Pricing private data

Abstract

We consider a market where buyers can access unbiased samples of private data by appropriately compensating the individuals to whom the data corresponds (the sellers) according to their privacy attitudes. We show how bundling the buyers’ demand can decrease the price that buyers have to pay per data point, while ensuring that sellers are willing to participate. Our approach leverages the inherently randomized nature of sampling, along with the risk-averse attitude of sellers in order to discover the minimum price at which buyers can obtain unbiased samples. We take a prior-free approach and introduce a mechanism that incentivizes each individual to truthfully report his preferences in terms of different payment schemes. We then show that our mechanism provides optimal price guarantees in several settings.

Appendix: A Proofs

We now provide the proofs of the results that were omitted from the main section.

Proof of Theorem 4.1

Consider some seller i and first suppose that $c_{i} < \max _{j \ne i} \hat {c}_{j}$. We observe that reporting any $\hat {c}_{i} < \max _{j \ne i} \hat {c}_{j}$ will not make a difference in the utility of seller i regardless of the outcome of the sampling; furthermore, he will derive a strictly positive utility every time seller i is sampled. On the other hand, if he reports $\hat {c}_{i} > \max _{j \ne i} \hat {c}_{j}$, then seller i will be excluded from the sampling and derive zero utility. The second case to consider is that $c_{i} > \max _{j \ne i} \hat {c}_{j}$. Then, by reporting $\hat {c}_{i} = c_{i}$, seller i’s utility is equal to zero. However, there is no way of getting positive utility in this case. In particular, by reporting $\hat {c}_{i} < \max _{j \ne i} \hat {c}_{j}$, seller i will get negative utility whenever he is sampled.

Thus, reporting $\hat {c}_{i} \ne c_{i}$ can never increase the utility of seller i but in some circumstances may actually decrease it. This shows that truthful reporting is a dominant strategy for each seller. To show ex-post individual rationality, we observe that by reporting $\hat {c}_{i} = c_{i}$ seller i gets a positive utility whenever sampled and zero utility otherwise. □

Proof of Lemma 4.3

Let Z _i be a random variable that denotes the number of times seller i is sampled. We have that ${\sum }_{i = 1}^{n} Z_{i} = {\sum }_{b \in B} d_{b}$ in order to meet the demand. Observe that the expected number of times that seller i is sampled under distribution ψ is $\mathbb {E}[Z_{i}] = {\sum }_{k=0}^{m} k p_{\psi }(k)$. Since ψ ∈ Ψ, this distribution produces unbiased samples, which implies that each seller is sampled the same expected number of times. Thus, summing over all sellers,

$$n^{\prime} \sum\limits_{k=0}^{m} k p_{\psi}(k) = \sum\limits_{i = 1}^{n} \mathbb{E}[Z_{i}] = \sum\limits_{b \in B} d_{b},$$

which concludes the proof. □

Proof of Theorem 4.2

If the payment will be determined by the function π ^max, Theorem 4.1 implies that it is a dominant strategy for seller i to report c _i truthfully and that we get ex-post individual rationality. For a seller i with $c_{i} < \max _{j \ne i} \hat {c}_{j}$, it is a dominant strategy to also report his certainty equivalent for (p _ψ , π ^max) truthfully in Step (2), because his report $\hat {e}_{i}$ does not affect the threshold $\overline {r}_{i}$. Finally, if the payment is determined to be $\pi _{i}(k) \equiv \overline {r}_{i} + \hat {c}_{i} (k - w)$ and seller i has reported $\hat {c}_{i}$ truthfully, we have ex-post individual rationality because $\pi _{i}(k) - c_{i} k = \overline {r}_{i} - c_{i} w > \hat {r}_{i} - c_{i} w = \hat {e}_{i} > 0$.

We now turn to the seller i with $c_{i} > \max _{j \ne i} \hat {c}_{j}$. By reporting any value $\hat {c}_{i} > \max _{j \ne i} \hat {c}_{j}$, the seller will not be sampled and gets utility zero. By reporting $\hat {c}_{i} < \max _{j \ne i} \hat {c}_{j} < c_{i}$, seller i gets a negative utility if assigned payment π ^max in Step 3. On the other hand, if assigned the payment $\pi _{i}(k) \equiv \overline {r}_{i} + \hat {c}_{i} (k - w)$, seller i derives utility $u_{i}(\overline {r}_{i} - \hat {c}_{i} w - (c_{i} - \hat {c}_{i})k)$ which may be positive for small values of k.

We now show that if seller i is risk-neutral or risk-averse, i.e., not risk-seeking, he derives negative utility in expectation. In particular, we have $\overline {r}_{i} - \hat {c}_{i} w - (c_{i} - \hat {c}_{i})k < (c_{i} - \hat {c}_{i}) (w-k)$. Thus, ${\sum }_{k} p_{\psi }(k) (\overline {r}_{i} - \hat {c}_{i} w - (c_{i} - \hat {c}_{i})k) < {\sum }_{k} p_{\psi }(k) (c_{i} - \hat {c}_{i}) (w-k) = 0$. And since u _i is concave or linear, ${\sum }_{k} p_{\psi }(k) u_{i}(\overline {r}_{i} - \hat {c}_{i} w - (c_{i} - \hat {c}_{i})k) < {\sum }_{k} p_{\psi }(k) u_{i}(c_{i} - \hat {c}_{i}) (w-k) = 0$.

To conclude the proof, we consider the case that the seller i with $c_{i} > \max _{j \ne i} \hat {c}_{j}$ is risk-seeking. Then, it is plausible that seller i is better off reporting $\hat {c}_{i} < c_{i}$ in order to get utility $u_{i}(\overline {r}_{i} - \hat {c}_{i} w - (c_{i} - \hat {c}_{i})k)$ which is positive for small values of k, but negative for large values. Even though such preferences are very unlikely, for the sake of completeness we describe how our CE mechanism can be extended to deal with this issue for the sake of completeness.

To avoid such situations, the mechanism can ask each seller j ≠ i to report his certainty equivalent for the lottery $(p_{\psi }, \pi _{-i}^{max})$, where $\pi _{-i}^{max} \equiv \max _{j\ne i} \hat {c}_{j}$, and use these values to determine the threshold $\overline {r}_{i}$ for seller i. Then, seller i will be included in the sampling only if $\hat {r}_{i} < \overline {r}_{i}$. This guarantees truthful reporting and individual rationality for selleri. Moreover, each seller j ≠ i has no reason to lie about his certainty equivalent for $(p_{\psi }, \pi _{-i}^{max})$.⁷ □

Proof of Theorem 4.5

Let S denote the set of all subsets of B (i.e., the powerset of B), and let S _b ⊆S denote the set of all such subsets that include buyer b. We sort all buyers in a non-increasing order of the sample sizes that they request, i.e., $d_{b^{\prime }} \geq d_{b}$ if b ^′<b. What follows is described from the perspective of some arbitrary seller i. Given a distribution ψ ∈ Ψ(N ^′), let q _ψ (s) denote the probability that seller i’s data is sold to all buyers in s but nobody else.⁸ Since q _ψ is a distribution over S, we have that ${\sum }_{s\in S} q_{\psi }(s)=1$. Since ψ ∈ Ψ(N ^′), the probability that buyer b gets seller i in his sample must be equal to d _b /n ^′, where n ^′ is the number of sellers in N ^′. Equivalently, for each buyer b, ${\sum }_{s\in S_{b}} q_{\psi }(s) = d_{b}/n^{\prime }$.

The ordering distribution ψ ^∗ = ψ ^∗(N ^′) satisfies the following simple predicate: If buyer b gets access to the data of seller i then so does buyer b − 1; equivalently, $q_{\psi ^{\ast }}(s)=0$ for every s∈(S _b ∖S _b−1). We will show that G(ψ) is minimized at ψ ^∗ over all ψ ∈ Ψ(N ^′) using proof by contradiction. Assume that G(ψ) < G(ψ ^∗) for some distribution ψ ∈ Ψ(N ^′) that does not satisfy the predicate. We will gradually modify distribution ψ until it satisfies the predicate without increasing the expected value G in the process (if g is strictly concave, then this modification leads to a decrease in G).

Let b be the first buyer in the ordering for which the predicate is not true, i.e., there exists some set s _A that contains b and does not contain b − 1 (s _A ∈ (S _b ∖S _b−1)) such that q _A ≡ q _ψ (s _A )>0. Since d _b−1≥d _b , there must also exist some s _B that contains b−1 but not b, and occurs with some positive probability q _B ≡ q _ψ (s _B ) > 0. Define q _min ≡ min(q _A , q _B )>0. Let s _I (resp., s _U ) denote the outcome that contains exactly the intersection (resp., union) of the buyers in s _A and s _B . We modify ψ by removing probability mass q _min from s _A and s _B and moving it to s _I and s _U . This leads to a new probability distribution $\hat {q}$ over S such that $\hat {q}(s_{A})=q_{A}-q_{\text {min}}$, $\hat {q}(s_{B})=q_{B}-q_{\text {min}}$, $\hat {q}(s_{I})=q(s_{I})+q_{\text {min}}$, and $\hat {q}(s_{U})=q(s_{U})+q_{\text {min}}$; for all other s ∈ S we have $\hat {q}(s)=q_{\psi }(s)$. $\hat {q}$ corresponds to some distribution $\hat {\psi } \in {\Psi }(N^{\prime })$.

We now show that $G(\hat {\psi }) \leq G(\psi )$. Let n _α be the number of buyers in s _α , and d the number of buyers in s _A ∖s _B ; thus, n _A = n _I +d, n _U = n _B +d, and $G(\psi ^{\prime }) - G(\hat {\psi })$ is equal to

$$\begin{array}{@{}rcl@{}} &&q_{\text{min}} g(n_{I}+d) + q_{\text{min}} g(n_{B}) - q_{\text{min}} g (n_{I}) - q_{\text{min}}g(n_{B}+d) \\ &=&q_{\text{min}} [(g(n_{I}+d)-g(n_{I})) - (g(n_{B}+d)-g(n_{B}))] \geq 0 \end{array} $$

The inequality holds because g is concave and n _B >n _I .

We can repeat the modification step for this same pair of buyers b and b−1 as long as the predicate is not satisfied. After every modification, either q(s _A ) or q(s _B ) becomes 0 and the probabilities of these sets are never raised again during the modification steps for this same pair of buyers. Thus, only a finite number of modifications is needed until the induced lottery satisfies the predicate. Since the expected value G does not increase at any point during this process, we conclude that G(ψ) is minimized at ψ ^∗. □

Proof of Theorem 5.1

Consider two instances (A and B) with n sellers and m≡n ²+n + 1 buyers. In both instances, buyers’ demand is the same: one buyer demands a sample of n (i.e., all of the sellers) and the remaining n ²+n buyers demand a sample of just one seller. The two instances differ with respect to the sellers’ cost functions and have different payment functions: $\pi ^{max}_{A}(k)=\min \{k,n\}$ and $\pi ^{max}_{B}(k)=\max \{k,n\}$; note that both can arise from concave c _i ’s.

Since we are interested in distributions that are oblivious to the payment function and the two instances differ only with respect to the payment functions, it suffices to show that if a distribution ψ ∈ Ψ(N) gives a 2-approximation for instance A, then the same distribution ψ cannot give a better than (2−𝜖)-approximation for instance B for 𝜖>0. More formally, we will show that if

$$ \sum\limits_{k=0}^{m} p_{\psi}(k) \pi_{A}^{max}(k) \leq 2 {\Pi}^{\text{OPT}}_{A}, $$

(2)

then

$$ \sum\limits_{k=0}^{m} p_{\psi}(k) \pi_{B}^{max}(k) > (2 - 6/n) {\Pi}^{\text{OPT}}_{B}. $$

(3)

Setting n>6/𝜖 will then conclude the proof.

Since $\pi _{A}^{max}$ is concave, ψ ^∗ is optimal for instance A and

$$\begin{array}{@{}rcl@{}} {\Pi}^{\text{OPT}}_{A} &=& \sum\limits_{k=0}^{m} p_{\psi^{\ast}}(k) \pi_{A}^{max}(k) \\ &=&\frac{1}{n} \min\{n^{2}+n+1,n\} + \frac{n-1}{n} \min\{1,n\} = 2 - \frac{1}{n}. \end{array} $$

Thus, (2) implies that ${\sum }_{k=0}^{m} p_{\psi }(k) \pi _{A}^{max}(k) < 4$. Then,

$$\sum\limits_{k=0}^{n} p_{\psi}(k)\pi^{max}_{A}(k) = \sum\limits_{k=0}^{n} k p_{\psi}(k)<4, $$

which, together with Lemma 4.3, implies

$$ \sum\limits_{k=n+1}^{m} k p_{\psi}(k) > n-2. $$

(4)

Moreover, ${\sum }_{k = n+1}^{m} p_{\psi }(k)\pi ^{max}_{A}(k) = n {\sum }_{k = n+1}^{m} p_{\psi }(k) < 4$, which implies that

$$ \sum\limits_{k = 0}^{n} p_{\psi}(k) > 1 - 4/n. $$

(5)

Now consider instance B. The buyer that requested a sample of size n will be given access to the data of all sellers. To determine what samples other buyers will get, suppose we randomly split the set of n ²+n buyers demanding a single seller into n groups of n + 1 buyers each. We label these groups {1,2,...,n}. We then assign seller i to the buyers of group i. This gives unbiased samples, because each buyer is equally likely to be in each group. Note that exactly n buyers get access to the data of a given seller, so ${\Pi }^{\text {OPT}}_{B} \leq n$.

To conclude the proof, we show that (4) and (5) imply (3). First note that in order to satisfy the demand of the buyer who requests n sellers, every seller will be sampled at least once and paid at least max{1,n}=n. Then, (5) implies that ${\sum }_{k=0}^{n} p_{\psi }(k) \pi ^{max}_{B}(k) > n-4$. On the other hand, (4) implies that ${\sum }_{k=n+1}^{m} p_{\psi }(k) \pi ^{max}_{B}(k) > n-2$. Summing these two inequalities, we conclude that ${\sum }_{k=0}^{m} p_{\psi }(k) \pi _{B}^{max}(k) > 2n-6$, which together with ${\Pi }^{\text {OPT}}_{B} \leq n$ implies (3). □

Proof Sketch for Theorem 5.2

Let ψ ∈ Ψ(N) be the distribution that achieves π^OPT; for notational simplicity, we write p≡p _ψ , $p_{o} \equiv p_{\psi ^{\ast }}$, and π≡π ^max for the remainder of the proof. We wish to show that

$$ \sum\limits_{k=0}^{m} p_{o}(k) \pi(k) \leq 2{\Pi}^{\text{OPT}} = 2 \sum\limits_{k=0}^{m} p(k) \pi(k). $$

(6)

Let $P(k)={\sum }_{k^{\prime } \leq k}{p(k^{\prime })}$ denote the cumulative distribution function of p, and P ⁻¹(t) = inf{k|P(k)≥t} denote its generalized inverse distribution function; the functions P _o (⋅) and $P_{o}^{-1}(\cdot )$ are defined similarly for p _o .

We decompose the [0,1] interval into subintervals (t _l , t _r ) for which we know that, for any two values t,t ^′∈(t _l , t _r ), we have P ⁻¹(t) = P ⁻¹(t ^′) and $P_{o}^{-1}(t)=P_{o}^{-1}(t^{\prime })$. In order to do so, we let T = {P(k)|(p(k)>0)∨(p _o (k)>0)} be the set of distinct values in [0,1] that either P(⋅) or P _o (⋅) takes. If 0<t ₁ < t ₂<...<t _|T| = 1 is an ordering of the values in T and t ₀ = 0, then we let I = {(t ₀, t ₁],(t ₁, t ₂],...,(t _|T|−1,1]}, which is a set of intervals that satisfy the property that we wanted. Given this property, (6) can be rewritten as follows:

$$\sum\limits_{t_{r}\in T}{(t_{r}-t_{r-1})\pi(P_{o}^{-1}(t_{r}))} \leq 2 \sum\limits_{t_{r}\in T}{(t_{r}-t_{r-1})\pi(P^{-1}(t_{r}))}. $$

Let $T_{A} = \{t_{r}\in T | P^{-1}(t_{r})>P_{o}^{-1}(t_{r})\}$ and T _B = T∖T _A . By definition of the set T _A , and using the fact that π(⋅) is an increasing function, it is easy to see that for any t _r ∈T _A we have $\pi (P^{-1}(t_{r})) \geq \pi (P_{o}^{-1}(t_{r}))$. Therefore

$$\begin{array}{@{}rcl@{}} \sum\limits_{t_{r}\in T_{A}}{(t_{r}-t_{r-1})\pi (P_{o}^{-1}(t_{r}))} &\leq &\sum\limits_{t_{r}\in T_{A}}{(t_{r}-t_{r-1}) \pi(P^{-1}(t_{r}))} \\ &\leq &\sum\limits_{t_{r}\in T}{(t_{r}-t_{r-1})\pi(P^{-1}(t_{r}))}. \end{array} $$

In order to prove the theorem (for a complete proof see (Gkatzelis et al. 2012)) we only need to show the following inequality; summing it with the previous one proves the theorem since T _A ∪T _B = T.

$$\sum\limits_{t_{r}\in T_{B}}{(t_{r}-t_{r-1})\pi(P_{o}^{-1}(t_{r}))}\leq \sum\limits_{t_{r}\in T}{(t_{r}-t_{r-1})\pi(P^{-1}(t_{r}))}. $$

Proving this inequality is significantly more demanding, but the main intuition behind why it holds is that, even though π may not be concave, the payment that a seller receives per buyer is decreasing in the total number of buyers that get access to his data. To verify that this is true, for some k, let i be the seller for which c _i (k) = π(k). Then, for k ^′≤k,

$$\begin{array}{@{}rcl@{}} \pi(k) = c_{i}(k) &\leq& \frac{k}{k^{\prime}}c_{i}(k^{\prime}) \leq \frac{k}{k^{\prime}}\pi(k^{\prime})\\ \Rightarrow \frac{\pi(k)}{k} &\leq& \frac{\pi(k^{\prime})}{k^{\prime}}, \end{array} $$

where the first inequality holds because of the concavity of c _i (⋅) and the second holds by definition of π(k ^′)= maxi c _i (k ^′).

We use this to upper bound the payment from distribution p _o due to t _r ∈T _B intervals; we show that for every expected buyer of p _o in the interval (t _l , t _r ] with t _r ∈T _B , there exists some other interval $(\bar {t}_{l},\bar {t}_{r}]$ with $\bar {t}_{r}\leq t_{r}$ and the same expected number of buyers for p. Since $\bar {t}_{r}\leq t_{r}$, the former interval leads to a smaller payment per buyer. □