Dissolving a partnership securely

Abstract

We characterize security strategies and payoffs for three mechanisms for dissolving partnerships: the Texas Shoot-Out, the $K+1$ auction, and the compensation auction. A security strategy maximizes a participant’s minimum payoff, and represents a natural starting point for analysis when a participant is either uncertain of the environment or uncertain of whether his rivals will play equilibrium. For the compensation auction, a dynamic dissolution mechanism, we introduce the notion of a perfect security strategy. Such a strategy maximizes a participant’s minimum payoff along every path of play. We show that while a player has many security strategies in the compensation auction, he has a unique perfect security strategy.

5 Appendix

Proof of Proposition 0

The proof is well known and is only included for completeness. It is easy to verify that for the given strategies the players each obtain a payoff of at least $x_{1}/2$ and $x_{2}/2$, respectively.

We show there is no strategy which guarantees player 1 more than $x_{1}/2$. Consider a strategy $\beta ^{1}$ such that $\beta ^{1}(x_{1})>x_{1}/2$ for some $x_{1}$. If player 1 with value $x_{1}$ sets a price $\beta ^{1}(x_{1})=p$ and player 2 chooses Sell, then player 1’s payoff is $x_{1}-p<x_{1}/2$. Likewise, if $\beta ^{1}(x_{1})=p<x_{1}/2$ for some $x_{1}$, then player 1 obtains a payoff less than $x_{1}/2$ if $\beta ^{2}(x_{2};p)=$ “Buy”.

Likewise, if $\beta ^{2}(x_{2},p)=$“Buy” for some $p>x_{2}/2$, then player 2 obtains a payoff $x_{2}-p<x_{2}/2$ if player 1 offers price p. $\square$

Proof of Proposition 1

For symmetric mechanisms, we can write $v_{i}(\hat{x}_{i},x_{-i},\hat{\beta }^{i},\beta ^{-i})$ as $v(\hat{x} _{i},x_{-i},\hat{\beta }^{i},\beta ^{-i})$.

Suppose to the contrary that bidder i with value $\hat{x}_{i}$ can guarantee more than $\hat{x}_{i}/N$, i.e., he has a (possibly mixed) strategy $\hat{\beta }^{i}$ such that

$$\begin{aligned} v(\hat{x}_{i},x_{-i},\hat{\beta }^{i},\beta ^{-i})\ge \bar{v}(\hat{x}_{i})> \frac{\hat{x}_{i}}{N}\quad \forall x_{-i},\beta ^{-i}. \end{aligned}$$

Since the inequality holds for all $x_{-i}$ and $\beta ^{-i}$, then it holds in particular for $\hat{x}_{-i}=(\hat{x}_{i},\ldots ,\hat{x}_{i})$ and $\hat{ \beta }^{-i}=(\hat{\beta }^{i},\ldots ,\hat{\beta }^{i})$, i.e.,

$$\begin{aligned} v(\hat{x}_{i},\hat{x}_{-i},\hat{\beta }^{i},\hat{\beta }^{-i})\ge \bar{v}( \hat{x}_{i})>\frac{\hat{x}_{i}}{N}. \end{aligned}$$

Summing across the N bidders, the total payoff when all bidders have a value of $\hat{x}_{i}$ and follow the same strategy $\hat{\beta }^{i}$ is $Nv( \hat{x}_{i},\hat{x}_{-i},\hat{\beta }^{i},\hat{\beta }^{-i})$, which is greater than $\hat{x}_{i}$. This contradicts that the mechanism is payoff feasible. $\square$

Proof of Proposition 2

We need to establish two facts: (i) $\bar{\beta }^{i}(x_{i})=x_{i}$ guarantees bidder i a payoff of at least $x_{i}/N$, and (ii) $\bar{\beta }^{i}(x_{i})=x_{i}$ is the unique security strategy.

Part (i): Suppose that $\bar{\beta }^{i}(x_{i})=x_{i}$. If bidder i wins, then he obtains a payoff of

$$\begin{aligned} x_{i}-\frac{N-1}{N}\left[ Kb_{\mathrm{s}}+(1-K)x_{i}\right] \ge x_{i}/N, \end{aligned}$$

where the inequality holds since $x_{i}\ge b_{\mathrm{s}}$ as $x_{i}$ is the winning bid. If bidder i loses, then he obtains

$$\begin{aligned} \frac{1}{N}\left[ Kb_{\mathrm{s}}+(1-K)b_{\mathrm{f}}\right] \ge x_{i}/N, \end{aligned}$$

where the inequality holds since $b_{\mathrm{f}}\ge b_{\mathrm{s}}\ge x_{i}$. His payoff, therefore, is at least $x_{i}/N$. By Proposition 00, $\bar{\beta } ^{i}(x_{i})=x_{i}$ is a security strategy.

Part (ii). Suppose $\bar{\beta }^{i}(x_{i})=x_{i}+\varDelta$ for $\varDelta >0$. Suppose all the other bidders bid $x_{i}+\varDelta /2$. Bidder i wins and obtains a payoff of

$$\begin{aligned} x_{i}-\frac{N-1}{N}\left[ K(x_{i}+\varDelta /2)+(1-K)(x_{i}+\varDelta )\right] <x_{i}/N, \end{aligned}$$

since $K(x_{i}+\varDelta /2)+(1-K)(x_{i}+\varDelta )>x_{i}$ for $K\in [0,1]$ . Suppose $\bar{\beta }^{i}(x_{i})=x_{i}-\varDelta$ for $\varDelta >0$. Suppose all the other bidders bid $x_{i}-\varDelta /2$. Bidder i loses and obtains a payoff of

$$\begin{aligned} \frac{1}{N}\left[ Kb_{\mathrm{s}}+(1-K)b_{\mathrm{f}}\right] =\frac{x_{i}-\varDelta /2}{N} <x_{i}/N. \end{aligned}$$

$\square$

Proof of Proposition 4

S $\bar{\beta }^{i}$ that satisfies the conditions of the Proposition. Let $x_{-i}$ and $\beta ^{-i}$ be arbitrary.

We first show that the bids of bidder i are in the interval $[\frac{x_{i}}{ N}+p_{k-1},\frac{kx_{i}}{N}]$ for rounds k where he remains active. In round 1, $p_{0}=0$ and $\bar{\beta }_{1}^{i}(x_{i};\mathbf {p}_{0})=x_{i}/N$ is determined. Suppose i does not drop at round $k-1$, i.e., $p_{k-1}\le \bar{\beta }_{k-1}^{i}(x_{i};\mathbf {p}_{k-2})$. Then, $\bar{\beta } _{k-1}^{i}(x_{i};\mathbf {p}_{k-2})\le (k-1)x_{i}/N$ implies

$$\begin{aligned} \frac{x_{i}}{N}+p_{k-1}\le \frac{x_{i}}{N}+\frac{(k-1)x_{i}}{N}=\frac{kx_{i} }{N}, \end{aligned}$$

and thus the interval $[\frac{x_{i}}{N}+p_{k-1},\frac{kx_{i}}{N}]$ is non-empty and $\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})\in [\frac{ x_{i}}{N}+p_{k-1},\frac{kx_{i}}{N}]$ at round k.

Next, we show that bidder i obtains a payoff of at least $x_{i}/N$ with $\bar{\beta }^{i}$. Suppose bidder i drops at some round k. Then, $\bar{ \beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})\ge x_{i}/N+p_{k-1}$ implies bidder i’s payoff is

$$\begin{aligned} \bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})-p_{k-1}\ge \frac{x_{i}}{N}. \end{aligned}$$

If bidder i wins the auction, then $p_{N-1}\le \bar{\beta }_{N-1}^{i}(x_{i}; \mathbf {p}_{N-2})\le \frac{(N-1)x_{i}}{N}$ and hence bidder i’s payoff is

$$\begin{aligned} x_{i}-p_{N-1}\ge x_{i}-\frac{(N-1)x_{i}}{N}=\frac{x_{i}}{N}. \end{aligned}$$

In either case, bidder i’s payoff is at least $x_{i}/N$. By Proposition 0, a bidder can guarantee at most $x_{i}/N$, and thus $\bar{\beta }^{i}$ is a security strategy.

We now establish that every security strategy satisfies the conditions of the proposition. Suppose to the contrary that $\bar{\beta }^{i}$ is a security strategy and $\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})<x_{i}/N+p_{k-1}$ for some k, $x_{i}$, and $\mathbf {p}_{k-1}$ such that $p_{m}\le$$\bar{\beta }_{m}^{i}(x_{i};\mathbf {p}_{m-1})$ for $m=1,\ldots ,k-1$. We show that there are bids for the other bidders such that player i’s payoff is less than $x_{i}/N$. Consider bids for the other bidders such that (i)$\ \mathbf {p}_{k-1}$ is the sequence of dropout prices, and (ii) the bids at round k exceed $\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})$. Then, bidder i drops at round k and his payoff is $\bar{\beta } _{k}^{i}(x_{i};\mathbf {p}_{k-1})-p_{k-1}<x_{i}/N$, which contradicts that $\bar{\beta }^{i}$ is a security strategy.

Suppose that $\bar{\beta }^{i}$ is a security strategy and $\bar{\beta } _{k}^{i}(x_{i};\mathbf {p}_{k-1})>kx_{i}/N$ for some k, $x_{i}$, and $\mathbf {p}_{k-1}$ such that $p_{m}\le$$\bar{\beta }_{m}^{i}(x_{i};\mathbf {p }_{m-1})$ for $m=1,\ldots ,k-1$. Consider bids of the other bidders such that (i)$\ \mathbf {p}_{k-1}$ is the sequence of dropout prices, (ii) some bidder $j\ne i$ drops at round k at price $p_{k}=kx_{i}/N+\varDelta$ for $\varDelta >0$ and $p_{k}<\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})$, and (iii) for rounds $k^{\prime }>k$ the remaining bidders bid $\frac{x_{i}}{N} +p_{k^{\prime }-1}$.

There are two cases. First, if bidder i drops at some round $k^{\prime }>k$ at a price strictly less than $\frac{x_{i}}{N}+p_{k^{\prime }-1}$, then his payoff is $\bar{\beta }_{k^{\prime }}^{i}(x_{i};\mathbf {p}_{k^{\prime }-1})-p_{k^{\prime }-1}<x_{i}/N$, which contradicts that $\bar{\beta }^{i}$ is a security strategy.¹³ Second, if bidder i wins the auction, then he pays

$$\begin{aligned} p_{N-1}=p_{k}+(p_{N-1}-p_{k})=\frac{kx_{i}}{N}+\varDelta +(N-k-1)\frac{x_{i}}{N} >\frac{(N-1)x_{i}}{N}, \end{aligned}$$

and hence his payoff is $x_{i}-p_{N-1}<x_{i}/N$, which contradicts that $\bar{\beta }^{i}$ is a security strategy. $\square$

Proof of Proposition 5

Using the same argument as in Proposition 1, it is immediate that each bidder i’s security payoff is at most $(x_{i}-p_{0})/n$. We show the given strategy achieves $(x_{i}-p_{0})/n$ , and therefore the given strategy is a security strategy in $\varGamma (n,p_{0})$ and the security payoff is $(x_{i}-p_{0})/n$.

Let $x_{-i}$ and $\beta ^{-i}$ be arbitrary, and let $p_{1},\ldots ,p_{n-1}$ be the sequence of dropout prices that results. Suppose that bidder i is not among the first $\hat{k}-1$ bidders to drop. We show, by induction, that for $k\in \{1,\ldots ,\hat{k}-1\}$ we have (i) $p_{k}-p_{0}\le k(x_{i}-p_{0})/n$ and (ii) $p_{k}-p_{k-1}\le (x_{i}-p_{k-1})/(n-k+1)$. Assume $x_{i}>p_{0}$. Since bidder i is not the first to drop, then

$$\begin{aligned} \bar{\beta }_{1}^{i}(x_{i};p_{0})=\frac{x_{i}-p_{0}}{n-1+1}+p_{0}\ge p_{1}, \end{aligned}$$

i.e.,

$$\begin{aligned} p_{1}-p_{0}\le \frac{x_{i}-p_{0}}{n}. \end{aligned}$$

Hence, (i) and (ii) hold for $k=1$.

Assume that (i) and (ii) hold for some $k^{\prime }<\hat{k}-1$. We show they hold for $k^{\prime }+1$. By the induction hypothesis, $p_{k^{\prime }}-p_{0}\le k^{\prime }(x_{i}-p_{0})/n$ and hence $k^{\prime }<n$ and $x_{i}>p_{0}$ imply $p_{k^{\prime }}-p_{0}\le x_{i}-p_{0}$, i.e., $p_{k^{\prime }}\le x_{i}$. Since bidder i did not drop at $k^{\prime }+1\le \hat{k}-1$, then

$$\begin{aligned} \bar{\beta }_{k^{\prime }+1}^{i}(x_{i};\mathbf {p}_{k^{\prime }})=\frac{ x_{i}-p_{k^{\prime }}}{n-(k^{\prime }+1)+1}+p_{k^{\prime }}\ge p_{k^{\prime }+1}, \end{aligned}$$

which establishes (ii) for $k=k^{\prime }+1$. This inequality can be rewritten as

$$\begin{aligned} p_{k^{\prime }+1}-p_{0}\le \frac{x_{i}+(n-k^{\prime }-1)p_{k^{\prime }}}{ n-k^{\prime }}-p_{0}. \end{aligned}$$

By the induction hypothesis, $p_{k^{\prime }}\le k^{\prime }(x_{i}-p_{0})/n+p_{0}$ and hence

$$\begin{aligned} p_{k^{\prime }+1}-p_{0}\le \frac{x_{i}+(n-k^{\prime }-1)(\frac{k^{\prime }(x_{i}-p_{0})}{n}+p_{0})}{n-k^{\prime }}-p_{0}=\frac{k^{\prime }+1}{n} (x_{i}-p_{0}), \end{aligned}$$

and (i) holds for $k=k^{\prime }+1$.

If bidder i drops in round $\hat{k}$, then his payoff is $(x_{i}-p_{\hat{k} -1})/(n-\hat{k}+1)$. Since $p_{\hat{k}-1}\le (\hat{k} -1)(x_{i}-p_{0})/n+p_{0}$, then

$$\begin{aligned} \frac{x_{i}-p_{\hat{k}-1}}{n-\hat{k}+1}\ge \frac{x_{i}-\left( \frac{\hat{k}-1}{n} (x_{i}-p_{0})+p_{0}\right) }{n-\hat{k}+1}=\frac{x_{i}-p_{0}}{n}. \end{aligned}$$

If bidder i is not among the first $n-1$ bidders to drop, then $p_{n-1}-p_{0}\le (n-1)(x_{i}-p_{0})/n$. He wins the auction and his payoff is

$$\begin{aligned} x_{i}-p_{n-1}\ge x_{i}-\left( \frac{n-1}{n}(x_{i}-p_{0})+p_{0}\right) =\frac{x_{i}-p_{0} }{n}\text {.} \end{aligned}$$

Hence, $\bar{\beta }^{i}$ guarantee’s bidder i a payoff of $(x_{i}-p_{0})/n$ and is therefore a security strategy.

If $x_{i}<p_{0}$, then bidder i’s payoff is negative if he wins the auction. We first show that $\bar{\beta }^{i}$ guarantees bidder i a payoff of a least $(x_{i}-p_{0})/n$. Since $\bar{\beta }_{1}^{i}$ calls for bidder i to drop immediately, his payoff is zero unless he wins the auction. The later occurs only if all $n-1$ other bidders drop immediately and ties are broken in bidder i’s favor. In this case, bidder i’s payoff is $x_{i}-p_{0}$. Since this occurs with at most probability 1 / n, his expected payoff is at least $(x_{i}-p_{0})/n$. $\square$

Proof of Proposition 6

Write $\bar{v} _{N-(k-1),p_{k-1}}(x_{i})$ for the security payoff of a bidder with value $x_{i}$ in the subauction $\varGamma (N-(k-1),p_{k-1})$. Suppose that $\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})<(x_{i}-p_{k-1})/(N-(k-1))+p_{k-1}$ for some k, $x_{i}$ and $\mathbf {p}_{k-1}$ such that $p_{0}\le p_{1}\le \cdots \le p_{k-1}$. We show that $\bar{\beta }^{i}$ is not a perfect security strategy. In particular, we show that $\bar{\beta }^{i}|_{\mathbf {p} _{k-1}}(x_{i})$ yields a payoff less than $\bar{v}_{N-(k-1),p_{k-1}}(x_{i})$ for some $x_{-i}$ and $\beta ^{-i}$.

From Proposition 5, the security payoff of bidder i in $\varGamma (N-(k-1),p_{k-1})$ is $\bar{v} _{N-(k-1),p_{k-1}}(x_{i})=(x_{i}-p_{k-1})/(N-(k-1))$. Let $x_{-i}$ and $\beta ^{-i}$ be such that the bids of the other $N-k$ bidders in round 1 of $\varGamma (N-(k-1),p_{k-1})$ are greater than $\bar{\beta }_{1}^{i}|_{\mathbf {p} _{k-1}}(x_{i})=\bar{\beta }_{k}^{i}(x_{i};\mathbf {p}_{k-1})$. Then, bidder i drops in round 1 and his payoff is

$$\begin{aligned} \bar{\beta }_{1}^{i}|_{\mathbf {p}_{k-1}}(x_{i})-p_{k-1}<\frac{x_{i}-p_{k-1}}{ N-(k-1)}+p_{k-1}-p_{k-1}=\bar{v}_{N-(k-1),p_{k-1}}(x_{i}). \end{aligned}$$

Hence, $\bar{\beta }_{i}$ is not a perfect security strategy.

Suppose that $\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})>(x_{i}-p_{k-1})/(N-k+1)+p_{k-1}$ for some k, $x_{i}$ and $\mathbf {p }_{k-1}$ such that $p_{0}\le p_{1}\le \cdots \le p_{k-1}$. Let $x_{-i}$ and $\beta ^{-i}$ be such that (i) one of the other $N-k$ bidders in $\varGamma (N-(k-1),p_{k-1})$ has a dropout price $\hat{p}_{k}$ satisfying

$$\begin{aligned} \bar{\beta }_{1}^{i}|_{\mathbf {p}_{k-1}}(x_{i})>\hat{p}_{k}>\frac{ x_{i}-p_{k-1}}{N-(k-1)}+p_{k-1}, \end{aligned}$$

and (ii) the remaining bidders’ dropout prices are above $\bar{\beta } _{1}^{i}|_{\mathbf {p}_{k-1}}(x_{i})=\bar{\beta }_{k}^{i}(x_{i};\mathbf {p} _{k-1})$. Then, bidder 1 does not drop out in round 1 of $\varGamma (N-(k-1),p_{k-1})$, but enters the subauction $\varGamma (N-k,\hat{p}_{k})$. From Proposition 5, the largest payoff he can guarantee himself in this subauction is $\bar{v}_{N-k,\hat{p}_{k}}(x_{i})=(x_{i}-\hat{p}_{k})/(N-k)$. We have that

$$\begin{aligned} \frac{x_{i}-\hat{p}_{k}}{N-k}<\frac{x_{i}-\left[ \frac{x_{i}-p_{k-1}}{N-(k-1) }+p_{k-1}\right] }{N-k}=\frac{x_{i}-p_{k-1}}{N-(k-1)}<\bar{v} _{N-(k-1),p_{k-1}}(x_{i}). \end{aligned}$$

Hence, $\bar{\beta }_{i}$ is not a perfect security strategy. $\square$