We begin by describing the general form of the problem faced by the profit-maximizing monopolist considered in the text. The incentive compatibility and individual rationality constraints associated with the seller's program are
$$B_\theta (x_\theta ) - t_\theta \ge B_\theta (x_{\hat \theta } ) - t_{\hat \theta }\quad \forall \theta ,\hat \theta$$
and
$$B_\theta (x_\theta ) - t_\theta \ge 0\quad\forall \theta ,$$
respectively. It is well known, that the seller's profit maximization program reduces to:
$$ \mathop {\max }\limits_{\{ x_1, {\ldots}, x_K \} } \sum\limits_{\theta = 1}^K {f_\sigma (\theta )B_\theta (x_\theta ) - } \sum\limits_{\theta = 1}^{K - 1} {(1 - F_\sigma (\theta ))\{ B_{\theta + 1} (x_\theta ) - B_\theta (x_\theta )\} } $$
(A1)
$$ s.t.\,\, x_{\theta + 1} \ge x_\theta \ge 0,$$
where
\(F_\sigma (\theta ) \equiv \sum\nolimits_{i = 1}^\theta {f_\sigma (i)}\).
41 The marginal benefit to the monopolist of increasing
x θ is thus proportional to
$$ b_\theta (x_\theta ) - \frac{{(1 - F_\sigma (\theta ))}}{{f_\sigma (\theta )}}\{ b_{\theta + 1} (x_\theta ) - b_\theta (x_\theta )\} $$
(A2)
(we need not worry about the definition of
b K +1(·) because 1 −
F σ(
K)=0). The firm increases
x θ as long as (A2) is positive, unless the
\(x_{\theta + 1} \ge x_\theta\) constraint binds.
Define
f λ(·) ≡ (1 −λ)
f m (·) + λ
f n (·), and define
F λ(·) analogously. Observe that, if
f m (θ) /
f n (θ) is monotonic in θ, then so too is
f λ(θ) /
f n (θ). Define
$$V({\bf x},\lambda ) \equiv \sum\limits_{\theta = 1}^K {B_\theta (x_\theta ) - } \sum\limits_{\theta = 1}^{K - 1} {\frac{{1 - F_\lambda (\theta )}}{{f_\lambda (\theta )}}\{ B_{\theta + 1} (x_\theta ) - B_\theta (x_\theta )\} } ,$$
where
x ≡ (
x 1, …,
x K ). Let
X be the subset of ℜ
K such that
\(x_{\theta + 1} \ge x_\theta \ge 0\quad\)for all
\(\theta \in \{ 1, \ldots K - 1\}\).
- (a)
V(·, λ) is supermodular for all λ.
- (b)
V(·, λ) has a unique maximizer in X for all λ.
- (c)
V(·,·) has decreasing first differences if f n (θ) / f m (θ) is increasing in θ.
- (d)
$$\Pi (x,\lambda ) = \sum\limits_{\theta = 1}^K {f_\lambda (\theta )B_\theta (x_\theta )\,\,-} \sum\limits_{\theta = 1}^{K - 1} {\{ 1 - F_\lambda (\theta )\} \{ B_{\theta + 1} (x_\theta ) - B_\theta (x_\theta )\} } .$$
Proof of Proposition 3:
(a) When σ=q is good news, f q (θ) / f 0(θ) is strictly increasing. Define f λ(·) ≡ (1−λ)f 0(·) + λf q (·). By Lemma A.2, we need only compare the consumption allocation that maximizes V(x, 0) to the one that maximizes V(x, 1). Denote the former by x 0 and the latter by x q . By Lemma A.1, Topkis's Monotonicity Theorem (Milgrom and Roberts, 1990) implies that the pointwise maximum of x 0 and x q also maximizes V(x, 0).42 Because V(x, 0) has a unique maximizer (Lemma A.1), the pointwise maximum of x 0 and x q equals x 0; that is, \(x_\theta ^0 \ge x_\theta ^q\) for all θ. As discussed in the text, there is no distortion at the top: \(x_K^0 = x_K^q = x_K^w\). Because there is no negative consumption, \(x_\theta ^0 = 0\) implies \(x_\theta ^q = 0\). To conclude our proof of part (a) we need to show that \(x_\theta ^0 > 0\) implies \(x_\theta ^0 > x_\theta ^q\) for θ < K. Suppose, counterfactually, that there exists at least one θ < K such that \(x_\theta ^0 = x_\theta ^q > 0\). Consider the smallest such θ, \(\tilde \theta\). Because \(\tilde \theta\) is the smallest such θ, it must be that \(\smash{x_{\tilde \theta }^q > x_{\tilde \theta - 1}^q }\) (without loss of generality, we can use the convention \(\smash{ x_0^\sigma = 0 }\) should \(\tilde \theta = 1\)). The fact that the \(x_{\tilde \theta }^q \ge x_{\tilde \theta - 1}^q\) constraint is not binding implies that
$$0 \le \frac{{\partial V({\bf x}^q ,1)}}{{\partial x_{\tilde \theta } }} = b_{\tilde \theta } (x_{\tilde \theta } ) - \frac{{1 - F_q (\tilde \theta )}}{{f_q (\tilde \theta )}}({b_{\tilde \theta + 1} (x_{\tilde \theta } ) - b_{\tilde \theta } (x_{\tilde \theta } )}).$$
As is well known, if
f σ(θ) /
f σ ′(θ) is increasing, then
$$ \frac{{f_\sigma (\theta )}}{{1 - F_\sigma (\theta )}} < \frac{{f_{\sigma '} (\theta )}}{{1 - F_{\sigma '} (\theta )}}. $$
(A6)
Hence,
$$\displaylines{
\frac{{\partial V({\bf x}^0 ,0)}}{{\partial x_{\tilde \theta } }} = b_{\tilde \theta } (x_{\tilde \theta } ) - \frac{{1 - F_0 (\tilde \theta )}}{{f_0 (\tilde \theta )}}( {b_{\tilde \theta + 1} (x_{\tilde \theta } ) - b_{\tilde \theta } (x_{\tilde \theta } )})\cr
> b_{\tilde \theta } (x_{\tilde \theta } )- \frac{{1 - F_q (\tilde \theta )}}{{f_q (\tilde \theta )}}( b_{\tilde \theta + 1} (x_{\tilde \theta } ) - b_{\tilde \theta } (x_{\tilde \theta } )) \ge 0.}$$
This expression implies that the constraint that
\(x_{\tilde \theta }^0 \le x_{\tilde \theta + 1}^0\) is binding when σ=0 (if not, then it would be profitable to increase
\(x_{\tilde \theta }^0\), contradicting the optimality of
x 0). Given
x 0 ≥
x q ,
\(x_{\tilde \theta }^q = x_{\tilde \theta }^0 = x_{\tilde \theta + 1}^0\) implies
\(x_{\tilde \theta }^q = x_{\tilde \theta + 1}^q\). Moreover, this argument can repeated inductively so that if
\(x_{\tilde \theta }^0 = x_{\tilde \theta + i}^0\) for
i=1, …,
I, then
\(x_{\tilde \theta }^q = x_{\tilde \theta + i}^q ( = x_{\tilde \theta }^0 )\) for
i=1, …,
I. Because it is never optimal to set
\(x_\theta > x_\theta ^w\), we know
\(\tilde \theta + I < K\); that is, there is a type
\(\tilde \theta + I + 1\) such that
\(x_{\tilde \theta }^0 = \cdots = x_{\tilde \theta + I}^0 < x_{\tilde \theta + I + 1}^0\). Substituting the constraint
\(x_{\tilde \theta }^q = \cdots = x_{\tilde \theta + I}^q\) in the maximization of
V(
x, 1) (recall λ=1 corresponds to σ=
q), the first-order condition with respect to
\(x_{\tilde \theta }^q\) implies
$$ \displaystyle\begin{array}{*{20}c}\hskip-60pt{0 \le \sum\limits_{\theta = \tilde \theta }^{\tilde \theta + I} {\left( {b_\theta (x_{\tilde \theta } ) - \displaystyle\frac{{1 - F_q (\theta )}}{{f_q (\theta )}}( {b_{\theta + 1} (x_{\tilde \theta } ) - b_\theta (x_{\tilde \theta } )})} \right)} } \hfill \\[12pt] \hskip20pt{ < \sum\limits_{\theta = \tilde \theta }^{\tilde \theta + I} {\left( {b_\theta (x_{\tilde \theta } ) -\displaystyle \frac{{1 - F_0 (\theta )}}{{f_0 (\theta )}}( {b_{\theta + 1} (x_{\tilde \theta } ) - b_\theta (x_{\tilde \theta } )})} \right)} = \sum\limits_{\theta = \tilde \theta }^{\tilde \theta + I} {\displaystyle\frac{{\partial V({\bf x}^q ,0)}}{{\partial x_\theta }}} ,} \hfill \\\end{array} $$
where the second line follows from (A6) and the fact that
\(\smash{ x_{\tilde \theta }^0 < x_{\tilde \theta + I + 1}^0 }\). But this implies that it would be feasible to increase profits when σ=0 by raising
\(\smash{ x_{\tilde \theta }^0 =\cdots = x_{\tilde \theta + I}^0 }\), which contradicts the optimality of
x 0. Hence, by contradiction, we've established the rest of part (a).
Part (b) has the identical proof, except that, now,f λ(·) ≡ λf 0(·) + (1−λ)f q (·).
Proof of Proposition 5:
As a preliminary, observe that, if (A2) is strictly increasing in θ for all x, then the associated order constraint (i.e., \(x_{\theta + 1} \ge x_\theta\)) is not binding and \(x_\theta\) is found by setting (A2) to zero and solving. Therefore, if (A2) is strictly increasing in θ for two distinct values of σ, say m and n, and if
$$\frac{{1 - F_m (\theta )}}{{f_m (\theta )}} < \frac{{1 - F_n (\theta )}}{{f_n (\theta )}},$$
then
\(x_\theta\) is greater when σ=
m than when σ=
n.
By assumptions (b) and (c), (A2) is strictly increasing in θ for all x when σ=0 (recall \(b_\theta (x)\) is strictly increasing in θ).
Suppose types are partitioned into J blocks, and let S j denote the maximal element in block j. Index the blocks so that \(1 \le S_1 < \cdots < S_J = K\). Let σ=j be that value of the indicator variable that indicates that a household with that realization of σ is in the j th block.
Under condition (i), any type in block
i is lower than any type in block
j if
i <
j. Observe that
f σ(θ) is equal to
\(f_0 (\theta ){\textstyle{{N_0 } \over {N_\sigma }}}\) if
θ is in the σ
th block and 0 otherwise. Hence,
$$\displaylines{
\frac{{1 - F_\sigma (\theta )}}{{f_\sigma (\theta )}} = \frac{{\sum\nolimits_{t > \theta } {f_\sigma (t)} }}{{f_\sigma (\theta )}} = \frac{{\sum\nolimits_{\scriptstyle t > \theta \hfill \atop\scriptstyle t \in \sigma \hfill} {f_0 (t)N_0 /N_\sigma } }}{{f_0 (\theta )N_0 /N_\sigma }}\cr
= \frac{{\sum\nolimits_{\scriptstyle t > \theta \hfill \atop\scriptstyle t \in \sigma \hfill} {f_0 (t)} }}{{f_0 (\theta )}} \le \frac{{\sum\nolimits_{t > \theta } {f_0 (t)} }}{{f_0 (\theta )}} = \frac{{1 - F_0 (\theta )}}{{f_0 (\theta )}}}$$
(A7)
for
θ in the σ
th block. The inequality is strict for all blocks except block
J. As noted before, there is no distortion for the top type within any population or sub-population; hence, each type
S enjoys efficient consumption. Except for
S J , this represents a strict increase in efficiency. For
S J (i.e.,
K), there is no change in efficiency. For the other types, recall that the marginal profit function is proportional to
$$b_\theta (x) - \frac{{1 - F_\sigma (\theta )}}{{f_\sigma (\theta )}}\{ b_{\theta + 1} (x) - b_\theta (x)\} .$$
From (A7), a switch from σ=0 (i.e., privacy) to σ ∈ {1, … ,
J}, cannot lower the marginal profit function and, for σ ∈ {1, … ,
J−1}, it strictly increases it. Hence, the consumption of these types increases, at least weakly. Because at least some types consume an amount strictly more efficient under the partition than under privacy, while no type consumes an amount that is less efficient, the result follows when condition (i) is satisfied.
Now assume condition (ii) is met. As above, the consumption of types {
S 1, …,
S J -1} is efficient and, thus, more efficient than under privacy. The consumption level of
S J =
K is equally efficient with or without privacy. Consider a type
θ that is not the maximal type within its block. Let type
θ+
k be the next higher type within that block. If
k=1 for all such types, then we have an ordered partition and the previous analysis applies. Restrict attention to the case
k ≥ 2. Now the marginal profit function is proportional to
$$b_\theta (x) - \frac{{1 - F_\sigma (\theta )}}{{f_\sigma (\theta )}}\{ b_{\theta + k} (x) - b_\theta (x)\} ,$$
which, by (A7) and the fact that
k ≥ 2, is weakly greater than
$$b_\theta (x) - \frac{{1 - F_0 (\theta + k - 1)}}{{f_0 (\theta )}}\{ b_{\theta + k} (x) - b_\theta (x)\} .$$
If we can show that term is greater than
$$b_\theta (x) - \frac{{1 - F_0 (\theta )}}{{f_0 (\theta )}}\{ b_{\theta + 1} (x) - b_\theta (x)\} ,$$
then we will have shown that marginal profit weakly increases for all non-maximal types, which completes the proof. A sufficient condition for that relation to hold is that
$$Z(n) \equiv (1 - F_0 (\theta + n - 1))\{ b_{\theta + n} (x) - b_\theta (x)\}$$
be decreasing in
n. Observe that
$$\displaylines{
Z(n + 1) - Z(n) = (1 - F_0 (\theta + n))\{ b_{\theta + n + 1} (x) - b_{\theta + n} (x)\}\cr
- f_0 (\theta + n)\{ b_{\theta + n} (x) - b_\theta (x)\} \cr
\le f_0 (\theta + n)\Delta (x)\left( {\frac{{1 - F_0 (\theta + n)}}{{f_0 (\theta + n)}} - n} \right),} $$
where
\(\Delta (x) \equiv b_{\theta + n + 1} (x) - b_{\theta + n} (x) > 0\) (because
\(x < x_{\theta + n + 1}^w\)) and the inequality follows from assumption (b). Assumption (c) and condition (ii) imply that the term in large parentheses is negative.
Turn now to our labor market example under the assumption that there is a continuum of household abilities. A necessary condition for a wage,
\(w_\sigma\), to be an equilibrium is that it equal the average productivity of the households willing to work at that wage, or
$$ w_\sigma = \nu\int_{\underline {\theta } }^{w_\sigma } {\frac{{\theta f_\sigma (\theta )}}{{F_\sigma (w_{\sigma})}}d\theta } . $$
(A8)
A necessary and sufficient condition for equilibrium is that the wage be the highest such wage satisfying Eq. (A8)—otherwise there would exist a wage such that the average productivity of job applicants would be higher than that wage and an employer would find it profitable to deviate by offering that wage. We know that such a value exists because both sides of the equation are continuous and the left-hand side ranges from 0 to ∞ while the right-hand side is bounded below by
\(\nu\underline \theta\) and above by
\(\nu\bar \theta\).
44