The “Italian Football Federation Auction” for Co-ownership Resolution

Abstract

The paper is inspired by the Italian Football Federation (FIGC) Auction. The auction was used by the FIGC from the 1950s until 2015 to resolve a 50% co-ownership between two teams in the same league with respect to the performance rights of a football player. Though FIGC banned such joint ownership in 2015, the resolution process is nevertheless interesting with respect to the received literature. In the FIGC Auction each co-owner submits a price offer for the whole asset. The highest bid wins the auction and obtains the asset, while the loser receives as payment the bid that is offered by the winner. Therefore, whether a bidder is a buyer, or a seller, can be established only after the auction has taken place—rather than before the auction. The main contribution of this paper is to investigate the set of pure strategy Nash Equilibria for two extensions of the FIGC Auction, with complete information. The first extension is the Extended Italian Football Federation Auction (EA), where asset shares can differ from 50%. The other extension is the Second Price Extended Italian Football Federation Auction (SEA), which is based on second price rather than first price. As compared to more standard auctions, though in the EA equilibria no bidder may have negative profits, in some circumstances we find that losing the auction may be more profitable than winning the auction. Finally, in the SEA we show that truthful bidding is not a weakly dominant strategy and that—somewhat surprisingly—the set of Nash Equilibria coincides with that of EA.

Appendix

Proof of Proposition 1

Suppose $v_{1} \ge \rho v_{2}$. From the agents’ payoff functions it follows that agent $1's$ best reply is given by

$$ b_{1} \left( {b_{2} } \right) = \left\{ {\begin{array}{ll} {b_{1} = \rho b_{2} } \hfill & {\quad if\;\;v_{1} > \rho b_{2} } \hfill \\ {b_{1} \in \left[ {0,\rho b_{2} } \right] } \hfill & {\quad if\;\;v_{1} = \rho b_{2} } \hfill \\ {b_{1} \in \left[ {0,\rho b_{2} } \right)} \hfill & {\quad if\;\;v_{1} < \rho b_{2} } \hfill \\ \end{array} } \right.; $$

while for agent $2$.

$$ b_{2} \left( {b_{1} } \right) = \left\{ {\begin{array}{ll} {b_{2} = \frac{{b_{1} }}{\rho } + \varepsilon } \hfill & {\quad if\;\;v_{2} > \frac{{b_{1} }}{\rho }} \hfill \\ {b_{2} \in \left[ {0,\frac{{b_{1} }}{\rho }} \right] } \hfill & {\quad if\;\;v_{2} = \frac{{b_{1} }}{\rho }} \hfill \\ {b_{2} \in \left[ {0,\frac{{b_{1} }}{\rho }} \right] } \hfill & {\quad if\;\;v_{2} < \frac{{b_{1} }}{\rho }} \hfill \\ \end{array} } \right., $$

with $\varepsilon > 0$ very small and such that $\frac{{b_{1} }}{\rho } + \varepsilon < v_{2}$. Then it follows immediately that $\left( {b_{1} = \rho b;b_{2} = b} \right)$ with $v_{2} \le b \le \frac{{v_{1} }}{\rho }$ are the price pairs where the two best replies meet: hence the set of pure strategy Nash Equilibria.□

Analogous reasoning holds for $v_{1} < \rho v_{2}$: where the two best replies meet at the following price pairs—$\left( {b_{1} = \rho b - \varepsilon ;\;b_{2} = b} \right)$ with $\frac{{v_{1} }}{\rho } < b \le v_{2}$—which then define the set of pure strategy Nash Equilibria, and the result is proved.

Proof of Proposition 2

Start by considering agent 2 and suppose $F_{1} \left( {b_{1} } \right)$ and $f_{1} \left( {b_{1} } \right)$ are agent 1’s mixed Nash Equilibrium distribution and probability density functions over the support $\left[ {\underline{v} ,\bar{v}} \right]$, with $\bar{v} > \underline{v}$. We assume $F_{1}^{'} \left( {b_{1} } \right) = f_{1} \left( {b_{1} } \right) > 0$ everywhere on $\left[ {\underline{v} ,\bar{v}} \right]$. Moreover, define

$$ E_{1} \left( {b_{1} } \right) = \mathop \smallint \limits_{{\underline{v} }}^{{\bar{v}}} b_{1} dF_{1} \left( {b_{1} } \right). $$

Then agent 2’s expected payoff is given by

$$ E\varPi_{2} \left( {b_{2} } \right) = \left\{ {\begin{array}{*{20}l} {E_{1} \left( {b_{1} } \right) - v_{2} } \hfill & {\quad if\;\;b_{2} \le \underline{v} } \hfill \\ { \mathop \smallint \limits_{{\underline{v} }}^{{b_{2} }} \left( {v_{2} - b_{2} } \right)dF_{1} \left( x \right) + \mathop \smallint \limits_{{b_{2} }}^{{\bar{v}}} \left( {x - v_{2} } \right)dF_{1} \left( x \right)} \hfill & {\quad if\;\;\underline{v} < b_{2} \le \bar{v}} \hfill \\ {v_{2} - b_{2} } \hfill & {\quad if\;\;b_{2} > \bar{v} } \hfill \\ \end{array} } \right.. $$

• (i) Suppose first $\underline{v} \ge v_{2}$. Hence for $b_{2} \le \underline{v}$ agent $2$ would lose the auction and obtain a constant, positive payoff $E_{1} \left( {b_{1} } \right) - v_{2} > 0$. If $\underline{v} < b_{2} \le \bar{v}$, then differentiating $E\varPi_{2} \left( {b_{2} } \right)$ over the interval $\underline{v} < b_{2} \le \bar{v}$ yields

  $$ \frac{{dE\varPi_{2} \left( {b_{2} } \right)}}{{db_{2} }} = - [F_{1} \left( {b_{2} } \right) - F_{1} \left( {\underline{v} } \right)] + 2\left( {v_{2} - b_{2} } \right)f_{1} \left( {b_{2} } \right), $$

  which, because $f_{1} \left( {b_{2} } \right) > 0$ over the interval $\left[ {\underline{v} ,\bar{v}} \right]$, is always negative for $b_{2} > v_{2}$. But since $\mathop {\lim }\limits_{{b_{2} \to \underline{v} }} (\int_{{\underline{v} }}^{{b_{2} }} {\left( {v_{2} - b_{2} } \right)dF_{1} \left( x \right)} + \int_{{b_{2} }}^{{\bar{v}}} {\left( {x - v_{2} } \right)dF_{1} \left( x \right))} = E_{1} \left( {b_{1} } \right) - v_{2}$, losing the auction would be preferable. Finally, if $b_{2} > \bar{v}$, agent $2$ wins the auction but obtains a negative payoff; therefore, his best choice would be to reply with any action in the interval $0 \le b_{2} \le \underline{v}$. However, against this agent $1$ would increase his profit by choosing $b_{1} \le \underline{v}$, which proves that in this case there cannot be a MSNE where agent $1$ randomizes over the interval $\left[ {\underline{v} ,\bar{v}} \right]$ with $\underline{v} \ge v_{2}$.

• (ii) Suppose $\underline{v} < v_{2} \le \bar{v}$. If $b_{2} \le \underline{v}$, then, again, agent $2$ loses the auction, and his payoff $E_{1} \left( {b_{1} } \right) - v_{2}$ is constant. If $\underline{v} < b_{2} \le \bar{v}$, the payoff $E\varPi_{2} \left( {b_{2} } \right)$ is non-negative at $b_{2} = v_{2}$ and decreasing for $v_{2} \le b_{2} \le \bar{v}$, which implies that $E\varPi_{2} \left( {b_{2} } \right) > E\varPi_{2} \left( {v_{2} } \right) \ge 0$ for at least one $b_{2} < v_{2}$. If $\bar{v} < b_{2}$, he would win the auction, but obtains negative profits. Therefore, for agent $2$ it is optimal to bid below $v_{2}$, which would imply that for agent $1$ too it is better to bid below $v_{2}$. Hence, also in this case, there cannot be a MSNE where agent $1$ randomizes over the interval $\left[ {\underline{v} ,\bar{v}} \right].$

• (iii) Finally, suppose $v_{2} > \bar{v}$. If $b_{2} \le \underline{v}$, agent $2$ loses the auction, with his expected payoff $E_{1} \left( {b_{1} } \right) - v_{2}$ being constant and negative. If $b_{2} > \bar{v}$, he wins the auction with expected profit $v_{2} - b_{2} ,$ which is decreasing in $b_{2}$ and is close to $v_{2} - \bar{v} > 0$ for values near to $\bar{v}$. Hence, he would never choose $b_{2} \le \underline{v}$; and, moreover, in a MSNE agent $2$ has to obtain at least $v_{2} - \bar{v}$. Therefore, in a MSNE agent $2$ would randomize over a closed and convex interval $\left[ {\underline{w} ,\bar{w}} \right]$, against agent $1$ randomizing on $\left[ {\underline{v} ,\bar{v}} \right]$, only if $\left[ {\underline{w} ,\bar{w}} \right] \subseteq \left[ {\underline{v} ,\bar{v}} \right]$. However, first notice that if $\bar{w} < \bar{v}$, then agent $1$ would not find it worthwhile to include $\bar{v}$ in his randomization interval, since in this case $E\varPi_{1} \left( {b_{1} = \bar{v}} \right) = v_{1} - \bar{v} < v_{1} - \bar{w} = E\varPi_{1} \left( {b_{1} = \bar{w}} \right)$, which contradicts the initial assumption that $b_{1} = \bar{v}$ is optimal. It follows that also $\bar{v} = \bar{w}$ is a necessary condition for agent $2$ to best-reply by randomizing over $\left[ {\underline{w} ,\bar{w}} \right]$. Then observe that $E\varPi_{2} \left( {b_{2} = \underline{v} } \right) = E_{1} \left( {b_{1} } \right) - v_{2} < 0$, while $E\varPi_{2} \left( {b_{2} = \bar{v}} \right) = v_{2} - \bar{v} > 0$. Hence, for agent $2$ to best-reply by randomizing over $\left[ {\underline{w} ,\bar{w}} \right]$ an additional necessary condition must be that $\underline{v} < \underline{w}$. But in this case for agent $1$ it will be $E\varPi_{1} \left( {\underline{v} \le b_{1} < \underline{w} } \right) = E_{2} \left( {b_{2} } \right) - v_{1} < 0$, with $E_{2} \left( {b_{2} } \right) = \int_{w}^{{\bar{w}}} {b_{2} dF_{2} \left( {b_{2} } \right),}$ while $E\varPi_{1} \left( {b_{1} = \bar{v}} \right) = v_{1} - \bar{v} > 0$, which implies that agent $1$ would not find it optimal to include offers $b_{1}$—such that $\underline{v} \le b_{1} < \underline{w}$—in his randomization interval, which proves the result.

□

Proof of Proposition 3

The proof is immediate. Consider agent $1^{\prime}s$ profit

$$ \varPi_{1} \left( {b_{1} } \right) = \left\{ {\begin{array}{ll} {\left( {v_{1} - \rho b_{2} } \right)} & {\quad if\;\;b_{1} \ge \rho b_{2} } \\ {\left( {\frac{{b_{1} - v_{1} }}{\rho }} \right)} & {\quad if\;\;b_{1} < \rho b_{2} } \\ \end{array} } \right., $$

and suppose that $v_{1} < \rho b_{2}$. Then if $b_{1} \ge$ $\rho b_{2}$, the payoff $\left( {v_{1} - \rho b_{2} } \right)$ is negative; while if $b_{1} < \rho b_{2}$, the payoff $\left( {\frac{{b_{1} - v_{1} }}{\rho }} \right)$ is positive, for $v_{1} < b_{1} < \rho b_{2}$ and for $b_{1} = v_{1}$ the payoff is equal to zero. Analogous considerations hold for agent $2$, and the result is proved.□

Proof of Proposition 4

(i) $v_{1} \ge \rho v_{2}$

Consider agent $1's$ payoff:

$$ \varPi_{1} \left( {b_{1} } \right) = \left\{ {\begin{array}{ll} {\left( {v_{1} - \rho b_{2} } \right)} & {\quad if\;\;b_{1} \ge \rho b_{2} } \\ {\left( {\frac{{b_{1} - v_{1} }}{\rho }} \right)} & {\quad if\;\;b_{1} < \rho b_{2} } \\ \end{array} } \right., $$

from which it follows immediately that his best reply is given by

$$ b_{1} \left( {b_{2} } \right) = \left\{ {\begin{array}{ll} {b_{1} \ge \rho b_{2} } & {\quad if\;\;v_{1} \ge \rho b_{2} } \\ {b_{1} = \rho b_{2} - \varepsilon } & {\quad if\;\;v_{1} < \rho b_{2} } \\ \end{array} } \right., $$

where $\varepsilon > 0$ is arbitrarily small.

For agent $2$ the payoff is

$$ \varPi_{2} \left( {b_{2} } \right) = \left\{ {\begin{array}{*{20}c} {\left( {v_{2} - \frac{{b_{1} }}{\rho }} \right)} & {\quad if\;\;b_{2} > \frac{{b_{1} }}{\rho }} \\ {\rho (b_{2} - v_{2} )} & {\quad if\;\;b_{2} \le \frac{{b_{1} }}{\rho }} \\ \end{array} } \right., $$

and his best reply given by

$$ b_{2} \left( {b_{1} } \right) = \left\{ {\begin{array}{ll} {b_{2} > \frac{{b_{1} }}{\rho }} \hfill & {\quad if\;\;v_{2} > \frac{{b_{1} }}{\rho }} \hfill \\ {b_{2} \ge \frac{{b_{1} }}{\rho }} \hfill & {\quad if\;\;v_{2} = \frac{{b_{1} }}{\rho }} \hfill \\ {b_{2} = \frac{{b_{1} }}{\rho }} \hfill & {\quad if\;\;v_{2} < \frac{{b_{1} }}{\rho }} \hfill \\ \end{array} } \right.. $$

It is then easy to check that the two best replies meet at price pairs $\left( {b_{1} = b\rho ;\;b_{2} = b} \right)$, with $v_{2} \le b \le \frac{{v_{1} }}{\rho }$, which represent the set of pure strategy Nash Equilibria. Analogous considerations hold for $v_{1} < \rho v_{2}$, and the result is proved.□