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 2 50 .5 Combo 2 2 1 50 .2 Kids menu 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: ρ(x, A) = u(x)

• y∈A u(y)

Average choice: ρ∗(A) =

• x∈A

xρ(x, A) Luce model iff

◮ Path independence ◮ Continuity

Ahn-Echenique-Saito Average choice

Results - II

Characterization of the (ordinally) linear Luce model: ρ(x, A) = u(x)

• y∈A u(y) =

f (v · x)

• y∈A f (v · y)

Average choice: ρ∗(A) =

• x∈A

xρ(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: ρ(x, A) = u(x)

• y∈A u(y) =

v · x + β

• y∈A(v · y + β)

Average choice: ρ∗(A) =

• x∈A

xρ(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 Rn, 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 Rn, 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 Rn, 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) ∈ convA.

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. ρ(x, A) = u(x)

• 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∈A

xρ(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∈A

xρ(x, A). i.e. if ∃ cont. u : X → R++ s.t. ρ∗(A) =

• x∈A
• u(x)
• y∈A u(y)
• x.

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

y x z

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

Path independence

y x z

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

Path independence

y x z

ρ∗({x, y}) ρ∗({x, y, z}) ρ∗({x, y, z})

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

z w y x

Path independence

z w y x

Ahn-Echenique-Saito Average choice

Violation of path independence

z w y x

ρ∗(A \ {y})

z w y x

ρ∗(A \ {x}) Ahn-Echenique-Saito Average choice

Violation of path independence

z w y x ρ∗(A) ?

Ahn-Echenique-Saito Average choice

Continuity

Let x / ∈ A. For any sequence xn in X, if x = limn→∞ xn, then ρ∗(A ∪ {x}) = lim

n→∞ ρ∗(A ∪ {xn}).

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: ρ∗(A) =

• x∈A
• u(x)
• y∈A u(y)
• x

Ahn-Echenique-Saito Average choice

Proof sketch

Necessity: ρ∗(A) =

• x∈A
• u(x)
• y∈A u(y)
• x

 

x∈A

u(x) +

• y∈B

u(x)   ρ∗(A ∪ B) =

• x∈A

u(x)x +

• x∈B

u(x)x

Ahn-Echenique-Saito Average choice

Proof sketch

Necessity: ρ∗(A) =

• x∈A
• u(x)
• y∈A u(y)
• x

 

x∈A

u(x) +

• y∈B

u(x)   ρ∗(A ∪ B) =

• x∈A

u(x)x +

• x∈B

u(x)x = ρ∗(A)

• x∈A

u(x) + ρ∗(B)

• x∈B

u(x);

Ahn-Echenique-Saito Average choice

Plott path independence

If convA ∩ convB = ∅ 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(x) + u(y) + u(z) = 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) + u(y) = u(ρ∗({x, y})) = u u(x)x + u(y)y u(x) + u(y)

• .

Choose y arbitrarily close to x while satisfying aff. indep. Then u(x) = 2u(x), a contradiction as u(x) > 0.

Ahn-Echenique-Saito Average choice

Plott path independence

Proposition

No average choice satisfies Plott path independence and (our) path independence.

Ahn-Echenique-Saito Average choice

Linear Luce

ρ is a linear Luce rule if

◮ ∃ v ∈ Rn; ◮ and a monotone and cont. f : R → R++

s.t. ρ(x, A) = f (v · x)

• y∈A f (v · y).

Ahn-Echenique-Saito Average choice

Independence

u(x) = u(y) iff ∀λ, z ρ∗({λx + (1 − λ)z, λy + (1 − λ)z}) = λρ∗({x, y}) + (1 − λ)z

Ahn-Echenique-Saito Average choice

x y z

ρ∗(λx + (1 − λ)z, λy + (1 − λ)z) ρ∗(x, y) Ahn-Echenique-Saito Average choice

x y z

ρ∗(λx + (1 − λ)z, λy + (1 − λ)z) ρ∗(λ′x + (1 − λ′)z, λ′y + (1 − λ′)z) ρ∗(x, y) Ahn-Echenique-Saito Average choice

Linear Luce

Theorem

An average choice is continuous linear Luce rationalizable iff it satisfies independence, continuity and path independence.

Ahn-Echenique-Saito Average choice

Linear Luce

Let ρ∗ be cont. Luce rationalizable.

Lemma

If ρ∗ satisfies independence then u(x) = u(y) iff u(λx + (1 − λ)z) = u(λy + (1 − λ)z) ∀λ, z

Ahn-Echenique-Saito Average choice

Linear Luce

Let ρ∗ be cont. Luce rationalizable.

Lemma

If ρ∗ satisfies independence, then u(x) ≥ u(y) iff u(λx + (1 − λ)z) ≥ u(λy + (1 − λ)z) ∀λ, z.

Ahn-Echenique-Saito Average choice

Strictly affine Luce

ρ is a strictly affine Luce rule if ∃

◮ v ∈ Rn; ◮ and β ∈ R

s.t. ρ(x, A) = v · x + β

• y∈A(v · y + β).

Ahn-Echenique-Saito Average choice

Calibration

ρ∗({λx + (1 − λ)y, λy + (1 − λ)x}) = ρ∗({x, y}) + 2λ(1 − λ) (x + y − 2ρ∗({x, y}))

Ahn-Echenique-Saito Average choice

Calibration

Interpret λx + (1 − λ)y and λy + (1 − λ)x as two perfectly correlated lotteries. ρ∗({λx + (1 − λ)y, λy + (1 − λ)x}) = (λ2 + (1 − λ)2)u(x)x + u(y)y u(x) + u(y) + (2λ(1 − λ))u(y)x + u(x)y u(x) + u(y)

Ahn-Echenique-Saito Average choice

Calibration

ρ∗ y

x

• y

x

• x

y

• x

y

• y

x

• x

y

• 1 − λ

1 − λ λ λ 1 − λ λ

Ahn-Echenique-Saito Average choice

Strictly affine Luce

Theorem

An average choice is strictly affine Luce rationalizable iff it satisfies calibration, independence, continuity and path independence.

Ahn-Echenique-Saito Average choice

On continuity and Debreu’s example

Debreu’s example: ρ(t, {t, b}) ρ(b, {t, b}) = ρ(t, {t, b, b′}) ρ(b, {t, b, b′})

Ahn-Echenique-Saito Average choice

On continuity and Debreu’s example

Let ρ∗ be continuous Luce rationalizable. zn → x then: ρ∗({x, y}) = 2u(x)x + u(y)y 2u(x) + u(y) = lim

n→∞ ρ∗({x, y, zn}).

Thus ρ∗ must be discontinuous.

Ahn-Echenique-Saito Average choice

Finite sample test

◮ Theory: ρ is observed. ◮ Reality: ρ is estimated.

Ahn-Echenique-Saito Average choice

Finite sample test

Fix A. Estimate ρ(x, A) by sampling from ρ. Luce (1959): “It is clear that rather large sample sizes are required from each subset to obtain sensitive direct tests of axiom 1.”

Ahn-Echenique-Saito Average choice

Finite sample test

Fix A. Population choices from A are given by a Luce rule p(A). Observe iid sample X1, . . . , Xk of choices: Xi ∈ A for i = 1, . . . , k. pk

x = |i : Xi = x|

k .

Ahn-Echenique-Saito Average choice

Finite sample test

Two possibilities to test the Luce model:

• 1. Use Luce’s IIA. Requires:

pk

x

pk

y

, for x, y ∈ A.

• 2. Use average choice:

µk =

• x∈A

xpk

x

Ahn-Echenique-Saito Average choice

√ k pk

x

pk

y

− px py

• d

− → N

• 0, 2p2

x

p2

y

• .

Recall that px py = u(x) u(y)

Ahn-Echenique-Saito Average choice

On the other hand, √ k(µk − µ) d − → N(0, Σ), where Σ = (σl,h) and |σl,h| ≤ max{xlxk : x ∈ A}.

Ahn-Echenique-Saito Average choice

Proposition

For any M, ∃ a Luce model s.t. asymptotic variance of pk

a /pk b

relative to max{σl,h}, the largest element of Σ, is greater than M. The inefficiency in using ratios relative to means can be arbitrarily large.

Ahn-Echenique-Saito Average choice

Proof sketch.

Ahn-Echenique-Saito Average choice

Recall:

Theorem

An average choice is continuous Luce rationalizable iff it satisfies continuity and path independence.

Ahn-Echenique-Saito Average choice

Proof sketch

Necessity: ρ∗(A) =

• x∈A
• u(x)
• y∈A u(y)
• x

Ahn-Echenique-Saito Average choice

Proof sketch

Necessity: ρ∗(A) =

• x∈A
• u(x)
• y∈A u(y)
• x

 

x∈A

u(x) +

• y∈B

u(x)   ρ∗(A ∪ B) =

• x∈A

u(x)x +

• x∈B

u(x)x

Ahn-Echenique-Saito Average choice

Proof sketch

Necessity: ρ∗(A) =

• x∈A
• u(x)
• y∈A u(y)
• x

 

x∈A

u(x) +

• y∈B

u(x)   ρ∗(A ∪ B) =

• x∈A

u(x)x +

• x∈B

u(x)x = ρ∗(A)

• x∈A

u(x) + ρ∗(B)

• x∈B

u(x);

Ahn-Echenique-Saito Average choice

Proof sketch

Sufficiency: First determine ρ on A with cardinality 2 and 3.

◮ A = {x, y} ◮ A = {x, y, z} with x, y, z affinely indep.

Ahn-Echenique-Saito Average choice

Proof sketch

Sufficiency: Determine ρ(x, {x, y}) and ρ(y, {x, y}) from ρ∗({x, y}) = xρ(x, {x, y}) + yρ(y, {x, y}).

Ahn-Echenique-Saito Average choice

Proof sketch

Sufficiency: Determine ρ(x, {x, y}) and ρ(y, {x, y}) from ρ∗({x, y}) = xρ(x, {x, y}) + yρ(y, {x, y}). For affinely indep. x, y, z, ρ(x, {x, y, z}), ρ(y, {x, y, z}) and ρ(z, {x, y, z}) are also determined from ρ∗({x, y, z}).

Ahn-Echenique-Saito Average choice

Proof sketch

By path independence, ∃ θ s.t. ρ∗(A) = θz + (1 − θ)ρ∗(A \ {z}) = θz + (1 − θ)[xρ(x, {x, y}) + yρ(y, {x, y})]. x, y and z are affinely indep., ⇒ ρ(x, A), ρ(y, A) and ρ(z, A) are unique; thus ρ(x, A) = (1 − θ)ρ(x, {x, y}) and ρ(y, A) = (1 − θ)ρ(y, {x, y}).

Ahn-Echenique-Saito Average choice

Proof sketch

ρ(x, A) = (1 − θ)ρ(x, {x, y}) and ρ(y, A) = (1 − θ)ρ(y, {x, y}). Hence ρ(x, A) ρ(y, A) = ρ(x, {x, y}) ρ(y, {x, y}). Luce’s IIA!

Ahn-Echenique-Saito Average choice

Proof sketch

We can define a Luce rule with utility u. Fix x∗ ∈ X. Let u(x∗) = 1, and u(x) = ρ(x, {x, x∗}) ρ(x∗, {x, x∗}). Then, u(x) u(y) = ρ(x, {x, y}) ρ(y, {x, y}) so u defines a Luce rule. Let ¯ ρ be the implied avg. choice. We need to show ¯ ρ = ρ∗.

Ahn-Echenique-Saito Average choice

Proof sketch

The proof is by induction on |A|. We know that ¯ ρ(A) = ρ∗(A) when |A| ≤ 3.

Ahn-Echenique-Saito Average choice

Proof sketch

Crucial lemma (roughly stated): Let A ∈ A be “generic” then there is x, y ∈ A s.t. conv0({x, ρ∗(A \ {x})}) ∩ conv0({y, ρ∗(A \ {y})}) is a singleton.

Ahn-Echenique-Saito Average choice

y x z w q r

ρ∗(A \ {y}) ρ∗(A \ {x})

ρ∗(A)

Ahn-Echenique-Saito Average choice

Proof sketch

¯ ρ(A \ {z}) = ρ∗(A \ {z}) So conv0({x, ¯ ρ(A \ {x})}) = conv0({x, ρ∗(A \ {x})}) conv0({y, ¯ ρ(A \ {y})}) = conv0({y, ρ∗(A \ {y})}) conv0({x, ρ∗(A \ {x})}) ∩ conv0({y, ρ∗(A \ {y})}) being a singleton, and path independence Implies ¯ ρ(A) = ρ∗(A).

Ahn-Echenique-Saito Average choice

Conclusion

◮ Plott’s path independent choice leads to an impossibility. ◮ A simple modification of path independence, allowing the

path to affect the weights of stochastic choice, avoid the impossibility.

◮ Our modified path independence and cont. pins down a

unique choice: Luce’s rule.

Ahn-Echenique-Saito Average choice