INFORMATION AND LEARNING
IN
ECONOMIC THEORY

(IN PROGRESS AND INCOMPLETE)
ANNIE LIANG

<EMAIL_ADDRESS>
Preface

These lecture notes are targeted towards graduate students. Several chapters
are also appropriate for an advanced undergraduate audience (e.g., Chapters
1-4 and Chapter 8). Exercises labeled U indicate an advanced undergraduate
level, exercises labeled G indicate graduate level, and exercises labeled G∗
are either more involved or rely on background knowledge that is not covered
in these notes.

Parts of these lecture notes are based on slides by Drew Fudenberg and
Xiaosheng Mu. Hershdeep Chopra, Andrei Iakovlev, and Tiago Cardoso Botelho
provided very helpful proofreading. I am also grateful to Kyohei Okumura, DJ
Thornton, and Jingyi Qiu for comments and corrections. All errors are mine
alone. Please feel free to email<EMAIL_ADDRESS>with any edits
and/or suggestions.

###### Contents

  1. 1 Foundations of Information
    1. 1 Information Partitions and Knowledge
      1. 1.1 Common Knowledge
      2. 1.2 Agreeing to Disagree
      3. 1.3 The Email Game
      4. 1.4 (Common) $p$-Belief
      5. 1.5 General State Spaces
      6. 1.6 Additional Exercises
    2. 2 Bayesian Updating and Beliefs
      1. 2.1 Preliminaries
      2. 2.2 Posterior Beliefs
        1. 2.2.1 Bayes’ Rule
        2. 2.2.2 Bayes’ Plausibility
        3. 2.2.3 Application of Bayes’ Rule: Incompatibility of Fairness Definitions
      3. 2.3 Gaussian Information
        1. 2.3.1 Formulae
        2. 2.3.2 Application 1: Career Concerns
        3. 2.3.3 Application 2: Linear-Quadatic Coordination Games
        4. 2.3.4 Application 3: Data Sharing
      4. 2.4 Additional Exercises
    3. 3 Properties of Information
      1. 3.1 Definitions
        1. 3.1.1 Monotone Likelihood Ratio Property
        2. 3.1.2 Affiliation
        3. 3.1.3 First-Order Stochastic Dominance
      2. 3.2 How They are Related
      3. 3.3 When These Conditions Fail
      4. 3.4 Additional Exercises
    4. 4 Comparing Information I: The Blackwell Order
      1. 4.1 Garblings
      2. 4.2 Decision Problems
        1. 4.2.1 Uniformly Better
        2. 4.2.2 Feasible Actions
      3. 4.3 Dispersion of Posterior Beliefs
        1. 4.3.1 Mean-Preserving Spreads
        2. 4.3.2 Convex Order
      4. 4.4 Blackwell’s Theorem and Proof
      5. 4.5 Additional Exercises
    5. 5 Comparing Information II: Cost of Information
      1. 5.1 Information Theoretic Preliminaries
        1. 5.1.1 Entropy
        2. 5.1.2 Kullback-Liebler Divergence
      2. 5.2 Prior-Dependent Costs
        1. 5.2.1 Uniform Posterior Separability
        2. 5.2.2 Decision-Theoretic Foundations
        3. 5.2.3 Posterior Separability
      3. 5.3 Prior-Independent Costs
      4. 5.4 Additional Exercises
  2. 2 Learning
    1. 6 Learning
      1. 6.1 Preliminaries
      2. 6.2 Binary Example
      3. 6.3 Doob’s Consistency Theorem
      4. 6.4 Merging of Beliefs
      5. 6.5 (Expected) Disagreement
      6. 6.6 Common Learning
      7. 6.7 Additional Exercises
    2. 7 Model Uncertainty and Misspecification
      1. 7.1 Model Uncertainty
        1. 7.1.1 Motivation
        2. 7.1.2 Expanded Framework
        3. 7.1.3 Failure of Asymptotic Agreement
      2. 7.2 Misspecified Learning
        1. 7.2.1 Role of KL Divergence
        2. 7.2.2 Berk Nash Equilibrium
      3. 7.3 Additional Exercises
    3. 8 Information Design
      1. 8.1 Bayesian Persuasion
        1. 8.1.1 Example
        2. 8.1.2 Model
        3. 8.1.3 Solution and Geometric Representation
        4. 8.1.4 Back to the Example
      2. 8.2 Additional Exercises

##  Part 1 Foundations of Information

###  Chapter 1 Information Partitions and Knowledge

Section 1.1 introduces the partitional model of information and three
definitions of common knowledge. Sections 1.2 presents Aumann (1976)’s result
that agents cannot agree to disagree. Section 1.3 presents Rubinstein (1989)’s
email game. Section 1.4 defines common $p$-belief.

We assume a finite state space in Sections 1.1-1.4 to ease exposition, and
discuss in Section 1.5 how these results extend more generally.

####  1.1 Common Knowledge

An unknown state $\omega$ takes values in the finite set $\Omega$. Agents
$i\in\mathcal{I}$ share a _common prior_ that the state $\omega$ is
distributed according $P\in\Delta(\Omega)$. Each agent $i$’s _information
partition_ $\Pi_{i}$ is a partition of $\Omega$, with the property that for
any realization of the state $\omega$, agent $i$ is informed that the state
belongs to $\Pi_{i}(\omega)$.

######  Assumption 1.

Every partition element has strictly positive probability under the prior;
that is, $P(\Pi_{i}(\omega))>0$ for every agent $i\in\mathcal{I}$ and state
$\omega\in\Omega$.

######  Definition 1.1 (Knowledge).

The set of states at which agent $i$ _knows_ the event $A\subseteq\Omega$ to
be true is

$K_{i}(A)=\\{\omega:\Pi_{i}(\omega)\subseteq A\\}.$

No agent can think that an event is true if it is not; that is,
$K_{i}(A)\subseteq A$ for every agent $i$ and event $A$.

######  Definition 1.2 (Mutual Knowledge).

The set of states at which the event $A\subseteq\Omega$ is _mutual knowledge_
is

$K(A)=\bigcap_{i\in\mathcal{I}}\\{\omega:\Pi_{i}(\omega)\subseteq A\\}$

i.e., all agents know $A$ to be true.

######  Example 1.1.

Suppose the set of states is $\Omega=\\{1,2,3,4,5,6\\}$, and there are two
agents with information partitions
$\Pi_{1}=\\{\\{1,2,3\\},\\{4,5\\},\\{6\\}\\}$ and
$\Pi_{2}=\\{\\{1,2\\},\\{3,4\\},\\{5\\},\\{6\\}\\}$. Let $A=\\{3,4,5,6\\}$.
Then, the set of states at which agent 1 knows $A$ to be true is
$K_{1}(A)=\\{4,5,6\\}$, the set of states at which agent 2 knows $A$ to be
true is $K_{2}(A)=\\{3,4,5,6\\}$, and the set of states at which both agents
know $A$ to be true is $K(A)=\\{4,5,6\\}$.

The knowledge operators $K_{i}$ and $K$ can be applied to events that
themselves represent knowledge or mutual knowledge of a state, thus building
up higher-order knowledge (agent 1 knows that agent 2 knows that…).

######  Exercise 1.1 (U).

Suppose there are two agents indexed to $i=1,2$.

  * (a)

Prove that $K_{1}(K_{2}(A)),K_{2}(K_{1}(A))\subseteq K(A)$ for every event
$A\subseteq\Omega$.

  * (b)

Provide an example in which $K(A)\nsubseteq K_{1}(K_{2}(A))$, demonstrating
that even if both players know an event to be true, either can fail to know
that the other knows it.

######  Exercise 1.2 (U).

Prove that $\neg K_{i}(\neg K_{i}(A))=K_{i}(A)$ for every event
$A\subseteq\Omega$ (where $\neg A$ denotes the complement of $A$.)

An implicit assumption is made that all agents know the state space $\Omega$
and the information partitions $(\Pi_{i})_{i\in\mathcal{I}}$. This assumption
is less strong than it might initially seem, since we can always redefine
states and expand the state space to accommodate uncertainty about other
players’ partitions, as in the following example.

######  Example 1.2.

Let $\Omega=\\{1,2\\}$, $\mathcal{I}=\\{1,2\\}$, and
$\Pi_{1}=\Pi_{2}=\\{\\{1\\},\\{2\\}\\}$. Suppose we want to model the
situation where agent 1 has uncertainty over whether agent 2’s information is
the complete partition $\Pi_{2}$ or the trivial partition
$\Pi_{2}^{\prime}=\\{\\{1,2\\}\\}$. One way to do this is to expand the state
space: Define
$\widetilde{\Omega}=\Omega\times\\{c,t\\}=\\{\\{1,c\\},\\{1,t\\},\\{2,c\\},\\{2,t\\}\\}$
and revise the agents’ information partitions to be

$\displaystyle\widetilde{\Pi}_{1}$
$\displaystyle=\\{\\{(1,c),(1,t)\\},\\{(2,c),(2,t)\\}\\}$
$\displaystyle\widetilde{\Pi}_{2}$
$\displaystyle=\\{\\{(1,c)\\},\\{(2,c)\\},\\{(1,t),(2,t)\\}\\}$

Then, for example, at state $(1,c)$ both agents know $\omega=1$ to be true,
but agent 1 does not know whether agent 2 knows it.

The event $A$ is _common knowledge_ at state $\omega$ if both agents know it
to be true, know the other to know it to be true, ad infinitum. There are at
least three equivalent ways to define this.

###### The First Definition.

The most direct approach is to recursively define higher-order levels of
knowledge.

######  Definition 1.3 (Common Knowledge, Definition 1).

For any event $A\subseteq\Omega$, define
$\mathscr{A}^{1}:=\bigcap_{i\in\mathcal{I}}K_{i}(A)$ to be the set of states
at which every agent knows $A$, and recursively define

$\mathscr{A}^{k}:=\bigcap_{i\in\mathcal{I}}K_{i}(\mathscr{A}^{k-1})$

for each $k\geq 2$. (For example, $\mathscr{A}^{2}$ is the set of states at
which every agent knows that every agent knows $A$.) The set of states at
which $A$ is _common knowledge_ is $\mathscr{A}^{\infty}:=\bigcap_{n\geq
1}\mathscr{A}^{n}$.

######  Exercise 1.3 (G).

Consider the informational environment of Example 1.1. Find the smallest value
of $k$ with the property that
$\mathscr{A}^{k^{\prime}}=\mathscr{A}^{k^{\prime}+1}$ for all $k^{\prime}\geq
k$.

###### The Second Definition.

Alternatively, we can define common knowledge using the meet of the players’
information partitions. If two partitions $\Pi$ and $\Pi^{\prime}$ satisfy

$\Pi^{\prime}(\omega)\subseteq\Pi(\omega)\quad\forall\omega\in\Omega$

then we say that $\Pi$ is a _coarsening_ of $\Pi^{\prime}$ (corresponding to
weakly less information at every state), and $\Pi^{\prime}$ is a _refinement_
of $\Pi$ (corresponding to weakly more information at every state). If
$\Pi^{\prime}$ a coarsening of both partitions $\Pi_{1}$ and $\Pi_{2}$, then
it is a _common coarsening_ of $\Pi_{1},\Pi_{2}$.

######  Definition 1.4.

Let $\Pi_{1}\wedge\Pi_{2}$ denote the finest common coarsening of
$\Pi_{1},\Pi_{2}$, i.e., the common coarsening of these partitions that is
moreover a refinement of every other common coarsening of $\Pi_{1}$ and
$\Pi_{2}$.

######  Definition 1.5.

For any sequence of information partitions
$(\Pi_{1},\dots,\Pi_{|\mathcal{I}|})$, let
$\mathscr{P}_{2}=\Pi_{1}\wedge\Pi_{2}$, and for each $k>2$, recursively define
$\mathscr{P}_{k}=\mathscr{P}_{k-1}\wedge\Pi_{k}$. The _meet_ of
$(\Pi_{1},\dots,\Pi_{|\mathcal{I}|})$ is
$\bigwedge_{i\in\mathcal{I}}\Pi_{i}\equiv\mathscr{P}_{|\mathcal{I}|}$.

######  Exercise 1.4 (G).

Prove that for any sequence of information partitions
$(\Pi_{1},\dots,\Pi_{|\mathcal{I}|})$, the meet $\mathscr{P}^{n}$ is invariant
to permutations of players indices.

######  Example 1.3.

Consider Example 1.1. Stack the two information partitions on top of one
another, and suppose an ant is placed on one of the states in an agent’s
partition (see Figure 1.1).

Figure 1.1: $\Pi_{1}\wedge\Pi_{2}(\omega)$ includes all states that an ant
seeded at $\omega$ can reach.

The ant’s movements obey two laws: The ant can move from side to side within
an information partition element, and it can jump across the players’
information partitions along the same state. The ant’s full range of motion
when seeded at any state $\omega$ then recovers the member of the meet that
includes that state. So in this example, we have
$\Pi_{1}\wedge\Pi_{2}=\\{\\{1,2,3,4,5\\},\\{6\\}\\}.$

######  Exercise 1.5 (G).

Formalize the statements in the example above by proving that two points
$x^{\prime}$ and $x^{\prime\prime}$ belong to the same element of
$\bigwedge_{i\in\mathcal{I}}\Pi_{i}$ if and only if there is a sequence
$(x_{0},x_{1},x_{2},…,x_{n},x_{n+1})$, with $x_{0}=x^{\prime}$ and
$x_{n+1}=x^{\prime\prime}$, such that for every $0\leq m\leq n$, $x_{m}$ and
$x_{m+1}$ belongs to the same element of $\Pi_{i}$ for some $i\in\mathcal{I}$.

######  Definition 1.6 (Common Knowledge, Definition 2).

An event $A\subseteq\Omega$ is common knowledge at state $\omega\in\Omega$ if
$\bigwedge_{i\in\mathcal{I}}\Pi_{i}(\omega)\subseteq A$.

######  Remark 1.1.

It is immediate that the set $\Omega$ is common knowledge at every
$\omega\in\Omega$.

###### The Third Definition.

Our final definition of common knowledge starts from the definition of an
_evident_ event which, upon its occurrence, is known to all agents.

######  Definition 1.7 (Evident Events).

The event $A\subseteq\Omega$ is _evident_ (or _public_) if $A\subseteq K(A)$.

######  Definition 1.8 (Common Knowledge, Definition 3).

The event $A\subseteq\Omega$ is common knowledge at $\omega$ if and only if
there is an evident event $E$ such that $\omega\in E$ and $E\subseteq K(A)$.

######  Exercise 1.6 (G).

Let $\mathcal{I}=\\{1,2\\}$. Prove that an event $E\subseteq\Omega$ is evident
if and only if it is a union of elements of the meet $\Pi_{1}\wedge\Pi_{2}$.

These three definitions of common knowledge are equivalent (see for example
Monderer and Samet (1989)).

####  1.2 Agreeing to Disagree

Often we are interested not only in agents’ knowledge (which depends only on
the agents’ information partitions) but also in agents’ posterior beliefs
(which depend additionally on the prior $P$). At any state $\omega$ and for
any event $A\subseteq\Omega$, agent $i$’s posterior probability of event $A$
is pinned down by Bayes’ rule (see Section 2.2):

$P(A\mid\Pi_{i}(\omega))=\frac{P(A\cap\Pi_{i}(\omega))}{P(\Pi_{i}(\omega))}$

Our assumption that every partition element has strictly positive prior
probability ensures that this expression is well-defined.

One event of interest is the one in which a player’s posterior belief takes on
a particular value. Fixing an event $A$ and a number $p\in[0,1]$, define
$A_{p}=\\{\omega:P(A\mid\Pi_{i}(\omega))=p\\}$ to be the set of states at
which player $i$ assigns posterior probability $p$ to the event $A$ being
true. If player 1 announces that he assigns probability $p$ to $A$, then all
other agents know that the state must belong to $A_{p}$.

######  Example 1.4.

Consider the informational environment of Example 1.1 with a uniform prior on
$\Omega$, and define the event $A=\\{2,3\\}$. Agent 2 has four partition
elements, $\\{1,2\\}$, $\\{3,4\\}$, $\\{5\\}$, $\\{6\\}$, and assigns to $A$ a
posterior probability of $1/2$, $1/2$, $0$, and $0$ (respectively) on these
partition elements. So the set of states $A_{1/2}$ at which agent 2 assigns
probability $1/2$ to event $A$ being true, is $A_{1/2}=\\{1,2,3,4\\}$.

The following theorem shows that whenever players’ posterior beliefs about an
event are common knowledge (e.g., because players have publicly announced
these beliefs), then these posterior beliefs must be identical. So
disagreement cannot be sustained whenever players’ beliefs are commonly known.

######  Theorem 1.1 (Aumann (1976)).

Suppose $\mathcal{I}=\\{1,2\\}$. Fix any state $\omega\in\Omega$ and event
$A\subseteq\Omega$. If it is common knowledge at $\omega$ that agent 1 assigns
(posterior) probability $q_{1}$ to event $A$, while agent 2 assigns
(posterior) probability $q_{2}$ to the same event, then $q_{1}=q_{2}$.

The result is stated for two agents, but the proof below directly extends for
an arbitrary finite number of players.

Proof. Let $\bf{P}$ be the element of $\Pi_{1}\wedge\Pi_{2}$ that contains
$\omega$. Then we can write $\textbf{P}=\cup_{k}\mathcal{P}^{k}$ where
$\mathcal{P}^{k}$ are elements of $\Pi_{1}$. Since the event $\\{$agent 1’s
posterior belief is $q_{1}\\}$ is common knowledge at $\omega$, agent 1 must
assign probability $q_{1}$ to event $A$ at every partition element
$\mathcal{P}^{k}$. So $q_{1}=P(A\cap\mathcal{P}^{k})/P(\mathcal{P}^{k})$ for
each $k$. This implies $q_{1}\cdot
P(\mathcal{P}^{k})=P(A\cap\mathcal{P}^{k})$. Summing over each of player 1’s
partition elements, we have
$q_{1}\sum_{k}P(\mathcal{P}^{k})=\sum_{k}P(A\cap\mathcal{P}^{k})$. Thus
$q_{1}\cdot P(\textbf{P})=P(A\cap\textbf{P})$. But repeating the same line of
logic for player 2, we obtain $q_{2}\cdot P(\textbf{P})=P(A\cap\textbf{P}).$
So it must be that $q_{1}=q_{2}$.

The following example explains why it is important that players’ posterior
beliefs are common knowledge and not simply mutual knowledge.

######  Example 1.5.

Let $\Omega=\\{1,2,3,4\\}$ with a uniform prior, and define
$\Pi_{1}=\\{\\{1,2\\},\\{3,4\\}\\}$ and $\Pi_{2}=\\{\\{1,2,3\\},\\{4\\}\\}$.
Choose $A=\\{1,4\\}$ and $\omega=2.$ Then agent 1 assigns posterior
probability $1/2$ to $A$ while agent 2 assigns posterior probability $1/3$.
Moreover, each agent knows one another’s posterior probability. But Theorem
1.1 is not violated: agent $2$ does not know that agent $1$ knows his
posterior probability to be $\frac{1}{3}$, so posterior beliefs are mutual
knowledge but not common knowledge.

The starting hypothesis of Theorem 1.1—that individuals have common knowledge
of one anothers’ beliefs—is strong. Geanakoplos and Polemarchakis (1982) show
that the same result obtains under a more realistic process: Communication of
posterior beliefs converges to common knowledge of identical posterior
beliefs, where this convergence occurs in fewer than $n_{1}+n_{2}$ steps with
$n_{i}$ the size of agent $i$’s partition.

######  Example 1.6.

Let $\Omega=\\{1,2,3,4,5,6,7,8,9\\}$ with all states equally likely. There are
two agents, Bob and Carly, with information partitions

$\Pi_{B}=\\{\\{1,2,3\\},\\{4,5,6\\},\\{7,8,9\\}\\}$

and

$\Pi_{C}=\\{\\{1,2,3,4\\},\\{5,6,7,8\\},\\{9\\}\\}.$

Suppose the true state is $\omega=1$, and the agents repeatedly communicate
their beliefs about the event $A=\\{3,4\\}$.

Round 1: Bob’s information partition reveals to him that the state belongs to
$\\{1,2,3\\}$, so he assigns posterior probability $1/3$ to the event $A$.
Carly’s information partition reveals to her that the state belongs to
$\\{1,2,3,4\\}$, so she assigns posterior probability $1/2$ to the event $A$.
The two agents announce these posterior beliefs.

Round 2: That Bob assigns probability $1/3$ to $A$ reveals to Carly that Bob
was either informed that the state belongs to $\\{1,2,3\\}$ or informed that
the state belongs to $\\{4,5,6\\}$. But Carly already knew in round 1 that
these were the two partition elements that Bob might have been informed of
(since she knew the state to be either 1, 2, 3, or 4), and so there is no
information for her in this announcement. That Carly assigns probability $1/2$
to $A$ reveals to Bob that agent 2 knows $\\{1,2,3,4\\}$, but again Bob knew
this in round 1. So both agents’ posterior beliefs are unchanged. They again
announce $1/3$ and $1/2$.

And now something interesting happens. That Bob sticks to his original belief
of $1/3$ tells Carly that Bob must have observed $\\{1,2,3\\}$. If instead Bob
observed $\\{4,5,6\\}$, then upon hearing that Carly’s belief was $1/2$ (and
thus learning that Carly observed $\\{1,2,3,4\\}$), Bob would have deduced
that the state was 4 with certainty, and hence revised his posterior belief of
$A$ to 1. So Carly now knows that the state is in $\\{1,2,3\\}$ and shares
Bob’s posterior belief, $\frac{1}{3}$. The two agents’ beliefs have converged,
and it is straightforward to show that these beliefs will not move after
subsequent communication.

Although agents’ beliefs must converge, the belief that they converge to need
not be the belief that agents would have held had they pooled their
information:

######  Example 1.7.

Let $\Omega=\\{1,2,3,4\\}$ with each state equally likely. Agents’ partitions
are given by $\Pi_{1}=\\{\\{1,2\\},\\{3,4\\}\\}$ and
$\Pi_{2}=\\{\\{1,3\\},\\{2,4\\}\\}$. Let $\omega=1$ and $A=\\{1,4\\}$. Both
posteriors are $1/2$ and the process of belief revision converges in one step.
But had agents shared their information, they would have learned that
$\omega\in\\{1,2\\}$ and also $\omega\in\\{1,3\\}$, leading to a (common)
posterior belief that $A$ is true with probability 1.

####  1.3 The Email Game

Common knowledge assumptions appear frequently in analyses of strategic
environments; for example, payoffs are assumed to be common knowledge in any
complete-information game. Do strategic predictions made under an assumption
of common knowledge approximately hold when we relax the assumption of common
knowledge? Rubinstein (1989)’s email game shows that for one formalism of what
“almost common knowledge” means, the answer is no: Strategic predictions can
change discontinuously when we move from common knowledge to almost common
knowledge.

In this game, two agents each choose an action from $\\{A,B\\}$. There are two
possible payoff matrices indexed to $\\{a,b\\}$ (depicted below with $a$ on
the left and $b$ on the right). The agents share a common prior that assigns
probability $1-p>\frac{1}{2}$ to the matrix indexed to $a$.

$\begin{array}[]{ccc}&A&B\\\ A&M,M&0,-L\\\ B&-L,0&0,0\end{array}\hskip
50.00008pt\begin{array}[]{ccc}&A&B\\\ A&0,0&0,-L\\\ B&-L,0&M,M\end{array}$

We assume throughout that $L>M>0$. Thus $(A,A)$ yields higher payoffs for both
agents when the payoff parameter is $a$ while $(B,B)$ yields higher payoffs
when the payoff parameter is $b$. The action $A$ is “safe,” in that it never
yields a negative payoff.

Communication Protocol. Both players have an automated email server, which is
the only means by which the players can communicate. Agent $1$ is informed of
the payoff parameter. If (and only if) the parameter is $b$, agent $1$’s email
server automatically sends an email to agent $2$ announcing that the parameter
is $b$. All emails are independently lost with probability $\varepsilon>0$, so
the agents’ email servers are set up to automatically send back confirmations
that emails have been received, and confirmations of confirmations, etc. Each
agent $i$’s type is the number of emails that agent $i$’s computer sends,
which is privately known to agent $i$.

In the special case $T_{1}=T_{2}=\infty$, there is common knowledge that the
parameter is $b$. But if for example $T_{1}=2$, then agent $1$ knows the
parameter is $b$, and agent 1 knows that agent $2$ knows that the parameter is
$b$, but agent $1$ does not know that agent $2$ knows that agent $1$ knows
that agent 2 knows that the parameter is $b$. In general, so long as $T_{1}$
and $T_{2}$ are finite, then higher-order knowledge of parameter $b$ must
break down at some stage.

######  Remark 1.2.

In the partitional framework of Section 1.1, we would model this information
environment as follows: The state space is

$\Omega=\left\\{(a,0,0),(b,1,0),(b,1,1),(b,2,1),(b,2,2),\dots\right\\}$

and agents’ information partitions are given by

$\displaystyle\Pi_{1}$
$\displaystyle=\left\\{\\{(a,0,0)\\},\\{(b,1,0),(b,1,1)\\},\\{(b,2,1),(b,2,2)\\},\dots\right\\}$
$\displaystyle\Pi_{2}$
$\displaystyle=\left\\{\\{(a,0,0),(b,1,0)\\},\\{(b,1,1),(b,2,1)\\},\dots\right\\}$

where, for example, $T_{1}=0$ reveals to player 1 the partition element
$\\{(a,0,0)\\}$, while $T_{2}=0$ reveals to player 2 the partition element
$\\{(a,0,0),(b,1,0)\\}$.

######  Proposition 1.

There is a unique Bayesian Nash equilibrium in which agent $1$ plays $A$ when
the payoff parameter is $a$. In this equilibrium, both agents play $A$
independently of the number of messages sent.

Proof. Let $s_{i}:T_{i}\rightarrow\Delta(\\{A,B\\})$ denote player $i$’s
equilibrium strategy. By assumption, $s_{1}(0)=A$. We will show that also
$s_{2}(0)=A$. Agent 2 of type $T_{2}=0$ knows that either agent $1$’s first
message was never sent (the state is $(a,0,0)$), or agent $1$’s first message
was sent but lost (the state is $(b,1,0)$). Unconditionally, the probabilities
of these states are $(1-p)$ and $p\varepsilon$. Conditional on $T_{2}=0$,
agent $2$ assigns a posterior probability of $\frac{1-p}{1-p+p\varepsilon}$ to
$(a,0,0)$, a posterior probability of $\frac{p\varepsilon}{1-p+p\varepsilon}$
to $(b,1,0)$ and zero probability to all other states.

So agent 2’s expected payoff from playing $A$ is at least

$M\cdot\left(\frac{1-p}{1-p+p\varepsilon}\right)+0\cdot\left(\frac{p\varepsilon}{1-p+p\varepsilon}\right)$
(1.1)

while agent 2’s expected payoff from playing $B$ is no more than

$(-L)\cdot\left(\frac{1-p}{1-p+p\varepsilon}\right)+M\cdot\left(\frac{p\varepsilon}{1-p+p\varepsilon}\right).$
(1.2)

Since $1-p>\frac{1}{2}$ and $L>M$ by assumption, (1.1) strictly exceeds (1.2),
and so agent 2’s strategy must satisfy $s_{2}(0)=A$.

Now suppose $s_{i}(T_{i})=A$ for $i=1,2$ and all $T_{i}<t$. We’ll argue that
$s_{1}(t)=s_{2}(t)=A$. Suppose first that agent 1’s computer sends $t$ emails
exactly, i.e., $T_{1}=t$. Since agent 1’s computer did not send a $(t+1)$-th
email, it must either be that agent $1$’s $t$-th message was lost (the state
is $(b,t,t-1)$), or that agent $1$’s $t$-th message was received, but its
confirmation was lost (the state is $(b,t,t)$). Agent 2’s posterior belief
conditional on $T_{1}=t$ then assigns probability
$z:=\frac{\varepsilon}{\varepsilon+(1-\varepsilon)\varepsilon}>\frac{1}{2}$ to
$(b,t,t-1)$ and probability $1-z$ to $(b,t,t)$. So the expected payoff to
playing $B$ is $z(-L)+(1-z)(M)<0$, while the payoff to playing $A$ is zero. We
conclude that agent 1’s strategy must satisfy $s_{1}(t)=A$, with nearly
identical reasoning yielding $s_{2}(t)=A$.

This result shows a sharp discontinuity in strategic predictions at common
knowledge. That is, $(B,B)$ is an equilibrium when agents have common
knowledge of the payoff parameter $b$, but fails to be an equilibrium when
players have knowledge of $b$ to arbitrarily high (finite) orders.

Whether this result is surprising depends on how natural we consider the
relaxation of common knowledge to be. Rubinstein (1989) argues that “high
$T_{i}$” is intuitively like common knowledge. Another view is that these are
substantially different, since for arbitrarily small but strictly positive
$\varepsilon$ the informational model is the one described in Remark 1.2, but
for $\varepsilon=0$ (corresponding to common knowledge of the state) the set
of states with positive ex-ante probability is
$\Omega=\\{(a,0,0),(b,\infty,\infty)\\}$ and the agents’ information
partitions are complete. So there is a discontinuity in the informational
environments as $\varepsilon\rightarrow 0$, and in this sense small
$\varepsilon$ may be quite unlike $\varepsilon=0$.

####  1.4 (Common) $p$-Belief

We now consider an alternative approach to formalizing almost common
knowledge, which defines common “almost-knowledge” in contrast to the above
“almost-common” knowledge.

######  Definition 1.9.

For any $p\in[0,1]$, say that agent $i$ _$p$ -believes_ $A$ at $\omega$ if
$P(A\mid\Pi_{i}(\omega))\geq p$. The set of states at which agent $i$
$p$-believes $A$ is

$\mathcal{B}_{i}^{p}(A)=\\{\omega:P(A\mid\Pi_{i}(\omega))\geq p\\}.$

######  Remark 1.3.

Is the case $p=1$ equivalent to knowledge? Suppose $\Omega=\\{1,2,3\\}$ and
the prior is $P=(0,1/2,1/2)$. Agent 1’s partition is $\\{\\{1,2\\},\\{3\\}\\}$
while agent 2’s partition is $\\{\\{1\\},\\{2\\},\\{3\\}\\}$. The state is
$\omega=2$. Then according to Definition 1.1, agent 2 knows $\\{2\\}$ but
agent 1 does not, while according to Definition 1.9, both agents have
$1$-belief of $\\{2\\}$. Whether knowledge and 1-belief represent distinct
modes of understanding is an interesting philosophical question, but we will
not have more to say on it here.

The following construction of _common $p$-belief_, due to Monderer and Samet
(1989), is parallel to Definition 1.3 for common knowledge.

######  Definition 1.10 (Common $p$-Belief).

For any $p\in[0,1]$ and event $A\subseteq\Omega$, define
$\mathscr{A}^{1}=\bigcap_{i\in\mathcal{I}}\mathcal{B}_{i}^{p}(A)$ to be the
set of states at which every agent $p$-believes $A$ to be true, and
recursively define
$\mathscr{A}^{k}=\bigcap_{i\in\mathcal{I}}\mathcal{B}^{p}_{i}(\mathscr{A}^{k-1})$
for every $k\geq 2$. Then $A$ is common $p$-belief at the set of states
$\mathscr{A}^{\infty}=\cap_{n\geq 1}\mathscr{A}^{n}$.

We can also define common $p$-belief by generalizing the definition of an
evident event (Definition 1.7) to events that are evident $p$-belief.

######  Definition 1.11.

For any $p\in[0,1]$, the event $A\subseteq\Omega$ is _evident $p$-belief_ if
$A\subseteq\bigcap_{i\in\mathcal{I}}\mathcal{B}_{i}^{p}(A).$

######  Definition 1.12.

For any $p\in[0,1]$, the event $A\subseteq\Omega$ is common $p$-belief at
$\omega$ if there exists an evident $p$-belief event $E$ such that

$\omega\in E\subseteq\bigcap_{i\in\mathcal{I}}\mathcal{B}_{i}^{p}(A).$

Definitions 1.10 and 1.12 are introduced in Monderer and Samet (1989) and
shown to be equivalent.

######  Exercise 1.7 (G∗).

Consider the email game of Rubinstein (1989). Let $P$ denote the common prior
on $\Omega$ (as defined in Remark 1.2), and define $\mathscr{C}^{p}$ to be the
event that agents have common $p$-belief in parameter $b$. For each
$\varepsilon\geq 0$, let

$\overline{p}(\varepsilon)=\sup_{p\in[0,1]}\\{p:P(\mathscr{C}^{p})>0\\}$

be the supremum of the set of values of $p$ such that $\mathscr{C}^{p}$ has
positive ex-ante probability. Is $\overline{p}(0)$ equal to the limit of
$\overline{p}(\varepsilon)$ as $\varepsilon\rightarrow 0$? Discuss your
answer.

####  1.5 General State Spaces

To show that the preceding insights do not require assumption of a finite
state space, we now briefly discuss two generalizations of these ideas. In
each case, we begin with a probability space $(\Omega,\Sigma,P)$ where
$\Omega$ is a set of states endowed with $\sigma$-algebra $\Sigma$, and
$P:\Sigma\rightarrow[0,1]$ is a probability measure.

###### The first generalization.

Let each information partition $\Pi_{i}$ be a partition of $\Omega$, where we
require that each partition element is $\Sigma$-measurable and has strictly
positive measure under $P$ (see e.g., Monderer and Samet (1989)). All of the
above definitions and proofs generalize as stated.

###### The second generalization.

Alternatively, we might model each agent $i$’s information as a sub
$\sigma$-algebra of $\Sigma$, denoted by $\Pi_{i}$.111That is, $\Pi_{i}$ is a
$\sigma$-algebra and $\Pi_{i}\subseteq\Sigma$. One foundation for this
approach (which we will examine in detail in subsequent chapters) is that each
agent $i$ privately observes a random variable
$X_{i}:\Omega\rightarrow\mathbb{R}$ that is measurable with respect to
$\Sigma$. In this case, each agent $i$’s $\sigma$-algebra is $\sigma(X_{i})$,
the $\sigma$-algebra generated by $X_{i}$, which is indeed coarser than
$\Sigma$.

The definition of knowledge can be extended as follows.

######  Definition 1.13.

Agent $i$ _knows_ the event $A\in\Sigma$ to be true at $\omega$ if there
exists some $B\in\Pi_{i}$ such that $\omega\in B\subseteq A$.

Common knowledge cannot in general be iteratively constructed (à la Definition
1.3) using this definition of $K_{i}$, since the set of states at which agent
$i$ knows $A$ to be true may not be $\Sigma$-measurable. Nevertheless, similar
to Definition 1.6, we can define $\bigwedge_{i\in\mathcal{I}}\Pi_{i}$ to be
the finest common coarsening of the $\sigma$-algebras $\Pi_{1},\dots,\Pi_{n}$,
and say that an event $A$ is common knowledge at $\omega$ if there is an
element $A$ of $\bigwedge_{i\in\mathcal{I}}\Pi_{i}$ such that $\omega\in A$.
We can also generalize Definition 1.8 as follows:

######  Definition 1.14.

The event $A\in\Sigma$ is _evident_ if $A\in\Pi_{i}$ for every
$i\in\mathcal{I}$, i.e., $A$ belongs to every agent’s $\sigma$-algebra.

######  Definition 1.15.

The event $A\in\Sigma$ is common knowledge at state $\omega$ if there is an
evident event $E$ such that $\omega\in E$ and $E\subseteq A$.

Theorem 1.1 can also be generalized, although the previous proof does not
extend (for example, there is no longer guaranteed to be a unique element of
$\Pi_{1}\wedge\Pi_{2}$ that contains $\omega$).

######  Proposition 2.

Let $X\in\mathcal{L}^{1}(\Omega,\Sigma,P)$, and define
$Y=\mathbb{E}(X\mid\Pi_{1})$, $Z=\mathbb{E}(X\mid\Pi_{2})$. If it is common
knowledge that $Y=y$ and $Z=z$ at a state $\omega$ with strictly positive
probability, then it must be that $y=z$.

Proof. If it is common knowledge that $Y=y$ and $Z=z$, there must exist an
event $E\in\Pi_{1}\cap\Pi_{2}$, where $Y$ takes the constant value $y$ on $E$,
and $Z$ takes the constant value $z$ on $E$. Let $\mathbbm{1}_{E}$ denote the
indicator variable that takes value 1 on $E$. Then

$\displaystyle y\cdot P(E)$ $\displaystyle=\mathbb{E}(Y\mathbbm{1}_{E})$
$\displaystyle=\mathbb{E}(X\mathbbm{1}_{E})$
$\displaystyle=\mathbb{E}(Z\mathbbm{1}_{E})=z\cdot P(E)$

using in the second and third equalities that $Y$ and $Z$ are conditional
expectations of $X$. Since $P(E)>0$ (by assumption that $\omega\in E$ has
strictly positive probability), it follows that $y=z$ as desired.

This result is in fact more general than Theorem 1.1, nesting the previous
result as a special case when we choose $X$ to be an indicator function on
some set.

######  Exercise 1.8 (G∗).

Generalize Proposition 2 by demonstrating that the conclusion still holds if
we assume that there is a measurable set of states $B\subseteq\Omega$ with
strictly positive probability, where at every $\omega\in B$ it is common
knowledge that $Y=y$ and $Z=z$.

####  1.6 Additional Exercises

######  Exercise 1.9 (G).

Two spies in an underground organization are stationed at remote locations.
Each spy privately observes whether the coast is clear at their location. The
spies share a common prior that the coast is clear at each location
independently with probability $1/2$.

Communication protocol. The spies communicate by email with a third party
electronic server at their home base. If and only if the coast is clear at a
spy’s location, that spy’s computer will automatically send a message to the
home base with the information that the coast is clear.

If the home base electronic server receives information from both spies
indicating that the coast is clear, then it will automatically send a message
to both spies indicating that it has received both messages. (Otherwise, it
will send no messages.) As these are dangerous times, each message has only a
$1-\varepsilon$ chance of being received (again independent). If either spy
receives a message from the home base, that spy will send a reply to the home
base confirming receipt. The reply is lost with probability $\varepsilon$,
independently of everything that’s happened before. So on and so forth.
Everything stated above is common knowledge.

Each spy observes the number of messages he has sent, and chooses an action in
$\\{A,B\\}$. If the coast is clear at both locations, then payoffs are given
by the right matrix below, and otherwise payoffs are given by the left matrix
below.

$\begin{array}[]{ccc}&A&B\\\ A&M,M&0,-L\\\ B&-L,0&0,0\end{array}\hskip
50.00008pt\begin{array}[]{ccc}&A&B\\\ A&0,0&0,-L\\\ B&-L,0&M,M\end{array}$

The payoff parameters satisfy $L>3M>0$.

  * (a)

Prove the following analogue of Rubinstein (1989)’s result: Let
$T_{1}=\mathbb{Z}_{+}$ and $T_{2}=\mathbb{Z}_{+}$ denote the two players’ type
spaces. There is a unique pure-strategy equilibrium in which both players
choose $A$ when the coast is not clear at their location, i.e.
$s_{1}(0)=s_{2}(0)=A$. In this equilibrium, players choose $A$ for any number
of messages sent, i.e. $s_{i}(t)=A$ for both players $i$ and all $t\in T_{i}$.

  * (b)

Suppose instead that $L=2$ while $M=1$, and demonstrate that the result in
Part (a) no longer holds by finding some $\varepsilon>0$ and a pair of
strategies $(s_{1},s_{2})$ that constitute a pure-strategy Bayesian Nash
equilibrium, where $s_{1}(0)=s_{2}(0)=A$ and $s_{i}(t)=B$ for some player $i$
and type $t\in T_{i}$.

######  Exercise 1.10 (G∗).

Let $X\in\mathcal{L}^{1}(\Omega,\Sigma,P)$, and define
$Y=\mathbb{E}(X\mid\Pi_{1})$, $Z=\mathbb{E}(X\mid\Pi_{2})$. Prove that if it
is common knowledge that $Y\in A$ and $Z\in B$ at a state $\omega$ with
strictly positive probability, then it must be that $A\cap B\neq\emptyset$.

###  Chapter 2 Bayesian Updating and Beliefs

Section 2.1 introduces the canonical Bayesian framework and the definition of
a signal. Section 2.2 reviews Bayes’ rule and key properties of Bayesian
posteriors. Section 2.3 provides closed-form expressions for posterior beliefs
in the special case of Bayesian updating to normal signals, with applications.

####  2.1 Preliminaries

There is a set of _parameters_ $\Theta$ endowed with a $\sigma$-algebra
$\Sigma$. An agent has a _prior_ $p\in\Delta(\Theta)$, where $\Delta(\Theta)$
denotes the set of $\Sigma$-measurable probability measures on $\Theta$. The
prior describes the agent’s belief at an “ex-ante” stage in the absence of any
information, where what is ex-ante is understood in the context of a specific
model.

The focus of this chapter is the object that we will call an _information
structure_ , _experiment_ , or a _signal_ , which can be formalized in either
of several ways:

  * (a)

We can define the signal to be a mapping $\sigma:\Theta\rightarrow\Delta(S)$
from the set of parameters to distributions over a set of signal realizations
$S$. See for example de Oliveira (2019).

  * (b)

We can define a signal to be an $(S,\mathcal{S})$-valued random variable $X$
on an underlying probability space $(\Omega,\Sigma,P)$, where
$\Omega=\Theta\times E$ for some set $E$. For example, we might define the
signal to be $X=\theta+\varepsilon$ for an $E$-valued noise term $\varepsilon$
that is independent of $\theta$, as we do in Section 2.3.

  * (c)

We can define a signal $S$ to be a finite partition of
$\Omega=\Theta\times[0,1]$, whose elements are non-empty and measurable with
respect to the Lebesgue $\sigma$-algebra on $\Omega$. Conditional on parameter
$\theta$, the probability of observing $s\in S$ is the Lebesgue measure of
$\\{x\in[0,1]\mid(\theta,x)\in s\\}$. See for example Frankel and Kamenica
(2019).

######  Remark 2.1.

It is straightforward to see that the first two formalisms nest one another
when all the relevant sets are finite. Suppose we are given a prior
$p\in\Delta(\Theta)$ and a signal $\sigma:\Theta\rightarrow\Delta(S)$. Define
the expanded state space to be $\Omega=\Theta\times S$ and let
$P(\theta,s)=p(\theta)\sigma(s\mid\theta)$. Then the random variable
$X:\Omega\rightarrow S$ satisfying $X(\theta,s)=s$ is equivalent to $\sigma$
in the sense that posterior beliefs about $\theta$ are the same whether we
condition on the realization of $X$ or the realization of $\sigma(\theta)$. In
the other direction, if we start with a random variable $X:\Theta\times
E\rightarrow S$ and a distribution $P\in\Delta(\Theta\times E)$, then we can
define $\sigma:\Theta\rightarrow\Delta(S)$ to satisfy
$\sigma(s\mid\theta)=P(X^{-1}(s)\mid\theta)$. The formalism in (c) is a
special case of (b), where $E=[0,1]$, the random variable $X:\Omega\rightarrow
S$ maps each $\omega$ into the partition element of $S$ to which it belongs,
and the probability distribution $P$ is the Lebesgue measure.

Example families of signals include:

######  Example 2.1 (Aumann (1976)’s Partitional Information Structures).

For each agent $i$, let $\Pi_{i}$ be a finite partition of $\Theta$ into
measurable elements of strictly positive measure. Index these partition
elements to $S=\\{1,\dots,n\\}$ where $n$ is the size of $\Pi_{i}$. Then let
$\sigma$ map each $\theta$ with probability 1 to the index of the partition
element to which $\theta$ belongs.

######  Example 2.2 (Finite Information Structures).

Suppose $\lvert\Theta\rvert,\lvert S\rvert<\infty$. Then we can express
$\sigma$ as a $|\Theta|\times|S|$ matrix where (1) all entries are
nonnegative, and (2) all rows sum to 1. For example, suppose a drug is either
good (g) or bad (b). The drug is administered to a patient who is either cured
(C) or not (N). The patient is cured with probability $3/4$ if the drug is
good and with probability $1/4$ if the drug is bad. Then $\Theta=\\{g,b\\}$
and $S=\\{C,N\\}$ and the information structure is

$\begin{array}[]{ccc}&C&N\\\\[-2.84526pt] g&3/4&1/4\\\ b&1/4&3/4\end{array}$

with each row depicting the probability over the signal realizations in the
associated state.

######  Example 2.3 (Gaussian Information).

The signal is $X=\theta+\varepsilon$, where
$\theta\sim\mathcal{N}(\mu_{\theta},\sigma_{\theta}^{2})$,
$\varepsilon\sim\mathcal{N}\left(0,\sigma_{\varepsilon}^{2}\right)$, and
$\theta\perp\\!\\!\\!\perp\varepsilon$.

####  2.2 Posterior Beliefs

#####  2.2.1 Bayes’ Rule

The agent updates his prior to the realization of the signal using Bayes’
rule.

######  Definition 2.1 (Bayes’ Rule, Finite Case).

Suppose $|\Theta|<\infty$. Fix any distribution $P\in\Delta(\Theta)$ and any
events $A,B\subseteq\Theta$ where $P(A),P(B)>0$. Then

$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}.$ (2.1)

######  Remark 2.2.

Rather than memorizing this formula, it is easier to remember that by the law
of total probability, we can rewrite $P(A\cap B)$ as $P(A\mid B)P(B)$ or as
$P(B\mid A)P(A)$, so

$P(A\mid B)P(B)=P(B\mid A)P(A).$

Dividing through by $P(B)$ yields (2.1).

######  Remark 2.3.

Applying (2.1) twice for the pairs of events $(A,E)$ and $(B,E)$, we have

$\frac{P(A\mid E)}{P(B\mid E)}=\frac{P(E\mid A)}{P(E\mid
B)}\cdot\frac{P(A)}{P(B)}$

so the relative conditional probabilities of events $A$ and $B$ is determined
by their relative probabilities under the prior, $\frac{P(A)}{P(B)}$, and the
_likelihood ratio_ of $E$ under $A$ and $B$, $\frac{P(E\mid A)}{P(E\mid B)}$.
_Base-rate neglect_ is the tendency to falsely equate $\frac{P(A\mid
E)}{P(B\mid E)}$ with $\frac{P(E\mid A)}{P(E\mid B)}$, neglecting the prior
distribution. This can lead to compelling but inaccurate statistical
conclusions.

For example, suppose $\Omega=\\{p,n\\}$, where $\omega=p$ indicates that an
individual is positive for a medical condition while $\omega=n$ indicates that
the individual is negative, with $P(\omega=p)=0.01$. Let $X\in\\{+,-\\}$ be
the outcome of a test where $P(X=+\mid\omega=p)=0.95$ and
$P(X=+\mid\omega=n)=0.05$. Since the likelihood of observing $X=+$ is much
higher when the individual has the condition than when he does not—indeed, the
likelihood ratio is $\frac{P(X=+\mid\omega=p)}{P(X=+\mid\omega=n)}=19$—it is
tempting to conclude from a positive test result that the individual has the
condition. But correctly applying Bayes’ rule yields that
$\frac{P(\omega=p\mid X=+)}{P(\omega=n\mid X=+)}<1$; that is, even with a
positive test it is more likely that the individual is negative for the
condition.

A useful rewriting of Bayes’ rule is

$P(\theta\mid
X=x)=\frac{P(X=x\mid\theta)P(\theta)}{\sum_{\theta^{\prime}\in\Theta}P(X=x\mid\theta^{\prime})P(\theta^{\prime})}\quad\forall\theta\in\Theta$
(2.2)

where the conditional distribution $P(\cdot\mid X=x)$ is precisely the agent’s
posterior belief upon observing $X=x$.

######  Example 2.4.

A drug is either effective ($\theta=A$) or not ($\theta=B$), where the prior
probability that the drug is effective is $p\in(0,1)$. The signal is

$\begin{array}[]{ccc}&a&b\\\ A&q&1-q\\\ B&1-q&q\end{array}$

for some $q\in(0,1)$. Then upon observing $a$, the agent assigns to $\theta=A$
a posterior probability of

$\frac{pq}{pq+(1-p)(1-q)}=\frac{1}{1+\frac{1-p}{p}\left(\frac{1-q}{q}\right)}$

which exceeds the prior belief of $p$ if and only if $q\geq\frac{1}{2}$.

More generally, when $\theta$ and $X$ are (not necessarily finite-valued)
random variables with densities $f_{\theta}$ and $f_{X}$ and conditional
densities $f_{\theta\mid X=x}$ and $f_{X\mid\theta=t}$, then the posterior
belief given $X=x$ is

$f_{\theta\mid
X=x}(t)=\frac{f_{X\mid\theta=t}(x)f_{\theta}(t)}{\int_{\theta^{\prime}\in\Theta}f_{X\mid\theta=t^{\prime}}(x)f_{\theta}(t^{\prime})dt^{\prime}}\quad\forall
t\in\Theta.$ (2.3)

Somewhat more generally, we may suppose that the joint distribution of
$(\theta,X)$ is such that for every realization $x$ of $X$, there is a
(measurable) function $q_{x}$ satisfying

$q_{x}(A)=\mathbb{E}(\mathbbm{1}_{A}\mid X=x)\quad\mbox{ for all events
$A\subseteq\Theta$}$

Then this $q_{x}$ is the posterior belief.

#####  2.2.2 Bayes’ Plausibility

Outside of special cases (such as the one we will cover in Section 2.3),
posterior beliefs often cannot be expressed in closed-form. Nevertheless,
there are certain properties they must satisfy. One important property is that
beliefs are a martingale, i.e., the expected posterior is equal to the prior.
Intuitively, if you expect to change your mind given more information, then
why haven’t you done so already?

######  Fact 2.1 (Beliefs are a martingale.).

Let $p\in\Delta(\Theta)$ denote the agent’s prior belief, and choose any event
$A$. Then the posterior probability assigned to this event conditional on the
realization of random variable $X$ is $\mathbb{E}(\mathbbm{1}_{A}\mid X)$. By
the law of iterated expectations,

$\mathbb{E}(\mathbb{E}(\mathbbm{1}_{A}\mid X))=\mathbb{E}(\mathbbm{1}_{A})$

so the expected posterior probability of $A$ is equal to the prior probability
of $A$. Since the event $A$ was arbitrarily chosen, we can conclude that the
expected posterior belief is equal to the prior belief. (In the case of a
finite state space $\Theta$, choosing $A=\\{\theta\\}$ yields
$\mathbb{E}(p(\theta\mid X))=p(\theta)$ for every $\theta$.)

Since any signal $X$ induces a distribution $\tau\in\Delta(\Delta(\Theta))$
over posterior beliefs, Fact 2.1 implies that this distribution must average
to the prior.

######  Definition 2.2.

Fixing a prior $p\in\Delta(\Theta)$, say that a distribution of posteriors
$\tau$ is _Bayes plausible_ if

$\int_{\Delta(\Theta)}qd\tau(q)=p$

i.e. the expected posterior is equal to the prior. We’ll use

$\mathcal{T}(p)\equiv\left\\{\tau\in\Delta(\Delta(\Theta))\,\mid\,\int
qd\tau(q)=p\right\\}$

to denote the set of Bayes plausible posterior distributions given prior $p$.

######  Exercise 2.1 (U).

The state space is $\Theta=\\{\theta_{1},\theta_{2}\\}$ and the prior is
$(\mu,1-\mu)$ for some $\mu\in[0,1]$. The signal structure is

$\begin{array}[]{cccc}&s_{1}&s_{2}\\\ \theta_{1}&p&1-p\\\
\theta_{2}&q&1-q\end{array}$

where $p,q\in[0,1]$. What is the distribution over posterior beliefs induced
by this signal structure? Verify that the expected posterior belief is equal
to the prior belief.

Not only are we guaranteed that any signal induces a Bayes-plausible
distribution over posterior beliefs, but also any Bayes-plausible distribution
over posterior beliefs can be induced by some signal.

######  Definition 2.3.

For any signal $X\sim P_{X}$, let $\tau_{X}\in\Delta(\Delta(\Theta))$ satisfy
$\tau_{X}(q)=P_{X}(\\{x:q_{x}=q\\})$. Say that $\tau\in\Delta(\Delta(\Theta))$
is _induced by $X$_ if $\tau=\tau_{X}$.

######  Proposition 3.

Suppose the prior $p$ belongs to the interior of the set $\Delta(\Theta)$.
Then every Bayes-plausible distribution $\tau\in\mathcal{T}(p)$ is induced by
some signal $X$.

The proof (demonstrated in Kamenica and Gentzkow (2011) and Shmaya and Yariv
(2016) among others) proceeds by construction. For any distribution $\tau$,
index the distinct posterior beliefs in the support of $\tau$ to be
$\\{q_{x}\\}_{x\in\mathcal{X}}$, where $\mathcal{X}$ may not be finite. Then
define $\sigma:\Theta\rightarrow\Delta(\mathcal{X})$ to satisfy

$\sigma(x\mid\theta)=\frac{q_{x}(\theta)\tau(q_{x})}{p(\theta)}$ (2.4)

We have constructed a signal $\sigma$ whose realizations $x$ are identified
with posterior beliefs $q_{x}$, where the conditional distribution over signal
realizations mimics Bayes’ rule $p(x\mid\theta)=\frac{p(\theta\mid
x)p(x)}{p(\theta)}$, setting $q_{x}(\theta)=p(\theta\mid x)$ and
$\tau(q_{x})=p(x)$ . This is a valid signal structure since

$\displaystyle\int_{\mathcal{X}}\sigma(x\mid\theta)dx=\int_{\mathcal{X}}\frac{q_{x}(\theta)\tau(q_{x})}{p(\theta)}dx=1$

by (2.4) and the definition of Bayes-plausibility. Moreover,

$\displaystyle\frac{\sigma(x\mid\theta)p(\theta)}{\int_{\Theta}\sigma(x\mid\theta)p(\theta)d\theta}$
$\displaystyle=\frac{\sigma(x\mid\theta)p(\theta)}{\tau(q_{x})\int_{\Theta}q_{x}(\theta)d\theta}=\frac{\sigma(x\mid\theta)p(\theta)}{\tau(q_{x})}=q_{x}(\theta)$

so $q_{x}(\cdot)$ is precisely the posterior belief when updating to the
signal $\sigma$.

Thus the probability that the posterior belief is $q_{x}$ is exactly the
probability that the realization of the constructed signal $\sigma$ is $x$, so
$\tau$ is induced by $\sigma$ as desired.

######  Exercise 2.2 (U).

Suppose the prior is over $\Theta=\\{\theta_{1},\theta_{2}\\}$ is $(1/3,2/3)$.
Provide a set $S$ and a signal structure $\sigma:\Theta\rightarrow\Delta(S)$
that induces the belief (0,1) with probability 1/3, and the belief (1/2,1/2)
with probability 2/3.

Together, Fact 2.1 and Proposition 3 imply:

######  Corollary 2.1.

Fix any prior belief $p\in Int(\Delta(\Theta))$. Then a distribution over
posteriors $\tau\in\Delta(\Delta(\Theta))$ is induced by some signal if and
only if it is Bayes-plausible, i.e., $\tau\in\mathcal{T}(p)$.

#####  2.2.3 Application of Bayes’ Rule: Incompatibility of Fairness
Definitions

Here we take a detour to demonstrate the power of Bayes’ rule. Individuals in
a population are each described by a covariate vector $C\in\mathcal{C}$, a
group membership $G\in\\{g_{1},g_{2}\\}$, and a type $\theta\in\\{0,1\\}$. For
example, we might interpret $\theta$ as the individual’s creditworthiness
(whether the individual would pay back a loan if approved), $G$ as a
demographic group, and $C$ as the individual’s credit history. Across
individuals, the random vector $(C,G,\theta)$ is distributed according to $P$,
and we use $p_{g}=P(\theta=1\mid G=g)$ for the base rate of $\theta=1$ in each
group $g$. A _scoring rule_ is any mapping $S:\mathcal{C}\rightarrow\\{0,1\\}$
that predicts the type given the covariate vector.

######  Definition 2.4 (Equality of False Positives).

A scoring rule $S$ has equal false positive rates if

$P(S=1\mid\theta=0,G=g_{1})=P(S=1\mid\theta=0,G=g_{2})$

In words, the probability of being incorrectly assessed to pay back the loan
is independent of group membership. Equivalently: $S\perp\\!\\!\\!\perp
G\mid\theta=0$, i.e., the score is conditionally independent of group
membership given type $\theta=0$.

######  Definition 2.5 (Equality of False Negatives).

A scoring rule $S$ has equal false negative rates if

$P(S=0\mid\theta=1,G=g_{1})=P(S=0\mid\theta=1,G=g_{2})$

In words, the probability of being incorrectly assessed to not pay back the
loan is independent of group membership. Equivalently: $S\perp\\!\\!\\!\perp
G\mid\theta=1$, i.e., the score is conditionally independent of group
membership given type $\theta=1$.

######  Definition 2.6 (Calibrated).

A score $S$ is calibrated if for each $s\in\\{0,1\\}$,

$P(\theta=1\mid S=s,G=g_{1})=P(\theta=1\mid S=s,G=g_{2})$

In words, among those assessed to pay back the loan (or, to not pay back the
loan), the probability of paying back the loan is independent of group
membership. Equivalently: $\theta\perp\\!\\!\\!\perp G\mid S$, i.e., type is
independent of group membership conditional on the score.

The following impossibility result demonstrates that (outside of edge cases)
these fairness criteria cannot be simultaneously satisfied.

######  Proposition 4 (Kleinberg, Mullainathan and Raghavan
(2017),Chouldechova (2017)).

Suppose $p_{g_{1}}\neq p_{g_{2}}$. Then no scoring rule $S$ can simultaneously
satisfy calibration, equal false positive rates, and equal false negative
rates.

Proof. Choose either group $g$ and define $FP_{g}=P(S=1\mid\theta=0,G=g)$,
$FN_{g}=P(S=0\mid\theta=1,G=g)$, and $PPV_{g}=P(\theta=1\mid S=1,G=g)$. We’ll
show that these quantities are related by the following identity:

$FP_{g}=\frac{p_{g}}{1-p_{g}}\times\frac{1-PPV_{g}}{PPV_{g}}\times(1-FN_{g}).$
(2.5)

To simplify notation, let $Q$ denote the joint distribution over $(C,G,S)$
after conditioning on $G=g$. Then, expanding (2.5), we have

$Q(S=1\mid\theta=0)=\frac{Q(\theta=1)}{Q(\theta=0)}\times\frac{Q(\theta=0\mid
S=1)}{Q(\theta=1\mid S=1)}\times Q(S=1\mid\theta=1)$

Multiplying both sides by $Q(\theta=0)$ and applying Bayes’ rule,

$\displaystyle Q(S=1,\theta=0)$ $\displaystyle=\frac{Q(\theta=0\mid
S=1)}{Q(\theta=1\mid S=1)}\times Q(S=1,\theta=1)$

Thus, (2.5) is equivalent to

$\frac{Q(S=1,\theta=0)}{Q(S=1,\theta=1)}=\frac{Q(\theta=0\mid
S=1)}{Q(\theta=1\mid S=1)}$ (2.6)

Again using Bayes’ rule, the RHS can be rewritten

$\frac{Q(\theta=0\mid S=1)}{Q(\theta=1\mid
S=1)}=\frac{Q(\theta=0,S=1)/Q(S=1)}{Q(\theta=1,S=1)/Q(S=1)}=\frac{Q(S=1,\theta=0)}{Q(S=1,\theta=1)}$

so (2.6) is equivalent to

$\frac{Q(S=1,\theta=0)}{Q(S=1,\theta=1)}=\frac{Q(S=1,\theta=0)}{Q(S=1,\theta=1)}$

and is therefore trivially true.

The identity (2.5) holds for both groups $g\in\\{g_{1},g_{2}\\}$. So if
$FP_{g_{1}}=FP_{g_{2}}$ (as required by equality of false positive rates),
$FN_{g_{1}}=FN_{g_{2}}$ (as required by equality of false negative rates), and
also $PPV_{g_{1}}=PPV_{g_{2}}$ (as required by calibration), it must also hold
that $p_{g_{1}}=p_{g_{2}}$.

####  2.3 Gaussian Information

Gaussian information environments are unusually tractable, since the posterior
belief can be expressed in closed-form. We’ll cover the main formulae for
Bayesian updating in these environments, and show how these can be used to
derive results in three applications.

#####  2.3.1 Formulae

We’ll start with the simplest case. The state is
$\theta\sim\mathcal{N}(\mu,\sigma_{\theta}^{2})$ and the signal is
$X=\theta+\varepsilon$, where
$\varepsilon\sim\mathcal{N}(0,\sigma_{\varepsilon}^{2})$,
$\theta\perp\\!\\!\\!\perp\varepsilon$, and
$\sigma_{\theta}^{2},\sigma_{\varepsilon}^{2}>0$. Then:

######  Fact 2.2.

The agent’s posterior belief about $\theta$ conditional on signal realization
$X=x$ is normally distributed with mean

$\mathbb{E}(\theta\mid
X=x)=\left(\frac{\sigma_{\varepsilon}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}\right)\mu+\left(\frac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}\right)x$

and variance

$Var(\theta\mid
X=x)=\frac{\sigma_{\theta}^{2}\sigma_{\varepsilon}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}.$

A key property worth remembering is that the posterior mean is a convex
combination of the prior mean $\mu$ and the signal realization $x$, where the
weights are proportional to prior precision and signal precision.
Additionally, while the posterior mean depends on the signal realization, the
posterior variance is a constant.

Fact 2.2 is also sometimes written as:

$(\theta\mid
X=x)\sim\mathcal{N}\left(\left(\frac{\tau_{\theta}}{\tau_{\theta}+\tau_{\varepsilon}}\right)\mu+\left(\frac{\tau_{\varepsilon}}{\tau_{\theta}+\tau_{\varepsilon}}\right)x\,,\,\frac{1}{\tau_{\theta}+\tau_{\varepsilon}}\right)$

where $\tau_{\theta}=1/\sigma_{\theta}^{2}$ is the precision of the prior
belief and $\tau_{\varepsilon}=1/\sigma_{\varepsilon}^{2}$ is the precision of
the signal. This restatement makes it apparent that the posterior precision is
the sum of the prior precision and signal precision.

We can use Fact 2.2 to derive the distribution of the posterior mean.

######  Exercise 2.3 (U).

Let $\theta\sim\mathcal{N}(\mu,1)$ and define two signals

$\displaystyle Y_{1}$ $\displaystyle=\theta+\varepsilon_{1}$ $\displaystyle
Y_{2}$ $\displaystyle=\theta+\varepsilon_{2}$

where $\theta$, $\varepsilon_{1}$, and $\varepsilon_{2}$ are all independent
of one another, and $\varepsilon_{1}\sim\mathcal{N}(0,1)$ while
$\varepsilon_{2}\sim\mathcal{N}(0,2)$.

  * (a)

Solve for the conditional distributions $\theta\mid Y_{1}=y$ and $\theta\mid
Y_{2}=y$.

  * (b)

What are the values of $(\mu,y)$ for which it the case that
$\mathbb{E}(\theta\mid Y_{1}=y)>\mathbb{E}(\theta\mid Y_{2}=y)?$ Provide
intuition for the condition you derive.

######  Exercise 2.4 (U).

Suppose we write the posterior belief as
$\mathcal{N}(\hat{\mu},\hat{\sigma}^{2})$, where $\hat{\mu}$ is a random
variable that depends on the realization of the signal $X$. Prove that

$\hat{\mu}\sim\mathcal{N}\left(\mu,\sigma_{\theta}^{2}-\frac{\sigma_{\theta}^{2}\sigma_{\varepsilon}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}\right),$

i.e. the expected posterior mean is the prior mean, and the variance of the
posterior mean is equal to the prior variance ($\sigma_{\theta}^{2}$), reduced
by the posterior variance,
$\left(\frac{\sigma_{\theta}^{2}\sigma_{\varepsilon}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}\right)$.

This characterization implies that the more informative the signal is, the
more variable the posterior mean is.

######  Remark 2.4.

More generally (i.e., for $\theta$ and $X$ that are not necessarily normally-
distributed), the law of total variance says that

$\operatorname*{\text{Var}}(\mathbb{E}[\theta\mid
X])=\operatorname*{\text{Var}}(\theta)-\mathbb{E}[\operatorname*{\text{Var}}(\theta\mid
X)]$

so the variance of the posterior mean is equal to the difference of the prior
variance and the expectation of the posterior variance.

Similar closed-forms exist for multivariate Gaussian states and signals.
Suppose $Z$ is a $1\times K$ vector distributed according to
$\mathcal{N}(\mu,\Sigma)$, where $\Sigma$ has full rank. Partition the vector
as follows:

$\left(\begin{array}[]{c}Z_{1}\\\
Z_{2}\end{array}\right)\sim\mathcal{N}\left(\left(\begin{array}[]{c}\mu_{1}\\\
\mu_{2}\end{array}\right),\left(\begin{array}[]{cc}\Sigma_{11}&\Sigma_{12}\\\
\Sigma_{21}&\Sigma_{22}\end{array}\right)\right)$

######  Fact 2.3.

The conditional distribution of $Z_{1}$ given $Z_{2}=z_{2}$ is
$\mathcal{N}(\hat{\mu},\widehat{\Sigma})$ where

$\displaystyle\hat{\mu}$
$\displaystyle=\mu_{1}+\Sigma_{12}\Sigma_{22}^{-1}(z_{2}-\mu_{2})$
$\displaystyle\widehat{\Sigma}$
$\displaystyle=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$

Again, the posterior mean depends on the signal realization, but the posterior
covariance matrix does not.

######  Example 2.5.

Let $\left(\begin{array}[]{c}Z_{1}\\\
Z_{2}\end{array}\right)\sim\mathcal{N}\left(\left(\begin{array}[]{c}\mu_{1}\\\
\mu_{2}\end{array}\right),\left(\begin{array}[]{cc}\sigma_{1}^{2}&\rho\sigma_{1}\sigma_{2}\\\
\rho\sigma_{1}\sigma_{2}&\sigma_{2}^{2}\end{array}\right)\right)$. Then
$(Z_{1}\mid Z_{2}=z_{2})\sim\mathcal{N}(\hat{\mu},\widehat{\Sigma})$ where

$\displaystyle\hat{\mu}$
$\displaystyle=\mu_{1}+\rho\frac{\sigma_{1}}{\sigma_{2}}(z_{2}-\mu_{2})$
$\displaystyle\widehat{\Sigma}$ $\displaystyle=\sigma_{1}^{2}(1-\rho^{2})$

######  Exercise 2.5 (U).

Let $Z_{1}=\theta$ and $Z_{2}=X$ where $\theta$ and $X$ are as defined at the
beginning of this section. Show that Fact 2.3 implies Fact 2.2.

Sections 2.3.2-2.3.4 demonstrate three applications of these Bayesian updating
formulae.

#####  2.3.2 Application 1: Career Concerns

Our first application is solving the two-period version of Holmstrom (1999)
model of career concerns.

There is a single agent and a manager. The agent has a type
$\theta\sim\mathcal{N}(\mu,\sigma_{\theta}^{2})$ that is unknown to both the
agent and the manager. In period 1, the agent chooses an effort level
$a\in\mathbb{R}_{+}$ at cost $c(a)=\frac{1}{2}a^{2}$. This effort is not
observed by the manager. The agent’s type and effort jointly determine the
realization of a performance signal

$X=\theta+a+\varepsilon$ (2.7)

where $\theta\perp\\!\\!\\!\perp\varepsilon$ and
$\varepsilon\sim\mathcal{N}(0,\sigma_{\varepsilon}^{2})$. In period 2, the
manager observes the realization of $X$ and forms an expectation about the
agent’s type. Since the manager does not observe $a$, this expectation is
taken with respect to the manager’s possibly misspecified perception about the
distribution of $X$ (more soon). The agent receives the manager’s expectation
of his type.

For arbitrary $a\in\mathbb{R}_{+}$, write $\mathbb{E}^{a}(\theta\mid X)$ for
the conditional expectation of $\theta$ with respect to
$X=\theta+a+\varepsilon$. If the manager expects the agent to choose effort
$a^{*}$ while the agent in fact chooses effort $a$, then the agent’s total
expected payoff is

$\mathbb{E}^{a}[\mathbb{E}^{a^{*}}(\theta\mid X)]-c(a),$

where the inner expectation $\mathbb{E}^{a^{*}}(\theta\mid X)$ is the
manager’s expectation of the agent’s type, and
$\mathbb{E}^{a}[\mathbb{E}^{a^{*}}(\theta\mid X)]$ is the agent’s expectation
of the manager’s expectation.

######  Claim 1.

There is a unique equilibrium in which the agent chooses effort
$a^{*}=\frac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}$.

######  Corollary 2.2.

Equilibrium effort $a^{*}$ is decreasing in $\sigma_{\varepsilon}^{2}$ (i.e.,
it is less valuable to manipulate a noisier signal) and is increasing in
$\sigma_{\theta}^{2}$ (i.e., it is more valuable to manipulate information
about a more uncertain unknown).

We’ll now prove Claim 1. Equilibrium effort $a^{*}$ must satisfy the first-
order condition

$\left.\frac{\partial\mathbb{E}^{a}[\mathbb{E}^{a^{*}}(\theta\mid
X)]}{\partial a}\right|_{a=a^{*}}=a^{*}$ (2.8)

equating the marginal value of increasing effort (over $a^{*}$) to the
marginal cost of increasing effort (over $a^{*}$). Applying Fact 2.2, the
manager’s expectation of $\theta$ with respect to the de-biased signal
$X-a^{*}=\theta+\varepsilon$ is

$\mathbb{E}^{a^{*}}(\theta\mid
X)=\frac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}(X-a^{*})+\frac{\sigma_{\varepsilon}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}\mu$

The agent’s expectation of this expectation (with respect to
$X=\theta+a+\varepsilon$) is

$\displaystyle\mathbb{E}^{a}\left[\mathbb{E}^{a^{*}}(\theta\mid
X)\right]=\mu+\frac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}(a-a^{*})$

So (2.8) implies that equilibrium effort is
$a^{*}=\frac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}$.
(Uniqueness follows from strict concavity of the agent’s payoff function.)

######  Exercise 2.6 (U).

Consider the model described in this section, and set
$\sigma_{\theta}^{2}=\sigma_{\varepsilon}^{2}=1$.

  * (a)

Suppose that in addition to the worker’s performance signal, the firm (through
collection of additional data about the worker’s type) is able to separately
observe a signal

$S=\theta+\delta$

where $\theta$ and $\delta$ are jointly normal and independent, and
$\delta\sim\mathcal{N}(0,1)$. The firm’s expectation about the worker’s type
is based both on $S$ as well as on the worker’s performance signal $X$ (as
defined in (2.7)). Solve for the worker’s equilibrium action and compare it
with the previous solution
$a^{*}=\frac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}$.
Does the worker exert more or less effort in equilibrium? Provide intuition
for your result.

  * (b)

Suppose that the firm instead acquires data that allows it to more accurately
monitor the performance shocks that the worker experiences (e.g., whether the
worker had a rough day, or had help at work). Formally, the firm observes

$S=\varepsilon+\delta$

where $\varepsilon$ and $\delta$ are jointly normal and independent, and
$\delta\sim\mathcal{N}(0,1)$. The firm’s expectation about the worker’s type
is based both on $S$ as well as on the worker’s performance signal $X$ (as
defined in (2.7)). Solve for the worker’s equilibrium action and compare it
with the previous solution
$a^{*}=\frac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}$.
Does the worker exert more or less effort in equilibrium? Provide intuition
for your result.

######  Exercise 2.7 (G).

Consider a variation on Holmstrom (1999)’s career concerns model, in which the
type $\theta$ and noise term $\varepsilon$ are correlated. Specifically, the
type is decomposed as $\theta=\theta_{1}+\theta_{2}$, the signal is
$X=\theta+\varepsilon+a$, and we suppose that

$\displaystyle\theta_{2}=\alpha\theta_{1}+z$
$\displaystyle\varepsilon=\beta\theta_{1}+w$

where $\alpha,\beta\in\mathbb{R}$ are known constants, and
$\theta_{1}\sim\mathcal{N}(\mu_{\theta},\sigma_{\theta}^{2})$,
$z\sim\mathcal{N}(0,\sigma_{z}^{2})$, and $w\sim\mathcal{N}(0,\sigma_{w}^{2})$
are mutually independent and unknown to both the agent and the manager.

  * (a)

Solve for equilibrium effort. How does this compare to Claim 1 in the special
case $\alpha=\beta=0$?

  * (b)

Suppose $\alpha,\beta>0$. How does equilibrium effort change in the parameters
$\alpha$ and $\beta$? Provide intuition.

#####  2.3.3 Application 2: Linear-Quadatic Coordination Games

Our second application is solving for equilibrium in a two-agent linear-
quadratic coordination game (Morris and Shin, 2002).

Let $\theta\sim\mathcal{N}(\mu,\sigma_{\theta}^{2})$ be an unknown state. Each
agent $i=1,2$ receives a private signal about the state

$X_{i}=\theta+\varepsilon_{i}$

where $\varepsilon_{i}\sim\mathcal{N}(0,\sigma_{\varepsilon}^{2})$ is
independent of the state and across agents. Each agent chooses an action
$a_{i}\in\mathbb{R}$ given their signal realization $x_{i}$. Agent $i$’s
payoff is

$U_{i}(a_{1},a_{2})=-(1-\beta)(a_{i}-\theta)^{2}-\beta(a_{i}-a_{j})^{2}$

where $\beta\in(0,1)$ controls how much the agent cares about matching the
state versus matching the other agent’s action.

We’ll solve for a symmetric linear Bayesian Nash equilibrium
$(a^{*}_{1},a^{*}_{2})$ in which each agent’s strategy satisfies

$a_{i}^{*}(x_{i})=cx_{i}+\kappa$ (2.9)

for some constants $c,\kappa\in\mathbb{R}$. Let’s first conjecture that such
an equilibrium exists. Given agent $j$’s strategy
$a_{j}(x_{j})=cx_{j}+\kappa$, agent $i$’s expected payoff (conditional on
$X_{i}=x_{i}$) is

$\mathbb{E}[-(1-\beta)(a_{i}-\theta)^{2}-\beta(a_{i}-(cX_{j}+\kappa))^{2}\mid
X_{i}=x_{i}]$

Taking a derivative with respect to $a_{i}$, agent $i$’s best reply is

$a_{i}^{*}(x_{i})=(1-\beta)\mathbb{E}(\theta\mid
X_{i}=x_{i})+\beta(c\mathbb{E}(\theta\mid X_{i}=x_{i})+\kappa).$

Plugging in the expression for $\mathbb{E}(\theta\mid X_{i}=x_{i})$ from Fact
2.2, and matching coefficients with (2.9), we have
$c=\frac{\sigma_{\theta}^{2}(1-\beta)}{\sigma_{\varepsilon}^{2}+\sigma_{\theta}^{2}(1-\beta)}$
and
$\kappa=\frac{\sigma_{\varepsilon}^{2}}{\sigma_{\varepsilon}^{2}+\sigma_{\theta}^{2}(1-\beta)}\mu$.
Thus a symmetric linear equilibrium exists in which each agent $i$ chooses

$a_{i}^{*}(x_{i})=\frac{\sigma_{\theta}^{2}(1-\beta)}{\sigma_{\varepsilon}^{2}+\sigma_{\theta}^{2}(1-\beta)}x_{i}+\frac{\sigma_{\varepsilon}^{2}}{\sigma_{\varepsilon}^{2}+\sigma_{\theta}^{2}(1-\beta)}\mu$
(2.10)

Morris and Shin (2002) further show that this is the unique pure-strategy
equilibrium.

Suppose we interpret the common prior $\mathcal{N}(\mu,\sigma_{\theta}^{2})$
as informed by a public signal, where a more informative signal implies a
smaller $\sigma_{\theta}^{2}$. Then we see from (2.10) that the more
informative the public signal is, the less weight agents place on their
private signal.

#####  2.3.4 Application 3: Data Sharing

Our final application is an example from Acemoglu et al. (2022) regarding why
online platforms don’t compensate users for the data that they give up.

There is a single platform and two agents $i=1,2$ with types distributed

$\begin{pmatrix}\theta_{1}\\\
\theta_{2}\end{pmatrix}\sim\mathcal{N}\left(\begin{pmatrix}0\\\
0\end{pmatrix},\begin{pmatrix}1&\rho\\\ \rho&1\end{pmatrix}\right)$

Each agent $i$ privately observes the realization of a signal
$X_{i}=\theta_{i}+\varepsilon_{i}$, where
$\varepsilon_{i}\sim\mathcal{N}(0,1)$ is independent across agents and
independent of both types.

The platform chooses a payment $p_{i}$ to offer to each agent $i$ for sharing
their data. After receiving these offers, each agent $i$ chooses whether to
share $(a_{i}=1)$ or withhold $(a_{i}=0)$ their signal realization. Write
$X_{\mathbb{a}}$ for the signals shared under action profile
$\mathbb{a}=(a_{1},a_{2})$. For example, if $\mathbb{a}=(1,0)$, then
$X_{\mathbb{a}}=X_{1}$, while if $\mathbb{a}=(1,1)$, then
$X_{\mathbb{a}}=(X_{1},X_{2})$.

Each agent $i$’s payoff is determined by the platform’s posterior uncertainty
about his type, a privacy parameter $v\in\mathbb{R}_{+}$, and his payment via

$u_{i}(\mathbb{a},\mathbb{p})=v\cdot\operatorname*{\text{Var}}(\theta_{i}\mid
X_{\mathbb{a}})+p_{i}\cdot\mathbbm{1}(a_{i}=1).$

The platform’s payoff is
$u_{P}(\mathbb{a},\mathbb{p})=-u_{1}(\mathbb{a},\mathbb{p})-u_{2}(\mathbb{a},\mathbb{p})$.
So the agents prefer for the platform to be more uncertain about their types,
while the platform prefers to be less uncertain.

We’ll show that when agent types are sufficiently correlated, i.e., $\rho$ is
large, then the platform can induce both agents to share their data at a lower
total payment than what is required to induce exactly one agent to share.

Let’s first solve for payment vectors $(p_{1},p_{2})$ given which it is an
equilibrium for both agents to share their signals. Suppose agent $j$ chooses
to share. Then if agent $i$ does not share, the platform’s belief about
$\theta_{i}$ is updated only to $X_{j}$. Since

$\begin{pmatrix}\theta_{i}\\\
X_{j}\end{pmatrix}\sim\mathcal{N}\left(\begin{pmatrix}0\\\
0\end{pmatrix},\begin{pmatrix}1&\rho\\\ \rho&2\end{pmatrix}\right)$

the platform’s posterior variance of $\theta_{i}$ is $1-\rho^{2}/2$ (by Fact
2.3). So agent $i$’s payoff is $v\cdot\left(1-\rho^{2}/2\right)$. If agent $i$
does share, then beliefs about $\theta_{i}$ are further updated to the signal
$X_{i}$, and (by Fact 2.2) the platform’s posterior variance of $\theta_{i}$
reduces to $\frac{2-\rho^{2}}{4-\rho^{2}}$. So agent $i$’s payoff is
$v\cdot\left(\frac{2-\rho^{2}}{4-\rho^{2}}\right)+p_{i}.$ Thus, agent $i$’s
best reply to $a_{j}=1$ is to share if and only if

$p_{i}\geq v\cdot\left(\frac{(2-\rho^{2})^{2}}{2(4-\rho^{2})}\right)$

and the action profile $(a_{1},a_{2})=(1,1)$ is an equilibrium if the above
display holds for both agents $i$. The minimum total payment is twice the
right-hand-side, i.e.,
$v\cdot\left(\frac{(2-\rho^{2})^{2}}{(4-\rho^{2})}\right)$.

Let’s now solve for payment vectors $(p_{1},p_{2})$ given which it is an
equilibrium for exactly one agent to share his data. Without loss, fix
$a_{2}=0$. If agent 1 chooses $a_{1}=0$, then the platform’s uncertainty about
$\theta_{1}$ is its prior uncertainty, 1, so agent $1$’s payoff is $v$. If
agent $1$ chooses $a_{1}=1$, then the platform’s belief about $\theta_{1}$
updates to the signal $X_{1}$. Applying Fact 2.2, the platform’s posterior
variance about $\theta_{1}$ is $1/2$ and so agent $1$’s payoff is
$v\cdot(1/2)+p_{1}.$ Thus, $a_{1}=1$ is a best reply to $a_{2}=0$ if and only
if $p_{1}\geq v/2.$ So the platform can induce (exactly) one agent to share if
it offers one agent a payment of at least $v/2$ (which is accepted) and
another a payment of no more than $v/2$ (which is rejected), at a total
payment of $v/2$.

When $\rho^{2}\geq\frac{7-\sqrt{17}}{4}\approx 0.71$, then
$v\cdot\left(\frac{(2-\rho^{2})^{2}}{(4-\rho^{2})}\right)<v/2$, so the
platform pays less to induce two users share than one. Intuitively, each
agent’s choice to share their data exerts a negative externality on other
agent: When both users share, each of their signals is less valuable in view
of the signal revealed by the other. Agents paid their marginal value thus
receive lower compensation, and in a limiting version of this model with a
growing number of agents, the amount of compensation needed to induce all
agents to share vanishes to zero.

####  2.4 Additional Exercises

######  Exercise 2.8 (U).

Suppose $\theta\sim\mathcal{N}(0,\sigma_{\theta}^{2})$ and

$\displaystyle Y_{1}$ $\displaystyle=\theta+b+\varepsilon_{1}$ $\displaystyle
Y_{2}$ $\displaystyle=b+\varepsilon_{2}$

where $\theta$, $b$, $\varepsilon_{1}$, and $\varepsilon_{2}$ are all
independent of one another, $b\sim\mathcal{N}(0,\sigma_{b}^{2})$,
$\varepsilon_{1}\sim\mathcal{N}(0,\sigma_{1}^{2})$, and
$\varepsilon_{2}\sim\mathcal{N}(0,\sigma_{2}^{2})$. We can interpret $Y_{1}$
as a biased signal about $\theta$ and $Y_{2}$ as a signal about the size of
the bias.

Your friend says: “The only value of $Y_{1}$ and $Y_{2}$ for learning about
$\theta$ is to provide information about the size of $b$. Since $Y_{1}-Y_{2}$
is an unbiased signal about $b$, it is equally valuable to learn the outcome
of $Y_{1}-Y_{2}$ as it is to learn the pair of signals $(Y_{1},Y_{2})$.”

Show that your friend is wrong: The distribution of $\theta\mid Y_{1},Y_{2}$
is different from the distribution of $\theta\mid Y_{1}-Y_{2}$. Also provide
an intuition explaining to your friend the error in their reasoning.

######  Exercise 2.9.

Consider the two-player game described in Section 2.3.4, but suppose that the
types are distributed

$\left(\begin{array}[]{c}X_{1}\\\
X_{2}\end{array}\right)\sim\mathcal{N}\left(\left(\begin{array}[]{c}0\\\
0\end{array}\right),\left(\begin{array}[]{cc}2&1\\\
1&1\end{array}\right)\right)$

As in Section 2.3.4, let $a_{1}$ and $a_{2}$ denote the actions of players 1
and 2, where an action of ‘0’ means that the player does not share their data,
while ‘1’ means that they do.

  * (a)

Suppose player 2 chooses $a_{2}=0$. What is player 1’s payoff from choosing
$a_{1}=0$ and what is player 1’s payoff from choosing $a_{1}=1$? Provide a
condition that characterizes when it is the case that player 1’s best reply is
to share data (i.e., $a_{1}=1$).

  * (b)

Suppose player 2 chooses $a_{2}=1$. What is player 1’s payoff from choosing
$a_{1}=0$ and what is player 1’s payoff from choosing $a_{1}=1$? Provide a
condition that characterizes when it is the case that player 1’s best reply is
to share data (i.e., $a_{1}=1$).

  * (c)

Suppose player 1 chooses $a_{1}=0$. What is player 2’s payoff from choosing
$a_{2}=0$ and what is player 2’s payoff from choosing $a_{2}=1$? Provide a
condition that characterizes when it is the case that player 2’s best reply is
to share data (i.e., $a_{2}=1$).

  * (d)

Suppose player 1 chooses $a_{1}=1$. What is player 1’s payoff from choosing
$a_{2}=0$ and what is player 2’s payoff from choosing $a_{2}=1$? Provide a
condition that characterizes when it is the case that player 2’s best reply is
to share data (i.e., $a_{2}=1$).

  * (e)

Suppose $v=1$. For what set of values of $(p_{1},p_{2})$ is it the case that:

    * (i)

$(a_{1},a_{2})=(1,1)$ is a Nash equilibrium?

    * (ii)

$(a_{1},a_{2})=(1,0)$ is a Nash equilibrium?

    * (iii)

$(a_{1},a_{2})=(0,1)$ is a Nash equilibrium?

    * (iv)

$(a_{1},a_{2})=(0,0)$ is a Nash equilibrium?

  * (f)

Again let $v=1$. What is the smallest total payment the firm must make to
induce an equilibrium where both players share their data? (That is, what is
the smallest sum $p_{1}+p_{2}$ such that $(a_{1},a_{2})=(1,1)$ is an
equilibrium given the payment profile $(p_{1},p_{2})$?) Comment on whether it
is the case that one player receives the higher payment, and why this answer
makes sense.

  * (g)

Again let $v=1$. Suppose there is a firm 1 and firm 2, where firm 1 only
interacts with player 1, and firm 2 only interacts with player 2. What is the
smallest amount $p_{1}$ that firm 1 must pay to induce player 1 to choose
$a_{1}=1$? What is the smallest amount $p_{2}$ that firm 2 must pay to induce
player 2 to choose $a_{2}=2$? Compare the sum of these values $p_{1}+p_{2}$ to
your answer from part (f).

######  Exercise 2.10 (G).

Suppose $\theta$ is normally distributed. For each $i=1,\dots,n$, let
$X_{i}=\theta+\varepsilon_{i}$ where $\varepsilon_{i}$ is independent of
$\theta$, the vector $(\varepsilon_{1},\dots,\varepsilon_{n})$ is jointly
normal, and the signals $X_{1},\dots,X_{n}$ are exchangeable. Define
$\overline{X}=\frac{1}{n}(X_{1}+\dots+X_{n})$. Prove that $\theta\mid
X_{1},\dots,X_{n}$ is identical in distribution to $\theta\mid\overline{X}$.

######  Hint.

Recall that $\mathbb{E}(\theta\mid X)$ minimizes
$\mathbb{E}[(\hat{\theta}-\theta)^{2}]$ among all $\sigma(X)$-measurable
random variables $\hat{\theta}$.

######  Exercise 2.11 (G).

Consider two processes of social learning about an unknown state
$\theta\sim\mathcal{N}(0,1)$.

Scenario 1: At $t=0$, a single agent privately observes the signal

$Y=\theta+\delta,\quad\delta\sim\mathcal{N}(0,1/\tau)$

where $\theta$ and $\delta$ are independent of one another, and the precision
$\tau\in\mathbb{R}_{+}$ is a known constant. The agent chooses an action $y$
and receives the payoff $-\mathbb{E}[(y-\theta)^{2}]$. At $t=1$, each of $n$
agents, indexed by $i$, privately observes a signal

$X_{i}=\theta+\varepsilon_{i},\quad\varepsilon_{i}\sim\mathcal{N}(0,1)$

as well as the action $y$ of the first agent. The error terms
$\varepsilon_{i}$ are independent across agents, and independent of $\theta$
and $\delta$. Each agent $i$ from this generation then takes an action $a_{i}$
to maximize the payoff $-\mathbb{E}[(a_{i}-\theta)^{2}]$. At $t=2$, you
arrive, observe the actions $(a_{1},\dots,a_{n})$ of the preceding generation
(but _not_ the action of the first agent), and choose an action $a^{*}$ with
payoff $-\mathbb{E}[(a^{*}-\theta)^{2}]$.

Scenario 2: At $t=1$, each of $m$ agents observes a private signal

$Z_{i}=\theta+\eta_{i},\quad\eta_{i}\sim\mathcal{N}(0,1)$

where the error terms $\eta_{i}$ are independent across agents and of
$\theta$. Each agent $i$ takes an action $b_{i}$ with payoff
$-\mathbb{E}[(b_{i}-\theta)^{2}]$. At $t=1$, you arrive, observe the actions
$(b_{1},\dots,b_{m})$ of the preceding generation, and choose an action
$a^{*}$ with payoff $-\mathbb{E}[(a^{*}-\theta)^{2}]$.

Characterize the function $h(n)$ such that your expected payoff is higher in
scenario 1 if and only if $m<h(n)$. As clearly as you can, write out an
intuition for this result.

######  Hint.

Use the fact given in Exercise 2.10.

###  Chapter 3 Properties of Information

Many economic settings involve an unknown type or quality $\theta$ and a
signal $X$ about $\theta$, where both $\theta$ and $X$ are ordered (i.e.,
there are “better” qualities $\theta$ and “higher” signal realizations $X$).
In these settings, we might think that higher realizations of $X$ are good
news about $\theta$—for example, that higher test scores suggest higher
ability or that better reviews for a product suggest higher quality. These
positive inferences are not in general justified, requiring assumptions on the
joint distribution of $(\theta,X)$.

Section 3.1 presents three useful definitions of positive dependence between
random variables, which are applied to our motivating problem (inference about
$\theta$ from observation of a signal $X$) in Section 3.2. Section 3.3
presents an example of the kind of counterintuitive result that can obtain
when these properties are not imposed on the informational environment.

####  3.1 Definitions

#####  3.1.1 Monotone Likelihood Ratio Property

Consider two random variables $Z$ and $\widetilde{Z}$ with distributions $F$
and $\widetilde{F}$ that admit densities $f$ and $\tilde{f}$. To simplify
exposition, all densities mentioned in this chapter are assumed to be
everywhere positive.

######  Definition 3.1.

The distribution $F$ _likelihood-ratio dominates_ the distribution
$\widetilde{F}$ if111The assumption that densities are everywhere strictly
positive allows us to define the monotone likelihood ratio property in terms
of likelihood ratios. More generally, we can consider a distribution $F$ to
likelihood-ratio dominate another distribution $\widetilde{F}$ if
$f(z)\tilde{f}(z^{\prime})\geq f(z^{\prime})\tilde{f}(z)$ for all
$z>z^{\prime}$.

$\frac{f(z)}{f(z^{\prime})}\geq\frac{\widetilde{f}(z)}{\widetilde{f}(z^{\prime})}\quad\quad\mbox{for
all $z>z^{\prime}$}$

Intuitively, moving up in the likelihood-ratio dominance order renders higher
realizations of $z$ more likely relative to lower realizations.

This definition is often specialized to conditional densities in the following
way. Suppose $\theta$ and $X$ are real-valued random vectors defined on the
same probability space with densities $f_{\theta}$ and $f_{X}$ and conditional
densities $f_{\theta\mid X}$ and $f_{X\mid\theta}$.

######  Definition 3.2.

The family of conditional densities
$\\{f_{X\mid\theta}(\cdot\mid\theta)\\}_{\theta\in\Theta}$ have the _monotone
likelihood ratio property_ (MLRP) if for every $x>x^{\prime}$ and
$\theta>\theta^{\prime}$,

$\frac{f_{X\mid\theta}(x\mid\theta)}{f_{X\mid\theta}(x^{\prime}\mid\theta)}\geq\frac{f_{X\mid\theta}(x\mid\theta^{\prime})}{f_{X\mid\theta}(x^{\prime}\mid\theta^{\prime})}.$
(3.1)

If the inequality above holds strictly at every $x>x^{\prime}$, then we say
that $\\{f_{X\mid\theta}(\cdot\mid\theta)\\}$ have the _strict_ monotone
likelihood ratio property.

######  Remark 3.1.

If $\\{f_{X\mid\theta}(\cdot\mid\theta)\\}$ satisfy MLRP, then
$\\{f_{\theta\mid X}(\cdot\mid X)\\}$ also satisfy MLRP. To see this, observe
that by Bayes’ rule, (3.1) can be rewritten

$\frac{f_{\theta\mid X}(\theta\mid
x)f_{X}(x)}{f_{\theta}(\theta)}\frac{f_{\theta}(\theta)}{f_{\theta\mid
X}(\theta\mid x^{\prime})f_{X}(x^{\prime})}\geq\frac{f_{\theta\mid
X}(\theta^{\prime}\mid
x)f_{X}(x)}{f_{\theta}(\theta^{\prime})}\frac{f_{\theta}(\theta^{\prime})}{f_{\theta\mid
X}(\theta^{\prime}\mid x^{\prime})f(x^{\prime})}$

which simplifies to the condition that $\\{f_{\theta\mid X}(\cdot\mid X)\\}$
have the monotone likelihood ratio property.

In the special case of an additive signal $X=\theta+\varepsilon$, where
$\varepsilon$ is independent of $\theta$ and has density $f_{\varepsilon}$,

$\frac{f_{\theta\mid X}(\theta\mid x)}{f_{\theta\mid X}(\theta^{\prime}\mid
x)}=\frac{f_{\varepsilon}(x-\theta)}{f_{\varepsilon}(x-\theta^{\prime})}$

so the MLRP condition in (3.1) becomes

$\frac{f_{\varepsilon}(x-\theta)}{f_{\varepsilon}(x^{\prime}-\theta)}\geq\frac{f_{\varepsilon}(x-\theta^{\prime})}{f_{\varepsilon}(x^{\prime}-\theta^{\prime})}\quad\quad\mbox{for
every $x>x^{\prime}$ and $\theta>\theta^{\prime}$},$

i.e., for every $\theta>\theta^{\prime}$, the function
$\frac{f_{\varepsilon}(x-\theta)}{f_{\varepsilon}(x-\theta^{\prime})}$ is
nondecreasing in $x$. It turns out that this is precisely the condition that
$f_{\varepsilon}$ is log concave.

######  Definition 3.3.

A function $f$ that maps a convex set into the positive reals is _log-concave_
if the function $\ln f$ is concave.

######  Proposition 5 (Saumard and Wellner (2014)).

A density function $f$ on $\mathbb{R}$ is log-concave if and only if for every
$\theta>\theta^{\prime}$, the ratio $\frac{f(x-\theta)}{f(x-\theta^{\prime})}$
is a non-decreasing function of $x$.

Thus, in any model where (1) $X=\theta+\varepsilon$, (2) the noise term
$\varepsilon$ is independent of $\theta$, and (3) $\varepsilon$ has a log-
concave density, we can be guaranteed that $\\{f_{\theta\mid X}(\cdot\mid
x)\\}$ has the monotone likelihood ratio property (no matter the distribution
of $\theta$).

Many distributions have log-concave densities—for example, normal
distributions, exponential distributions, the uniform distribution over any
convex set, the logistic distribution, and the extreme value distribution. But
others do not—for example, the Pareto distribution and Cauchy distribution.
See Saumard and Wellner (2014) or Bagnoli and Bergstrom (2005) for other
examples and properties of log-concave distributions.

#####  3.1.2 Affiliation

Let $Z_{1},\dots,Z_{n}$ be real-valued random variables taking values in
$\mathbb{R}^{n}$ and admitting joint density $f$, which again we’ll assume to
be everywhere strictly positive. For any $z,z^{\prime}\in\mathbb{R}^{n}$, let
$z\wedge z^{\prime}$ (“z meet z’) denote the component-wise minimum of $z$ and
$z^{\prime}$, and $z\vee z^{\prime}$ (“z join z’) denote the component-wise
maximum, i.e.,

$\displaystyle z\vee z^{\prime}$
$\displaystyle=(\max(z_{1},z^{\prime}_{1}),\dots,\max(z_{n},z^{\prime}_{n}))$
$\displaystyle z\wedge z^{\prime}$
$\displaystyle=(\min(z_{1},z^{\prime}_{1}),\dots,\min(z_{n},z^{\prime}_{n}))$

######  Definition 3.4.

The variables $Z_{1},\dots,Z_{n}$ are _affiliated_ if

$f(z\vee z^{\prime})f(z\wedge z^{\prime})\geq f(z)f(z^{\prime})$ (3.2)

for all $z,z^{\prime}\in\mathbb{R}^{n}.$

This condition loosely says that larger realizations of any one variable make
larger realizations of the other variables more likely. Figure 3.1 depicts
this relationship for two binary variables.

Figure 3.1: Two binary variables with joint density $f$ are affiliated if
$f(1,1)f(0,0)\geq f(1,0)f(0,1)$.

######  Remark 3.2.

If $Z_{1},\dots,Z_{n}$ are mutually independent, then they are affiliated.

Besides Definition 3.4, there are several equivalent ways to characterize
affiliation. The first follows by taking logs of both sides of (3.2).

######  Proposition 6.

$Z_{1},Z_{2},\dots,Z_{n}$ are affiliated if and only if $f$ is log-
supermodular, i.e.

$\log f(z\vee z^{\prime})+\log f(z\wedge z^{\prime})\geq\log f(z)+\log
f(z^{\prime})$

for all $z,z^{\prime}$.

######  Proposition 7.

Suppose the joint density $f$ is twice-differentiable. Then
$Z_{1},Z_{2},\dots,Z_{n}$ are affiliated if and only if
$\frac{\partial^{2}\log f}{\partial z_{i}z_{j}}\geq 0$.

We show the only if direction below, leaving the if direction for an exercise.

Proof. Without loss let $i=1$ and $j=2$. Choose any
$z_{1},z^{\prime}_{1},z_{2},z_{2}^{\prime}\in\mathbb{R}$ where
$z_{1}>z_{1}^{\prime}$ and $z_{2}>z_{2}^{\prime}$. Suppose
$Z_{1},Z_{2},\dots,Z_{n}$ are affiliated. Then by definition

$\log f(z_{1},z_{2},z_{-12})-\log f(z_{1}^{\prime},z_{2},z_{-12})\geq\log
f(z_{1},z_{2}^{\prime},z_{-12})-\log f(z_{1}^{\prime},z_{2}^{\prime},z_{-12})$

where $z_{-12}$ is shorthand for $(z_{3},\dots,z_{n})$. Rewrite $z_{1}$ as
$z_{1}^{\prime}+\varepsilon$ and divide both sides by $\varepsilon$. Taking
the limit as $\varepsilon\rightarrow 0$, we have

$\displaystyle\lim_{\varepsilon\rightarrow 0}$ $\displaystyle\left(\frac{\log
f(z_{1}^{\prime}+\varepsilon,z_{2},z_{-12})-\log
f(z_{1}^{\prime},z_{2},z_{-12})}{\varepsilon}\right)$
$\displaystyle\quad\quad\quad\quad\quad\geq\lim_{\varepsilon\rightarrow
0}\left(\frac{\log f(z_{1}^{\prime}+\varepsilon,z_{2}^{\prime},z_{-12})-\log
f(z_{1}^{\prime},z_{2}^{\prime},z_{-12})}{\varepsilon}\right)$

so $\frac{\partial\log f}{\partial z_{1}}$ is increasing in $z_{2}$, as
desired.

######  Exercise 3.1 (G).

Prove the ‘if’ direction of Proposition 7: If the joint density $f$ is twice-
differentiable and satisfies $\frac{\partial^{2}\log f}{\partial
z_{i}z_{j}}\geq 0$, then $Z_{1},Z_{2},\dots,Z_{n}$ are affiliated.

The next characterization simplifies (3.2) to a pairwise condition.
Specifically, for any $(Z_{i},Z_{j})$ and any realization of the remaining
variables $Z_{-ij}$, higher realizations of $Z_{i}$ must imply higher
realizations of $Z_{j}$.

######  Proposition 8.

$Z_{1},\dots,Z_{n}$ are affiliated if and only if

$f(z_{i},z_{j},z_{-ij})f(z_{i}^{\prime},z_{j}^{\prime},z_{-ij})\geq
f(z^{\prime}_{i},z_{j},z_{-ij})f(z_{i},z_{j}^{\prime},z_{-ij})$ (3.3)

for every pair of distinct indices $i$, $j$, and every $z_{i}>z_{i}^{\prime}$,
$z_{j}>z_{j}^{\prime}$, and $z_{-ij}\in\mathbb{R}^{n-2}$.

######  Exercise 3.2 (G).

Prove Proposition 8.

This pairwise characterization immediately implies the following
characterization, which says that $(Z_{1},\dots,Z_{n})$ are affiliated if and
only if for every pair of variables $i,j$, and every realization of $z_{-ij}$,
the family of conditional densities $\\{f(\cdot\mid
z_{j},z_{-ij})\\}_{z_{j}\in\mathbb{R}}$ has the monotone-likelihood ratio
property.

######  Proposition 9.

$Z_{1},\dots,Z_{n}$ are affiliated if and only if

$f(z_{i}\mid z_{j},z_{-ij})f(z^{\prime}_{i}\mid z^{\prime}_{j},z_{-ij})\geq
f(z_{i}\mid z^{\prime}_{j},z_{-ij})f(z^{\prime}_{i}\mid z_{j},z_{-ij})$ (3.4)

for every pair of distinct indices $i$, $j$, and every $z_{i}>z_{i}^{\prime}$,
$z_{j}>z_{j}^{\prime}$, and $z_{-ij}\in\mathbb{R}^{n-2}$.

Proof. The displays in (3.3) and (3.4) are equivalent to one another by Bayes’
rule, so Proposition 8 implies Proposition 9.

Operations that preserve affiliation include:

######  Proposition 10 (Monotone Functions).

Suppose $Z_{1},\dots,Z_{n}$ are affiliated, and the functions
$g_{i}:\mathbb{R}\rightarrow\mathbb{R}$, $1\leq i\leq n$, are either all
nondecreasing or all nonincreasing. Then the variables
$g_{1}(Z_{1}),\dots,g_{n}(Z_{n})$ are affiliated.

######  Proposition 11 (Subsets).

Suppose $Z_{1},\dots,Z_{n}$ are affiliated and let $A\subseteq\\{1,\dots,n\\}$
be any subset of these variables. Then the variables $(Z_{i})_{i\in A}$ are
affiliated.

######  Proposition 12 (Order Statistics).

For each $1\leq i\leq n$, let $z^{(i)}$ denote the $i$-th largest realization
among $(z_{1},\dots,z_{n})$. Then the variables $(Z^{(1)},\dots,Z^{(n)})$ are
affiliated.

######  Exercise 3.3 (G).

Show that affiliation is not preserved under arbitrary linear combinations of
affiliated variables by constructing an example of random variables
$Z_{1},Z_{2},Z_{3}$ where $(Z_{1},Z_{2},Z_{3})$ are affiliated but
$(Z_{1}+Z_{2},Z_{3})$ are not.

#####  3.1.3 First-Order Stochastic Dominance

Again consider two real-valued random variables, a parameter $\theta$ and a
signal $X$, defined on the same probability space with joint distribution $F$.
In many applications we may expect a higher signal realization to lead to a
higher inference about the unknown parameter. We now formalize ‘higher
inference’ as a first-order stochastic dominance shift in the posterior
belief.

######  Definition 3.5.

A distribution $F$ _first-order stochastically dominates_ $\widetilde{F}$,
which we denote by $F\geq_{FOSD}\widetilde{F}$, if

$\int u(\theta)dF(\theta)\geq\int u(\theta)d\widetilde{F}(\theta)$

for every nondecreasing function $u:\mathbb{R}\rightarrow\mathbb{R}$.
Equivalently, $F(\theta)\leq\tilde{F}(\theta)$ at every $\theta\in\Theta$.

If $u$ is interpreted as a utility function over money, then a monetary gamble
distributed according to $F$ is preferred over one distributed according to
$\tilde{F}$ by every agent that prefers more money over less, regardless of
the specific shape of the agent’s utility function. We can use this definition
to compare conditional beliefs about $\theta$.

######  Definition 3.6.

Say that $F$ has the _FOSD property_ if $F_{\theta\mid X}(\cdot\mid
X=x)\geq_{FOSD}F_{\theta\mid X}(\cdot\mid X=x^{\prime})$ for all
$x>x^{\prime}$.

Milgrom (1981) proposed a closely related property, which is imposed on
conditional distributions $F_{X\mid\theta}$ rather than joint distributions
$F$. (This is analogous to considering a signal
$\sigma:\Theta\rightarrow\Delta(S)$ without fixing a prior on $\Theta$.)

######  Definition 3.7.

Say that a signal realization $x$ is _more favorable than_ signal realization
$x^{\prime}$ if for every prior distribution $F_{\theta}\in\Delta(\Theta)$,
the posterior distribution $F_{\theta\mid X}(\cdot\mid x)$ first-order
stochastically dominates the posterior distribution $F_{\theta\mid
X}(\cdot\mid x^{\prime})$.

That is, $x$ is more favorable than $x^{\prime}$ if observing the realization
$x$ leads to a FOSD-higher posterior belief about $\theta$ (compared to
observing $x^{\prime}$). If $x$ is more favorable than $x^{\prime}$ for all
$x>x^{\prime}$, then we have a stronger version of the FOSD property (given in
(3.6)) that holds not only for the specific joint distribution $F$, but for
all joint distributions $F$ that are generated by $F_{X\mid\theta}$ and some
choice of prior $F_{\theta}$.

######  Example 3.1.

Recall that in the normal-updating setting with
$\theta\sim\mathcal{N}(\mu,\sigma_{\theta}^{2})$, $X=\theta+\varepsilon$,
$\varepsilon\sim\mathcal{N}(0,\sigma_{\varepsilon}^{2})$, and
$\theta\perp\\!\\!\\!\perp\varepsilon$, the agent’s posterior belief about
$\theta$ conditional on $X$ is

$\mathcal{N}\left(\frac{\sigma_{\theta}^{2}}{\sigma_{\varepsilon}^{2}+\sigma_{\theta}^{2}}X+\frac{\sigma_{\varepsilon}^{2}}{\sigma_{\varepsilon}^{2}+\sigma_{\theta}^{2}}\mu,\frac{\sigma_{\varepsilon}^{2}\sigma_{\theta}^{2}}{\sigma_{\varepsilon}^{2}+\sigma_{\theta}^{2}}\right).$

This distribution is increasing (in the FOSD order) in the realization of $X$
for all parameters $\mu$ and $\sigma_{\theta}^{2}$. So $x$ is more favorable
than $x^{\prime}$ for every pair $x>x^{\prime}$.

####  3.2 How They are Related

Let $\theta$ and $X$ be real-valued random vectors defined on the same
probability space. We’ll use $F$ to denote their joint distribution, and
assume throughout that the densities $f_{\theta}$ and $f_{X}$ and conditional
densities $f_{\theta\mid X}$ and $f_{X\mid\theta}$ exist. In this setting, our
main properties from above are:

  * A: $(X,\theta)$ are affiliated.

  * MLRP: $\\{f_{X\mid\theta}(\cdot\mid\theta)\\}$ satisfies MLRP.

  * FOSD: For all $x>x^{\prime}$, $F(\cdot\mid X=x)\geq_{FOSD}F(\cdot\mid X=x^{\prime})$

  * MF: For all $x>x^{\prime}$, $x$ is more favorable than $x^{\prime}$

These properties are related in the following way:

$\mbox{{(A)}}\quad\Longleftrightarrow\quad\mbox{{(MLRP)}}\quad\Longleftrightarrow\quad\mbox{{(MF)}}\quad\Longrightarrow\quad\mbox{{(FOSD)}}$

where the one-directional implication from (MF) to (FOSD) is strict. See de
Castro (2009) for an example of a distribution satisfying (FOSD) but not
(MLRP).

######  Remark 3.3.

(MLRP) is equivalent to (MF) but strictly stronger than (FOSD). Thus if a
joint distribution $F$ satisfies (MLRP) then it must satisfy (FOSD), but $F$
can satisfy (FOSD) and fail (MLRP). On the other hand, a conditional
distribution $F_{X\mid\theta}$ that satisfies (FOSD) for every completion to a
joint distribution $F$ (i.e., for every choice of prior $F_{\theta}$) must
also satisfy (MLRP). So ”FOSD for every prior” is equivalent to MLRP, while
“FOSD for some prior” is weaker.

We’ve already established the equivalence between (A) and (MLRP) in
Proposition 9. Since (FOSD) is necessary for (MF), clearly (MF) implies
(FOSD). The following result proves equivalence of (MLRP) and (MF).

######  Proposition 13 (Milgrom (1981)).

$x$ is more favorable than $x^{\prime}$ if and only if for every
$\theta>\theta^{\prime}$,

$\frac{f_{X\mid\theta}(x\mid\theta)}{f_{X\mid\theta}(x^{\prime}\mid\theta)}\geq\frac{f_{X\mid\theta^{\prime}}(x\mid\theta^{\prime})}{f_{X\mid\theta^{\prime}}(x^{\prime}\mid\theta^{\prime})}$
(3.5)

Proof. We will first show that if (3.5) is satisfied at every
$\theta>\theta^{\prime}$, then $x$ must be more favorable than $x^{\prime}$.
Fix any prior $F_{\theta}$ and parameter $\theta^{*}\in\Theta$. If
$F_{\theta}(\theta^{*})\in\\{0,1\\}$ then the conclusion is trivially reached.
So suppose $F_{\theta}(\theta^{*})\in(0,1)$.

For any $\theta\leq\theta^{*}$ and $\tilde{\theta}>\theta^{*}$, (3.5) implies

$\frac{f(x\mid\tilde{\theta})}{f(x\mid\theta)}\geq\frac{f(x^{\prime}\mid\tilde{\theta})}{f(x^{\prime}\mid\theta)}$

where we omit subscripts on the densities here and elsewhere in the proof to
ease notation. Integrating over all $\tilde{\theta}$ such that
$\tilde{\theta}>\theta^{*}$ (with respect to the prior distribution
$F_{\theta}$), we obtain

$\frac{\int_{\tilde{\theta}>\theta^{*}}f(x\mid\tilde{\theta})dF_{\theta}(\tilde{\theta})}{f(x\mid\theta)}\geq\frac{\int_{\tilde{\theta}>\theta^{*}}f(x^{\prime}\mid\tilde{\theta})dF_{\theta}(\tilde{\theta})}{f(x^{\prime}\mid\theta)}$

or equivalently

$\frac{f(x\mid\theta)}{\int_{\tilde{\theta}>\theta^{*}}f(x\mid\tilde{\theta})dF_{\theta}(\tilde{\theta})}\leq\frac{f(x^{\prime}\mid\theta)}{\int_{\tilde{\theta}>\theta^{*}}f(x^{\prime}\mid\tilde{\theta})dF_{\theta}(\tilde{\theta})}.$

Integrating over all $\theta$ such that $\theta\leq\theta^{*}$, we obtain

$\frac{\int_{\theta\leq\theta^{*}}f(x\mid\theta)dF_{\theta}(\theta)}{\int_{\tilde{\theta}>\theta^{*}}f(x\mid\tilde{\theta})dF_{\theta}(\tilde{\theta})}\leq\frac{\int_{\theta\leq\theta^{*}}f(x^{\prime}\mid\theta)dF_{\theta}(\theta)}{\int_{\tilde{\theta}>\theta^{*}}f(x^{\prime}\mid\tilde{\theta})dF_{\theta}(\tilde{\theta})}.$

Recall that $f(x\mid\theta)f(\theta)=f(\theta\mid x)f(x)$, so the above
display implies

$\frac{\int_{\theta\leq\theta^{*}}f(\theta\mid
x)d\theta}{\int_{\tilde{\theta}>\theta^{*}}f(\tilde{\theta}\mid
x)d\tilde{\theta}}\leq\frac{\int_{\theta\leq\theta^{*}}f(\theta\mid
x^{\prime})d\theta}{\int_{\tilde{\theta}>\theta^{*}}f(\tilde{\theta}\mid
x^{\prime})d\tilde{\theta}}$

or more simply

$\frac{F(\theta^{*}\mid x)}{1-F(\theta^{*}\mid x)}\leq\frac{F(\theta^{*}\mid
x^{\prime})}{1-F(\theta^{*}\mid x^{\prime})}$

Since $\frac{y}{1-y}$ is a strictly increasing function in $y$, we have
$F(\theta^{*}\mid x)\leq F(\theta^{*}\mid x^{\prime})$ as desired.

In the other direction, we will show that if $x$ is more favorable than
$x^{\prime}$, then (3.5) holds everywere. Consider any two parameter values
$\theta>\theta^{\prime}$, and let $F_{\theta}$ be a prior distribution
supported on these two points with equal probability on each.

Since by assumption $x$ is more favorable than $x^{\prime}$, then
$F(\theta^{\prime}\mid x)\leq F(\theta^{\prime}\mid x^{\prime})$, implying

$\frac{F(\theta^{\prime}\mid x)}{1-F(\theta^{\prime}\mid
x)}\leq\frac{F(\theta^{\prime}\mid x^{\prime})}{1-F(\theta^{\prime}\mid
x^{\prime})}$

or equivalently

$\frac{f(\theta^{\prime}\mid x^{\prime})}{f(\theta\mid
x^{\prime})}\geq\frac{f(\theta^{\prime}\mid x)}{f(\theta\mid x)}.$

Applying Bayes’ rule again, we can rewrite the above as
$\frac{f(x\mid\theta)}{f(x^{\prime}\mid\theta)}\geq\frac{f(x\mid\theta^{\prime})}{f(x^{\prime}\mid\theta^{\prime})}$,
which is the desired conclusion.

######  Remark 3.4.

Milgrom (1981)’s result is not precisely the proposition above, but instead
the equivalence between strict MLRP (as defined in Definition 3.2) and a
definition of “more favorable” that replaces FOSD with strict FOSD.

Specifically, say that $F$ strictly first-order stochastically dominates
$\widetilde{F}$ if $F(\theta)\leq\widetilde{F}(\theta)$ everywhere with strict
inequality at some $\theta$. (Equivalently, $\int u(\theta)dF(\theta)>\int
u(\theta)d\widetilde{F}(\theta)$ for every strictly increasing function
$u:\mathbb{R}\rightarrow\mathbb{R}$.) Say that $x$ is strictly more favorable
than $x^{\prime}$ if for every prior distribution $F_{\theta}$, the posterior
distribution $F_{\theta\mid X}(\cdot\mid x)$ strictly first-order
stochastically dominates $F_{\theta\mid X}(\cdot\mid x^{\prime})$. Then, by
substituting strict inequalities in place of weak inequalities in the proof
above where appropriate, we can conclude that
$\\{f_{X\mid\theta}(\cdot\mid\theta)\\}$ satisfies strict MLRP if and only if
$x$ is strictly more favorable than $x^{\prime}$.222Indeed, the same proof
demonstrates a stronger (if slightly more cumbersome to state) result: If and
only if $\\{f_{X\mid\theta}(\cdot\mid\theta)\\}$ satisfies strict MLRP, then
$F_{\theta\mid X}(\theta\mid x)<F_{\theta\mid X}(\theta\mid x^{\prime})$ at
every $\theta$ such that $0<F(\theta)<1$.

We conclude by briefly summarizing other notions of positive dependence and
placing the above properties relative to these.

  * Positive covariance (C): $Cov(X,\theta)\geq 0$

  * Positive quadrant dependence (QD): $Cov(g(X),h(\theta))\geq 0$ for all non-decreasing functions $g$ and $h$

  * Association (As): $Cov(g(X,\theta),h(X,\theta))\geq 0$ for all non-decreasing functions $g$ and $h$

  * Left-Tail Decreasing (LT): For all $x$, $F_{X\mid\theta}(X\leq x\mid\theta\leq t)$ is non-increasing in $t$, and for all $t$, $F_{\theta\mid X}(\theta\leq t\mid X\leq x)$ is non-increasing in $x$.

  * Inverse Hazard Rate Decreasing (IHR): For all $x$, $F_{X\mid\theta}(x\mid t)/f_{X\mid\theta}(x\mid t)$ is non-increasing in $t$, and for all $t$, $F_{\theta\mid X}(t\mid x)/f_{\theta\mid X}(t\mid x)$ is non-increasing in $x$.

These properties are extensively studied in, for example, Lehmann (1966),
Esary, Proschan and Walkup (1967), de Castro (2009), and Chapter 3 of
Balakrishna and Lai (2009). The following chain of implications is summarized
in de Castro (2009):

######  Theorem 3.1.

_(A)_ $\Longleftrightarrow$ _(MLRP)_ $\Longrightarrow$ _(IHR)_
$\Longrightarrow$ _(FOSD)_ $\Longrightarrow$ _(LT)_ $\Longrightarrow$ _(As)_
$\Longrightarrow$ _(QD)_ $\Longrightarrow$ _(C)_

Thus the standard properties of affiliation and MLRP are in fact strong,
implying all of the other properties but not in general implied by them. These
properties are equivalent to one another in the special case in which the two
variables are jointly normal.

######  Exercise 3.4 (G).

Suppose $(X_{1},\dots,X_{n})$ are jointly normal and exchangeable, where
$\sigma^{2}=\operatorname*{\text{Var}}(X_{i})$ for each $i$, and
$\rho=Cov(X_{i},X_{j})$ for each pair of indices $i,j$. Prove that these
variables are affiliated if and only if $\rho\geq 0$.

######  Hint.

Use the fact given in Exercise 2.10.

####  3.3 When These Conditions Fail

An example from Lagziel and Lehrer (2019) demonstrates the kind of
counterintuitive result that can hold in settings where (A) and (MLRP) fail.

An editor chooses which papers to publish. Papers have unknown quality graded
on a 9-point scale (A+, A, A-, B+, B, B-, C+,C,C-), whose prior distribution
is given in Figure 3.2.

Figure 3.2: Distribution of Papers’ Quality

The editor learns about quality via a noisy refereeing process, which
generates an unbiased signal $X$ about the paper. The realization of $X$ is
equal to the true quality with probability 0.8, and otherwise exactly two
levels higher or lower than the true quality (each with probability 0.1). The
distribution of $X$ is reported in Figure 3.3:

Figure 3.3: Distribution of Referee Signal

The editor chooses a threshold and accepts all papers whose expected quality
(given the referee’s report) exceeds this threshold. Intuitively, we may
expect that the editor faces a tradeoff between publishing more papers versus
publishing higher quality papers, where a higher threshold corresponds to
publishing fewer but higher quality papers.

But observe that if the editor chooses to publish only papers with an expected
quality that (weakly) exceeds $A$ (i.e., the top-rated 5% of papers), then the
expected value of the published work is close to $B+$. If the editor lowers
the bar to $A-$ (i.e., the top-rated 13%), then the expected value of the
published work _increases_ to $A-$. Not only are more papers published, but
their expected quality is higher.

In this example, we have $\mathbb{E}(\theta\mid X=x)<\mathbb{E}(\theta\mid
X=x^{\prime})$ even while $x>x^{\prime}$, so clearly the posterior belief at
$x^{\prime}$ does not first-order stochastically dominate the posterior belief
at $x$. Chambers and Healy (2011) demonstrate an even stronger reversal by
constructing signals such that the posterior belief at the lower signal
realization first-order stochastically dominates the posterior belief at the
higher signal realization. Notably, their result relies on natural-seeming
signals that satisfy various reasonable properties.

######  Theorem 3.2.

For every non-degenerate, bounded $\theta$ there exists a signal structure $X$
and two signal realizations $x^{\prime}>x$ such that $f(\theta\mid
X=x^{\prime})$ is strictly first-order stochastically dominated by
$f(\theta\mid X=x)$. Furthermore, $X$ can be chosen to have the following
properties: i) $X$ is an additive signal structure, and ii) $e:=X-\theta$ is
mean-zero, symmetric, quasiconcave, and has bounded support.

See Heinsalu (2020) for a strengthening of Lagziel and Lehrer (2019)’s example
using this result, in which lowering the threshold not only increases the
expected quality, but results in a quality distribution for published papers
that first-order stochastically dominates the one that would obtain at the
higher threshold.

####  3.4 Additional Exercises

######  Exercise 3.5 (G).

Let $Z_{1},\dots,Z_{n}$ be affiliated and let
$h:\mathbb{R}^{n}\rightarrow\mathbb{R}$ be any function that is nondecreasing
in each of its coordinates. Prove that the function

$\mathbb{E}(h(Z_{1},\dots,Z_{n})\mid Z_{1}=z_{1})$

is nondecreasing in $z_{1}$.

######  Exercise 3.6 (G).

Let $X$ be any real-valued random variable and let
$f:\mathbb{R}\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ be
bounded nondecreasing functions. Prove that $Cov(f(X),g(X))\geq 0$. (Do not
apply the FKG inequality.)

######  Hint.

There are at least two short proofs, one that uses Fubini’s theorem and the
fact that $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)$ for any independent
random variables $X$ and $Y$, and another which relies entirely on elementary
(if not obvious) arguments.

######  Exercise 3.7 (G).

Ann and Bob share the same prior $p$ over an unknown real-valued state
$\theta$, and observe a common realization of the signal $X$, but disagree
about the distribution of $S$. Ann believes that $X=\theta+\varepsilon$, where
$\theta\perp\\!\\!\\!\perp\varepsilon$ and $\varepsilon$ is a real-valued
noise term with density $f_{\varepsilon}$. Bob believes that
$X=\theta+\varepsilon+\Delta$ for some $\Delta>0$. That is, Ann perceives Bob
as adding $\Delta$ to the realization of the signal, while Bob perceives Ann
as subtracting $\Delta$ from the realization of the signal.

Let $f_{A}$ denote the joint density of $(\theta,X)$ according to Ann’s model
and $f_{B}$ denote the joint density according to Bob’s model, with
$\mathbb{E}^{A}$ and $\mathbb{E}^{B}$ denoting their respective expectation
operators. Impose the monotone likelihood-ratio property on
$\\{f_{A}(\cdot\mid\omega)\\}$, that is,

$\frac{f_{A}(x^{\prime}\mid\theta^{\prime})}{f_{A}(x\mid\theta^{\prime})}\geq\frac{f_{A}(x^{\prime}\mid\theta)}{f_{A}(x\mid\theta)}\quad\forall
x^{\prime}>x,\theta^{\prime}>\theta$

  * (a)

Prove that $\\{f_{B}(\cdot\mid\theta)\\}$ also satisfies MLRP.

  * (b)

Prove that $\mathbb{E}^{A}[\mathbb{E}^{B}[\theta\mid X]]$ is decreasing in
$\Delta$, and interpret this result.

  * (c)

Suppose that Ann and Bob now additionally observe a common vector of iid
signals $(Y_{1},Y_{2},\dots,Y_{N})$ where each $Y_{i}=\theta+\delta_{i}$ with
$\theta\perp\\!\\!\\!\perp\delta_{i}$ and $\delta_{i}$ are iid across signals.
Prove that

$\mathbb{E}^{A}[\mathbb{E}^{B}[\theta\mid
X,Y_{1},\dots,Y_{N}]]\leq\mathbb{E}^{A}[\mathbb{E}^{B}[\theta\mid
X,Y_{1},\dots,Y_{N},Y_{N+1}]]$

for every $N\geq 1$. Again, interpret the result.

###  Chapter 4 Comparing Information I: The Blackwell Order

When an agent has access to a choice between multiple signals, we may desire
to order these signals based on how informative they are. Intuition can guide
us on how to define such an ordering in specific cases, for example:

  * •

Adding noise to a signal decreases its informativeness.

  * •

Observing the realization of $(X,Y)$ is more informative than observing the
realization of $X$ alone.

Any informativeness ordering should satisfy these properties, but there are
different ways to generalize from here. One approach is to fix a decision
problem and characterize the instrumental value of the signal for that
decision problem. Alternatively, we could look for a universal informativeness
ordering over signals that holds for all decision problems (as will be the
focus of this chapter). Yet another approach is to quantify the “signal
content” contained within the signal based on the physical difficulty of
producing or processing that information (see Chapter 5).

In this chapter we introduce the Blackwell partial order on signals, which
considers one signal to be more informative than another if it is more useful
for all decision problems. If $\sigma$ dominates $\sigma^{\prime}$ in this
Blackwell order, we will say that _$\sigma$ is more informative than
$\sigma^{\prime}$_ or that _$\sigma$ Blackwell-dominates $\sigma^{\prime}$_.
The following sections demonstrate five perspectives on this order,
culminating in Blackwell’s theorem (establishing their equivalence) and the
proof of this theorem.

####  4.1 Garblings

We may consider a signal to be more informative than another if the latter is
a noised-up version of the former.

######  Definition 4.1 (Markov matrix).

A matrix $M$ is a _Markov matrix_ if its entries are nonnegative and its rows
sum to 1.

Recall that when the set of states and the set of signal realizations are
finite, we can represent any signal as a Markov matrix.

######  Definition 4.2 (Garblings, Finite Version).

Markov matrix $P$ is a _garbling_ of Markov matrix $Q$ if there exists a
Markov matrix $M$ s.t. $QM=P$

######  Example 4.1.

Let $\Theta=\\{\theta_{1},\theta_{2}\\}$ and consider the signals

$P=\left(\begin{array}[]{cc}3/4&1/4\\\
1/4&3/4\end{array}\right)\quad\quad\quad
Q=\left(\begin{array}[]{cccc}9/16&3/16&3/16&1/16\\\
1/16&3/16&3/16&9/16\end{array}\right)$

where as usual the rows are indexed to states and the columns are indexed to
signal realizations. Then since

$\underbrace{\left(\begin{array}[]{cccc}9/16&3/16&3/16&1/16\\\
1/16&3/16&3/16&9/16\end{array}\right)}_{Q}\underbrace{\left(\begin{array}[]{cc}1&0\\\
1&0\\\ 0&1\\\ 0&1\\\
\end{array}\right)}_{M}=\underbrace{\left(\begin{array}[]{cc}3/4&1/4\\\
1/4&3/4\end{array}\right)}_{P}$

where $M$ is a Markov matrix, we can conclude that $P$ is a garbling of $Q$.

This example has a particularly nice intuition. Label the possible
realizations of the first information structure $P$ as $s_{1}$ and $s_{2}$,
and consider the signal which is two independent realizations of $P$. The set
of possible realizations of this new signal is then
$\\{s_{1}s_{1},s_{1}s_{2},s_{2}s_{1},s_{2}s_{2}\\}$ with the conditional
distributions over these realizations given precisely by $Q$. So observing $P$
is statistically equivalent to observing $Q$ and forgetting the second
realization. Clearly then $P$ is less informative than $Q$.

######  Exercise 4.1 (U).

The state space is $\Theta=\\{\theta_{1},\theta_{2}\\}$ and two signals are
described by the following signal structures

$P=\left(\begin{array}[]{cccc}&s_{1}&s_{2}&s_{3}\\\ \theta_{1}&2/3&1/3&0\\\
\theta_{2}&0&1/3&2/3\end{array}\right)$

$Q=\left(\begin{array}[]{ccc}&\tilde{s}_{1}&\tilde{s}_{2}\\\
\theta_{1}&1/6&5/6\\\ \theta_{2}&5/6&1/6\end{array}\right)$

Show that $Q$ is a garbling of $P$. (How does this exercise relate to Example
4.1?)

######  Exercise 4.2 (G).

For any $q\in[1/2,1]$, define the matrix

$S_{q}=\left(\begin{array}[]{cc}q&1-q\\\ 1-q&q\end{array}\right)$

Suppose $q,q^{\prime}\in[1/2,1]$ with $q>q^{\prime}$. Prove that
$S_{q^{\prime}}$ is a garbling of $S_{q^{\prime}}$.

More generally, we can replace the Markov matrix $M$ in Definition 4.2 with a
Markov kernel.

######  Definition 4.3 (Garblings, General Version).

The signal $\sigma^{\prime}:\Theta\to\Delta(S^{\prime})$ is a _garbling_ of
the signal $\sigma:\Theta\to\Delta(S)$ if there exists a Markov kernel
$\gamma:S\to\Delta(S^{\prime})$ such that

$\sigma^{\prime}(s^{\prime}\mid\theta)=\int_{s\in S}\gamma(s^{\prime}\mid
s)\sigma(s\mid\theta)ds$

######  Example 4.2.

Let $\theta$, $\varepsilon$, and $\delta$ be independent real-valued random
variables with densities $f_{\theta}$, $f_{\varepsilon}$, and $f_{\delta}$.
Then the signal $X=\theta+\varepsilon+\delta$ is a garbling of
$Y=\theta+\varepsilon$, since

$f_{X\mid\theta}(x\mid
t)=\int_{y\in\mathbb{R}}f_{\delta}(x-y)f_{Y\mid\theta}(y\mid t)dy$

where $f_{\delta}$ is a Markov kernel.

######  Example 4.3.

Consider an arbitrary finite set $\Theta$ and let $I$ be the
$|\Theta|\times|\Theta|$ identity matrix. Then for any set of signal
realizations $S$ and any $|\Theta|\times|S|$ Markov matrix $Q$, we have
$IQ=Q$, so $Q$ is a garbling of $I$.

######  Exercise 4.3 (U).

Is it possible for $P$ and $Q$ to both be garblings of one another if $P\neq
Q$? Provide an example if so, and otherwise prove that it is not possible.

######  Remark 4.1.

Let $X$ and $X^{\prime}$ respectively denote the random realizations of the
signals $\sigma$ and $\sigma^{\prime}$. Then $\theta$, $X$, and $X^{\prime}$
are random variables which can be defined on a common probability space. The
property that $\sigma^{\prime}$ is a garbling of $\sigma$ does not however pin
down the joint distribution of $(\theta,X,X^{\prime})$. What it guarantees is
that there is a way of generating these variables such that $\theta$ is
independent of $X^{\prime}$ conditional on $X$, in which case $\theta\mid X$
is identical in distribution to $\theta\mid X,X^{\prime}$.111First draw the
state $\theta$, then draw $X$ according to its conditional distribution, and
finally draw $X^{\prime}$ according to the garbling kernel $\gamma$,
independent of $\theta$. Other ways of generating these variables—still
consistent with $\sigma^{\prime}$ being a garbling of $\sigma$—can yield
different relationships.

For example, suppose $\theta\sim\mathcal{N}(0,1)$ while

$\displaystyle X$ $\displaystyle=\theta+\varepsilon_{1}$ $\displaystyle
X^{\prime}$ $\displaystyle=\theta+\varepsilon_{2}$

where $\varepsilon_{1}\sim\mathcal{N}(0,1)$ and
$\varepsilon_{2}\sim\mathcal{N}(0,2)$ are both independent of $\theta$. Then
clearly the latter signal is a garbling of the former. If we further assume
that $\varepsilon_{2}=\varepsilon_{1}+\delta$ where
$\delta\sim\mathcal{N}(0,1)$ is an independent noise term, then the following
statements are true:

  * •

$X^{\prime}$ is independent of $\theta$ conditional on $X$.

  * •

$X^{\prime}$ is not independent of $X$ conditional on $\theta$ (since they are
further related through the common component $\varepsilon_{1}$).

On the other hand, if we assume that $\varepsilon_{1}$ and $\varepsilon_{2}$
are independent, then the statements above are reversed:

  * •

$X^{\prime}$ is not independent of $\theta$ conditional on $X$ (since
$X^{\prime}$ provides additional information about $\theta$ beyond what is
revealed by $X$).

  * •

$X^{\prime}$ is independent of $X$ conditional on $\theta$.

Thus in general, the assumption that two signals are related by a garbling
does not imply either conditional independence statement given above.

####  4.2 Decision Problems

Our next two definitions are based on the instrumental value of the signal for
decision problems.

######  Definition 4.4.

A decision problem is any pair $\mathbb{D}=(A,u)$ where $A$ is an action set
and $u:A\times\Theta\rightarrow\mathbb{R}$ is a payoff function.

The full decision problem is described as follows. Fix a prior
$p\in\Delta(\Theta)$ and a signal $\sigma:\Theta\rightarrow\Delta(S)$.

  1. 1.

The agent chooses a strategy $\alpha:S\rightarrow A$.

  2. 2.

The state $\theta\sim p$ and signal realization $s\sim\sigma(\cdot\mid\theta)$
are realized, and the agent takes action $\alpha(s)$. The agent’s payoff is
$u(\alpha(s),\theta)$.

Without the benefit of further information, the best expected payoff the agent
can achieve is

$\sup_{a\in A}\mathbb{E}\left[u(a,\theta)\right]$ (4.1)

With the benefit of the signal, the agent can achieve an expected payoff of

$\sup_{\alpha:S\rightarrow
A}\mathbb{E}\left[u(\alpha(s),\theta)\right]=\mathbb{E}\left[\sup_{a\in
A}\mathbb{E}\left[u(a,\theta)\mid s\right]\right]$ (4.2)

where we abuse notation on the LHS by using $s$ to denote the random variable
which is the realization of the signal. On the RHS, the inner expectation is
with respect to uncertainty about $\theta$ (conditional on the realization of
$s$) and the outer expectation is with respect to uncertainty about $s$. One
measure of the value of the signal is the difference in these expected
payoffs, i.e.,

$\displaystyle V_{\mathbb{D},p}(\sigma)\equiv\mathbb{E}\left[\sup_{a\in
A}\mathbb{E}\left[u(a,\theta)\mid s\right]\right]-\sup_{a\in
A}\mathbb{E}\left[u(a,\theta)\right]$

where $\mathbb{D}=(A,u)$ is the decision problem and $p\in\Delta(\Theta)$ is
the agent’s prior.

######  Remark 4.2.

It is without loss to assume the use of pure strategies above, but in the
subsequent development of the Blackwell order it will be useful to replace $a$
with a mixed strategy $\alpha\in\Delta(A)$ in (4.1) and $\alpha$ with a
stochastic map $\alpha:S\rightarrow\Delta(A)$ in (4.2).

######  Example 4.4.

Suppose $\Theta=\\{\theta_{1},\theta_{2}\\}$ with a uniform prior $p$. The
decision problem is $(A,u)$ where $A=\\{a_{1},a_{2}\\}$ and the utility
function $u:A\times\Theta\rightarrow\mathbb{R}$ assigns a payoff of 1 when the
action matches the state, and zero otherwise. The signal $\sigma$ is

$\begin{array}[]{ccc}&s_{1}&s_{2}\\\ \theta_{1}&q&1-q\\\
\theta_{2}&1-q&q\end{array}$

where $q>1/2$. Then the agent’s ex-ante payoff is maximized by choosing the
strategy $\alpha$ that maps $s_{1}$ to action $a_{1}$ and $s_{2}$ to action
$a_{2}$, with an expected payoff of $q$. In the absence of information the
agent’s best payoff is $1/2$, so $V_{\mathbf{D},p}(\sigma)=q-1/2$.

######  Example 4.5.

Suppose $\Theta=\mathbb{R}$ with a prior
$\theta\sim\mathcal{N}(0,\sigma_{\theta}^{2})$. The decision problem is
$(A,u)$ where $A=\mathbb{R}$ and $u(a,\theta)=-(a-\theta)^{2}$. The signal is
$X=\theta+\varepsilon$ where
$\varepsilon\sim\mathcal{N}(0,\sigma_{\varepsilon}^{2})$ is independent of
$\theta$. Then the agent’s ex-ante payoff is maximized by choosing the
strategy $\alpha(x)=\mathbb{E}(\theta\mid X=x)$, with an expected payoff of

$\mathbb{E}_{X}\left[-(\mathbb{E}(\theta\mid
X)-\theta)^{2}\right]=\mathbb{E}_{X}\left[-\operatorname*{\text{Var}}(\theta\mid
X)\right]=-\frac{\sigma_{\theta}^{2}\sigma_{\varepsilon}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}$

using Fact 2.2 in the final equality (and in particular, the property that
posterior variance is independent of the signal realization). In the absence
of information the agent’s best payoff is
$-\operatorname*{\text{Var}}(\theta)=-\sigma_{\theta}^{2}$, so
$V_{\mathbf{D},p}(X)=\frac{\sigma_{\theta}^{2}\sigma_{\varepsilon}^{2}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}-\sigma_{\theta}^{2}=\frac{\sigma_{\theta}^{4}}{\sigma_{\theta}^{2}+\sigma_{\varepsilon}^{2}}$.

In any specific decision problem, a signal that is informative (in the sense
of moving the agent’s beliefs about $\theta$) may nevertheless have no
instrumental value, as demonstrated in the following exercise.

######  Exercise 4.4 (U).

Suppose $\Theta=\\{1,2\\}$ and let the prior $p$ assign equal probability to
either state. Consider the decision problem $(A,u)$ with $A=\\{1,2\\}$ and
$u(a,\theta)=\mathbbm{1}(a=\theta)$. Let $\sigma_{P}$ and $\sigma_{Q}$
respectively be the two signals described by $P$ and $Q$ in Example 4.1. Show
that $V_{\mathbb{D},p}(\sigma_{P})=V_{\mathbb{D},p}(\sigma_{Q})$. That is, the
second independent observation of signal $P$ has no value to the agent over
the first.

#####  4.2.1 Uniformly Better

We’ll say that a signal is more informative than another if it is more useful
in every decision problem and for every prior belief.

######  Definition 4.5.

The signal $\sigma$ is more informative than $\sigma^{\prime}$ if
$V_{\mathbb{D},p}(\sigma)\geq V_{\mathbb{D},p}(\sigma^{\prime})$ for every
decision problem $\mathbb{D}$ and every prior $p$.

This is a strong condition, and we generally won’t be able to order signals in
this way.

######  Exercise 4.5 (U).

Let $\Theta=\\{\theta_{1},\theta_{2},\theta_{3}\\}$ with a uniform prior $p$.
Let $A=\\{a_{1},a_{2}\\}$ and consider two utility functions: Let
$u:A\times\Theta\rightarrow\mathbb{R}$ take value 1 if
$(a,\theta)\in\\{(a_{1},\theta_{1}),(a_{2},\theta_{2}),(a_{2},\theta_{3})\\}$,
and value 0 otherwise. Let $u^{\prime}:A\times\theta\rightarrow\mathbb{R}$
take value 1 if
$(a,\theta)\in\\{(a_{1},\theta_{1}),(a_{2},\theta_{2}),(a_{1},\theta_{3})\\}$,
and value 0 otherwise. Consider the two information structures

$\sigma$:  $s_{1}$ $s_{2}$ $\theta_{1}$ $1$ $0$ $\theta_{2}$ $0$ $1$
$\theta_{3}$ $0$ $1$ $\sigma^{\prime}$:  $s_{1}$ $s_{2}$ $\theta_{1}$ $1$ $0$
$\theta_{2}$ $0$ $1$ $\theta_{3}$ $1$ $0$

Show that $V_{\mathbb{D},p}(\sigma)>V_{\mathbb{D},p}(\sigma^{\prime})$ where
$\mathbb{D}=(A,u)$, but
$V_{\mathbb{D}^{\prime},p}(\sigma)>V_{\mathbb{D}^{\prime},p}(\sigma^{\prime})$
where $\mathbb{D}^{\prime}=(A,u^{\prime})$, i.e. the agent prefers the first
information given payoffs $u$ and the second given payoffs $u^{\prime}$.

The definition of uniformly better varies both the decision problem and also
the prior, but the additional flexibility due to arbitrary priors is not
substantial:

######  Exercise 4.6 (G).

Prove that if there is a full-support prior $p_{0}\in\Delta(\Theta)$ such that

$V_{\mathbb{D},p_{0}}(\sigma)\geq
V_{\mathbb{D},p_{0}}(\sigma^{\prime})\quad\mbox{for every decision problem
$\mathbb{D}$}$

then $\sigma$ is more informative than $\sigma^{\prime}$.

#####  4.2.2 Feasible Actions

Our third definition says that a signal is more informative if observing the
realization of the signal allows the agent to more effectively tailor his
action to the state.

######  Definition 4.6.

Fix any action set $A$. A conditional distribution over actions
$d:\Theta\to\Delta(A)$ is _feasible under $\sigma:\Theta\rightarrow\Delta(S)$_
if there exists a mapping $\alpha:S\to\Delta(A)$ such that

$d(a\mid\theta)=\int_{s\in S}\alpha(a\mid s)\sigma(s\mid\theta)ds$

We’ll use $\Lambda_{\sigma}(A)$ to denote the set of all feasible
distributions under $\sigma$ given action set $A$.

When $\sigma$ is a fully revealing signal (e.g.,
$\sigma:\Theta\rightarrow\Delta(\Theta)$ satisfying
$\sigma(\theta\mid\theta)=1$ for every $\theta$), then every mapping
$d:\Theta\rightarrow\Delta(A)$ is feasible under $\sigma$. (Simply set
$\alpha=d$.) When $\sigma$ is uninformative—for example, a constant—then
$\Lambda_{\theta}(A)$ consists of all mappings $d:\Theta\rightarrow\Delta(A)$
that take each state into the same distribution over actions. Larger sets
$\Lambda_{\sigma}(A)$ allow the agent more flexibility in tailoring his action
to the state, and in this sense are more valuable.

######  Remark 4.3.

Observe that $\alpha$ is itself a Markov kernel, so $d$ can be interpreted as
a garbling of $\sigma$ where $A$ is the set of signal realizations.

####  4.3 Dispersion of Posterior Beliefs

Our final perspective adopts the view on a signal introduced in Section 2.2,
where a signal is identified with the distribution over posterior beliefs that
it induces. We consider the dispersion of these posterior beliefs. Given an
uninformative signal, the agent’s posterior is deterministically equal to the
agent’s prior, so there is no dispersion. And if the signal reveals the state
directly, then the posterior belief is a point mass on the true state, which
“maximally varies” depending on the realization of the signal.

We may expect more informative signals to be associated with more dispersed
beliefs, but the measure of dispersion is important. For example, using
variance to measure dispersion yields a complete order on signals, which
cannot possibly be equivalent to the (strict) partial order described in the
previous definitions. Below we define two alternative measures for
dispersion—mean-preserving spreads and dominance in the convex order—which
will turn out to again characterize the previous partial order on signals.

#####  4.3.1 Mean-Preserving Spreads

######  Definition 4.7.

A distribution of posterior beliefs $F\in\Delta(\Delta(\Theta))$ is a _mean-
preserving spread_ of another distribution $\widetilde{F}$ if there exist
$\Delta(\Theta)$-valued random variables $Z,\widetilde{Z}$ satisfying the
following conditions:

  1. 1.

$Z\sim F,\widetilde{Z}\sim\widetilde{F}$

  2. 2.

$\mathbb{E}(Z\mid\widetilde{Z})=\widetilde{Z}$ (thus in particular
$\mathbb{E}(Z)=\mathbb{E}(\widetilde{Z})$)

The name “mean-preserving spread” reflects that each realization of
$\widetilde{Z}$ is spread out into a random $Z$ with the same mean. When $Z$
and $\widetilde{Z}$ are both real-valued, then the second condition can also
be stated as $Z=\widetilde{Z}+\varepsilon$ for some random variable
$\varepsilon$ satisfying $\mathbb{E}(\varepsilon\mid\widetilde{Z})=0$.

######  Example 4.6.

Consider the two signals

$P=\left(\begin{array}[]{cc}3/4&1/4\\\
1/4&3/4\end{array}\right)\quad\quad\quad
Q=\left(\begin{array}[]{cccc}9/16&3/16&3/16&1/16\\\
1/16&3/16&3/16&9/16\end{array}\right)$

from Example 4.1, where the set of states is
$\Theta=\\{\theta_{1},\theta_{2}\\}$. Let the agent’s prior be uniform over
these states. Then the agent has two possible posterior beliefs after
observing $P$, $(3/4,1/4)$ and $(1/4,3/4)$, which are equally likely. We will
write the distribution of posterior beliefs as

$F_{P}=1/2\cdot(3/4,1/4)+1/2\cdot(1/4,3/4).$

Under $Q$, the distribution of posterior beliefs is instead

$F_{Q}=5/16\cdot(9/10,1/10)+3/8\cdot(1/2,1/2)+5/16\cdot(1/10,9/10).$

We will now show that $F_{Q}$ is a mean-preserving spread of $F_{P}$. Let
$\widetilde{Z}$ be a random variable satisfying $\widetilde{Z}\sim F_{P}$ and
construct the random variable $Z$ given $\widetilde{Z}$ as follows:

  * •

If $\widetilde{Z}=(1/4,3/4)$ then $Z=(1/10,9/10)$ with probability $5/8$ and
$Z=(1/2,1/2)$ with probability $3/8$.

  * •

If $\widetilde{Z}=(3/4,1/4)$ then $Z=(9/10,1/10)$ with probability $5/8$ and
$Z=(1/2,1/2)$ with probability $3/8$.

Then $\mathbb{E}(Z\mid\widetilde{Z})=\widetilde{Z}$ and also $Z\sim F_{Q}$, so
$F_{Q}$ is a mean-preserving spread of $F_{P}$ as desired. This construction
is depicted in Figure 4.1.

Figure 4.1: Depiction of the mean-preserving spread, where the numbers
represent the probability of state $\theta_{1}$.

######  Exercise 4.7 (G).

Let $\Theta=\\{\theta_{1},\theta_{2}\\}$ and consider the two signals

$P=\left(\begin{array}[]{cc}2/3&1/3\\\
1/4&3/4\end{array}\right)\quad\quad\quad
Q=\left(\begin{array}[]{cccc}1/3&1/2&1/6\\\ 1/8&1/2&3/8\end{array}\right)$

Define $F_{P}$ to be the distribution of posterior beliefs induced by $P$ and
$F_{Q}$ to be the distribution of posterior beliefs induced by $Q$. Prove that
$F_{P}$ is a mean-preserving spread of $F_{Q}$.

######  Exercise 4.8 (G).

Suppose $Y_{1},Y_{2},\dots,Y_{n}$ are independent and identically distributed
random variables, and define $\overline{Y}_{n}=\frac{1}{n}\sum_{i=1}^{n}Y_{i}$
to be their sample average. Let $n^{\prime}<n$ and define
$\overline{Y}_{n^{\prime}}=\frac{1}{n^{\prime}}\sum_{i=1}^{n^{\prime}}Y_{i}$.
Prove that the distribution of $\overline{Y}_{n^{\prime}}$ is a mean
preserving spread of the distribution of $\overline{Y}_{n}$.

#####  4.3.2 Convex Order

Another partial order of dispersion is the following:

######  Definition 4.8.

A distribution of posterior beliefs $F\in\Delta(\Delta(\Theta))$ _dominates_
another distribution $G$ _in the convex order_ if for every continuous convex
function $h:\Delta(\Theta)\rightarrow\mathbb{R}$,

$\int_{\Delta(\Theta)}h(p)dF(p)\geq\int_{\Delta(\Theta)}h(p)dG(p)$

This implies that $F$ and $G$ have the same mean (choosing $h(p)=p$) and that
$F$ has the larger variance (choosing $h(p)=\|p\|^{2}$).

You may recall the concept of _second order stochastic dominance_ :

######  Definition 4.9.

For any lotteries $F$ and $G$, $F$ _second-order stochastically dominates_ $G$
if and only if

$\int_{\Delta(\Theta)}u(p)dF(p)\geq\int_{\Delta(\Theta)}u(p)dG(p)$

for every nondecreasing and concave utility function $u$.

Dominance in the convex order is stronger than SOSD.

######  Exercise 4.9 (G).

Prove that if $F$ dominates $G$ in the convex order, then $G$ second order
stochastically dominates $F$.

The converse is not in general true.

######  Example 4.7.

Let $G$ be a distribution uniform on $[1,2]$ and let $F$ be a point mass at
zero. Then $G$ second order stochastically dominates $F$ but $F$ does not
dominate $G$ in the convex order.

Intuitively, second-order stochastic dominance confounds changes in the
dispersion of the distribution with shifts in the distribution, while
dominance in the convex order isolates the former comparison.

####  4.4 Blackwell’s Theorem and Proof

We now state and prove Blackwell (1951)’s theorem, which demonstrates
equivalence of these five definitions. For the proof we will work with finite
sets (in particular assuming finite $\Theta$) but several parts of the proof
extend more generally.

######  Theorem 4.1.

The following are equivalent:

  1. 1.

$\sigma^{\prime}$ is a garbling of $\sigma$.

  2. 2.

$\sigma$ is more informative than $\sigma^{\prime}$.

  3. 3.

$\Lambda_{\sigma}(A)\supseteq\Lambda_{\sigma^{\prime}}(A)$ for every finite
action set $A$.

  4. 4.

For any prior on $\Theta$, if we define $F$ and $F^{\prime}$ to be the
distributions of posterior beliefs induced by $\sigma$ and $\sigma^{\prime}$
(under this prior), then $F$ is a mean-preserving spread of $F^{\prime}$.

  5. 5.

For any prior on $\Theta$, if we define $F$ and $F^{\prime}$ to be the