Choice Under Certainty Bruno Salcedo Cornell University Decision Theory Spring 2017 1 / 1
decision theory Decision theory is about making choices Normative aspect: what rational people should do Descriptive aspect: what people do do Not surprisingly, it s been studied by economists, psychologist, and philosophers More recently, computer scientists have looked at it too How should we design robots that make reasonable decisions What about software agents acting on our behalf Agents bidding for you on ebay Managed health care Algorithmic issues in decision making This course will focus on normative aspects, informed by a computer science perspective 2 / 1
axiomatic decision theory Standard (mathematical) approach to decision theory: Give axioms characterizing reasonable decisions Ones that any rational person should accept Then show show that these axioms characterize a particular approach to decision making For example, we will discuss Savage s axioms that characterize maximizing expected utility An issue that arises frequently: How to represent a decision problem 3 / 1
expressed preferences LetX be a set of alternatives Typical elements ofx are denoted byx,y,z... For each pairx,y X we ask a subject whether he prefersx overy, y overx, or neither Notation: x y means the subject strictly prefers x over y The relation is a binary relation onx Example:X={a,b,c},b a,a c, andb c What if the individual also saida b? 4 / 1
some axioms What are some (arguably) reasonable properties for a preference order? A binary relation on X satisfies Completeness ifx y impliesy x Asymmetry ifx y impliesy x Acyclicity ifx 1 x 2... x n impliesx 1 x n Transitivity if[x y andy z] impliesx z Negative transitivity if[x y andy z] impliesx z Example:X={a,b,c},a b,b c,c a Are such cyclic preferences reasonable? How would this person choose an alternative fromx? 5 / 1
preference relations Definition: A binary relation is a strict preference relation if it is asymmetric and negatively transitive Example: X ={a, b, c}, and is a strict preference relation withb a, anda c. What can we tell aboutb vs.c? Asymmetry impliesc a anda b Ifb c, then NT would implyb a Hence, it must be thatb c Asymmetry then impliesc b 6 / 1
negative transitivity Proposition: The binary relation is NT iffx z implies that for ally X,y z orx y Proposition: If is NT then[x y andy x] imply that for allz,[z x iffz y] and[x z iffy z] Example: LetX={1,2,3,...} and suppose that1is not ranked by relative to any other alternative If satisfies NT, then there are no alternatives which can be ranked 7 / 1
weak preference and indifference Definition: Given a relation onx 1.x is weakly preferred toy,x y, iffy x 2.x is indifferent toy,x y, iffy x andx y 8 / 1
Proposition: Given a relation onx 1. is asymmetric if and only if is complete 2. is negatively transitive if and only if is transitive Proof of 1. Asymmetry implies we cannot have bothx y andy x Hence, at least one ofx y andy x holds 2. Ifx y z, thenz y x Therefore, by NT,z z, i.e.,x z Proof of will be on homework 1 9 / 1
transitivity Why do we care about transitivity? People s preferences sometimes fail transitivity However, if this is pointed out, most people think they should change their preferences Example:X={a,b,c},a b,b c,c a No undominated alternative inx a b andb c, buta c Susceptible to money pumps Starting froma, pay a penny to switch fromatoc Then pay a penny to switch fromc tob Then pay a penny to switch frombtoa... 10 / 1
transitivity and cycles Proposition: If is a strict preference relation, then it is transitive and acyclic Proof for transitivity: Supposex y z, we need to showx z Asymmetry impliesz y Ifx z, NT would implyx y Hence we must havex z Proof for acyclicity: Suppose towards a contradictionx 1 x 2... x n andx 1=x n Transitivity would implyx 1 x n=x 1, which contradicts asymmetry 11 / 1
examples Example: X ={a, b, c}, a b, b c, no other strict preference rankings The relation is clearly acyclic (and asymmetric) It is not transitive becausea b c buta c Can you come up with a relation that is acyclic and complete but not transitive? Example: X ={a, b, c}, a c, no other strict preference rankings The relation is clearly transitive and asymmetric It is not NT becausea b c buta c Can you come up with an example of a relation that is NT but not transitive? What if we also require thatx y ory x wheneverx y? 12 / 1
no sugar in my coffee X={1,2,3,...}, number of grains of sugar in my coffee I do not like sugar, but I cannot distinguish less than one teaspoon If x y 10,000, thenx y andy x Ifx y>10,000, theny x The relation is transitive and asymmetric Ifx y, thenx<y and thusy x Ifx y z, thenz x=(z y)+(y x)>20,000, and thusx z The induced indifference relation is not transitive 1 2 3... 9,999 10,000 10,001, but1 10,001 13 / 1
choice functions Hard to measure preferences beyond introspection and surveys Choices can be observed to some extent Maintained assumption: X is finite LetP(X) denote the set of all non-empty subsets ofx Typical elements ofp(x) are denoted bya,b,... Definition: A choice function is a functionc:p(x) P(X) such thatc(a) A for everya P(X) 14 / 1
examples X={a,b,c} By definitionc({a})={a},c({b})={b} andc({c})={c} A {x,y} {x,z} {y,z} {x,y,z} c(a) {x} {x,z} {z}?? 15 / 1
examples X={a,b,c} By definitionc({a})={a},c({b})={b} andc({c})={c} A {x,y} {x,z} {y,z} {x,y,z} c(a) {x} {x,z} {z} {x,z} c (A) {x} {x,z} {y} {y,z} c (A) {y} {z} {y} {y} c (A) {x,y} {x,z} {y,z} {x,y,z} 15 / 1
to some extent Knowledge of a choice function requires observing Choices from different alternative sets ceteris paribus Choices from all elements ofp(x) All possibly chosen alternatives from each element ofp(x) 16 / 1
choices induced by preferences How would a decision maker with preferences make choices? Presumably it would never choose dominated alternatives Any undominated alternative would be acceptable Definition: The choice function induced by a preference onx is the functionc(, ):P(X) P(X) given by c(a, )= { x A for ally A, y x } Example:X={a,b,c},a b,b c, anda c c({a,b}, )={a},c({a,c}, )={a},c({a,b,c}, )={a} c({b,c}, )={b} 17 / 1
acyclicity and choices Proposition: If is acyclic andx is finite, thenc(, ) is a choice function Proof thatc(a, ) for alla P(X) Letm< be the number of alternatives ina Suppose towards a contradiction thatc(a, )= and fix somex 1 A There would existx 2 A such thatx 2 x 1 andx 2 x 1 (why?) There would existx 3 A such thatx 3 x 2 x 1 andx 3 x 1,x 2. There would existx m+1 A such thatx m+1... x 3 x 2 x 1 and x m+1 x 1,x 2,...x m IfX is finite andc(, ) is a choice function, then is acyclic 18 / 1
rationalization What are the implications of preferences in terms of choices? Ifc is an arbitrary choice function, need there be a preference such that c( )=c(, )? Example:X={a,b,c},c({a,b,c})=c({a})={a} andc({a,b})={b} c({a})={a} requiresa a Thenc({a,b})={b} requiresb a But then c({a, b, c}) ={a} is not possible Is there a context where these choices would be reasonable? 19 / 1
Where do we stand? If is a preference order (negatively transitive + asymmetric) on a setx, then we can partition the elements ofx into indifference classes X 1,...,X k such that X 1 X 2... X k more precisely, ifx,y X i, thenx y (i.e.,x y andy x) ifx X i andy X j wherei<j, thenx y Now suppose that po is such thatx po z, andy is incomparable tox andz ( po = partial order ) then po can t be a preference order (x po y,y po z, butx po z) But po is acyclic So there is a choice functionc induced by po : c({x,z})=x, c({x,y})={x,y}, c({y,z})={y,z}, c({x,y,z})={x,y} What properties does a choice functionc induced by a preference order have? We ll consider two properties of a choice functionc that are necessary and sufficient forc to be induced by a preference order every choice function induced by a preference order has these properties. if a choice function c has these properties, then c( ) = c(, ) for some preference order 20 / 1
Sen sα A B x Sen s axiomα: Ifx B Aandx c(a), thenx c(b) If the smartest student in the U.S. is from Cornell, then he/she is also the smartest student at Cornell. 21 / 1
Sen sα Proposition: For any preference relation, the induced choice functionc(, ) satisfies axiomα Proof: Supposex B A X andx c(a, ), butx c(b, ) x c(b, ) implies thaty x for somey B Sincey A, thenx c(a, ) Actually, even if is acyclic,c(, ) satisfies axiomα The same proof applies without change! So if is a preference order, there must be additional properties thatc(, ) satisfies. 22 / 1
Sen sβ A B x y Sen s axiomβ: IfB A,x,y c(b), andy c(a), then x c(a) If the smartest student in the U.S. is from Cornell, then all the smartest students at Cornell are among the smartest students in the U.S. 23 / 1
Sen sβ Proposition: For any preference relation, the induced choice functionc(, ) satisfies axiomβ Proof: Supposex,y B A X,x c(b, ), andy c(a, ) x,y c(b, ) implies thaty x y c(a, ) implies thatz y for allz A NT then implies thatz x for allz A Hence,x c(a, ) c=c(, po ) does not satisfy axiomβ: z c({y,z}),y c({x,y,z}), butz/ c({x,y,z}) 24 / 1
If c = c(, ) and is a preference order, then it satisfies axioms α and β Are there any more requirements? No Theorem: (a) If is a preference relation on a finite setx, then c( )=c(, ) is a choice function satisfying axiomsα andβ (b) Ifc is a choice function onx satisfying axiomsαandβ, thenc( )=c(, ) for a unique preference relation We ve already shown part (a). But note that we needx to be finite. Why? To ensure that is acyclic, and thus to ensure thatc(a, ) is nonempty if A is nonempty. We need to work a little to prove part (b)... 25 / 1
revealed preference Given a choice functionc satisfying axiomsαandβ, we want to find a preference relation such thatc( )=c(, ) Givenx andy, how can we figure out ifx y ory x? Look atc({x,y})! Definition: Given a choice function c, x is revealed preferred toy ifx y andc({x,y})={x} Let denote the revealed preferences fromc, we need to show 1. c(a) c(a, ) for alla P(X) 2. c(a, ) c(a) for alla P(X) 3. is asymmetric 4. is negatively transitive 26 / 1
proof 1.c(A) c(a, ) for alla P(X) Letx c(a) By Axiomα,x c({x,y}) for ally A Hence,y x for ally A 2.c(A, ) c(a) for alla P(X) Letx c(a, ) and take anyy c(a) Axiomαimpliesy c({x,y}) Sincex c(a, ),y x and thusc({x,y}) {y} Thereforec({x,y})={x,y} Axiomβthen impliesx c(a) 27 / 1
proof 3. is asymmetric x x by construction Ifx y thenc({x,y})={x}, soc({x,y}) {y}, hencey x 4. is negatively transitive Suppose thatz y x; we want to show thatz x; that is,x c({x,z}) By Axiomα, it suffices to show thatx c({x,y,z}) Ify c({x,y,z}) Sincey x, we must havex c({x,y}) Axiomβ then impliesx c({x,y,z}) Ifz c({x,y,z}) then an analogous argument impliesy c({x,y,z}) and thus,x c({x,y,z}) Ify c({x,y,z}) andz c({x,y,z}) thenc({x,y,z})={x} 28 / 1