erm; Slusky called he ex ane aroach (layers choose sraeges before knowng her ye +: Fewer layers : More comlcaed sraeges (mus address sraeges for each layer ye Examle - 3 yes of layer ; yes of layer Naure 3 Dfferen? - many quesons on wheher hese wo mehods are dfferen n equlbra or mehod (e.g. easer o solve Subgame Perfecon - he layers aroach guaranees subgame erfec; some eole argue he sraeges aroach could have sraeges ha aren' subgame erfec Harsany - argued he wo were dfferen; dea of mmedae vs. delayed commmen (here "commmen" means ckng a sraegy Immedae Commmen - a sar of game before knowng ye (sraeges aroach of 7

Delayed Commmen - choose sraegy afer knowng ye (layers aroach Slusky - "The ermnology s no longer used n ar because 's no useful." Problem - Harsany argued hese are dfferen and sad delayed commmen s beer bu hs examle roblem combned cooerave and non-cooerave elemens Slusky - no sure wha equlbrum noon s n a mxed cooerave and noncooerave game Same - n fully non-cooerave game sraeges and layers aroach are THE SAME; some felds (e.g. regulaon use he layers aroach because "'s rgh" bu hey're ncorrecly aly Harsany concluson o fully non-cooerave games Hsory - Blackwell Nash Von-Neumann Kuhn all dd wo-layer oker examles usng he sraeges aroach 5 years before Harsany; hey usually used a smlfed verson lke layers each wh 3 yes (hgh medum and low hand; n ha caseraeges aroach s easer hen layers aroach ( layer 8 sraegy game vs. 6 layer sraegy Players Aroach Easer - for nfne yes wh connuous sraegy saces he "easy" way s o arameerze ye and solve usng he layers aroach (e.g. rncal agen roblem; consumer maxmzaon; frm rof maxmzaon Examle - smles case: layers yes of each sraeges: a a and b b Players Aroach - game shown here s Player for layer (only shows hs ayoffs b b even hough we need o know layer a x x 's o solve he game; a a mnmum we also need anoher ar of ayoff a x 3 x ables for layer ; ha wll conan all he nformaon alhough Slusky refers o have a ar for each layer ye (oal of four of hese Noaon: = Pr[ lays a ] (so = Pr[ lays a ] = Pr[ beleves s hs oonen] (so = Pr[ beleves s hs oonen] So he robables always defaul o he frs sraegy or he frs ye of he oonen Execed Uly - E[ u ] = x + ( x + ( x3 + ( ( x + ( y + ( ( y + ( ( y + 3 ( ( ( y Ths s lnear n o we can wre roblem as max f ( x - x y - y + c Prob n box (a b vs. Payoff n box (a b Player Prob n box (a b vs. Player Chosen by naure Payoff n box (a b Player b b a y y a y 3 y... only & are from oher layers Bes Rely - f f ( > 0 = 0 If f ( < 0 (0 f f ( = 0 There wll be a bes rely for he oher hree layer yes resulng n equaons wh unknowns ( 5 of 7

Sraeges Aroach - wll show hs s he same as he layers aroach; assume Eq s he ure sraegy equlbrum New Noaon - = Pr[ ] (robably layer s ye Same as Players Aroach - use execed ayoffs o show ex os ex ane Look a Eq frs: Player ye (rob s lays a ; hs oonen could be ye of layer (rob Pr[ ] and lays b or ye of layer (rob Pr[ ] and lays b Player ye (rob s lays a ; hs oonens are he same and lay he same sraeges (bu he condonal robables are dfferen E[Eq] = [ Pr[ ] u ( a b ; + Pr[ ] u ( a b ; + [ ] u ( a b ; + Pr[ ] u ( a b ; Pr[ Snce hs s an equlbrum we know: E[Eq] E[c] c has he followng sraeges ( a a bb so bascally only he sraegy for layer s changed (from a o a ; usng he same logc as above we have E[r] = [ Pr[ ] u ( a b ; + Pr[ ] u ( a b ; + [ ] u ( a b ; + Pr[ ] u ( a b ; Pr[ The corresondng crcled erms cancel ou so E[Eq] E[c] becomes [ Pr[ ] u ( a b ; + Pr[ ] u ( a b ; [ ] u ( a b ; + Pr[ ] u ( a b ; Pr[ Tha s he execed ayoff of a once layer knows he's ye s o he execed ayoff of a once he knows he's ye : E[ u ( a E[ u ( a We can use hs same logc o ge E[Eq] E[c] yelds same condon as layng a over a E[Eq] E[r] yelds same condon as layng b over b E[Eq] E[r] yelds same condon as layng b over b Combnng all four comarson yelds he same equlbrum sraegy as he ex os game: s = a = a = b = b equlbrum n ex ane game s he same equlbrum n he ex os game Now assume s = a = a = b = b s an equlbrum n he ex os game; we wan o show hs s also an equlbrum n he ex ane game; usng he work above we can work backwards o show Eq = ( aa bb has a hgher execed ayoff han r r c c; now consder E[Eq] vs. E[r];.e. comare ( s = a = a = b = b o ( s = a = a = b = b so boh yes of layer are changng her sraegy; changng he noaon a lle for convenence: E[Eq] = E u ( a + E[ u ( [ a E[ u ( a E[ u ( a E[r] = + Player b b b b b b b b a a c a a c a a r r Eq r a a Player 's sraegy f he's: Tye Tye Player c 6 of 7

From he equlbrum assumon we know E u ( a E[ u ( and [ a E[ u ( a E[ u ( a (.e. each erm n E[Eq] s 's resecve erm n E[r]... ha means E[Eq] E[r] We can reea hs logc o show E[Eq] E[c] ( s = a = a = b = b n he ex os game s he same as ( a a bb n he ex ane game Why? - hs works because he yes are ndeenden of each oher; we can' have and a he same me so here s no neracon (.e. can change boh her sraeges a he same me and 's equvalen o change one a a me Mxed Sraeges - frs look a ex os; suose lays (// and lays (/3/3 n he ex ane game ha's he same as: ex ane a a /3 /6 a a / /3 /6 a a / /3 /6 a a /3 /6 Problem? - gong he oher way 's ossble o come u wh a mxed sraegy n he ex ane game ha can' be relcaed by mxed sraeges n he ex os game (e.g. (0//0; hs s ar of he nuon why Harsany sad hey were dfferen No Problem - won' ge hs as a mxed sraegy because here's no gan n correlaon beween he ye yes; (0//0 mgh as well be (//// whch also has each ye layng a 50-50 mxed sraegy; n a real layer game here could be correlaed sraeges so hs sn' he case 7 of 7