Monopolistic Signal Provision^†

7 Appendix: Omitted proofs

Proof of lemma 1. () Let x be an arbitrary nondecreasing function, and let . Define, recursively, for each n = 0, 1, 2, …, two functions, , and three finite collections of intervals, , , and , as follows:

• 

• equals the collection of maximal intervals in (θ₁, θ₂) over which ordered so that .

• equals the complementary set of maximal intervals in (θ₁, θ₂), also ordered so that .

  (Notice that the intervals in these two sets must alternate and, since h satisfies NC, the left-most interval must be , and the right-most interval is an element of only when over such interval.)

• , or if is the right-most interval.

• And, for every n ≥ 1 i, and ,

  

  (Notice that every hⁿ also satisfies NC.)

Since every step merges at least two intervals, there exists an n such that , and therefore . Let N be the smallest integer such that this holds. We now have

()Suppose condition NC is violated for some θ. Then, any “one-step”function x that has a constant value over (θ₁, θ), and a higher constant value over [θ, θ₂), has a positive covariance with h over (θ₁, θ₂). ■

Proof of remark 3. Notice that the collection of all open intervals over which h is weakly decreasing satisfies conditions (a) and (b). Denote this collection . Now, for each interval D ɛ let P(D) be the largest open interval containing D while satisfying the condition NC. Notice that the collection {P(D): D ɛ } satisfies the conditions (a)-(c).¹⁵

For uniqueness, suppose toward a contradiction that both and ^′ ≠ satisfy (a)-(c). Since the complement of ∪P consists of intervals over which h is increasing, it cannot be the case that some P^′ ≠ ^′ lies outside ∪P, or vice versa (otherwise condition (b) would not hold). Therefore, there must exist a pair of intervals P ∈ , and P^′ ≠ ^′, such that P^∩′ ≠ Ø, and P ≠ ^′. Suppose without loss that . I now show that satisfies condition NC, a contradiction to (c). Let θ ɛ P ∪ P′. We have two cases to consider, according to whether or not θ ɛ P ∩ P′. Suppose first that θ ɛ P ∩ P′. From condition (b), applied to both P and P′, one obtains the desired inequality:

The case in which θ ∉ P ∩ P′ is similar, and left to the reader.

Finally, the claim that all intervals in are disjoint and never adjacent follows from the same reasoning as above, namely, the union of two intervals in that intersect, or are adjacent, also satisfies NC, a contradiction to (c). ■

Proof of theorem 1. Suppose is non-empty, otherwise the theorem follows from proposition 1. Let denote the collection of maximal intervals complementary to , i.e., the largest intervals in , and let (which constitutes a partition of [θ_L, θ_H]). Also let Φ denote the set of truthful filters satisfying the following condition: for every P ∈, θ ∈ P, and θ^′ ∉ cl (P) φ is such that φ(θ) φ(θ^′).

I proceed in two steps. I first show that every truthful filter φ^′ outside Φ is dominated by some truthful filter within this set, and then show that the truthful filter φ* in the theorem, which belongs to Φ, weakly dominates every other member of Φ.

Let φ^′∉Φ, and suppose it maximally pools a collection of intervals . Now suppose one replaces with , which, instead, maximally pools the collection of smaller intervals ∩: (T∩I)_T∈,I∈ and is otherwise equal to θ. Notice that belongs to Also, from the properties of, for every T∈, both the average and the expectation must be increasing in I. The change in the objective is given by

where the inequality follows from the fact that each term in the outside sum, which is proportional to the covariance between and across intervals I, is always non-negative, and any term in this sum becomes positive whenever T intersects more than one interval I with positive probability (which must occur for at least one interval T ∈ This concludes the first step of the proof.

For the second step, let be any element of other than . We can write

[6]

Consider the first sum. Since , it follows that, for every P ∈ , , and therefore each term in the sum is proportional to

But, from lemma 1, this quantity is non-negative.

Consider now the second sum. Notice that, for any S ∈ , the problem of maximizing the integral , subject to , is identical to the original problem [I] with [θ_L, θ_H] = S. From proposition 1, this problem is solved by setting φ^′(θ) = θ for all θ ∈ S As a result, each term in this second sum is also non-negative. ■

Proof of lemma 2. Let P = (θ₁,θ₂) ∈ , and suppose x is non-constant and nondecreasing. Also, let be such that either for all , or x(θ)≥(<)x(θ)^* for all .

Suppose towards a contradiction that From the algorithm in the proof of lemma 1, this requires that, for every is constant over for all i. Therefore, for some and as a result This implies in turn that xⁿ is non-constant over (θ₁, θ₂) for all . So that, in particular, x^N−1 is non-constant over , and implying that and . From the latter equalities for all θ<θ*, and for all θ<θ* for all . But, since h is generic, and , the boundary conditions for (θ₁,θ₂)∈ imply that the first expectation must be larger than the second, i.e., . By combining this with the fact that x^N−1 is non-constant, one obtains a strict inequality in the final step of the algorithm, which is a contradiction:

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Proof of Corollary 1. Consider the proof of theorem 1 for a generic h. By invoking lemma 2, at least one of the summations in [6] must be positive when and differ over a positive-measure subset. It follows that when a given truthful filter differs from over a positive-measure subset, it is dominated by whether or not . ■

Proof of lemma 3. Suppose φ, q satisfy constraints [3] and [4]. I show that there exist schedules , with a truthful filter, that satisfy constraints [3] and [4] and deliver profits weakly higher than . For any given type θ, let A(θ) denote the largest type interval (possibly a singleton) such that for every , and set

By construction, the new schedules satisfy constraints [3] and [4], and is a truthful filter (indeed, and [ for all θ. Moreover, if one replaces with profits change by

where the inequality follows from the convexity of c. ■

Proof of lemma 5. Consider schedules φ, q satisfying the constraints in problem [III], such that q satisfies lemma 4, but is not constant over some P ∈ . I consider a specific change in this schedule, resulting in a new schedule , such that q′ is constant over every interval in , and then show that this change is weakly profitable.

Let denote the collection of intervals T over which is constant. Suppose the new schedule satisfies

and q′ satisfies

Notice that these new schedules satisfy all the constraints in problem [III].

Consider now the change in the objective. In order to measure this change, it is useful to decompose the shift from to into a shift from to an intermediate regime , plus a shift from to . Define the intermediate schedule φ″ as follows:

In other words, the intermediate regime is equal to the original one, , except for the fact that, within each interval in , both and q are replaced with their expected values. ( need not satisfy the constraints as it is only a device for calculating profits.)

Consider first the change in the objective when shifting from to . This change only affects the intervals in . For each P ∈ , the change is equal to

The inequality follows because both terms on the right are non-negative: The first term is non-negative because is nondecreasing and h satisfies condition NC over P (lemma 1). The second term is non-negative because c is convex.

Consider now the change in the objective when shifting from to . After some algebra, this change is given by

where μ(A) := ∫_AdF (θ). Every term in this sum is non-negative. To see this, fix T ∈ and define a new function such that

Notice from the boundary conditions for the intervals in that g(θ) is nondecreasing. Each term in brackets in the above sum over intervals T ∈ can now be written as

which is proportional to the covariance between and across intervals I ∈, and is therefore non-negative. ■