Acquiring High Quality Customer Data with Low Cost

Abstract

This work concerns optimal customer data acquisition and selection problem. We first propose using a divide-and-conquer technique to find empirical distribution of the customer data. We then formulate a data acquisition problem as an optimization problem that maximizes the quality of the acquired data while keeping the cost of acquisition as low as possible. We propose using generalized second-price (GSP) auction to acquire customer data and show that when the number of bidders is large, GSP is a truth-telling mechanism. We derive the analytical solution for the optimization problem, which finds a set of data that best represents the probability distribution of the target population relative to the acquisition cost. An experimental study is conducted to demonstrate the effectiveness of the proposed approach.

Appendix

Proof of Theorem 1 (GSP is asymptotically truth-telling).

Let v be the true value of a bidder and $\beta (v)$ be the bidder’s bid. For a GSP with N bidders (large) and S slots (fixed), Gomes and Sweeney [5] show the following relationship:

$$ \beta (v) = v - \sum\nolimits_{s = 2}^{S} {\gamma_{s} (v)\int_{0}^{v} {(v - \beta (x))F^{N - s - 1} (x)f(x)dx} } , $$

where F (with the density f) represents the distribution of bidders’ values and $\gamma_{s} (v)$ is a function of v with a complicated expression. We have derived the result $\beta (v) = v$ by showing that

$$ \mathop {\lim }\limits_{N \to \infty } \sum\nolimits_{s = 2}^{S} {\gamma_{s} (v)\int_{0}^{v} {(v - \beta (x))F^{N - s - 1} (x)f(x)dx} } = 0. $$

The details are omitted due to space limitations. Instead, we provide a sketch of the proof here in a Nash equilibrium context. First, suppose that the bidder underbid with $\beta^{ - } (v) < v$ and wins. He will be paid an amount next to his price $\beta * > \beta^{ - } (v)$. If $\beta * > v$, then the bidder would win with bid v anyway. If $\beta^{ * } \le v$, then the payoff to the bidder would be $\beta^{ * } - v \le 0$. So, underbidding does not improve the bidder’s payoff. Now, suppose that the bidder overbid with $\beta^{ + } (v) > v$. Since N is large and S is fixed, there will be a large number of bids within $[v,\beta^{ + } (v)]$ and this number will be larger than S. Consequently, the bidder will not win by overbidding.

Proof of Theorem 2 (Solutions to the optimization problem (1)).

We standardize the minimization problem (1) to

$$ { \hbox{min} }\,\sum\limits_{i = 1}^{m} {\frac{{\left( {x_{i} - np_{i} } \right)^{2} }}{{np_{i} }}} + \frac{1}{2}w\sum\limits_{i = 1}^{m} {d_{i} x_{i}^{2} } , $$

(A.1)

$$ {\text{s}}.{\text{t}}. { 1} - x_{i} \le 0, \, i = 1, \ldots ,m, $$

(A.2)

$$ \sum\limits_{i = 1}^{m} {x_{i} } - n = 0. $$

(A.3)

Applying the KKT conditions to this problem, we have for Lagrange multipliers $\lambda_{i} , \, i = 1, \ldots ,m$,

$$ \begin{aligned} \lambda_{i} \ge 0, \, i = 1, \ldots ,m, \hfill \\ \lambda_{i} ( - x_{i} + 1) = 0, \, i = 1, \ldots ,m, \hfill \\ \sum\limits_{i = 1}^{m} {x_{i} } - n = 0. \hfill \\ \end{aligned} $$

(A.4)

It follows from (A.1), (A.2) and (A.3) that

$$ \frac{d}{{dx_{i} }}\left( {\sum\limits_{i = 1}^{m} {\frac{{\left( {x_{i} - np_{i} } \right)^{2} }}{{np_{i} }}} + \frac{1}{2}w\sum\limits_{i = 1}^{m} {d_{i} x_{i}^{2} } + \sum\limits_{i = 1}^{m} {\lambda_{i} ( - x_{i} + 1)} + \mu (\sum\limits_{i = 1}^{m} {x_{i} } - n)} \right) = 0, \, i = 1, \ldots ,m, $$

or

$$ \frac{{2\left( {x_{i} - np_{i} } \right)}}{{np_{i} }} + wd_{i} x_{i} - \lambda_{i} + \mu = 0, \, i = 1, \ldots ,m. $$

For non-boundary solutions, $x_{i} > 1$. It then follows from (A.4) that $\lambda_{i} = 0, \, i = 1, \ldots ,m$. Thus,

$$ \begin{aligned} 2(x_{i} - np_{i} ) + wnp_{i} d_{i} x_{i} + np_{i} \mu = 0, \, i = 1, \ldots ,m, \hfill \\ x_{i} (2 + wnp_{i} d_{i} ) - 2np_{i} + np_{i} \mu = 0, \, i = 1, \ldots ,m, \hfill \\ \end{aligned} $$

i.e.,

$$ x_{i} - \frac{{2np_{i} }}{{(2 + wnp_{i} d_{i} )}} + \mu \frac{{np_{i} }}{{(2 + wnp_{i} d_{i} )}} = 0, \, i = 1, \ldots ,m, $$

(A.5)

or

$$ \sum\limits_{i = 1}^{m} {x_{i} } - \sum\limits_{i = 1}^{m} {\frac{{2np_{i} }}{{(2 + wnp_{i} d_{i} )}}} + \mu \sum\limits_{i = 1}^{m} {\frac{{np_{i} }}{{(2 + wnp_{i} d_{i} )}}} = 0. $$

Substituting (A.3) into the above equation, we have

$$ n - \sum\limits_{i = 1}^{m} {\frac{{2np_{i} }}{{(2 + wnp_{i} d_{i} )}}} + \mu \sum\limits_{i = 1}^{m} {\frac{{np_{i} }}{{(2 + wnp_{i} d_{i} )}}} = 0. $$

Solving it for μ, we get $\mu = 2 - \frac{1}{{\sum\limits_{i = 1}^{m} {\frac{{p_{i} }}{{(2 + wnp_{i} d_{i} )}}} }}$. Substituting it into (A.5), we have

$$ x_{i} = {{\frac{{np_{i} }}{{2 + wnp_{i} d_{i} }}} \mathord{\left/ {\vphantom {{\frac{{np_{i} }}{{2 + wnp_{i} d_{i} }}} {\sum\limits_{j = 1}^{m} {\frac{{p_{j} }}{{2 + wnp_{j} d_{j} }}} }}} \right. \kern-0pt} {\sum\limits_{j = 1}^{m} {\frac{{p_{j} }}{{2 + wnp_{j} d_{j} }}} }}, \, i = 1, \ldots ,m. $$

This solution satisfies KKT condition and the objective function is convex, so it is the unique solution.