Average choice David Ahn Federico Echenique Kota Saito - PowerPoint PPT Presentation

Average choice David Ahn Federico Echenique Kota Saito Harvard-MIT, March 10, 2016

Path independence This paper: An exploration of path independence in stochastic choice. Ahn-Echenique-Saito Average choice

Plott path independence Plott (Ecma 1973) in response to Arrow’s impossibility theorem: c ( A ∪ B ) = c ( c ( A ) ∪ c ( B )) For example: if c ( x , y ) = x , then c ( x , y , z ) = c ( c ( x , y ) ∪ c ( z )) = c ( x , z ) c = “choice” Ahn-Echenique-Saito Average choice

Plott path independence ◮ Kalai-Megiddo (Ecma 1980) ◮ Machina-Parks (Ecma 1981) NO stochastic* choice can be continuous and Plott Path Indep. Restore the impossibility. Primitive: stochastic* choice. (I’ll explain what I mean by stochastic*). Ahn-Echenique-Saito Average choice

Path independence We: allow the path to affect choice. Choice from A ∪ B is a lottery between choice from A and choice from B. Who A and B are may affect the lottery. Ahn-Echenique-Saito Average choice

Our main result Theorem A stochastic* choice is cont. and path independent iff it is a cont. Luce (or Logit) rule. Ahn-Echenique-Saito Average choice

Our main result Kalai-Megiddo and Machina-Parks Impossibility thm.: NO stochastic* choice can be cont. and PPI. Our paper: tweaking PPI avoids impossibility and characterizes the Luce model. Ahn-Echenique-Saito Average choice

Stochastic choice Stochastic choice: for each A , given prob. of choosing x out of A . Average (= stochastic*) choice: given the average (or mean) stochastic choice from A . Ahn-Echenique-Saito Average choice

Stochastic choice at McDonalds Burger Cheese burger Fries Drink prob Combo 1 1 0 2 50 .5 Combo 2 0 2 1 50 .2 Kids menu 0 1 .5 25 .3 avg. .5 .7 1.35 42.5 For ex. standard IO models (Berry-Levinsohn-Pakes). Ahn-Echenique-Saito Average choice

Why average choice? ◮ Aggregate data can be available when choice frequencies are not. ◮ Aggregate data can be more reliably estimated. ◮ Allows us to understand how utility depends on object characteristics. This is how economists use the Logit model. Ahn-Echenique-Saito Average choice

Main thm. Luce or Logit: u ( x ) ρ ( x , A ) = � y ∈ A u ( y ) Average choice: � ρ ∗ ( A ) = x ρ ( x , A ) x ∈ A Luce model iff ◮ Path independence ◮ Continuity Ahn-Echenique-Saito Average choice

Results - II Characterization of the (ordinally) linear Luce model: f ( v · x ) u ( x ) ρ ( x , A ) = y ∈ A u ( y ) = � � y ∈ A f ( v · y ) Average choice: � ρ ∗ ( A ) = x ρ ( x , A ) x ∈ A An avg. choice is cont. PI, and independent iff it is a linear Luce rule. Ahn-Echenique-Saito Average choice

Results - III Characterization of the (cardinally) affine Luce model: v · x + β u ( x ) ρ ( x , A ) = y ∈ A u ( y ) = � � y ∈ A ( v · y + β ) Average choice: � ρ ∗ ( A ) = x ρ ( x , A ) x ∈ A An avg. choice is cont. PI, independent, and calibrated iff it is an affine Luce rule. Ahn-Echenique-Saito Average choice

Small sample advantage. Luce’s IIA ρ ( x , { x , y } ) ρ ( y , { x , y } ) = ρ ( x , { x , y , z } ) ρ ( y , { x , y , z } ) ◮ Theory: ρ is observed. ◮ Reality: ρ is estimated. Ahn-Echenique-Saito Average choice

Small sample advantage. Estimating frequencies can require large samples. Luce (1959): need 1000s of observations to test his model. “It is clear that rather large sample sizes are required from each subset to obtain sensitive direct tests of axiom 1.” Average choice avoids the problem. Ahn-Echenique-Saito Average choice

Primitive ◮ Let X be a compact and convex subset of R n , with n ≥ 2. For ex. X = ∆( P ) and P set of prizes Ahn-Echenique-Saito Average choice

Primitive ◮ Let X be a compact and convex subset of R n , with n ≥ 2. For ex. X = ∆( P ) and P set of prizes ◮ Let A be the set of all finite subsets of X . Ahn-Echenique-Saito Average choice

Primitive ◮ Let X be a compact and convex subset of R n , with n ≥ 2. For ex. X = ∆( P ) and P set of prizes ◮ Let A be the set of all finite subsets of X . ◮ An average choice is a function ρ ∗ : A → X , such that, for all A ∈ A , ρ ∗ ( A ) ∈ conv A . Ahn-Echenique-Saito Average choice

Luce model A stochastic choice is a function ρ : A → ∆( X ) s.t. ρ ( A ) ∈ ∆( A ). ρ : A → ∆( X ) is a continuous Luce rule if ∃ a cont. u : X → R ++ s.t. u ( x ) ρ ( x , A ) = y ∈ A u ( y ) . � Ahn-Echenique-Saito Average choice

Luce rationalizable ρ ∗ is continuous Luce rationalizable if ∃ a cont. Luce rule ρ s.t. � ρ ∗ ( A ) = x ρ ( x , A ) . x ∈ A Ahn-Echenique-Saito Average choice

Luce rationalizable ρ ∗ is continuous Luce rationalizable if ∃ a cont. Luce rule ρ s.t. � ρ ∗ ( A ) = x ρ ( x , A ) . x ∈ A i.e. if ∃ cont. u : X → R ++ s.t. � � u ( x ) � ρ ∗ ( A ) = x . � y ∈ A u ( y ) x ∈ A Ahn-Echenique-Saito Average choice

Path independence If A ∩ B = ∅ then ρ ∗ ( A ∪ B ) = λρ ∗ ( A ) + (1 − λ ) ρ ∗ ( B ) , for some λ ∈ (0 , 1). Ahn-Echenique-Saito Average choice

Path independence Contrast with Plott P.I.: ρ ∗ ( A ∪ B ) = ρ ∗ ( { ρ ∗ ( A ) , ρ ∗ ( B ) } ) . Ahn-Echenique-Saito Average choice

Path independence Contrast with Plott P.I.: ρ ∗ ( A ∪ B ) = ρ ∗ ( { ρ ∗ ( A ) , ρ ∗ ( B ) } ) . Let ρ ∗ ( A ) = ρ ∗ ( A ′ ). Then PPI demands: ρ ∗ ( A ∪ B ) = ρ ∗ ( A ′ ∪ B ). We allow for the “path” to matter through the weights on ρ ∗ ( A ) = ρ ∗ ( A ′ ) and ρ ∗ ( B ). Ahn-Echenique-Saito Average choice

Path independence x ρ ∗ ( { x , y , z } ) z ρ ∗ ( { x , y } ) y Ahn-Echenique-Saito Average choice

Path independence x ρ ∗ ( { x , y , z } ) z ρ ∗ ( { x , y } ) y Ahn-Echenique-Saito Average choice

Path independence x ρ ∗ ( { x , y , z } ) ρ ∗ ( { x , y , z } ) z ρ ∗ ( { x , y } ) y Luce’s IIA: ρ ( x , { x , y } ) ρ ( y , { x , y } ) = ρ ( x , { x , y , z } ) ρ ( y , { x , y , z } ) Ahn-Echenique-Saito Average choice

Path independence x y z w

Path independence x y z w Ahn-Echenique-Saito Average choice

Violation of path independence x x y y ρ ∗ ( A \ { x } ) ρ ∗ ( A \ { y } ) z w z w Ahn-Echenique-Saito Average choice

Violation of path independence x y ρ ∗ ( A ) ? z w Ahn-Echenique-Saito Average choice

Continuity Let x / ∈ A . For any sequence x n in X , if x = lim n →∞ x n , then ρ ∗ ( A ∪ { x } ) = lim n →∞ ρ ∗ ( A ∪ { x n } ) . Ahn-Echenique-Saito Average choice

Theorem An average choice is continuous Luce rationalizable iff it satisfies continuity and path independence. Ahn-Echenique-Saito Average choice

Proof sketch Necessity: � � u ( x ) � ρ ∗ ( A ) = x � y ∈ A u ( y ) x ∈ A Ahn-Echenique-Saito Average choice

Proof sketch Necessity: � � u ( x ) � ρ ∗ ( A ) = x � y ∈ A u ( y ) x ∈ A   � �  ρ ∗ ( A ∪ B ) = � � u ( x ) + u ( x ) u ( x ) x + u ( x ) x x ∈ A y ∈ B x ∈ A x ∈ B Ahn-Echenique-Saito Average choice

Proof sketch Necessity: � � u ( x ) � ρ ∗ ( A ) = x � y ∈ A u ( y ) x ∈ A   � �  ρ ∗ ( A ∪ B ) = � � u ( x ) + u ( x ) u ( x ) x + u ( x ) x x ∈ A y ∈ B x ∈ A x ∈ B � � = ρ ∗ ( A ) u ( x ) + ρ ∗ ( B ) u ( x ); x ∈ A x ∈ B Ahn-Echenique-Saito Average choice

Plott path independence If conv A ∩ conv B = ∅ then ρ ∗ ( A ∪ B ) = ρ ∗ ( { ρ ∗ ( A ) , ρ ∗ ( B ) } ) . Ahn-Echenique-Saito Average choice

Plott path independence Kalai-Megiddo ( Ecma 1980) and Machina-Parks ( Ecma 1981): Theorem If ρ ∗ is continuous then it cannot satisfy Plott path independence. Ahn-Echenique-Saito Average choice

Plott path independence Proposition If an average choice is continuous Luce rationalizable, then it cannot satisfy Plott path independence. Ahn-Echenique-Saito Average choice

Plott path independence Let x , y , z ∈ X be aff. indep.. PPI ⇒ ρ ∗ ( { x , y , z } ) = ρ ∗ ( ρ ∗ ( { x , y } ) , { z } ) = u ( ρ ∗ ( { x , y } )) ρ ∗ ( { x , y } ) + u ( z ) z . u ( ρ ∗ ( { x , y } )) + u ( z ) Ahn-Echenique-Saito Average choice

Plott path independence Let x , y , z ∈ X be aff. indep.. PPI ⇒ ρ ∗ ( { x , y , z } ) = ρ ∗ ( ρ ∗ ( { x , y } ) , { z } ) = u ( ρ ∗ ( { x , y } )) ρ ∗ ( { x , y } ) + u ( z ) z . u ( ρ ∗ ( { x , y } )) + u ( z ) But ρ ∗ ( { x , y , z } ) = u ( x ) x + u ( y ) y + u ( z ) z . u ( x ) + u ( y ) + u ( z ) Ahn-Echenique-Saito Average choice

Plott path independence By aff. indep.: u ( z ) u ( z ) u ( x ) + u ( y ) + u ( z ) = u ( ρ ∗ ( { x , y } )) + u ( z ) . ⇒ u ( ρ ∗ ( { x , y } )) = u ( x ) + u ( y ) . Ahn-Echenique-Saito Average choice

Plott path independence So: � u ( x ) x + u ( y ) y � u ( x ) + u ( y ) = u ( ρ ∗ ( { x , y } )) = u . u ( x ) + u ( y ) Choose y arbitrarily close to x while satisfying aff. indep. Then u ( x ) = 2 u ( x ), a contradiction as u ( x ) > 0. Ahn-Echenique-Saito Average choice