Demand reduction and preemptive bidding in multi-unit license auctions

Abstract

Multi-unit ascending auctions allow for equilibria in which bidders strategically reduce their demand and split the market at low prices. At the same time, they allow for preemptive bidding by incumbent bidders in a coordinated attempt to exclude entrants from the market. We consider an environment where both demand reduction and preemptive bidding are supported as equilibrium phenomena of the ascending auction. In a series of experiments, we compare its performance to that of the discriminatory auction. Strategic demand reduction is quite prevalent in the ascending auction even when entry imposes a (large) negative externality on incumbents. As a result, the ascending auction performs worse than the discriminatory auction both in terms of revenue and efficiency, while entrants’ chances are similar across the two formats.

Appendix B

In this Appendix we provide the analytic solution for the incumbent’s equilibrium bid function in the ascending auction as determined by (3). To solve this first-order condition it will prove useful to consider a related differential equation:

$$ zG^{\prime}(z)=2\bigl(1+G(z)-z^{2}G(z)^{2}\bigr), $$

(11)

with general solution

$$ G_{\alpha_{p,x}}(z)=\frac{1}{z}\frac{I_{1}(2z)+\alpha_{p,x}K_{1}(2z)}{I_{2}(2z)-\alpha_{p,x}K_{2}(2z)} $$

(12)

for z≥0. Here I _n (K _n ) is the n ^th modified Bessel function of the first (second) kind and α _p,x is a constant chosen such that the boundary condition B _I (p)=p is satisfied. In the proof of Theorem 1 below we show that this boundary condition is met if $\alpha_{p,x}=\alpha (\frac{1-p}{x/3} )$ where we define

$$ \alpha(z)=\frac{I_{2}(2z)-(1-1/z)I_{1}(2z)}{K_{2}(2z)+(1-1/z)K_{1}(2z)} $$

(13)

Lemma 1

$G_{\alpha_{p,x}}(\cdot)$ and α(⋅) satisfy the following properties:

• (i) α(z) is strictly increasing in z with α(0)=0.

• (ii) $G_{\alpha_{p,x}}(z)$ has an asymptote at z=z ^∗ where z ^∗ solves I ₂(2z ^∗)/K ₂(2z ^∗)=α _p,x .

• (iii) The inverse $G_{\alpha_{p,x}}^{(-1)}(z)$ is well defined for $z\in\mathcal{D}$ where

  $$\mathcal{D}=\left \{ \begin{array}{l@{\quad}l} \lbrack\frac{x/3}{1-x/3},\frac{x/3}{1-x/3-p}] & \mbox{\emph{if}}\ p<1-x/3\\[2mm] \lbrack\frac{x/3}{1-x/3},\infty)\cup(-\infty,\frac{x/3}{1-x/3-p}] & \mbox{\emph{if}}\ p>1-x/3 \end{array} \right . $$

  an d $G_{\alpha_{p,x}}^{(-1)}(z)$ for $z\in\mathcal{D}$ is minimized at $z=\frac{x/3}{1-x/3-p}$ with $G_{\alpha _{p,x}}^{(-1)}(\frac{x/3}{1-x/3-p})=\frac{1-p}{x/3}$.

• (iv) For $z\in\mathit{Int}({\mathcal{D}})$ we have

  $$ \frac{1}{z} \biggl(\frac{1}{z}+1 \biggr)<G_{\alpha_{p,x}}^{(-1)}(z)^{2}< \biggl(\frac{1}{z}+1 \biggr)^{2} $$

  (14)

Proof

Properties (i) and (ii) can be verified by plotting the right side of (13) and the ratio of modified Bessel functions I ₂(2z)/K ₂(2z). To establish (iii) note that if $G_{\alpha_{p,x}}(z)\geq0$ then (12) implies that $G_{\alpha_{p,x}}(z)\geq G_{0}(z)\equiv I_{1}(2z)/(zI_{2}(2z))$ and using standard properties of the modified Bessel functions we have

$$\frac{1}{G_{0}(z)} \biggl(\frac{1}{G_{0}(z)}+1 \biggr)= \biggl( \frac {I_{0}(2z)I_{2}(2z)}{I_{1}(2z)I_{1}(2z)} \biggr)z^{2}<z^{2}$$

Since $G_{\alpha_{p,x}}(z)\geq G_{0}(z)$ we conclude that

$$ \frac{1}{G_{\alpha_{p,x}}(z)} \biggl(\frac{1}{G_{\alpha_{p,x}}(z)}+1 \biggr)<z^{2}$$

(15)

when $G_{\alpha_{p,x}}(z)\geq0$. Moreover, inequality (15) is (trivially) satisfied when $G_{\alpha_{p,x}}(z)\leq-1$ since then the left side is non-positive. Combined with (11) inequality (15) implies that $G_{\alpha_{p,x}}(z)$ is strictly decreasing, and, hence, invertible when $G_{\alpha_{p,x}}(z)\leq-1$ or $G_{\alpha_{p}}(z)\geq0$. This proves that $G^{(-1)}_{\alpha_{p,x}}(z)$ is well defined for $z\in\mathcal{D}$ since $\mathcal{D}\subseteq[0,\infty)\cup(-\infty,-1]$ for all 0≤p≤1 and x>0. If p<1−x/3 then $G^{(-1)}_{\alpha_{p,x}}(z)$ is strictly decreasing on $[\frac{x/3}{1-x/3},\frac{x/3}{1-x/3-p}]$ and is thus minimized at $z=\frac{x/3}{1-x/3-p}$. If p>1−x/3 then $G^{(-1)}_{\alpha_{p,x}}(z)$ declines on $[\frac{x/3}{1-x/3},\infty)$ and it declines on $(-\infty ,\frac{x/3}{1-x/3-p}]$ with $G^{(-1)}_{\alpha_{p,x}}(\pm\infty)=z^{*}$, so the minimum is again attained at $z=\frac{x/3}{1-x/3-p}$. A direct computation verifies that $G_{\alpha_{p,x}}(\frac{1-p}{x/3})=\frac{x/3}{1-x/3-p}$, or, equivalently $G^{(-1)}_{\alpha_{p,x}}(\frac{x/3}{1-x/3-p})=\frac{1-p}{x/3}$. Finally, the left inequality of (iv) follows from (15) and the fact that $G^{(-1)}_{\alpha_{p,x}}(z)$ is well defined for $z\in\mathcal{D}$. To show the right inequality of (iv), note that $G_{\alpha_{p,x}}^{(-1)}(z)<1/z+1$ is equivalent to $z<1/G_{\alpha_{p,x}}(z)+1$, which using (12) can be written as

$$1-\frac{1}{z}<\frac{I_{2}(2z)-\alpha_{p,x} K_{2}(2z)}{I_{1}(2z)+\alpha_{p,x} K_{1}(2z)}$$

Using $\alpha_{p,x}=\alpha (\frac{1-p}{x/3} )$ and the definition of α(⋅) in (13) this can be rewritten as $\alpha (z)>\alpha (\frac{1-p}{x/3} )$. Since α(⋅) is increasing, the right inequality in (iv) follows if $z> (\frac{1-p}{x/3} )$ for all z such that $G_{\alpha_{p,x}}(z)\in\mbox{Int}({\mathcal{D}})$. Since $\frac{1-p}{x/3}=G^{(-1)}_{\alpha_{p,x}}(\frac{x/3}{1-x/3-p})$ this is true if $G^{(-1)}_{\alpha_{p,x}}(z)>G^{(-1)}_{\alpha_{p,x}}(\frac{x/3}{1-x/3-p})$ for all $z\in\mbox{Int}({\mathcal{D}})$, which holds since $G^{(-1)}_{\alpha_{p,x}}(z)$ is minimized at $z=\frac{x/3}{1-x/3-p}$. □

With Lemma 1 we are able to characterize B _I (v) in Proposition 2.

Theorem 1

The incumbents’ equilibrium bid function in the ascending auction is given by:

$$ B_{I}(v)=1-\frac{(x/3)^{2}}{1-v}G_{\alpha_{p,x}}^{(-1)} \biggl( \frac {x/3}{1-x/3-v} \biggr)^{2} $$

(16)

B _I (v) is strictly increasing for 0≤v<p with B _I (p)=p, satisfies v<B _I (v)<v+x/3 for 0≤v<p, and lim _x→0 B _I (v)=v.

Proof

Note that for 0≤v≤p and 0≤p≤1, the ratio (x/3)/(1−x/3−v) lies in the set (−∞,−1]∪[0,∞) on which $G_{\alpha_{p,x}}^{(-1)}$ is well defined (see proof of Lemma 1). We first verify the necessary (first-order and boundary) conditions. Differentiating (16) with respect to v yields

Using (16) we can rewrite this as

$$B_{I}^{\prime}(v)=-\frac{1-B_{I}(v)}{1-v} \biggl(1+ \frac{(1-v)(x/3)}{(1-x/3-v)^{2}}\frac{2}{G_{\alpha_{p,x}}^{(-1)} (\frac{x/3}{1-x/3-v})G_{\alpha_{p,x}}^{\prime} (G_{\alpha_{p,x}}^{(-1)}(\frac {x/3}{1-x/3-v}) )} \biggr) $$

which can be rewritten using (11)

which, using (16), further simplifies to

$$B_{I}^{\prime}(v)=-\frac{1-B_{I}(v)}{1-v} \biggl(1+ \frac{(1-v)(x/3)}{(1-x/3-v)^{2}}\frac{1}{1+\frac{(x/3)}{(1-x/3-v)}-\frac{(1-v)(1-B_{I}(v))}{(1-x/3-v)^{2}}} \biggr) $$

and, after rearranging terms

$$B_{I}^{\prime}(v)=\frac{(1-B_{I}(v))(B_{I}(v)-v)}{(1-v)(v-B_{I}(v)+x/3)}$$

which is the first-order condition in (3). To verify the boundary condition, note that B _I (p)=p can be rewritten as

$$G_{\alpha_{p,x}}^{(-1)} \biggl(\frac{x/3}{1-x/3-p} \biggr)= \frac{1-p}{x/3}$$

or, equivalently,

$$\frac{x/3}{1-x/3-p}=G_{\alpha_{p,x}} \biggl(\frac{1-p}{x/3} \biggr) $$

which yields a linear equation in α _p,x , see (12), that can readily be solved to yield $\alpha_{p,x}=\alpha (\frac{1-p}{x/3} )$, with α(⋅) defined in (13).

To show that B _I (v)<v+x/3, use the left inequality of (14) in property (iv) of Lemma 1

$$B_{I}(v)<1-\frac{(x/3)^{2}}{1-v}\frac{1}{\frac{x/3}{1-x/3-v}}\biggl( \frac{1}{\frac{x/3}{1-x/3-v}}+1 \biggr)=v+x/3 $$

Similarly, the right inequality in (14) implies

$$B_{I}(v)>1-\frac{(x/3)^{2}}{1-v} \biggl(\frac{1}{\frac{x/3}{1-x/3-v}}+1 \biggr)^{2}=v $$

Note that v<B _I (v)<v+x/3 for 0≤v<p implies that $B_{I}^{\prime }(v)>0$ for all 0≤v<p, see the first-order condition (3).

Finally, the limiting property of B _I (v) follows from the fact that $G_{\alpha_{p,x}}^{(-1)}(z)\sim1/z$ for z small. Hence, as x tends to zero, B _I (v) limits to

$$\lim_{x\rightarrow0}B_{I}(v)=\lim_{x\rightarrow0}1-\frac{(x/3)^{2}}{1-v} \biggl(\frac{1-x/3-v}{x/3} \biggr)^{2}=1-(1-v)=v $$

 □