Privacy, property rights and efficiency: The economics of privacy as secrecy

Benjamin E. Hermalin; Michael L. Katz

doi:10.1007/s11129-005-9004-7

Quantitative Marketing and Economics

September 2006, Volume 4, Issue 3, pp 209–239 | Cite as

Privacy, property rights and efficiency: The economics of privacy as secrecy

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Benjamin E. Hermalin
Michael L. Katz

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First Online: 08 August 2006

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Abstract

There is a long history of governmental efforts to protect personal privacy and strong debates about the merits of such policies. A central element of privacy is the ability to control the dissemination of personally identifiable data to private parties. Posner, Stigler, and others have argued that privacy comes at the expense of allocative efficiency. Others have argued that privacy issues are readily resolved by proper allocation of property rights to control information. Our principal findings challenge both views. We find: (a) privacy can be efficient even when there is no “taste” for privacy per se, and (b) to be effective, a privacy policy may need to ban information transmission or use rather than simply assign individuals control rights to their personally identifiable data.

Keywords

Privacy Property rights Personal data Asymmetric information

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Notes

Acknowledgment

The authors would like to thank Thomas Davidoff, Giancarlo Spagnolo, Jean Tirole, Hal Varian, Michael Waldman, and Luc Wathieu, as well as participants in the 2004 “Quantitative Marketing and Economics Conference” and 2004 “Conference on the Economics of Electronics Communications Markets,” for helpful discussions of these issues. We are particularly grateful to an anonymous referee. An earlier version of this paper was titled “Is Privacy Efficient? The Economics of Privacy as Secrecy.”

Appendix

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(·) + λf _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 $\theta \in \{ 1, \ldots K - 1\}$.

Lemma A.1.

Suppose that, for all $x < x_\theta ^w$, $b_{\theta + 1} (x) - b_\theta (x)$ is non-decreasing in x for all $\theta \in \{ 1,2,\ldots,K - 1\}$. Then V(x, λ) and X have the following properties:

(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)

X is a lattice.

Proof:

(a)

By Topkis's Characterization Theorem (Milgrom and Roberts, 1990), property (a) is implied by the fact that $\frac{{\partial ^2 V}}{{\partial x_\theta \partial x_{\hat \theta } }} \ge 0\quad$for $\theta \ne \hat \theta$.
(b)

V(·, λ) attains a maximum over X because V(·, λ) is continuous and bounded and there would be no loss of generality in limiting its domain to $X \cap [0,x_K^w ]^K$, which is closed and bounded. $b_\theta (x)$ is decreasing and $b_{\theta + 1} (x) - b_\theta (x)$ non-decreasing in x for all $\theta \in \{ 1,2,\ldots ,K - 1\}$. Thus, V(·, λ) is concave. Given that X is convex, the maximizer of V(·, λ) is unique.
(c)

The additive separability of V(x, λ) with respect to the components of x implies it is sufficient to show decreasing differences with respect to a single x _θ, which can be shown by demonstrating that the cross-partial derivative of V with respect to λ and x _θ,
$$ \frac{{(F_n - F_m )f_\lambda + (1 - F_\lambda )(f_n - f_m )}}{{f_\lambda ^2 }}(b_{\theta + 1} - b_\theta ), $$

(A3)
is negative, where arguments have been suppressed for readability. Straightforward algebra reveals that the sign of (A3) equals the sign of F _n(θ) − F _m(θ). As is well known, f _n(θ) / f _m(θ) increasing in θ implies that F _m(θ) > F _n(θ) for all θ < K, which establishes (c).
(d)

X is a lattice if the meet (pointwise minimum) and join (pointwise maximum) of any two elements of X are in X. Suppose the join of x ¹ and x ² was not in X for two elements x ¹ and x ² of X. Then there must exist a θ such that $\max \{ x_\theta ^1 ,x_\theta ^2 \} > \max \{ x_{\theta + 1}^1 ,x_{\theta + 1}^2 \}$, which is impossible given that $x_\theta ^i < x_{\theta + 1}^i$ for i=1, 2. The proof that the meet is in X is similar.

Lemma A.2.

Let z be the value of x ∈ X that maximizes V(x, λ). Then z uniquely maximizes Π(x, λ) subject to x ∈ X, where

$$\Pi (x,\lambda ) = \sum\limits_{\theta = 1}^K {f_\lambda (\theta )B_\theta (x_\theta )\,\,-} \sum\limits_{\theta = 1}^{K - 1} {\{ 1 - F_\lambda (\theta )\} \{ B_{\theta + 1} (x_\theta ) - B_\theta (x_\theta )\} } .$$

Proof:

In maximizing either V or Π, there are K − 1 constraints of the form $x_{\theta + 1} - x_\theta \ge 0$. Form the Lagrangian

$$\displaylines{ L_\Pi = \Pi ({\bf x},\lambda ) + \sum\limits_{\theta = 1}^{K - 1} {\mu _\theta (x_{\theta + 1} - x_\theta ) + \mu _0 x_1}\cr = {\Pi ({\bf x},\lambda ) + \sum\limits_{\theta = 1}^K {(\mu _{\theta - 1} - \mu _\theta )x_\theta ,} } }$$

where μ₀ is the Lagrange multiplier on the restriction that sales be non-negative and μ_K ≡ 0. Making the change of variables $\alpha _\theta = \mu _\theta - \mu _{\theta - 1}$, the Lagrangean is

$$L_\Pi = \Pi ({\bf x},\lambda ) - \sum\limits_{\theta = 1}^K {\alpha _\theta x_\theta } .$$

We have a solution if there exist x and $\{ \alpha _\theta \}$ such that

$$\sum\limits_{k = \theta + 1}^K {\alpha _k \le 0} \quad{\rm for}\;{\rm all}\;\theta \in\left\{ {0, \ldots ,K - 1} \right\},$$

$$ \frac{{\partial L_\Pi }}{{\partial x_\theta }} = f_\lambda (\theta )b_\theta (x_\theta ) - \left( {1 - F_\lambda (\theta )} \right)\left( {b_{\theta + 1} (x_\theta ) - b_\theta (x_\theta )} \right) - \alpha _\theta = 0, $$

(A4)

and

$$(x_{\theta + 1} - x{}_\theta )\sum\limits_{k = \theta + 1}^K {\alpha _k = 0} .$$

We can similarly write the Lagrangean for the V problem as

$$L_V = V({\bf x},\lambda ) - \sum\limits_{\theta = 1}^K {\tilde \alpha _\theta x_\theta } .$$

The problem is unchanged if we scale each Lagrange multiplier by defining $\hat \alpha _\theta \equiv \tilde \alpha _\theta f_\lambda (\theta )$. We then have a solution for the V problem if there exist x and $\{ \hat \alpha _\theta \}$ such that

$$\sum\limits_{k = \theta + 1}^K {\hat \alpha _k \le 0} \quad{\rm for}\;{\rm all}\;\theta \in\left\{ {0, \ldots ,K - 1} \right\},$$

$$ \frac{{\partial L_V }}{{\partial x_\theta }} = b_\theta (x_\theta ) - \frac{{1 - F_\lambda (\theta )}}{{f_\lambda (\theta )}}\left( {b_{\theta + 1} (x_\theta ) - b_\theta (x_\theta )} \right) - \frac{{\hat \alpha _\theta }}{{f_\lambda (\theta )}} = 0. $$

(A5)

and

$$(x_{\theta + 1} - x{}_\theta )\sum\limits_{k = \theta + 1}^K {\hat \alpha _k = 0}.$$

Multiplying (A5) by f _λ(θ), we see that, if x and $\{ \hat \alpha _\theta \}$ are a solution to (A5), then they also satisfy (A4). V and Π are both concave (the proof of the latter parallels the proof that V is concave) and X is convex. Hence, both V and Π have unique maximizers in X. Therefore the x and $\{ \hat \alpha _\theta \}$ that solve the V problem are the unique solution to the Π problem.

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 4:

Let –s be the common slope. Let b ₂(x) − b ₁(x)=δ, where δ > 0. If x ₁ is an interior solution, then it follows from (A2) that

$$x_1 = \frac{{\beta - \delta H_\sigma }}{s},$$

where H _σ=f _σ(2) / f _σ(1). An interior solution exists only if and only if $\beta > \delta H_\sigma$. The resulting deadweight loss triangle per low-type is ${\textstyle{s \over 2}}(x_1^w - x_1 )^2$.

Consider each case in turn:

(a)

If $\beta \le \delta H_0$, then $x_1 = 0$ under privacy and the deadweight loss is maximized. If β > min{δH ₁, δH ₂}, then $x_1 > 0$ for one sub-population absent privacy, and privacy is less efficient.
(b)

When β > max{δH ₁, δH ₂}, direct calculations show that privacy reduces deadweight loss.⁴³
(c)

The per-capita deadweight loss under dissemination tends to 0 as the sorting of household types becomes perfect and thus dissemination is then more efficient than privacy. Instances in which the information improvement lowers total surplus can be constructed by making use of (b) and the continuity of average deadweight loss with respect to the proportion of low types, and considering values of max{δH ₁, δH ₂}that are just above $\beta$.

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

Proof of Proposition 7:

It suffices to show that $\frac{{f_i (\theta )}}{{f_j (\theta )}}$ monotonically decreasing implies w $_i < {\it w}_j$. Define $g_\sigma (\theta ,{\it w}) = \frac{{f_\sigma (\theta )}}{{F_\sigma ({\it w})}}$. By the definitions of $f_\sigma (\theta )$ and $F_\sigma (\theta )$,

$$\int_{\underline {\theta } }^{w} {\left[ {\frac{{g_i (\theta ,w)}}{{g_j (\theta ,w)}} - 1} \right]g_j (\theta ,w)\, d\theta } = 0.$$

Moreover, $\frac{{g_i (\theta ,w)}}{{g_j (\theta ,w)}}$ is monotonically decreasing, so it and θ covary negatively. Hence, their covariance,

$$ \int_{\underline {\theta } }^{w} {\theta \left[ {\frac{{g_i (\theta ,w)}}{{g_j (\theta ,w)}} - 1} \right]g_j (\theta ,w)\, d\theta } , $$

(A9)

is negative for all $w > \underline \theta$. Therefore,

$$v\int_{\underline {\theta } }^{w_i } {\frac{{\theta f_j (\theta )}}{{F_j (w_i )}}d\theta } > v\int_{\underline {\theta } }^{w_i } {\frac{{\theta f_i (\theta )}}{{F_i (w_i )}}d\theta } = w_i .$$

Both the integral and w are continuous in w, and $\nu\int_{\underline {\theta } }^{w} {\theta {\textstyle{{f_j (\theta )} \over {F_j (w)}}}d\theta - w}$ goes to −∞ as $w \to \infty$. Hence, there exists $w_j > w_i$ such that $v\int_{\underline {\theta } }^{w_j } {\frac{{\theta f_j (\theta )}}{{F_j (w_j )}}d\theta } = w_j$.

Proof of Proposition 10:

Consider an equilibrium that arises when firms can compel households to reveal the values of their indicator variables. Let $ - t_n (\sigma )$ be the equilibrium wage offered by firm n to any household that has value σ. Because there are no matching benefits, the industry is competitive, and no firm would, in equilibrium, offer a wage that leads it to lose money, it must be that $t_n (\sigma ) = t_m (\sigma )$ for all n and m. Without loss of generality, order the σs so that $t(\sigma _i ) \ge t(\sigma _j )$ if i < j. Observe that $\sigma _1$ is the least-favored indicator variable because $u( {\theta ,x(\theta ,t(\sigma _1 )),t(\sigma _1 )} ) \le u( {\theta ,x(\theta ,t(\sigma _j )),t(\sigma _j )})$ for all θ and all j > 1, where $x\left( {\theta ,t} \right)$ is the best-response labor supply decision of a type-θ household when offered t. The result then follows from Proposition 2.

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Benjamin E. Hermalin
- 1
Email author
Michael L. Katz
- 1

1.Haas School of BusinessUniversity of California, BerkeleyBerkeleyUSA

Privacy, property rights and efficiency: The economics of privacy as secrecy

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