Computing Profit-Maximizing Bid Shading Factors in First-Price Sealed-Bid Auctions

Abstract

Computational methods are used to determine a profit-maximizing shading factor by which rational bidders shade their bid in first price sealed bid auctions for a broad range of realistic scenarios when the prior is diffuse. Bidders’ valuations may have both common value and firm-specific components, and the accuracy of their estimates of the common value component may differ. In addition, we allow for a subset of “naive” rivals, defined as bidders who do not account for the Winners’ Curse. Our computations show that profit-maximizing shading is greatly impacted by asymmetries in the bidding population and, in particular, by the presence of naive bidders. Failing to account for the presence of naive bidders results in underbidding only in one case, when facing a single rival who is naive, and in overbidding in all other cases. Overbidding is particularly severe when the population of naive competitors is large.

Appendices

Appendix A Effect of naive competitors on the Winner’s Curse

See Fig. 8.

Fig. 8

Bias factor for an increasing number of bidders. In one group only one of the bidders is rational (solid) and in the other, all bidders are rational (dashed)

Appendix B Proof of Lemma 1

As stated in the main text, the Shading Factors satisfy the following:

$$\begin{aligned} c_i\in \arg \max _{{\hat{c}}} \int _{\mathbb {R}} f_i(\epsilon _i) ({\hat{c}}-\epsilon _i)\left( \prod _{j\ne i} F_j(\varDelta _i-{\hat{c}}-(\varDelta _j-c_j)+\epsilon _i)\right) d\epsilon _i \end{aligned}$$

(7)

Take the derivative with respect to $c_i$ and set it equal to zero:

$$\begin{aligned} \frac{\partial }{\partial c_i}\left[ \int _{\mathbb {R}} f_i(\epsilon _i) (c_i-\epsilon _i)\left( \prod _{j\ne i} F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\right) d\epsilon _i\right] =0 \end{aligned}$$

(8)

Noting that $f_i(\epsilon _i)$ does not depend on $c_i$, solve:

$$\begin{aligned} \int _{\mathbb {R}} f_i(\epsilon _i) \frac{\partial }{\partial c_i}\left[ (c_i-\epsilon _i)\left( \prod _{j\ne i} F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\right) \right] d\epsilon _i=0 \end{aligned}$$

(9)

The derivative corresponds to the addition of two terms:

$$\begin{aligned} \frac{\partial }{\partial c_i}\left[ (c_i-\epsilon _i)\right] \left( \prod _{j\ne i} F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\right) + \ldots \end{aligned}$$

(10)

$$\begin{aligned} (c_i-\epsilon _i)\frac{\partial }{\partial c_i}\left[ \left( \prod _{j\ne i} F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\right) \right] \end{aligned}$$

(11)

It is clear that (10) is equal to:

$$\begin{aligned} \left( \prod _{j\ne i} F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\right) \end{aligned}$$

Finally, to solve (11) let’s focus for now on the term:

$$\begin{aligned}&\frac{\partial }{\partial c_i}\left[ \prod _{j\ne i} F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\right] = \\&\quad -\sum _{j\ne i} \left( f_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\prod _{k\ne j, i} F_k(\varDelta _i-c_i-(\varDelta _k-c_k)+\epsilon _i)\right) \end{aligned}$$

Multiplying and dividing by $F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)$ we obtain

$$\begin{aligned} -\sum _{j\ne i} \left( \frac{f_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)}{F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)}\prod _{k\ne i} F_k(\varDelta _i-c_i-(\varDelta _k-c_k)+\epsilon _i)\right) \end{aligned}$$

Note that the product does not depend on j, so it can go out of the summation:

$$\begin{aligned} -\left( \prod _{k\ne i} F_k(\varDelta _i-c_i-(\varDelta _k-c_k)+\epsilon _i)\right) \left( \sum _{j\ne i} \frac{f_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)}{F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)}\right) \end{aligned}$$

So the third term, replacing k with j, corresponds to:

$$\begin{aligned} -(c_i-\epsilon _i)\left( \prod _{j\ne i} F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)\right) \left( \sum _{j\ne i} \frac{f_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)}{F_j(\varDelta _i-c_i-(\varDelta _j-c_j)+\epsilon _i)}\right) \end{aligned}$$

Finally, to obtain the full expression, apply the change of variables $z_i = \varDelta _i-c_i-(\varDelta _j-c_j)$, to obtain:

$$\begin{aligned} \left( \prod _{j\ne i} F_j(z_{ij}+\epsilon _i)\right) -(c_i-\epsilon _i)\left( \prod _{j\ne i} F_j(z_{ij}+\epsilon _i)\right) \left( \sum _{j\ne i} \frac{f_j(z_{ij}+\epsilon _i)}{F_j(z_{ij}+\epsilon _i)}\right) \end{aligned}$$

Replacing back into the integral in (9) we have:

$$\begin{aligned} \int _{\mathbb {R}}f_i(\epsilon _i)\left( \prod _{j\ne i} F_j(z_{ij}+\epsilon _i)\right) \left( 1-(c_i-\epsilon _i)\sum _{j\ne i} \frac{f_j(z_{ij}+\epsilon _i)}{F_j(z_{ij}+\epsilon _i)}\right) d\epsilon _i = 0 \end{aligned}$$

(12)

This is the first order condition.

Appendix C Monte Carlo method

See Figures 9, 10, 11 and 12

We benchmarked the Shading Factor in some cases using Monte Carlo simulations, as set out below:

Generate signals randomly for the number of bidders, in a matrix of size $n\times m$. Each column represents a bidder, and each row a draw. Define an interval in which to look for the equilibrium. Apply the shading to the competitors and test shading within that interval and up to 2 decimal places, and pick the one with the highest expected profit.

This procedure has limitations. First, we are taking a fixed number of shadings within a bounded and fixed interval; however, given the results obtained using the first order condition, we can use intervals easily wide enough to give reasonable assurances that it includes the optimum. Second, this procedure is inefficient, as it tries every single possibility within that set. However, this procedure also has important advantages. It can deal with any trouble embedded functions might have and it is agnostic about the curvature of the expected profits functions (does not assume differentiability or a unique maximum within the interval).

We plot the histograms of the estimates of specific simulations.²⁸ Each SF in the sample, consists on the equilibrium shading that maximizes expected profits using 100.000 hypothetical draws for each bidder. We do this 500 times to obtain the sample of optimal factors we include in the histograms shown below. We use a tolerance of 0.01 for each optimal factor and a maximum of iterations (over best responses) of 50 times. In all these samples, the algorithm converged (within tolerance) every time.

Fig. 9

Two sophisticated bidders, both with signals N(0, 1). The SF is 1.77 for both

Fig. 10

Four sophisticated bidders, with signals N(0, 1). The $SF=1.51$ for all

Fig. 11

Two sophisticated bidders, one with signals N(0, 1), the other with signals N(1, 1). The true corrections are $SF=1.62$ and $SF=2.14$ respectively

Fig. 12

Two sophisticated bidders, one with signals N(0, 1), the other with signals N(0, 0.5). The true correction is 1.40 for both