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CHAPTER 7
Stochastic Choice
We now study the empirical content of individual rational choice when choice
is stochastic. There are two possible interpretations of this exercise.
The ﬁrst is that we lack data on individual choices. There is instead
a population of agents, and we observe the distribution of choices in the
population. For example we may know how many people purchased Italian
wine in a wine store, and how many purchased French cheese in a cheese
store, but we do not know if those who bought the French product in one store
are the same people who bought Italian in the other. The theory to be tested
is that of rational agents with stable preferences. Thus we want to know when
an observed distribution of choices is consistent with a population of rational
agents with potentially different, but stable, preferences.
The second interpretation is that we observe an individual who literally
randomizes among different alternatives. We might observe this individual
agent over time, enough to infer a stochastic rule that he uses to select an
element at random when faced with a given set of available choices.
7.1
STOCHASTIC RATIONALITY
The model of stochastic choice can be described as follows. A system of choice
probabilities is a pair (X,P), where X is a ﬁnite set of alternatives and P is
a function with domain contained in 2X\{∅} × X, where P(A,x) = PA(x) is
a non-negative number for each nonempty A ⊆X and x ∈A, and such that

x∈A PA(x) = 1. In other words, PA(x) deﬁnes a probability distribution over
the set A.
In our ﬁrst interpretation of stochastic choice, there would be an underlying
large population of agents: PA(x) is the fraction of agents who choose x when
the set A of possible alternatives is available. In the second interpretation of
stochastic choice, an individual agent’s choices really are random, and we
know that PA(x) is her probability of choosing x from A.
We assume here that all nonempty A ⊆X are possible sets of alternatives
to choose from (they are budgets in the terminology of Chapters 2 and 3). We


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96
Stochastic Choice
explain below (Remark 7.5) which results hold true when this assumption is
relaxed.
Given a ﬁnite set X, we consider the set  of all strict preferences on X. We
can identify each element of  with a one-to-one function π : X →{1,...,|X|};
such a function is a speciﬁc utility representation of the preference in question.
It is convenient in what follows to describe preferences using utility functions.
We use the following notational simpliﬁcation. Denote by π(A) the set
{π(x) : x ∈A} and write π(x) ≥π(A) to mean that π(x) ≥π(y) for all y ∈A.
Recall that the preferences in  are strict, so if x ∈A then π(x) ≥π(A) means
that x gives the highest utility in A for utility function π.
A probability distribution on  is a function ν :  →R+ with 
π∈ ν(π)
= 1. Denote by () the set of all probability distributions on . When ν ∈
() and E ⊆ then we write ν(E) for 
π∈E ν(π).
A system of choice probabilities (X,P) is rationalizable if there is ν ∈()
such that for all nonempty A ⊆X and all x ∈A,
PA(x) =

π∈
ν(π)1{π(x)≥π(A)} = ν ({π ∈ : π(x) ≥π(A)}).
Rationalizability has different interpretations, depending on how we interpret
the stochastic choice and the system (X,P). If we interpret stochastic choice
as the choices of a population of agents, and PA(x) as the fraction of agents
that choose x from A, then rationalizability means that there is a population
distribution ν over the possible preferences that agents can have. Then the
fraction of agents choosing x is the fraction of agents for whom x is best in the
set A. If, on the other hand, PA(x) is the result of individual random choices,
then rationalizability means that the individual’s preferences change. Given
a choice problem A, the individual agent draws a utility at random from ,
and chooses x with probability equal to the probability of drawing a utility for
which x is best in A. Given this interpretation, the model is often called a model
of random utility.
Before we go any further, it is worth setting down a very basic implication
of rationalizability:
Observation 7.1
If (X,P) is rationalizable, then for any x ∈X and nonempty
A,A′ ⊆X such that x ∈A ⊆A′, we have
PA(x) ≥PA′(x).
The monotonicity property in Observation 7.1 is the stochastic counterpart
of Sen’s α (discussed in Chapter 2). The property is called regularity in the
literature on stochastic choice. Regularity is clearly too weak to characterize
rationalizable systems of choice probabilities. We turn instead to two stronger
properties: the axiom of revealed stochastic preference and the non-negativity
of the Block–Marschak polynomials.
A system of choice probabilities (X,P) satisﬁes the axiom of revealed
stochastic preference if, for all sequences (x1,A1),...(xn,An), with xi ∈Ai for


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7.1 Stochastic rationality
97
i = 1,...,n, we have that
n

i=1
PAi(xi) ≤max
π∈
n

i=1
1{π(xi)≥π(Ai)}.
Note that the sequence (x1,A1),...(xn,An) may repeat the same term (xi,Ai)
many times. As a consequence, testing the satisfaction of the axiom of revealed
stochastic preference is problematic because one must verify that an inﬁnite
number of sequences have the property above. This is never an issue when
using other revealed preference axioms, for example when testing for SARP
or GARP. It turns out, however, that there is an algorithm that inﬁnitely many
steps determines if the axiom is satisﬁed. The algorithm amounts to checking
the existence of a solution of a system of linear inequalities.1
In addition to the axiom of revealed stochastic preference, a certain system
of polynomials turns out to characterize stochastic rationality. For all A ⊊X
and x ∈Ac = X \ A, deﬁne the number Kx,A by
Kx,A =
|A|

i=0
(−1)|A|−i

{C⊆A:|C|=i}
PCc(x).
The collection of all Kx,A comprise the Block–Marschak polynomials for the
system of choice probabilities (X,P).
The meaning of the Block–Marschak polynomials is made clear by
Proposition 7.3 below. They are in principle difﬁcult to interpret. Note,
however, that a simple calculation gives:
Kx,{y} = PX\{y}(x) −PX(x),
for x ̸= y. So Observation 7.1 means that Kx,{y} ≥0 is necessary for
rationalizability. It turns out that, not only Kx,{y}, but all the Block–Marschak
polynomials must be non-negative for rationalizability; and conversely that
if they are all non-negative, then the system of choice probabilities is
rationalizable.
The following result collects two theorems, one due to McFadden and
Richter, and one due to Falmagne.
Theorem 7.2
Let (X,P) be a system of choice probabilities, where X is a
ﬁnite set. The following statements are equivalent:
I) (X,P) is rationalizable.
II) (X,P) satisﬁes the axiom of revealed stochastic preference.
III) The Block–Marschak polynomials for (X,P) are non-negative.
The role of the Block-Marschak polynomials may seem obscure. The
following result gives them a natural interpretation. We present this result
1 The existence of such solutions lies at the heart of many problems studied in this book: see the
discussion in Chapter 12.


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98
Stochastic Choice
before the proof of Theorem 7.2 because it turns out to play a crucial rule
in the proof.
For C ⊆X, and any x ∈Cc, let
Mx,C = {π ∈ : π(C) > π(x) ≥π(Cc)}
be the set of all utilities in  that rank any member of C above x, and x at the
top of Cc. Proposition 7.3 says that, if (X,P) is rationalizable, then Kx,A is the
probability that all the elements in A are ranked above x, and that x is at the top
of Ac; put differently, Kx,A is the probability that A is the upper contour set of x.
Proposition 7.3
The system of choice probabilities (X,P) is rationalized by
ν ∈() iff Kx,A = ν(Mx,A) for all (x,A).
Proof. The proof is an application of a combinatorial technique called
M¨obius inversion; this speciﬁc type of M¨obius inversion is called the
inclusion–exclusion principle. The technique lets us invert variables which are
deﬁned by cumulative sums of real-valued functions deﬁned on a lattice. For
a set A and x ∈A, ν rationalizes the system of choice probabilities iff PA(x)
is the probability that the strict upper contour set of π at x is contained in Ac;
formally
PA(x) = ν({π ∈ : {y : π(y) > π(x)} ⊆Ac}).
Moreover, ν(Mx,B) is by deﬁnition the probability that the strict upper contour
set of x is exactly B. Consequently, ν rationalizes the system of choice
probabilities iff PA(x) = 
B⊆Ac ν(Mx,B), or inverting the role of A and Ac,
PAc(x) =

B⊆A
ν(Mx,B).
(7.1)
Equation (7.1) shows that PAc is deﬁned by a cumulative sum; namely, one
deﬁnes it by summing ν(Mx,B) across all B ⊆A. M¨obius inversion tells us
that, conversely, if we know the value of the left-hand side of equation (7.1)
for every A, we can recover ν(Mx,B) for all B. An explicit formula for this
inversion in this case is well known, and is called the inclusion–exclusion
principle. The application of this principle in this environment gives exactly
ν(Mx,A) = Kx,A. (See Proposition 2 of Rota (1964) and the Corollary (Principle
of Inclusion–Exclusion) on p. 345.)
7.1.1
Proof of Theorem 7.2
The proof requires the following technical lemma, which we state here without
proof. Note that the ﬁrst part of the lemma is an alternative inductive deﬁnition
of the Block–Marschak polynomials.
Lemma 7.4
For all A ⊆X:
I) Kx,A = PAc(x) −
C⊊A Kx,C if x ∈Ac;
II) 
x∈Ac Kx,A = 
x∈A Kx,A\{x}.


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7.1 Stochastic rationality
99
We start by proving the equivalence of (I) and (III). The proof uses
Proposition 7.3 in important ways. First, note that the implication (I) ⇒
(III) is immediate from Proposition 7.3. We shall prove that (III) ⇒(I) by
constructing a rationalizing ν ∈().
The structure of this proof is quite involved, so we divide it into steps.
Here is the basic structure. Our ultimate goal is to construct ν ∈() which
rationalizes P. We do this by ﬁrst recursively constructing a set function ν∗on
a collection of “cylinders” in . Let us consider π∗∈, and suppose x1 is the
π∗-maximal element, x2 is the second highest ranked element, and so forth. We
will ﬁrst deﬁne ν∗, the probability of the set of all π for which x1 is maximal.
Using this number, we then ﬁnd the probability of the set of all π for which x1
is maximal and x2 comes second. Ultimately, this will allow us to construct the
probability of π∗.
We use the notation ν∗simply because the function is not deﬁned on all
subsets of , but rather on the set of cylinders described in the previous
paragraph. But ν∗will then be deﬁned on atoms (on each singleton {π});
and will thus have a natural extension from the set of atoms to a probability
measure. This probability measure is ν. Of course, ν and ν∗will coincide on all
cylinders. Along the way we will simply need to show that ν(π) ≥0 for every
π, that the associated numbers add to one, and that we have rationalization.
Note that the cylinders referred to in the previous paragraph will be exactly the
type we need to consider in order to discuss rationalization.
Step 1: Deﬁning the cylinders.
We use the term d-sequence for a sequence (x1,...,xk) such that all its terms
are (pairwise) distinct. For any d-sequence (x1,...,xk) let
S(x1,...,xk) = {π ∈ : π(x1) > π(x2) > ··· > π(xk) > π(X \ {x1,...,xk})}.
The set S(x1,...,xk) is the cylinder associated to (x1,...,xk). Let S be the collection
of all cylinders: the subsets of  of the form S(x1,...,xk), for some d-sequence
(x1,...,xk). We shall deﬁne a function ν∗on S.
Step 2: Constructing ν∗, and verifying two important additivity properties.
We want to deﬁne ν∗on S by induction. Let A = {x1,...,xk} for some
d-sequence (x1,...,xk) and let RA be the set of d-sequences of length k with
elements in A. The key properties required of ν∗will be the following.

(x′
1,...,x′
k)∈RA
ν∗(S(x′
1,...,x′
k,x)) = Kx,{x1,...,xk}.
(7.2)
and

x∈Ac
ν∗(S(x1,...,xk,x)) = ν∗(S(x1,...,xk))
(7.3)
We proceed to deﬁne ν∗. The guiding principle in the deﬁnition of ν∗is the
interpretation of Kx,A obtained from Proposition 7.3. We seek to construct ν so
that Kx,A is the probability that A is the strict upper contour set of x. The same
guiding principle suggests properties (7.2) and (7.3).


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100
Stochastic Choice
First, for every x ∈X, let ν∗(S(x)) = Kx,∅. Second, for every x,y ∈X,
with x ̸= y, let ν∗(S(y,x)) = Kx,{y}. Suppose now that ν∗(S(y1,...,yl)) has
been deﬁned for all d-sequences (y1,...,yl) with l ≤k. Fix a d-sequence
(x1,...,xk,xk+1). Let A = {x1,...,xk} and RA be the set of all d-sequences in
A. If 
(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k)) = 0, let ν∗(S(x1,...,xk,xk+1)) = 0. Otherwise, let
ν∗(S(x1,...,xk,xk+1)) = ν∗(S(x1,...,xk))Kxk+1,{x1,...,xk}

(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k)).
(7.4)
This deﬁnes ν∗on S by induction.
We shall prove that (7.2) and (7.3) are satisﬁed. The proof is by induction on
the length of the d-sequence deﬁning A. It is easy to see by a direct calculation
that (7.2) and (7.3) are satisﬁed for all sequences of length 0 and 1, where a
sequence of length 0 is associated to A = ∅, and we deﬁne ν∗(S∅) = 1.
Suppose that (7.2) holds for all d-sequences of length l ≤k −1. Let A =
{x1,...,xk} for some d-sequence (x1,...,xk), of length k. Then we know that

(x′
1,...,x′
k)∈RA
ν∗(S(x′
1,...,x′
k)) =

x∈A
⎛
⎜⎝

(x′
1,...,x′
k−1)∈RA\{x}
ν∗(S(x′
1,...,x′
k−1,x))
⎞
⎟⎠
(7.5)
=

x∈A
Kx,A\{x}
(7.6)
=

x∈Ac
Kx,A.
(7.7)
Here, (7.5) follows from the inductive hypothesis and (7.3) (by adding over the
right-hand-side of (7.3)). Furthermore, (7.6) is a consequence of the inductive
hypothesis and (7.2); and (7.7) follows from the second part of Lemma 7.4.
Step 2a: Verifying the two additivity properties ((7.2) and (7.3)) in the case

(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k)) = 0.
There
are
two
cases
to
consider.
Suppose
ﬁrst
that

(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k)) = 0. Then (7.7) implies that for any x ∈Ac,
we have Kx,A = 0, as all Block-Marschak polynomials are non-negative (the
hypothesis). Therefore, as by deﬁnition of ν∗, 
(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k,x)) = 0,
we have 
(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k,x)) = Kx,A. This veriﬁes (7.2).
Further, Equation (7.3) is clearly satisﬁed when 
(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k))
= 0, because, by deﬁnition, ν∗(S(x1,...,xk,x)) = 0.
Step 2b: Verifying the two additivity properties ((7.2) and (7.3)) in the case

(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k)) > 0.
The second case to consider is when 
(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k)) > 0.
By deﬁnition of ν∗(S(x1,...,xk,x)), Equation (7.2) is always satisﬁed when


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7.1 Stochastic rationality
101

(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k)) > 0. To see that (7.3) also holds, note that:

x∈Ac
ν∗(S(x1,...,xk,x)) = ν∗(S(x1,...,xk))
x∈Ac Kx,{x1,...,xk}

(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k))
= ν∗(S(x1,...,xk))
x∈A Kx,A\{x}

(x′
1,...,x′
k)∈RA ν∗(S(x′
1,...,x′
k))
= ν∗(S(x1,...,xk)).
The second equality above follows from the second property in Lemma 7.4;
the third equality follows from Equation (7.6).
This ﬁnishes the proof that ν∗satisﬁes (7.2) and (7.3).
Step 3: Deﬁning ν from the ν∗(π), and verifying that they coincide on
cylinders.
Now, ν can be deﬁned on  in the following way. If we ﬁx π ∈
then S(x1,...,x|X|) = {π}, where (x1,...,x|X|) is the sequence deﬁned by π(xl) >
π(xl+1), for l = 1...,|X| −1. We write ν(π) for ν∗(S(x1,...,x|X|)) and let ν be
the obvious extension of this measure to all subsets of , namely ν(E) =

π∈E ν(π). Equation (7.3) establishes that ν∗= ν on S. Moreover, ν ≥0,
and it is easily veriﬁed that 
π∈ ν({π}) = 1, so it is a probability measure.2
Finally, the fact that ν(Mx,A) = Kx,A is the content of Equation (7.2). By
Proposition 7.3, the proof is complete.
We proceed to prove that (I) is equivalent to (II). The proof is a direct
application of a version of Farkas’ Lemma.
Note ﬁrst that (I) implies (II) because, for any sequence (x1,A1),...,(xn,An),
rationalizability implies n
i=1 PAi(xi) ≤maxν∈()
n
i=1 ν({π : π(xi) ≥
π(Ai)}). The latter expression exhibits a linear function being maximized over
a compact and convex set. Hence, there is a maximizer at an extreme point;
in this case, there is a maximizer at some δπ∗∈(), where δπ∗is the point
mass on {π∗}. Then δπ∗({π : π(xi) ≥π(Ai)}) = 1{π∗(xi)≥π∗(Ai)}, concluding this
direction.
Conversely, let W be a matrix that has one column for every π ∈ and
one row for every pair (x,A) with x ∈A. In the entry corresponding to row
(x,A) and column π we have a zero if there is y ∈A with π(y) > π(x) and a
one otherwise; so the entry is 1{π(x)≥π(A)}. The matrix W can be represented as
follows:
⎡
⎢⎢⎣
π1
···
π|X|!
...
...
···
...
(x,A)
1{π1(x)≥π1(y)∀y∈A}
···
1{π|X|(x)≥π|X|!(y)∀y∈A}
...
...
···
...
⎤
⎥⎥⎦
2 Indeed, using (7.6) we obtain that (−1)ν(π) = |X|−1
i=0
(−1)|X|−i|X|
i

= −1, by the
Binomial Theorem.


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102
Stochastic Choice
Let p be the vector with as many entries as there are pairs (x,A), whose entries
of p are arranged in the same order as the rows of W, so that PA(x) is the entry
in position (x,A) of the vector p.
Then we can represent a probability distribution ν ∈() as a vector with
one entry for every π ∈. The existence of a rationalizing ν is the same as the
existence of a solution ν to the system
p = W · ν,
such that ν ≥0 and ν · (1,...,1) = 1.
By Farkas’ Lemma (Lemma 1.14), there is no solution to the system iff there
is a vector η and a scalar θ such that
η · W + θ(1,...,1) ≤0
(7.8)
η · p + θ > 0.
(7.9)
We proceed to show ﬁrst that a violation of the axiom of revealed stochastic
preference follows from the existence of a solution to (7.8)–(7.9) in which
the entries of η are non-negative integers (in fact the two statements are
equivalent).
Let η and θ be a solution to (7.8)–(7.9) in which the entries of η are
non-negative integers. Deﬁne a sequence (x1,A1),...(xn,An) by including (in
any order) η(x,A) times the term (x,A); the sequence must have at least one term
since a solution η to (7.8)–(7.9) cannot be the null vector. Then n
i=1 PAi(xi) =

(x,A) η(x,A)PA(x) and, for any π, n
i=1 1{π(xi)≥π(Ai)} = 
(x,A) η(x,A)1{π(x)≥π(A)}.
Since η and θ solve (7.8)–(7.9) we obtain that
n

i=1
PAi(xi) + θ > 0 ≥
n

i=1
1{π(xi)≥π(Ai)} + θ,
for all π. Thus the sequence (x1,A1),...(xn,An) presents a violation of the
axiom.
Finally, we prove that if there is a solution to (7.8)–(7.9) then we can take
the entries of η to be non-negative integers. We show how to reduce the system
to a collection of strict inequalities in η, by substituting out θ. Then we can
take η to have rational entries and satisfy the system. After multiplying η by a
large enough positive integer, we can assume that its entries are integers. We
shall prove that they can be assumed to be non-negative.
To see how to substitute out θ, note that Equations (7.8) and (7.9) imply
that, for every π,

(x,A)
η(x,A)1{π(x)≥π(A)} + θ ≤0 <

(x,A)
η(x,A)PA(x) + θ.
Hence for all π,

(x,A)
η(x,A)1{π(x)≥π(A)} <

(x,A)
η(x,A)PA(x).
(7.10)


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7.2 Luce’s model
103
(In fact (7.10) being true for every π is necessary and sufﬁcient for the
existence of θ that satisﬁes Equations (7.8) and
(7.9), by setting θ =
−maxπ∈

(x,A) η(x,A)1{π(x)≥π(A)} for a solution to (7.10).)
So it is without loss of generality to assume that η is integer-valued.
We show that we can take η ≥0 in (7.10) by showing that whenever η(ˆx,ˆA) < 0
then Equation (7.10) holds for some η′ with η′
(ˆx,ˆA) = 0 and η ≤η′.
Suppose
η(ˆx,ˆA) < 0.
Note
that
for
any
π ∈,
1{π(ˆx)≥π(ˆA)} =
1 −
z∈ˆA\{ˆx} 1{π(z)≥π(ˆA)} and PˆA(ˆx) = 1 −
z∈ˆA\{ˆx} PˆA(z). Consequently
we get
η(ˆx,ˆA)1{π(ˆx)≥π(ˆA)} = η(ˆx,ˆA) + (−η(ˆx,ˆA))

z∈ˆA\{ˆx}
1{π(z)≥π(ˆA)}
and
η(ˆx,ˆA)PˆA(ˆx) = η(ˆx,ˆA) + (−η(ˆx,ˆA))

z∈ˆA\{ˆx}
PˆA(z).
So now ﬁnd η′ as follows. Add −η(ˆx,ˆA) to both sides of Equation (7.10) for all π
and make the preceding substitutions. Hence, η′ coincides with η everywhere
except that η(ˆx,ˆA) = 0 and for all z ∈ˆA\{ˆx} η′
(z,ˆA) = η(z,ˆA) −η(ˆx,ˆA), we obtain that
η′ satisﬁes (7.10) for all π ∈ while η ≤η′ and η′
(ˆx,ˆA) = 0.
Remark 7.5
The proof that (I) is equivalent to (II) in the preceding does
not rely on the ability to observe the entire system of choice probabilities. In
particular, the axiom of revealed stochastic preference is also necessary and
sufﬁcient for rationalization by ν ∈() for environments as in Chapter 2,
whereby we only observe PA for A in some set of budgets  ⊆2X\{∅}.
7.2
LUCE’S MODEL
We now analyze the special class of systems (X,P) introduced by Duncan
Luce. The model is heavily used in applied work, and it lies at the foundation of
statistical and econometric studies of discrete choice. We proceed to describe
the model, and study its relation to stochastic rationality.
A system of choice probabilities (X,P) satisﬁes Luce’s independence of
irrelevant alternatives (LIIA) if for any A, and any x,y ∈A, PA(x)P{x,y}(y) =
P{x,y}(x)PA(y). The LIIA axiom is easiest to interpret when the probabilities
involved are non-zero, so we can divide and obtain
PA(x)
PA(y) = P{x,y}(x)
P{x,y}(y).
So the LIIA axiom says that the likelihood of choosing x relative to y is
independent of what other alternatives may be available in A.
An example illustrates that LIIA may be unreasonable. Consider an agent
facing the set of alternatives {car,bus1}, who chooses each with probability
1/2. If the agent faces instead the set {car,bus1,bus2}, where the two buses only


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104
Stochastic Choice
differ in their color (they go to the same place at the same speed), then we might
expect him to choose the car or either of the two buses with probability 1/2.
LIIA, however, implies that he must choose each alternative with probability
1/3.
LIIA has a clear implication. For notational simplicity, write qx,y for P{x,y}(x).
Suppose that PA(x) > 0 for all x ∈A, and for all nonempty A. Fix an element
z ∈X. Then we can deﬁne u(x) = qx,z/qz,x.
Note that LIIA implies that 1 = 
x∈A PA(x) = PA(y)
x∈A
qx,y
qy,x . Note also
that
qx,y
qy,x
= Px,y,z(x)
Px,y,z(y) = Px,y,z(z)qx,z/qz,x
Px,y,z(z)qy,z/qz,y
= u(x)
u(y).
Then,
PA(y) =
u(y)

x∈A u(x).
(7.11)
We say that (X,P) conforms to the Luce model if there is a function u : X →
R+ such that (7.11) holds for all A and y. We can make the interpretation of u
a bit more precise. Say that x ⪰∗y if qx,y ≥1/2. If (X,P) conforms to Luce’s
model then x ⪰∗y iff u(x) ≥u(y). So ⪰∗is a preference relation represented
by u.
The numbers u(x) can be thought of as utility intensities. In previous
chapters, the utility functions have played a purely ordinal role in choice. But in
the Luce model, an alternative that has a higher utility than another alternative
has a higher probability of being chosen. So the utility function u conveys a
meaning above and beyond how it orders the different objects of choice.
The “revealed preference” problem of testing whether (X,P) conforms to
Luce’s model is very simple to solve: set u(x) = PX(x) and verify whether the
resulting u satisﬁes the deﬁnition. Instead of testing Luce’s model, we focus
on the relation between the model and stochastic rationality, as described in
Section 7.1.
Theorem 7.6
If (X,P) conforms to Luce’s model, then it is rationalizable.
Before proving Theorem 7.6 we show that Luce’s model does not exhaust
all the rationalizable systems of choice probabilities. Consider the following
example.
Let X = {a,b,c}. Suppose that the utility of a is given by a random variable
˜a; and that there are random variables ˜b and ˜c that deﬁne the utilities of b and
c. Suppose that the three random variables, ˜a, ˜b and ˜c, are independent and
distributed on {1,...,6} according to the following table:
1
2
3
4
5
6
˜a
0
0
1/2
1/2
0
0
˜b
0
0.6
0
0
0
0.4
˜c
0.4
0
0
0
0.6
0.


--- Page 11 ---
7.2 Luce’s model
105
The distributions of ˜a, ˜b, and ˜c describe a probability distribution on , as
any speciﬁcation of random utilities is equivalent to a probability distribution
on .
Then qa,b = 0.6, qb,c = 0.4 + 0.6 × 0.4 = 0.64, and qc,a = 0.6. So a ≻∗b,
b ≻∗c and c ≻∗a. Then ⪰∗cannot have any utility representation, let alone
one that allows (X,P) to conform with the Luce model.
7.2.1
Proof of Theorem 7.6
Suppose that (X,P) conforms to Luce’s model with a corresponding function
u; by a normalization we can suppose that 
x∈X u(x) = 1. We use the same
notation as in the proof of Theorem 7.2.
For any π ∈ and j, let xπ
j denote the alternative in X with the jth highest
value in π. Thus, π(xπ
1 ) = |X|, π(xπ
2 ) = |X| −1, . . . π(xπ
|X|) = 1.
The proof proceeds by constructing a probability space in which one can
calculate the probability that a sequence of random draws will correspond to a
preference π. Consider a probability space deﬁned as follows. Draw inﬁnite
sequences in X at random by drawing independently (with replacement)
elements from X such that each z is drawn with probability u(z). Let μ be
the associated probability measure on X. It should be clear that for any A ⊆X
and x ∈A, the probability that x is drawn before any other element in A is equal
to u(x)/
y∈A u(y).3
For any sequence j1 < ... < j|X| with j1 = 1, let Dj1,...,j|X|(π) denote
the event that xπ
k is drawn for the ﬁrst time at draw number jk. Then

j1<...<j|X| Dj1,...,j|X|(π) is the event C(π) that the draws will conform to π: so
C(π) = 
j1<...<j|X| Dj1,...,j|X|(π) is the event where xπ
1 is drawn ﬁrst; followed
by xπ
2 (possibly after several repeated draws of xπ
1 ); followed by xπ
3 (possibly
after several repeated draws of xπ
1 and xπ
2 ), and so on. The sets Dj1,...,j|X|(π) are
disjoint, so the probability that the draws will conform to π is
μ(C(π)) =

j1<...<j|X|
μ(Dj1,...,j|X|(π))
=

j1<...<j|X|
u(xπ
1 )j2−j1u(xπ
2 )

u(xπ
1 ) + u(xπ
2 )
j3−j2−1
· u(xπ
3 )

u(xπ
1 ) + u(xπ
2 ) + u(xπ
3 )
j4−j3−1 ···u(xπ
|X|).
The events C(π) form a partition. Deﬁne ν({π}) to be the probability of
C(π). We can explicitly calculate ν as follows. Let hk = jk+1 −jk −1 and
3 To see this: Let E be the event that x is drawn before any other element in A. Then E occurs
if either x is obtained in the ﬁrst draw, which has probability u(x), or else an element of Ac is
obtained in the ﬁrst draw (which has probability 1 −
y∈A u(y)) and then E occurs. Then the
probability q of E obeys the equation q = u(x) + (1 −
y∈A u(y))q.


--- Page 12 ---
106
Stochastic Choice
vπ
k = k
l=1 u(xπ
l ). Then,
ν(π) = μ(C(π)) =
"
z∈X
u(z)

h1≥0,...h|X|−1≥0
(vπ
1 )h1(vπ
2 )h2 ···(vπ
|X|−1)h|X|−1
=
"
z∈X
u(z)
|X|−1
"
k=1
1
1 −vπ
k
=
"
z∈X
u(z)
|X|−1
"
k=1
1
|X|
l=k+1 u(xπ
l )
=
|X|−1
"
k=1
u(xπ
k )
|X|
l=k u(xπ
l )
;
where the next-to-last equality follows by distributing the product 
z∈X u(z)
appropriately, and using that |X|
l=1 u(xπ
l ) = 1.
Finally, we have already observed that the probability of drawing x before
any other alternative in A is equal to u(x)/
y∈A u(y) = PA(x). Note that this
probability also equals 
{π:π(x)≥π(A)} μ(C(π)) = ν({π : π(x) ≥π(A)}). Hence
ν rationalizes (X,P).
7.2.2
Luce’s model and the logit model
Luce’s model can be interpreted as a random utility model in which the
“average” utility of x is some known quantity v(x), but where the actual utility
is v(x) + ε(x), where ε(x) is unknown and random.
The following calculation shows a method of determining v, and motivates
why we can think of Luce’s model as the “logit model.” Suppose that (X,P)
conforms to Luce’s model, with utility index u. Note that
qx,y =
u(x)
u(x) + u(y) =
u(x)/u(y)
u(x)/u(y) + 1 =
ev(x)−v(y)
1 + ev(x)−v(y) ,
where v(x) = log(u(x)). Then the probability of choosing x over y is qx,y =
ϕ(v(x) −v(y)). The function ϕ is the logistic distribution function. Then
PA(x) =
ev(x)

y∈A ev(y) .
(7.12)
In particular, Equation (7.12) alternatively derives from a particular random
utility model speciﬁcation, which is called the logit model. Suppose that PA(x)
equals the probability that v(y) + ε(y) ≤v(x) + ε(x) for all y ∈A, y ̸= x.
That is, the probability that ε(y) −ε(x) ≤v(x) −v(y) for all y ∈A, y ̸= x.
Let v : X →R be an arbitrary function, and let G be a cdf on R, from
which the terms ε(x) are drawn independently across x. In particular, if G is
a Gumbel distribution, then qx,y, the probability of choosing x over y, is equal
to the probability that a logistic random variable is below v(x) −v(y). The


--- Page 13 ---
7.2 Luce’s model
107
choice of a Gumbel distribution is suggested by the function ϕ being logistic.
The following proposition establishes that this speciﬁcation implies the choice
probabilities in Equation (7.12). It can therefore be viewed as a counterpart to
Theorem 7.6.
Proposition 7.7
Let v : X →R, and suppose that G(α) = exp(−exp(−α)).
Let ε be a random vector drawn from RX according to |X| independent draws
of G. Let (X,P) be deﬁned by
PA(x) = Pr(v(y) + ε(y) ≤v(x) + ε(x) for all y ∈A4).
Then (X,P) conforms to Luce’s model with index u(x) = ev(x).
Proof. Fix A and x ∈A. Let u(y) = ev(y) and δ(y) = e−ε(y). Note that the form
of G implies that the distribution function of δ(y) is Pr(δ(y) ≤α) = 1 −e−α
(the exponential distribution). Then, by the deﬁnition of (X,P),
PA(x) = Pr(u(x)/δ(x) ≥u(y)/δ(y) for all y ∈A)
=
, ∞
0
Pr(δ(y) ≥¯δu(y)/u(x) for all y ∈A)e−¯δd¯δ
=
, ∞
0
⎛
⎝"
y∈A\{x}
e−¯δu(y)/u(x)
⎞
⎠e−¯δd¯δ
=
, ∞
0
exp
⎧
⎨
⎩−¯δ
⎛
⎝1 +

y∈A\{x}
u(y)
u(x)
⎞
⎠
⎫
⎬
⎭d¯δ
=
u(x)

y∈A u(y).
Proposition 7.7 relies on a particular distribution for the random utility term ε.
One may ask if there are other distributions for ε that lead to the Luce model.
We provide a partial answer in the next result (which is due to McFadden),
where we show that if we insist on G being translation complete and the utility
index being onto, then the distribution giving rise to the Luce model is unique.
Consider now a system of choice probabilities (X,P), where we depart from
the assumptions in this chapter by allowing that X may be inﬁnite. Suppose that
PA(x) is only deﬁned for ﬁnite sets A. As before, x →PA(x) is a probability
distribution on A.
Assume that PA(x) is obtained from a random utility model, as above.
Speciﬁcally, suppose that there is a utility v deﬁned on A, and random
variables ε(x) such that x is chosen from A if v(y) + ε(y) < v(x) + ε(x) for all
y ∈A. Suppose that the random variables ε(x) are independent with identical
distribution function G on R. Say that (X,P) is generated from v : X →R
and G.
4 Here, Pr is probability calculated according to independent draws from G.


--- Page 14 ---
108
Stochastic Choice
The distribution function G is translation complete if, for any function f
such that
0 = lim
x→−∞f(x) = lim
x→+∞f(x),
0
f(x + t)dG(x) = 0 for all t implies that f = 0 a.s. The property of translation
completeness is shared by many common distribution functions.
Proposition 7.8
Suppose that (X,P) is generated from v and G, and that
(X,P) conforms to Luce’s model with utility index u(x) = ev(x). Suppose that
v : X →R is such that v(X) = R, and that G is translation complete. Then there
is α > 0 such that G(x) = exp(−α exp(−x)).
Proof. Fix w ∈R and x ∈X. Let A = {x,y1,...,yK}, and let δi be such that
v(yi) = w + δi for i = 1,...,K. Importantly, K is an arbitrary positive integer.
Note that
PA(x) =
exp(v(x))
exp(v(x)) + K
i=1 exp(w + δi)
=
,  K
"
i=1
(G(v(x) + t −w −δi))

dG(t).
Because v(X) = R, we can let all the δi →0 from below to obtain that (using
the right continuity of G)
exp(v(x))
exp(v(x)) + K exp(w) =
,
(G(v(x) + t −w))KdG(t).
On the other hand, if we choose z ∈X such that v(z) = w + log(K) then
P{x,z}(x) =
0
G(v(x) + t −w −log(K))dG(t). Now,
P{x,z}(x) =
exp(v(x))
exp(v(x)) + exp(v(z)) =
exp(v(x))
exp(v(x)) + K exp(w),
hence,
(G(v(x) + t −w))KdG(t) =
,
(G(v(x) + t −w −log(K)))dG(t).
Since, w and x were arbitrary, we obtain that for all w,
, 
(G(v(x) + t −w))K −G(v(x) + t −w −log(K))dG(t)

= 0.
Since x was arbitrary, we can choose v(x) = 0 and note that by translation
completeness, (G(t))K = G(t −log(K)) for all t. We claim that there is α > 0
such that G(t) = exp(−α exp(−t)).
First, note that G(t −log(K)) = (G(t))K implies, setting t = 0, that
G(−log(K)) = (G(0))K for any positive integer K. Likewise, for any positive
integer L, we have, by setting t = log(K/L), G(−log(L)) = (G(log(K/L)))K.
But G(−log(L)) = (G(0))L, so (G(0))L = (G(log(K/L)))K, or G(log(K/L)) =
(G(0))
L
K . G ◦log is therefore continuous on the strictly positive rational
numbers, and by deﬁnition, it is right-continuous, so that for any strictly
positive real number r, G(log(r)) = (G(0))1/r. Now, let x ∈R. We have


--- Page 15 ---
7.3 Random expected utility
109
G(x) = G(log(exp(x))) = (G(0))exp(−x). Finally, since G(x) is a cumulative
distribution function, we infer that 0 < G(0) < 1, so we may set α =
−log(G(0)) > 0, and obtain G(x) = exp(−α exp(−x)).
7.3
RANDOM EXPECTED UTILITY
The previous discussion has not sought to limit the structure of rationalizing
preferences in any way. We shall now study the random choice of lotteries, and
consider only von Neumann–Morgenstern expected utility preferences.
Let Y be a ﬁnite set of “prizes.” The objects of choice will be lotteries over
Y. A lottery is a probability distribution over Y. Let X be the set of all lotteries
over Y. An alternative x in X indicates the probability xi of the ith element
of Y.
In the present setting, a (von Neumann–Morgenstern) utility function is a
vector u ∈RY. An expected utility maximizing agent with utility function u
weakly prefers a lottery x over y iff u · x ≥u · y.
We have as before a system of choice probabilities, but where X is inﬁnite
and we deﬁne the choice only over ﬁnite nonempty sets. So a system of choice
probabilities is a pair (X,P), where PA is a probability distribution over A, for
all ﬁnite nonempty sets A. More speciﬁcally, PA is a Borel probability measure
on X with PA(A) = 1.
We say that (X,P) is expected-utility rationalizable if there is a probability
measure μ on RY such that for every ﬁnite nonempty A and every x ∈A,
PA(x) = μ

{u ∈RY : u · x ≥u · y for all y ∈A}

.
The domain of μ is required to be the σ-algebra generated by all sets of the
form {u ∈RY : u · x ≥u · y for all y ∈A}. Such sets are discussed in a bit more
detail below. Further, μ is required to be regular, in the sense that for every
possible A, with probability 1, u has a unique maximizer.
The system (X,P) is monotonic if A ⊆A′ and x ∈A, then PA(x) ≥PA′(x).
The property of monotonicity, or regularity, was discussed earlier in 7.1. By
Observation 7.1, any rationalizable system of choice probabilities must be
monotonic.
We say that the system (X,P) is linear if
PλA+(1−λ)y(λx + (1 −λ)y) = PA(x).
The notion of linearity is analogous to the property of independence in
expected-utility theory.5
The convex hull of a ﬁnite set A, denoted by conv(A) is the set of all convex
combinations of the elements in A. Write ext(A) for the extreme points of the
5 If ⪰is a preference relation over lotteries, independence says that x ⪰y iff λx + (1 −λ)z ⪰
λy + (1 −λ)z, for all z and λ ∈(0,1). In the present setup, the probability of selecting x is
unaffected by a mixture with y because any of the intended rationalizing preferences should be
unaffected by the mixture.


--- Page 16 ---
110
Stochastic Choice
convex hull of A: these are the points in the convex hull of A which cannot be
written as a convex combination of other points in the convex hull of A.
A system (X,P) is extreme if PA(ext(A)) = 1.
Finally, say that (X,P) is mixture continuous if PλA+(1−λ)A′ is continuous as
a function of λ, for all ﬁnite nonempty sets A and A′.
The next result is due to Gul and Pesendorfer.
Theorem 7.9
A system of choice probabilities is expected-utility rationaliz-
able iff it is monotone, linear, extreme, and mixture continuous.
We proceed to discuss the main ideas in the proof of Theorem 7.9. Let
N(A,x) = {u ∈RY : u · x ≥u · y
∀y ∈A}.
Thus N(A,x) is the set of utilities rationalizing the choice of x from the set A.
Note that the rationalizability of (X,P) by a probability measure μ means that
PA(x) = μ(N(A,x)).
The proof of Theorem 7.9 uses basic ideas from convex analysis in
Euclidean spaces. Part of the difﬁculty in the proof is due to the domain of
P being subsets of the simplex, not more general subsets of a Euclidean space.
So the ﬁrst step is to recast the problem using ﬁnite subsets of Rn as the domain
of P. The way to do that is to assume that Y has n+1 elements, and observe that
the (n + 1)-dimensional simplex is contained in an n-dimensional hyperplane.
This hyperplane is isomorphic to Rn. Then use linearity to extend P to the
domain of all ﬁnite subsets of Rn. We omit the details.
Suppose then that we have deﬁned PA for ﬁnite sets A ⊆Rn. Linearity can
be more simply recast in this case as stating that PA(x) = PA+{y}(x + y) for all
x ∈A.
For P to be rationalizable, we need a probability measure μ for which
PA(x) = μ(N(A,x)). So one can deﬁne a function μ on all sets of the form
N(A,x) by setting μ(N(A,x)) = PA(x). The proof proceeds by ﬁrst showing
that μ is well deﬁned, and then that μ is additive on the family of sets of the
form N(A,x).
Note that if u,v ∈N(A,x) then αu+βv ∈N(A,x) for any α,β ≥0. So N(A,x)
is a convex cone. In fact, 0 ∈N(A,x) so it is a pointed cone, and A is ﬁnite so
it is a polyhedral cone (one deﬁned by the intersection of a ﬁnite number of
halfspaces).
To see that μ is well deﬁned on sets of this kind (pointed polyhedral cones),
we need to establish two things. The ﬁrst is a basic result from convex analysis:
if K is any pointed cone, then there is always a ﬁnite set A and x ∈A such that
K = N(A,x). The result is geometrically intuitive, and illustrated in Figure 7.1.
We do not provide a formal proof of the existence of A and x with K = N(A,x),
but hope that the ﬁgure suggests one. The dotted lines in the ﬁgure are obtained
as perpendicular to the extreme rays of the cone.
The second fact we need to establish says that if K = N(A,x) = N(A′,x′),
then PA(x) = PA′(x′). To prove this fact, note ﬁrst that linearity implies that
PA(x) = PA−{x}(0). In an abuse of notation, let A and A′ denote A −{x} and


--- Page 17 ---
7.3 Random expected utility
111
Κ
x
A pointed cone Κ.
(a)
(b)
A set A = (x, x′, x′′)
such that Κ  = N (A, x).
x
x′
x′′
Fig. 7.1 Cones and decision problems.
Κ′
Κ = Κ′ ∪ Κ′′.
(a)
(b)
Κ = Κ′ ∪ Κ′′.
Κ′
x
Aλ
yλ
zλ
Κ′′
Κ′′
Fig. 7.2 Finite additivity.
A′ −{x′}. Let K = N(A,0) = N(A′,0). Then we know that pos(A) = N(K,0)
and pos(A′) = N(K,0), so A′ ⊆pos(A).6 Since A′ is ﬁnite and 0 ∈A, there is
λ > 0 such that λA′ ⊆conv(A). Then monotonicity and linearity imply that
PA′(0) = PλA′(0) ≥PA(0).
The reverse argument establishes PA′(0) ≤PA(0).
To show that μ is (ﬁnitely) additive we need to show that if K, K′, and
K′′ are pointed polyhedral cones (meaning sets of the form N(A,0)) such that
K = K′∪K′′, and the cones have disjoint interior, then μ(K) = μ(K′)+μ(K′′).7
We give a sketch of the argument based on Figure 7.2. On the left of the ﬁgure
we have a cone K that is the union of K′ and K′′, two cones with disjoint
interior. By the previous graphical argument, one can ﬁnd A′ and A′′ such that
6 The notation posA stands for the set of all positive linear combinations of elements in A.
7 The intersection of these cones must have probability zero, if we are to obtain a regular
probability measure.


--- Page 18 ---
112
Stochastic Choice
K′ = N(y,A′) and K′′ = N(z,A′′) for some points y and z. Consider x chosen on
the line connecting y and z. Let A be such that K = K′ ∪K′′ = N(x,A).
Let Aλ = (1−2λ)A+λA′ +λA′′, and let yλ and zλ be as in Figure 7.2(a). By
linearity, PAλ(yλ) = PA′(y) = μ(K′) and PAλ(zλ) = PA′′(z) = μ(K′′).
Let B be any closed ball such that the intersection of B with the extreme
points of A is {x}. Then by choosing λ small enough we can guarantee that the
only extreme points of Aλ in B are yλ and zλ. Hence the extremeness property
of P implies that
PAλ(B) = PAλ(yλ) + PAλ(zλ) = μ(K′) + μ(K′′).
By mixture continuity, PA(B) = limλ→0 PAλ(B). Since B was arbitrary, μ(K) =
PA(x) = μ(K′) + μ(K′′).
The general argument for additivity is more sophisticated. It relies on
basic results from convex analysis and the idea that N(
i βiDi,
i βiyi) is
independent of βi so long as they are positive (think of the set of all rectangles
with sides parallel to the axes).
Once μ has been established to be an additive measure of the set of all
pointed cones, an extension argument is required to show that it is a probability
measure on vectors u (utility indices).
7.4
CHAPTER REFERENCES
The two interpretations of random choice outlined at the beginning of
the chapter are very standard. Many econometric applications of these
models assume a population distribution of preferences (so-called individual
unobserved heterogeneity). There is also substantial evidence that agents
choose different alternatives when faced with the same choice problem: see,
for example, Mosteller and Nogee (1951), Papandreou (1953), and Chipman
(1960) for early discussions of this fact. Agranov and Ortoleva (2013) conduct
an experiment in which subjects seem to deliberately randomize.
Theorem 7.2 is due to McFadden and Richter (1971, 1990) and Falmagne
(1978). In particular, the equivalence between (I) and (II) is due to McFadden
and Richter, while the equivalence between (I) and (III) is due to Falmagne.
The earlier work of Block and Marschak (1960) had established that
the non-negativity of the Block-Marschak polynomials was necessary for
rationalization. Lemma 7.4 is from Falmagne (1978) (see his Theorem 3).
Proposition 7.3 appears in Barber´a and Pattanaik (1986). The idea of applying
M¨obius inversion to these problems ﬁrst appears in Colonius (1984), see also
Fiorini (2004) and Billot and Thisse (2005). Rota (1964) provides a classic
discussion of M¨obius inversion and the inclusion–exclusion principle.
Luce’s model is presented in Luce (1959). The example of the buses and the
car is due to Debreu (1960b), while a related example attributed to Savage
is presented in Luce and Suppes (1965). This example motivates a model
in which objects are ﬁrst categorized before applying a Luce-style model,


--- Page 19 ---
7.4 Chapter references
113
see Gul, Natenzon, and Pesendorfer (2014). There is plenty of experimental
evidence where subjects exhibit violations of LIIA. One of the best-known
such experiments is reported in Huber, Payne, and Puto (1982): it is called the
attraction effect.
Theorem 7.6 is due to Block and Marschak (1960). The proof presented here
follows Debreu (1960a). The example showing that there are rationalizable
systems that do not conform to Luce’s model is attributed by Block and
Marschak (1960) to Paul Halmos.
Propositions 7.7 and 7.8 appear in McFadden (1974). He attributes
Proposition 7.7 to Holman and Marley, who never published their result.
Proposition 7.8 is part of a collection of uniqueness results regarding models
of random utility: see Yellott (1977). Hausman and Wise (1978) investigate the
related random utility model where error terms are normally distributed.
Theorem 7.9 is due to Gul and Pesendorfer (2006). Our informal discussion
of the proof is largely taken from their paper.
There are other natural approaches to stochastic choice which we have
not discussed here. For example, Machina (1985) suggests that the choice
probabilities attributed to any set may be generated by maximizing a convex
preference relation on the space of lotteries over that set.
