Information Hierarchies
Abstract
If experiment A is Blackwell more informative than experiment B, it is always possible that A and B are induced by signals A′ and B′ such that A′ is a refinement of B′, that is, A′ entails observing B′ plus some additional information. We first show that this result does not extend beyond pairs of experiments: There exist collections of experiments that cannot be induced by a collection of signals so that whenever two experiments are Blackwell ordered, the associated signals are refinement ordered. In other words, sometimes it is impossible for more informed agents to know everything that less informed agents know. More broadly, define an information hierarchy to be a partially ordered set that ranks experiments in terms of informativeness. Is it the case that for any choice of experiments indexed on the hierarchy such that higher experiments are Blackwell more informative, there are signals that induce these experiments with higher signals being refinements of lower signals? We show that the answer is affirmative if and only if the undirected graph of the information hierarchy is a forest.
1 Introduction
There are two distinct things we might mean when we say that A is more informed than B. One is that A's information about some state of the world is more accurate than B's. The other is that A knows everything that B knows; that is, A observes what B observes, plus something else. The first notion is typically formalized by representing a source of information as an experiment, that is, a distribution of posterior beliefs (Blackwell (1953)).1 We then say that A is Blackwell more informed than B if A's beliefs are a mean-preserving spread of B's beliefs. The second notion is formalized by representing a source of information as a partition of some expanded state space (Green and Stokey (1978)). This formulation distinguishes payoff-relevant states (Ω) from the realizations of signals conditional on those states (X). We then say that A knows everything that B knows if A's partition is a refinement of B's partition.
These two formalisms are closely related. Every signal, that is, partition of the expanded state space, induces an experiment. If one signal is a refinement of another, then the experiment it induces is Blackwell more informative. Moreover, Green and Stokey (1978) established a partial converse: given two Blackwell-ordered experiments, there exist refinement-ordered signals that induce those experiments. In other words, if A's information is more accurate than B's, it is always possible that A knows everything that B knows.
A natural question is whether Green and Stokey's result extends beyond pairs of experiments. Suppose we have some collection of experiments. Is it always the case that we can construct a collection of signals that induce those experiments, and whenever two experiments are Blackwell ordered, the two corresponding signals are refinement ordered?
We show that the answer is negative. We construct an example where A is Blackwell more informed than B and C, who are in turn more informed than D, and yet it cannot be that A knows everything that B and C know and that all three know everything that D knows.
To examine this issue more generally, we introduce the notion of an information hierarchy, which is simply a partially ordered set. We consider experiment allocations that assign an experiment to each element of the hierarchy, with the property that a higher element's experiment is Blackwell more informative than a lower element's experiment. We say an experiment allocation is constructible if it is possible to assign a signal to every element of the hierarchy so that each signal induces the corresponding experiment, and a higher element's signal is a refinement of a lower element's signal. If every monotone experiment allocation is constructible, we say that the information hierarchy is universally constructible.
Our main theorem characterizes the set of information hierarchies that are universally constructible. A partially ordered set (and thus an information hierarchy) is associated with an undirected graph, whose nodes are the elements of the set and whose edges are determined by the partial order.2 We establish that an information hierarchy is universally constructible if and only if this graph is a forest.
We also establish a generalization of this result. We consider a class of other economically meaningful orders on signals, and we show that, if we replace refinement with any one of these orders in our notion of constructibility, our theorem continues to hold.
We discuss two applications of our theorem. First, suppose there is an agent who observes information from several sources and an econometrician who does not know the data generating process behind these sources. We analyze the conditions under which the econometrician can rationalize the agent's reactions to this information. In particular, we show that two necessary conditions, Bayes plausibility and Blackwell monotonicity,3 do not suffice for rationalizability when there are more than two sources of information. Second, we study information design in organizations under the constraint that managers must have access to the information of their subordinates.
The rest of this paper proceeds as follows. Section 2 establishes basic definitions. Section 3 presents motivating examples. Section 4 contains the remaining definitions and our main result. Section 5 discusses extensions. Section 6 provides two applications. We discuss related literature in probability theory and information economics in Section 7. Section 8 is a conclusion.
2 States, Signals, and Experiments
There is a finite state space Ω and a prior .
A signal π is a finite partition of s.t.
, where S is the set of non-empty Lebesgue-measurable subsets of
(Green and Stokey (1978), Gentzkow and Kamenica (2017)). An element
is a signal realization. The interpretation of this formalism is that a random variable x, drawn uniformly from
, determines the signal realization conditional on the state. Thus, the conditional probability of s given ω is
where
denotes the Lebesgue measure. Importantly, if we know that one agent observed some signal, say
, and another agent observed some signal
, we can deduce not only how informed each agent is about ω, but also the joint distribution of their signal realizations (and thus their higher-order beliefs).
In contrast, a (Blackwell) experiment only specifies how informed an agent is about ω, without specifying her information about the beliefs of other agents. The standard definition of an experiment is a map from Ω to distributions over signal realizations, but for ease of exposition we identify each experiment with the distribution of beliefs it induces. Thus, we define an experiment, denoted by τ, as an element of that has finite support and satisfies
.
We write if π is a refinement of
, that is, every element of π is a subset of some element of
.4 If
, an agent who observes π has access to all the information available to the agent who observes
(and thus knows those agent's beliefs). We write
if τ is Blackwell more informative than
, that is, τ is a mean-preserving spread of
. If
, an agent who observes τ obtains a higher payoff than the agent who observes
in any decision problem.
We denote the set of all signals by Π. The pair is a lattice and we let ∨ denote the join, that is,
is the coarsest refinement of both π and
. Note that
is the signal that is equivalent to observing both π and
.








3 Illustrative Examples
There are four experiments, ,
,
, and
, some of which might be Blackwell more informative than others. We wish to know whether there exist signals
for
such that
- (C1)
and
- (C2)
implies
.
As we discussed in the Introduction, the answer to this question will shed light on (i) whether Green and Stokey's (1978) result on the equivalence of representing information as experiments versus signals generalizes beyond pairwise comparisons, and (ii) whether it is always possible that a more informed agent knows what all less informed agents know.
As we will see, for some Blackwell rankings of the experiments, the answer is yes, no matter what the the exact experiments are. For other rankings, this is not the case; there exist experiments for which there do not exist signals satisfying (C1) and (C2). We illustrate these results through three examples, depicted in Figure 1.

Example hierarchies.
Chain
Suppose . When experiments are ordered in this way, it is always possible to construct signals that satisfy (C1) and (C2), regardless of the particular experiments. This follows from Green and Stokey (1978): For any
and τ with
, there exists a π such that (i)
and (ii)
.7 In other words, take any signal that induces some experiment; there is a refinement of this signal that induces any particular more-informative experiment.
Now, consider any that induces
. Since
, there exists some signal
that induces
. Similarly, there is a
that induces
. Finally, there is a
that induces
. Hence, we constructed signals that satisfy (C1) and (C2).
Tree
Next, suppose ,
, and
is not comparable with
or
. In this case, it is again always possible to construct signals so that
and
implies
. The argument, however, is more subtle than for the chain. To see the issue, suppose we first go “up the tree” and proceed as in the previous example: we construct
and then
that induce
and
. Now, to assign a signal
, we have to go “down the tree.” But there is no guarantee that, given the
we constructed, there exists a
that is coarser than
and induces
.
We establish a result (Lemma 1), however, that tells us we can construct that induces
and has the property that
. (To do so, we build a
whose realizations are independent of the state given the realization of
.) Then, we replace the “provisional”
with
(which is of course also finer than
). We then apply a similar procedure to assign
and update the provisional
and
. Hence, we constructed signals that satisfy (C1) and (C2). This algorithm is discussed in greater detail in Section 4.
Diamond
Finally, suppose ,
, and
and
are not (necessarily) comparable. We now give an example of four experiments ordered in this manner such that there do not exist signals that satisfy (C1) and (C2).8
Let with a uniform prior. Since the state space is binary, we associate each belief with
. The rows in Figure 2 depict the four experiments, with the number of dots in a circle proportional to the probability mass on that belief. It is easy to see that these four experiments satisfy the given Blackwell ordering, with arrows (solid or dashed) showing the unique way to obtain a higher-ranked distribution as a mean-preserving spread of a lower-ranked distribution.

The diamond hierarchy is not universally constructible.
Now, to obtain a contradiction, suppose that the signals ,
,
, and
satisfy (C1) and (C2). First, note that these signals do not merely induce the respective marginal distributions of beliefs (i.e., the experiments), but they also determine the joint distribution of these beliefs. In particular, as noted earlier, whenever
, we must have
.
We use this fact to show that when , (i)
implies that
, while (ii)
implies that
with a strictly positive probability. To see (i), note that when
, we must have
and thus
(since
if and only if
). To see (ii), note that when
, there is a strictly positive probability that
(since
if and only if
); and when
, there is a strictly positive probability that
(since all of the other beliefs in the support of
are strictly lower than
). Thus, we have reached a contradiction.
4 Universal Constructibility
4.1 Information Hierarchies and Constructibility
These examples motivate the following issue. Partially order a collection of experiments so that higher experiments are Blackwell more informative than lower experiments. Is it possible to find signals that induce these experiments, and that satisfy the property that signals inducing higher experiments are refinements of signals inducing lower ones? We argued that this was in fact always possible when the ranking on experiments corresponded to the chain or the tree example above but was not necessarily possible when experiments were ranked as a diamond. The main result of this paper is an if-and-only-if characterization of the rankings of experiments such that suitable refinement-ordered signals exist for any experiments whose Blackwell order is consistent with the ranking.
We now provide the remaining definitions needed to state our result. An information hierarchy H is a finite partially ordered set . We refer to elements of N as nodes. Given
, we say that n covers
if
and there does not exist
with
. The graph of H, denoted
, is an undirected graph whose nodes are the elements of N with an edge between two nodes if one covers the other.
is a forest if there is at most one path between any two nodes.9
Fix an information hierarchy H and . An experiment allocation on H is a map that assigns an experiment to every node; experiment allocation β is monotone if
implies that
, that is, higher nodes are Blackwell more informed than lower nodes. A signal allocation on H is a map that assigns a signal to every node; signal allocation σ is monotone if
implies that
, that is, signals associated with higher nodes refine signals associated with lower nodes. A signal allocation σ induces an experiment allocation β if, for all n,
. A (monotone) experiment allocation on H is constructible if there exists a monotone signal allocation that induces it.10
We say that H is universally constructible if, for every Ω and , every monotone experiment allocation is constructible.
4.2 Main Result
We now present the main result of the paper:
Theorem 1.An information hierarchy is universally constructible if and only if its graph is a forest.
A rigorous proof of Theorem 1 is in the Appendix. The remainder of this section sketches the argument.
4.3 Proof Sketch: if
We begin by explaining why, if is a forest, then any monotone experiment allocation can be constructed. We illustrate this for the special case when
is a tree. One can then obtain the general case by applying the same argument on each of the disjoint trees that make up the forest.
The proof relies on the following result, which is of independent information-theoretic interest. We say that is statistically redundant given
if
, that is, observing
and
yields the same beliefs as observing
only.11
Lemma 1.Fix signals and π, with
, and an experiment
that is Blackwell less informative than
. There exists a signal
that induces
and is statistically redundant given any signal that is between π and
in the refinement order.












Why do we need the upper bound ? Once we fix any
and π, we can construct
that is statistically redundant given
, π, and anything in between. But, suppose we only fix π and consider an arbitrary
. As long as
is not coarser than π, there is always some refinement of π, say
, such that
is not statistically redundant given
.
The proof of Lemma 1 is constructive. To see how the construction works, consider the example in Figure 3. The state is either L or R. Each of the four rows represents a signal, that is, a partition of . The two rectangles in each row represent the unit interval crossed with the two states. Each signal realization is indicated by its pattern. Note that
, and while
, it is not the case that
nor that
.

Illustration of Lemma 1.
To establish the claim in Lemma 1, we need to construct a signal that induces the same beliefs as
, but is statistically redundant given π, and given any refinement of π up to
. The bottom row illustrates such a construction. Each signal realization of
corresponds to a signal realization of
, with the same likelihood in each state. However, the “locations” of the signal realizations in
are rearranged so that the conditional probability of each signal realization of
in state ω given
is (i) the same for
and
, and (ii) the same for any elements of
that refine the same element of π. Property (i) ensures that
is statistically redundant given
, while (i) and (ii) together ensure that it is also redundant given any
s.t.
. The proof in the Appendix generalizes this construction.
With Lemma 1 in hand, we can now establish the universal constructibility of trees. Consider the tree hierarchy described in Section 3. Take some monotone experiment allocation β on this hierarchy. We construct a sequence of monotone signal allocations on progressively larger subsets of N, adding one node at a time. The signal assigned to a given node may be revised as the sequence progresses.
In Step 1, we pick an arbitrary node, say node D. Given , let
be any signal that induces
. In Step 2, we pick a node connected to D, node A. Node A is above node D and so we follow the same procedure as in the chain example to tentatively assign a suitable signal
to A, with
and
; the signal to D is unchanged in this step, that is, we let
. In Step 3, we pick an unassigned node connected to A or D, which here must be node B. Since B is below A, a complication in assigning a signal to B is that there may not exist a signal
such that
and yet
. This is where Lemma 1 comes into play. Applying Lemma 1 with
and
, we conclude that there exists a signal
that induces
and is statistically redundant with
, so that
. We then set
, we replace the initial assignment to A with
, and we leave the signal at D unchanged at
. Finally, in Step 4, we move to the final node C, below the previously assigned node B. We now apply Lemma 1 with
,13
, and
to obtain a new signal
that induces
and is statistically redundant with the previously assigned signals at nodes above C. We then set
and replace the signals of the previously assigned nodes above C as in Step 3.
This completes the construction of a monotone signal allocation inducing experiment allocation β. The full details of this procedure, applied to any hierarchy whose graph is a forest, are the heart of the formal proof of the if direction.
Which properties of a tree are used in this argument? When we apply this procedure to trees (or forests more generally), we can add the nodes in a way so that the new node n always covers or is covered by exactly one node to which we have previously assigned a signal. When n covers
, we can apply Green and Stokey's (1978) result to obtain the new signal at n. When n is covered by
, we can apply Lemma 1 instead. If the hierarchy is not a forest, however, no matter how we add the nodes, we will inevitably find ourselves in a situation where the new node covers or is covered by more than one node, in which case neither Green and Stokey's (1978) result nor Lemma 1 is of use.
4.4 Proof Sketch: Only if
We now sketch the proof of the other direction of Theorem 1: a hierarchy is universally constructible only if its graph is a forest. The proof consists of three main steps. Step 1 introduces a notion of a closed subhierarchy and establishes that a hierarchy is universally constructible only if its closed subhierarchies are as well. Step 2 shows that every hierarchy that is not a forest contains a closed subhierarchy that is one of two types. Step 3 shows that both of these types are not universally constructible, thus completing the proof.
Step 1: Given a hierarchy , a subset of nodes
induces the hierarchy
, which we refer to as a subhierarchy of H, with the partial order being the restriction of ≥ to
. Say that
is closed if, for every
and
,
implies
. In other words,
contains all the nodes from N that are “between” the nodes of
. Lemma 2 in the Appendix shows that if H is universally constructible, then every closed subhierarchy of H is also universally constructible. The argument is as follows. Fix a closed subhierarchy
. We show that any monotone experiment allocation β on
can be extended to a monotone experiment allocation on all of H. This step uses the hypothesis that
is closed, since it means that in extending β, we never have to fill in beliefs for “in between” nodes in
. As H is universally constructible, there is a monotone signal allocation on H that induces the extension of β, and the restriction of this signal allocation to
is a monotone signal allocation on
that induces β. Since β is arbitrary,
is universally constructible.
Step 2: Say that a hierarchy H is cyclic if is not a forest. Suppose H is cyclic. Given that N is finite and H is cyclic, it follows that H contains a minimal cyclic closed subhierarchy (MCC), that is, a subhierarchy
such that:
is cyclic, closed, and there does not exist
such that
is cyclic and closed. Any MCC
must fall into one of the following categories:
is a union of non-comparable paths (UNP): It contains a maximal node
and a minimal node
, and its graph consists of at least two paths between
and
. Moreover, nodes in
are comparable only if they are in the same path. The diamond is an example of a UNP; another example is depicted in Figure 4(a).
is a crown: It contains exactly four nodes, which we label F(ather), M(other), S(on), D(aughter), and the partial order consists of
,
,
, and
. A crown is depicted in Figure 6(a).

For adjacent nodes, the node that is higher in the figure covers the node that is lower. (a) A UNP. Nodes A, B, and C are in different paths, and therefore are not comparable. (b) A between set which is not an MCC. There are two overlapping (directed) paths that go from to A to
, one that goes through B and one that does not. There is a smaller cyclic closed subhierarchy, namely the nodes between A and
.
This taxonomy takes considerable effort to prove formally, but the high level argument is as follows. An MCC contains maximal nodes, which are not covered by any other node, and minimal nodes, which do not cover any other nodes. We distinguish two cases on these maximal and minimal nodes.
In the first case, every maximal node in covers every minimal node. Then, the fact that
is minimal, together with the existence of a cycle, implies that
contains exactly four nodes and is in fact a crown.
Alternatively, there is a maximal node that does not cover a minimal node
. Then, we show that
is a UNP. To see this, note the following two subcases. First, it may be that
is simply the set of nodes that are between
and
, that is, a between set. Then,
consists of a series of directed paths between
and
. Now, if nodes in distinct paths were comparable, then it would be possible to find a smaller cyclic closed subhierarchy, as illustrated in Figure 4(b), violating the fact that
is minimal. Thus, nodes must not be comparable across paths and
is a UNP. The second subcase is that
is not a between set. Lemma 4 in the Appendix shows that then every cycle in
must contain every node in
. Any such spanning cycle can be decomposed into two undirected paths between a maximal node and a minimal node. If any nodes in these two paths were comparable, we could find a smaller cycle that does not contain every node in
, which would contradict Lemma 4. As a result,
is a UNP.
Step 3: Finally, we show that UNPs and crowns are not universally constructible. The former is demonstrated in Lemma 6 in the Appendix using a similar construction as we used for the diamond in Section 3. Indeed, the diamond is an example of a UNP in which there are exactly two undirected paths from to
.
For the crown, we construct a particular monotone experiment allocation that is not constructible: Let and
,
, and
. We represent each belief as a pair
in the triangle
, where
and
. Consider the experiment allocation β that assigns to F, M, S, and D the experiments indicated in Figure 5(a). Each belief realization is a circle, the letters inside the circle indicate the nodes for which this belief is in the support, and the area of the circle is the likelihood of the belief (which is the same for all nodes attached to the belief). For example, the belief
can be realized for F, S, and D, with likelihood of 1/2 in each case. It is easy to see from Figure 5 that β assigns higher nodes more informative signals. For example, going from D to M (Figure 5(b)), the mass on the east and west beliefs stays the same, but the mass on the central belief is spread out in the north-south directions.

The crown is not universally constructible.
Lemma 7 in the Appendix formally defines this experiment allocation and shows that it is not constructible. To see what goes wrong, suppose there is a monotone signal allocation that induces β. Such a signal allocation induces a joint distribution over beliefs at all four nodes. Consider first the joint distribution of beliefs between D, M, and S. The center D belief occurs if and only if the M belief is north or south (Figure 5(b)). The center belief of S occurs if and only if the M belief is east or west (Figure 5(c)). Hence, the probability of both D and S having the center belief is zero. Now consider the joint distribution of beliefs between D, F, and S. The belief at D is in the center if and only if the belief at F is in the center (Figure 5(d)), and the belief at S is in the center if and only if the belief at F is in the center (Figure 5(e)). So, the probability that both D and S have the center belief is positive, yielding a contradiction.
5 Discussion
5.1 Binary States
A key step in our proof of the only-if direction of Theorem 1 shows that the crown is not universally constructible. Specifically, Figure 5 presents a particular experiment allocation involving three states that is not constructible.
It is natural to ask whether there is an example of a non-constructible allocation on the crown using two states. No such example exists: when the state space is binary, every monotone experiment allocation on the crown is constructible. The reason is as follows. Take an arbitrary monotone experiment allocation β on . It is known that the set of experiments under the Blackwell order is a lattice when the state space is binary (Kertz and Rösler (2000), Müller and Scarsini (2006)). Thus, there exists an experiment
such that
,
, and
for any
. We can therefore expand the crown hierarchy by adding a T(utor) whom F(ather) and M(other) hired to oversee the children:
,
,
, and
. We refer to this hierarchy as the cross, which is depicted in Figure 6(b). Let
be the experiment allocation on the cross that sets
for
and
. The allocation
is monotone by construction. The cross is a forest, and hence by Theorem 1, it is universally constructible. Thus, there is a monotone signal allocation
that induces
. Restricting
to
yields a monotone signal allocation on the crown that induces β.

The crown and cross hierarchies.
This example illustrates that the binary state space is special. With three or more states, every monotone experiment allocation is constructible if and only if the hierarchy is a forest. With binary states, being a forest is sufficient for constructibility, but it is not necessary. A natural direction for future research is to characterize necessary and sufficient conditions for every monotone experiment allocation to be constructible for the case of binary states.
5.2 Beyond the Refinement Order
As we mentioned at the outset, there are various things one might mean by “A is more informed than B.” One is that A's signal is Blackwell more informative than B's, that is, . Another is that A has observed all of B's information, that is,
. These, however, are not the only economically relevant comparisons of signals. For instance, it might be that A knows B's belief about the state. Or, it might be that if A observed B's information, A's belief about the state of the world would not change, that is, that
is statistically redundant given
. In a companion paper (Brooks, Frankel, and Kamenica (2022)), we explore these and other relations on signals in more detail. In this section, we explain how our results extend to a large class of notions of “more informed,” including those just mentioned.





Many economically relevant comparisons of signals are “in between” refinement and the belief-martingale relation, that is, they are weaker than the former and stronger than the latter. For example, the two aforementioned comparisons—A knowing B's belief or B's signal being statistically redundant given A's—fall in this in-between category. Formally, we say a binary relation on Π is proper if
.
Our main result can be extended by replacing the refinement order with any proper relation. In particular, say a signal allocation σ on H is -monotone if
implies that
. The experiment allocation β on H is
-constructible if β is induced by some
-monotone signal allocation on H. A hierarchy is said to be
-universally constructible if every monotone experiment allocation is
-constructible for any Ω and
. With this terminology, we have the following:
Theorem 2.Fix any proper relation . A hierarchy H is
-universally constructible if and only if its graph is a forest.








6 Applications
6.1 Rationalizing Reaction to Unknown Sources of Information
Consider an agent who obtains information from multiple sources. If we do not know the information-generating process, what restrictions does the agent's rationality impose on her potential reactions to this information? Concretely, suppose a decision maker has access to a set of information sources . Suppose further that our data set
tells us the distribution of beliefs conditional on observing any non-empty subset of information sources. When can we rationalize a given data set
in the sense that we can associate each information source
with some signal (i.e., an element of Π) and conclude that belief formation is consistent with Bayes's rule?
To be rationalized, belief distributions (i.e., experiments) in have to satisfy two obvious properties. First, there is Bayes plausibility: the average belief cannot differ across sets of information sources, that is,
for any two subsets S and
. Second, there is Blackwell monotonicity: observing a superset of sources necessarily induces a more dispersed distribution of beliefs, that is,
is a mean-preserving spread of
if
. A natural question is whether these are the only properties imposed by Bayesian updating.
Theorem 1 tells us that, when there are three or more sources, the answer is no.15 In this case, Bayesian updating requires more than just Bayes plausibility and Blackwell monotonicity. To see why, consider the set-inclusion information hierarchy H where each non-empty collection of sources is associated with a node
and the partial order is the superset order:
if
. As illustrated in Figure 7, the graph of this information hierarchy is not a forest. By Theorem 1, this means that there is some monotone experiment allocation on H, call it β, that cannot be induced by any monotone signal allocation on H. Now, we can associate with this β a data set
by setting
. Note that
necessarily satisfies Bayes plausibility and Blackwell monotonicity (since β is monotone). If we could rationalize
by associating each
with some signal
, then the signal allocation
would induce β and be monotone (since
implies
), contradicting Theorem 1. Thus, we know that there are data sets that satisfy Bayes plausibility and Blackwell monotonicity, yet cannot be rationalized.

The set-inclusion hierarchy.
A fruitful direction for future research would be to fully characterize which reactions to unknown sources of information are rationalizable.
6.2 Information Design
Suppose a sender provides information to a set of agents. Moreover, she faces certain types of monotonicity constraints such as some agents must know the beliefs of some other agents or some agents must have access to others' information. Then, we can think of agents as elements of an information hierarchy, and the information design problem consists of selecting a (suitably monotone) signal allocation on this hierarchy. Our results shed light on how such monotonicity constraints affect the information design problem.
For example, consider organizations. A long literature in organization economics emphasizes the importance of the hierarchical structure of managerial relationships (Williamson (1967)). One important aspect of organizational design is deciding how much information to provide—about individuals' prospects for promotion, about the overall performance of the organization, etc.—to each member of the organization. It is often suboptimal to provide full transparency and share full information with everyone (Fuchs (2007), Jehiel (2015), Smolin (2017)).
A natural constraint that an information designer might face is that anyone in the organization ought to have access to the information that is available to her subordinates, that is, that the allocation of signals within the organization should be monotone with respect to the management structure, so that the signal of a superior refines the signals of their subordinates. This constraint interacts with the organization structure. Theorem 1 implies that, if an organization has the feature that every subordinate has at most one superior, the aforementioned constraint can always be satisfied as long as individuals who are higher up in the organization are more informed in the Blackwell sense. Furthermore, if the information designer's objectives only depend on each agent's experiment, then one could reformulate the information design problem in terms of the choice of monotone experiment allocation, rather than choosing monotone signal allocations directly. With richer managerial relationships,16 however, Theorem 1 also tells us there could be desirable allocations of information which are incompatible with the monotonicity constraint, even though they provide (Blackwell) more information to those higher up in the organization.
7 Related Literature
We now discuss the connection between our results and related literature in probability theory and in information economics.17
7.1 Connections to Probability Theory
Mathematically, our results are close to a literature in probability theory on the relationship between orders on probability distributions and stochastic processes. Because the connection to this literature is technical, our discussion is somewhat formal and detailed.
Fix two distributions . We say that τ is greater than
in the convex order if, for every convex function
,
. The Sherman–Stein theorem says that the following are equivalent: (i) τ is greater than
in the convex order; (ii) τ can be obtained from
via mean-preserving spreads. Blackwell (1953, Section 3) observed that his result is implied by the Sherman–Stein theorem, since we can interpret τ as an experiment and the convex function ϕ as the maximum utility obtained in a decision problem given a belief. Importantly, the mean-preserving spread from τ to
is a probability transition kernel which defines a martingale on
. Thus, an equivalent way of stating (ii) is that there exists a two-period martingale on
such that the marginals are τ and
.
This equivalence was later generalized by Strassen (1965, Theorem 8): the sequence is increasing in the convex order if and only if there exists an M-period martingale with marginals
. This result is closely related to Green and Stokey (1978) and, more broadly, whether experiments can be represented as partitions: When the set of realizations of the martingale is finite, the martingale generates a filtration
, where each
is generated by a finite partition
of the underlying probability space, and the partitions are increasing in the refinement order. We can therefore view the probability space itself as an expanded state space, so that the
are refinement-ordered signals that induce the given experiments. In our terminology, Strassen's theorem implies that the chain is universally constructible.
There is a substantial literature that further generalizes and strengthens Strassen's theorem. See Hirsch, Profeta, Roynette, and Yor (2011) for a survey.18 To our knowledge, Juillet (2016) is the only paper in this literature to study collections of distributions that are indexed on partially ordered sets. He presented examples that illustrate various ways in which Strassen's theorem does not generalize. Most closely related to the present study, Juillet (2016, Section 4.1) constructed a collection , where X is the diamond and
, such that the collection is increasing in the convex order but there does not exist a real-valued martingale on X with the given marginals. The distributions in Juillet's example are similar in spirit to (but distinct from) our example of a non-constructible belief allocation given in Section 3.19 Our Theorem 1 implies that the equivalence between
increasing in the convex order and induced by a martingale holds if and only if X is a forest.
Also related is Fill, Allen, and Machida (2001), who studied an analogue of the problem of Juillet (2016), but where the convex order is replaced by a generalization of first-order stochastic dominance. Specifically, Fill, Allen, and Machida (2001) considered collections of distributions indexed on a partially ordered set , where each
is a distribution on a different partially ordered set S. They defined a notion of stochastic monotonicity for distributions on S that reduces to first-order stochastic dominance when S is totally ordered. A pair
is monotonicity equivalent if the following condition holds: a collection
is stochastically monotone if and only if there is a distribution over non-decreasing functions
such that
is the marginal distribution of
. Monotonicity equivalence is analogous to universal constructibility, but with stochastic monotonicity in place of the Blackwell order. In the special case that S contains a sub-poset that is either a cycle or a crown, Fill, Allen, and Machida (2001) showed that
is monotonicity equivalent if and only if X is a forest. The cases where X is a diamond or a crown also feature prominently in their proof, which follows a similar strategy of showing that if X is not a forest, then it contains a sub-poset on which monotonicity equivalence fails, and then extending the counterexample to all of X.
7.2 Connections to Information Economics
Our paper connects to various threads in the economics literature studying information and beliefs. In addition to the aforementioned classical literature on information orders, Bergemann and Morris (2016) studied an extension of the Blackwell order to type spaces, and Mu, Pomatto, Strack, and Tamuz (2019) considered comparisons of repeated experiments.
We contribute to the literature on higher order beliefs (Harsanyi (1967), Mertens and Zamir (1985), Brandenburger and Dekel (1993)). Specifically, we characterize when restrictions of the form “player i knows player j's type” place constraints on i's and j's first-order beliefs, beyond the obvious constraint that i must be Blackwell more informed than j.20
We also study how signals can be combined to produce more informative signals. Gentzkow and Kamenica (2017) studied this issue in the context of a communication game with a receiver who combines information provided by multiple senders. Börgers, Hernando-Veciana, and Krähmer (2013) studied the interaction between signals from the perspective of whether signals are substitutes or complements.
We also contribute to the growing literature on information design (Kamenica and Gentzkow (2011), Bergemann and Morris (2016)). Arieli, Babichenko, Sandomirskiy, and Tamuz (2021) characterized feasible joint belief distributions of a group of agents in a binary state case. Mathevet and Taneva (2020) analyzed the implications of information design for organizational structure.
Finally, Shmaya and Yariv (2016) studied behavior by subjects in lab experiments when the experimental protocol potentially generates information. Their analysis led them to a question that is related to the result we present in Section 6.1. In particular, given data on how an agent behaves following any sequence of signal realizations, Shmaya and Yariv derived necessary and sufficient conditions for the behavior to be consistent with Bayesian updating if the data generating process, the agent's preferences, and the agent's prior are not known. On a similar topic, Molavi (2021) studied which belief dynamics are consistent with Bayesian updating, but allowed for the possibility that the agent has an incorrect model of the data generating process.
8 Conclusion
We conclude with a brief discussion of an important direction for future work. Our analysis has focused on the question of when an information hierarchy is universally constructible. That condition requires that every monotone experiment allocation is constructible. Of course, even when not every monotone experiment allocation is constructible, some are. A natural goal would therefore be to characterize, for a given information hierarchy, the set of constructible monotone experiment allocations. Such a characterization would be of particular interest in the case of the set-inclusion hierarchy (Figure 7), which would identify the exact sufficient and necessary conditions for rationalizability of reactions to unknown sources of information, as discussed in Section 6.1.
























































































Appendix: Proofs
A.1 Proof of Lemma 1
Proof of Lemma 1.Let be a signal s.t.
. Since
, there exists a garbling
such that
, and
. For every
, let
denote the element of π s.t.
. (This element exists since
.) Now,
, let
be a partition of
s.t. ∀ω,
, where
. Such a partition exists because
for all
. Let
with
. We now show that
satisfies (i) and (ii). To show (i), it suffices to show that
for every
and ω. We have
































A.2 Proof of Theorem 1: if
Let H be an information hierarchy and suppose is a forest. Let β be a monotone experiment allocation on H. We will construct a monotone signal allocation that induces β. To do so, we construct a sequence of subhierarchies of H, adding nodes of H one by one, until we reach the full hierarchy H. At each step, we assign a signal to the newly added node and potentially reassign the signals allocated to the previously added nodes.








Now, consider a construction procedure f of the following form. Let be any node in N. For
, let
be an arbitrary element of
. Note that for any
,
is not empty.
Claim 1.For each , there is at most one edge in
between
and nodes in
.
Proof of Claim 1.Suppose toward contradiction that has an edge in
with distinct n,
. Since n and
both have an edge with
, they must belong to the same tree in
. Moreover, there must be a path between n and
in
. To see this, let
be the node that was added earliest to
among the nodes in the tree to which n and
belong. For every other node
from this tree, we must have
, which in turn means that there is a path from
to
in
and thus in
. Hence, there is a path from both n and
to
and thus a path between n and
in
. So, there must be a path between n and
in
that does not go through
. But, because
has an edge with both n and
, there is another path from n to
that goes through
. However,
is a forest, so there cannot be multiple paths between two nodes; we have reached a contradiction. Q.E.D.









First, to node , we assign an arbitrary signal
such that
. Note we are vacuously satisfying the base case of the induction argument: the signal allocation to the single node in
induces appropriate experiments. For
, there are three cases:
,
, and
.
We first consider the case
. Note that, by Claim 1,
covers exactly one node in
(call this node
) and is not covered by any nodes in
. Since
, there exists some
such that
(which follows from Theorem 1 of Green and Stokey (1978)). We set
and we keep the signal allocation for nodes in
unchanged, that is,
for all
. It is clear that
induces appropriate experiments (by the inductive hypothesis for
and by construction for
). We also need to show that this signal allocation is monotone on
. Consider any
such that
. Since
, either
or
. In the former case, we know
by the inductive hypothesis. If
, we know
. By the inductive hypothesis,
and thus
. That completes the proof for this case.
Now consider the case where . Let
be the node in
that covers
. Denote
,
, and
. By Lemma 1, we know
such that (i)
, and (ii)
s.t.
,
. We set
. For
, if
, we set
; otherwise, we set
. We need to show that
is monotone on
and induces appropriate experiments. We have that
. For
, first consider cases where
, so
. Since
(recall that
covers
), by the inductive hypothesis,
; moreover,
; hence,
. For
s.t.
,
. Since by the inductive hypothesis,
, we have established that
for all
. We now need to show that
is monotone. Consider any
s.t.
. There are three cases. First, suppose
. In that case, we know that
and
. Since (by the inductive hypothesis)
, we know that
, and hence
. The second case is where
and
. Then,
and
. Since (by the inductive hypothesis)
, we have that
. Finally, suppose that
and
. Then,
and
. Since (by the inductive hypothesis)
, we have that
.
Finally, suppose . We assign an arbitrary signal
to
such that
, and we keep the signal allocation to nodes in
unchanged, that is,
for all
. It is clear that
induces appropriate experiments (by the inductive hypothesis for
and by construction for
). Since
is not comparable to any node in
, the fact that
is a signal allocation on
implies that
is monotone on
. This completes the proof. □
A.3 Proof of Theorem 1: Only if
Lemma 2.If an information hierarchy H is universally constructible and is a closed subhierarchy of H, then
is universally constructible.
Proof of Lemma 2.Suppose H is an information hierarchy that is universally constructible under . Fix a state space Ω and a prior
. Suppose
is a closed subhierarchy of H. Let
be a monotone experiment allocation on
. We need to construct a monotone signal allocation on
that induces
. Let
defined as follows: (i) if
, let
; (ii) if
and
such that
, let
; and (iii) if
and
such that
, let
.
Claim 2.β is monotone on H.
Proof of Claim 2.Consider with
. We show that
by considering four exhaustive cases:
If n and are both in
, this follows from the fact that
is monotone on
.
If n and are both not
, consider two subcases. If
such that
, then
. Otherwise, since
and
such that
, it must be that
such that
, so
.
If and
,
.
Finally, if and
, then there cannot exist an
with
. If such an
did exist, then since
is closed and
, we would have that
, a contradiction. Thus,
. Q.E.D.








Lemma 3.Fix a hierarchy and a closed subhierarchy
. Let
be the set of edges in
. Then
, where
.
Proof of Lemma 3.Fix . We need to show that
if and only if
. If
, then n covers
in H, that is,
and there is no
with
. A fortiori, there is no
with
; hence, n covers
in
, so
. If
, then n covers
in
. As a result,
. If there exists
such that
, then, since
is closed,
, so n must not cover
in
, a contradiction. As a result, n covers
in H as well, so
. Q.E.D.
















Lemma 4.Fix a hierarchy and a subhierarchy
. Suppose
is an MCC of H. Then, either (i)
is a simple between set in H, or (ii) every cycle in
is a spanning cycle.
Proof of Lemma 4.For this proof, all between sets are defined relative to H, and we simply write for
. Similarly, by closed we mean closed in H. Note that by Lemma 3,
is the subgraph of
obtained by dropping edges with nodes that are not in
. In particular,
is cyclic if and only if H contains a cycle whose nodes are in
. This fact is used freely below.
We consider two cases. First, suppose there are two nodes such that
and there are two distinct paths from n to
in the directed graph
. Note that
is closed, so that these paths are in
as well, so that
is cyclic. Moreover, since
is closed, we have that
. Hence, since
is an MCC, we must have that
. It remains to show that the between set
is simple. Suppose to the contrary there is some node
such that
belongs to two distinct paths from n to
in
. Then, there are either two distinct directed paths from n to
or two distinct directed paths from
to
; thus, either
or
must be cyclic. Since both
and
are closed and strict subsets of
,
must not be an MCC, so we have reached a contradiction. Thus, we have established that
must be a simple between set.
Now consider the second case where for every , there is at most one path from n to
in the directed graph
. Given a path P (either directed or undirected), let
denote the set of nodes that appear in P. Since
is an MCC,
contains a cycle
where
,
. We will argue that
is closed. The fact that
will then follow directly from the hypothesis that
is an MCC.
Let us then suppose that is not closed, in order to reach a contradiction. Given a directed path
, its undirected analogue is the undirected path
where
. Say that a directed path P in
only contains edges in C if every edge in the undirected analogue of P is in C. A directed path P in
is an external directed connection (EDC) from n to
if (i) P is a directed path from n to
; (ii)
; and (iii) P does not only contain edges in C. Say that
are an externally connected pair (ECP) if there is an external directed connection from n to
or from
to n. An ECP
is said to be minimally close if, for every
,
is an ECP only if
and
.
Claim 3.Given any two nodes , if P is the unique directed path in
from n to
, then
.
Proof of Claim 3.If there are two non-comparable nodes in , there would be two distinct directed paths from n to
. Hence, all nodes in
are comparable. Therefore, there is a directed path from n to
whose nodes are
. Since there is a unique directed path from n to
, the set of nodes in P is
. Q.E.D.
Claim 4.There exist such that
is a minimally close ECP.
Proof of Claim 4.We know there is a pair of nodes in that are an ECP. Otherwise,
would be closed. Moreover, since L is finite, there is a pair of nodes in
that are a minimally close ECP. Q.E.D.


















Claim 5. is cyclic.
Proof of Claim 5.It suffices to show there are two distinct undirected paths from to
in
. One path is
. The other path is the undirected analogue of the external directed connection
. Since
is external, these two undirected paths must be distinct. Q.E.D.
Claim 6.S is closed.
Proof of Claim 6.Let . We will show that Y is closed and that
.
First we show that Y is closed. Consider any and
. By definition of Y,
and
, where
for
. Hence,
and thus
.
It remains to show that . Given
,
. Moreover,
. Hence,
.
Now, consider some . We need to show that
. If
, then we are done. Otherwise,
. We know
for some
. If
,
. Suppose instead that
. We will reach a contradiction. Let P denote the directed path from
to
whose nodes include n. Because
is a minimally close ECP, path P must only include edges in C. Since
, the nodes in path P cannot be a subset of
. Thus, the nodes in P contain the nodes in
, including
and
. The sequence of nodes and edges in P between
and
is a directed path between those nodes, and thus is equal to
(by uniqueness of the directed path). Since
contains an edge which is not in C, we have contradicted the hypothesis that P only contains edges in C. Thus, we have established that
. Q.E.D.










Lemma 5.If the hierarchy is such that every cycle in
is a spanning cycle, then for any pair of nodes
, there exist two undirected paths from n to
such that the union of the nodes in the two paths is N and the intersection of the nodes in the two paths is
.
Proof of Lemma 5.Since there exists a spanning cycle in , for any pair of nodes
, there exist two paths
and
in
such that
and
, where
and
. We need to show that
. Suppose to the contrary that there exists
with
. We know there exist
and
such that
. Now, consider the path
in
. Since
, this is a well-defined path. But it is a cycle that is not spanning and thus we have reached a contradiction. Q.E.D.












Lemma 6.If the hierarchy H is a UNP, then H is not universally constructible.
Proof of Lemma 6.Let . We will present a monotone experiment allocation based on the allocation on the diamond in Section 3 and show that it is not constructible. Consider the experiment allocation given by21
;
,
for
; and
for all
and
. These are the experiments depicted in Figure 2, where
;
; for all
,
; and for all
and
,
. Monotonicity of β is immediate from the mean-preserving spreads depicted in Figure 2.22
Toward a contradiction, suppose σ is monotone signal allocation that induces β. We begin by establishing that it must be the case that, along any undirected path from to
, the realized beliefs must be equal across the interior nodes on that path.
Claim 7.For all , we have
for all
.
Proof of Claim 7.It suffices to establish that for any , we have
. We know
and
are comparable. Assume that
; the case where
is analogous and omitted. Since σ is monotone, we have that
. As the refinement order implies the belief-martingale property, we have that
. Now suppose toward contradiction that
. We would conclude that
is a strict mean-preserving spread of
.23 But that would mean that
is not equal to
in distribution and thus that
. Q.E.D.

Step 1: We show that . First, note that
since
and the support of
lies above 0. Moreover, since
, we must also have
. A similar argument establishes that
and
. Claim 7 tells us that
. Hence,
.
Step 2: We show that . It suffices to show that (a)
, and (b)
.
Arguments analogous to the ones in Step 1 yield . Because
, this in turn implies that
. Thus, Claim 7 tells us that
. Therefore,
, establishing part (a).
Now, note that implies
. Moreover, since
, we have
, establishing (b). Q.E.D.
Lemma 7.The crown is not universally constructible.
Proof.Consider the experiment allocation given by ,
,
, and
. These experiments are depicted in Figure 5. To see that
is a mean-preserving spread of
, note that we can obtain
from
by (i) spreading the realization
in
to
in
, (ii) leaving the realization
in
unchanged, and (iii) spreading the realization
in
to
. To see that
is a mean-preserving spread of
, note that we can obtain
from
by (i) leaving the realization
in
unchanged, (ii) spreading the realization
in
to
in
, and (iii) leaving the realization
unchanged. The argument for why
and
are mean-preserving spreads of
is symmetric. Hence, β is monotone.
We claim that β is not constructible. As before, consider the joint distribution of beliefs on induced by any monotone signal allocation σ, and recall that
implies
. Specifically, consider the conditional probability of
given
. Since
, we have that
. Since
, we have that
. Hence, the joint probability of
and
must be zero, that is,
. But,
implies
, and
implies
. Hence,
, a contradiction. Q.E.D.
Lemma 8.If the hierarchy is cyclic and N is a simple between set in H, then H is not universally constructible.
Proof of Lemma 8.Since is cyclic and N is a simple between set, H must be a UNP. Therefore, Lemma 6 establishes that H is not universally constructible. Q.E.D.
Lemma 9.If the hierarchy H is cyclic and not a crown, and every cycle in is a spanning cycle, then H is not universally constructible.
Proof of Lemma 9.Suppose that is cyclic and not a crown, and that every cycle in
is a spanning cycle. We say an element
of N is maximal if there is no
such that
. We say an element
of N is minimal if there is no
such that
.
Claim 8.There exist and
in N that are maximal and minimal, respectively, such that
does not cover
.
Proof of Claim 8.Suppose toward contradiction that every maximal element covers every minimal element. Hence, is a complete bipartite graph of maximal and minimal elements. If there were only one maximal element or only one minimal element, there could not be a cycle. So, there must be at least two of each. Take any
consisting of exactly two maximal and two minimal elements, and let
.
is a crown and therefore is cyclic. Moreover,
is clearly closed, so that by Lemma 3, the nodes in
are part of a cycle in
as well. Since every cycle in
is a spanning cycle, we must have
, and thus H is a crown subhierarchy, and we have reached a contradiction. Q.E.D.








Proof of Theorem 1: Only if.Suppose is not a forest, that is, it contains a cycle. Since N is finite, H contains a subhierarchy
that is an MCC of H. By Lemma 4, either (i)
is a simple between set in H, or (ii) every cycle in
is a spanning cycle. Consider case (i). Because
is closed and
is a simple between set in H,
is also a simple between set in
. Thus, Lemma 8 implies that
is not universally constructible. Now consider case (ii). If
is not a crown, then Lemma 9 implies that
is not universally constructible. If
is a crown, Lemma 7 implies it is not universally constructible. Therefore, by Lemma 2, H is not universally constructible. Q.E.D.