Noncooperative games, coupling constraints, and partial efficiency

Abstract

Many noncooperative settings require sharing of aggregate holdings—be these of natural resources, production tasks, or pollution permits. This paper considers instances where the shared items eventually become competitively priced. For that reason, the solution concept incorporates features of Nash and Walras equilibria. Focus is on how the concerned agents, by themselves, may reach an outcome of such sort. A main mechanism is direct bilateral exchange, repeated time and again.

Appendix 1: Proofs

Proof of Proposition 3.1

The “best response” correspondence

$$\begin{aligned} x\in X\rightrightarrows \mathcal {B}(x):=\arg \max _{\hat{x}\in X}\pi (\hat{x} ,x)\subseteq X \end{aligned}$$

has non-empty convex values. If a convergent sequence $(\hat{x} ^{k},x^{k})\rightarrow (\hat{x},x)$ satisfies $\hat{x}^{k}\in \mathcal {B} (x^{k}),$ with $x^{k}\in X,$ it holds for any $\chi \in X$ that

$$\begin{aligned} \pi (\hat{x},x)\ge \lim \sup _{k}\pi (\hat{x}^{k},x^{k})\ge \lim \sup _{k}\pi (\chi ,x^{k})\ge \pi (\chi ,x). \end{aligned}$$

Hence, $\hat{x}\in \mathcal {B}(x)$. This tells that $\mathcal {B}$ has closed graph. By Kakutani’s theorem, there is a fixed point $x\in \mathcal {B} (x)$. Any such point is a normalized equilibrium. $\square$

Proof of Proposition 3.2

x is a normalized equilibrium iff

$$\begin{aligned} x\in \arg \max \left\{ \pi (\cdot ,x)+\mathcal {\iota }_{X}(\cdot )\right\} \Longleftrightarrow 0\in \partial \left\{ \pi (\cdot ,x)+\mathcal {\iota } _{X}(\cdot )\right\} (x)=M(x)-N(x). \end{aligned}$$

The last equality, which derives from the above constraint qualification, holds iff there exists some $m\in M(x)\cap N(x).$ This proves the first bullet. For the second, note that any $m\in N(x)$ yields $\left\langle m,X-x\right\rangle \le 0$ by definition (3). For the third bullet, recall Moreau’s orthogonal decomposition, saying that $m=\tau +n,$ with $\tau \in T(X,x),$ $n\in N(x)$ and $\left\langle n,\tau \right\rangle =0$; see Prop.0.6.3 in Aubin and Cellina (1984). In fact, $\tau =P_{T(X,x)}\left[ m\right]$ so that $\tau =0\Longleftrightarrow m\in N(x).$ $\square$

Proof of Proposition 4.1

Let $x(\cdot )$ be a solution to (7). Use the norm $\left\| x\right\| :=\left\langle x,x\right\rangle ^{1/2}$ to define a Lyapunov function $\mathcal {L} (t):=\left\| x(t)-\bar{x}\right\| ^{2}/2$ for $t\ge 0.$ Under (8), as long as $x=x(t)\in X$ differs from $\bar{x},$ it holds almost everywhere

$$\begin{aligned} \begin{array}{rl} \mathcal {\dot{L}}= &{} \left\langle x-\bar{x},\dot{x}\right\rangle \in \left\langle x-\bar{x},P_{T(x)}\left[ M(x)\right] \right\rangle \\ \subseteq &{} \left\langle x-\bar{x},M(x)-N(x)\right\rangle \le \left\langle x-\bar{x},M(x)\right\rangle <0. \end{array} \end{aligned}$$

The inclusion $\subseteq$ stems from the Moreau’s decomposition theorem, mentioned above. Further, the first inequality followed from (3) which tells that $\left\langle \bar{x}-x,N(x)\right\rangle \le 0.$ This proves the global asymptotic stability of $\bar{x}$ as a steady state of ( 7). That is, $x(t)\rightarrow \bar{x}.$ Then clearly, $\bar{x}$ must be unique.

Quite likewise, the relaxed system yields

$$\begin{aligned} \mathcal {\dot{L}}=\left\langle x-\bar{x},\dot{x}\right\rangle \in \left\langle x-\bar{x},M(x)-N(x)\right\rangle , \end{aligned}$$

and the same arguments apply. $\square$

Lemma 1

(On outer semicontinuity of margins). The correspondence $x\in X\rightrightarrows M(x)$ has closed graph. Hence, M is outer (upper) semicontinuous, and M(X) must be bounded.

Proof of Lemma 1

If some convergent sequence $(m^{k},x^{k}) \rightarrow (m,x)$ satisfies $m^{k}\in M(x^{k})$ with $x^{k}\in X,$ it holds for any $\chi \in \mathbb {X}$ that

$$\begin{aligned} \pi (\chi ,x^{k})\le \pi (x^{k},x^{k})+\left\langle m^{k},\chi -x^{k}\right\rangle . \end{aligned}$$

Passing to the limit, we obtain

$$\begin{aligned} \pi (\chi ,x)\le \pi (x,x)+\left\langle m,\chi -x\right\rangle , \end{aligned}$$

hence $m\in M(x).$ Since X is compact, M is outer semicontinuous, and M(X) must be bounded. $\Diamond$

Proof of Proposition 4.2

Let $\bar{x}$ be the unique normal equilibrium. By Lemma 1, it holds for any number $\varepsilon >0$ that

$$\begin{aligned} \sup \left\{ \left\langle M(x),x-\bar{x}\right\rangle :\left\| x-\bar{x} \right\| \ge \varepsilon , x\in X\right\} <0. \end{aligned}$$

(23)

From here on the argument is known, but included for completeness.⁷ Pick $m^{k}\in M(x^{k})$ such that $x^{k+1}=P_{X} \left[ x^{k}+s_{k}m^{k}\right] .$ Since orthogonal projection $P_{X}$ onto X is non-expansive,

$$\begin{aligned} \left\| x^{k+1}-\bar{x}\right\| ^{2}= & {} \left\| P_{X}\left[ x^{k}+s_{k}m^{k}\right] -P_{X}\left[ \bar{x}\right] \right\| ^{2}\le \left\| x^{k}+s_{k}m^{k}-\bar{x}\right\| ^{2} \\= & {} \left\| x^{k}-\bar{x}\right\| ^{2}+2s_{k}\left\langle x^{k}-\bar{x} ,m^{k}\right\rangle +s_{k}^{2}\left\| m^{k}\right\| ^{2}. \end{aligned}$$

(8) tells that $\left\langle x^{k}-\bar{x},m^{k}\right\rangle \le 0.$ Upon introducing non-negative numbers

$$\begin{aligned} A_{k}:=\left\| x^{k}-\bar{x}\right\| ^{2}, B_{k}:=0, \quad C_{k}:=2s_{k}\left\langle \bar{x}-x^{k},m^{k}\right\rangle , \quad D_{k}:=s_{k}^{2}\sup \left\| M(X)\right\| ^{2}, \end{aligned}$$

the string, stated just here above, takes the form

$$\begin{aligned} A_{k+1}\le A_{k}(1+B_{k})-C_{k}+D_{k} \end{aligned}$$

with $\sum _{k=0}^{\infty }B_{k}=0$ and $\sum _{k=0}^{\infty }D_{k}<+\infty .$ From the Robbins–Siegmund Lemma, restated as another Lemma below, it follows that $A_{k}$ converges to some $A\ge 0,$ and $\sum _{k=0}^{\infty }C_{k}<+\infty$. It remains to show that $A=0.$ Otherwise, take $\varepsilon :=A/2>0$ to have from (23), that $\left\langle x^{k}-\bar{x},m^{k}\right\rangle <0$ is bounded away from 0 for sufficiently large k. This gives the contradiction $\sum _{k=0}^{\infty }C_{k}=+\infty$. $\square$

Proof of Theorem 7.1

At any stage $k=0,1,2, \ldots ,$ consider the prevailing profile $x^{k}\in X,$ and two agents $i^{k},j^{k}$ just when these are about to update their actual endowments $z_{i}^{k}$ and $z_{j}^{k}$ . For simpler notation, temporarily omit mention of k. Note that bilateral trade (17) amounts to posit

$$\begin{aligned} x^{+1}=x+sd \end{aligned}$$

(24)

with $d_{i}:=(0,d_{ij})\in \mathbb {Y}_{i}\times \mathbb {Z}$, $d_{j}:=(0,-d_{ij})\in \mathbb {Y}_{j}\times \mathbb {Z},$ all other components of $d\in \mathbb {X}$ being nil. Since M is bounded on X, one may assume that each such d be bounded in norm by some constant $\delta >0.$

Let $\bar{x}$ denote the unique normal equilibrium. From (24) it follows that

$$\begin{aligned} \left\| x^{+1}-\bar{x}\right\| ^{2}&=\left\| x+sd-\bar{x}\right\| ^{2}=\left\| x-\bar{x}\right\| ^{2}+2s\left\langle x-\bar{x},d\right\rangle +s^{2}\left\| d\right\| ^{2} \\&\le \left\| x-\bar{x}\right\| ^{2}+2s\left\{ \left\langle x_{i}-\bar{x} _{i},(0,d_{ij})\right\rangle -\left\langle x_{j}-\bar{x}_{j},(0,d_{ij}) \right\rangle \right\} +s^{2}\delta ^{2}. \end{aligned}$$

Invoke (4), $y^{+1}\in Nash(z),$ and (18), (21) to see that, for agent i, 

$$\begin{aligned} (0,d_{ij})=\left[ m_{i}-n_{i}\right] -\left[ m_{j}-n_{j}\right] \in \left[ M_{i}(x)-N(X_{i},x_{i})\right] -\left[ M_{j}(x)-N(X_{j},x_{j})\right] , \end{aligned}$$

and likewise for his interlocutor j.

Let $\mathcal {\nu }:=\#I\ge 2$ denote the number of agents. Any pair (i, j) of distinct agents is selected with probability $1/\left( {\begin{array}{c}\mathcal {\nu }\\ 2\end{array}}\right) = \frac{2}{\mathcal {\nu }(\mathcal {\nu }-1)}.$ Repeated draws are independent of each other. In the last string of inequalities take expectation E with respect to drawing the agent pair (i, j). By (21) this gives

$$\begin{aligned} E\left\| x^{+1}-\bar{x}\right\| ^{2}\le \left\| x-\bar{x} \right\| ^{2}+\frac{2}{( \begin{array}{c} \mathcal {\nu } \\ 2 \end{array} )}\sum _{i}\sum _{j}s\left\langle x_{i}-\bar{x}_{i},m_{i}-n_{i}-m_{j}+n_{j} \right\rangle +s^{2}\delta ^{2} \end{aligned}$$

(25)

To grasp the nature of the last inequality, let

$$\begin{aligned} \mathbb {D}:=\left\{ d=(d_{i})\in \mathbb {X}:\sum _{i\in I}d_{i}=0\right\} , \end{aligned}$$

denote the space of “redistributions,” and consider the orthogonal projection $P_{\mathbb {D}}\left[ \cdot \right]$ from the ambient space $\mathbb {X}$ onto $\mathbb {D}$. Given any vector $v=(v_{i})\in \mathbb {X},$ it holds for each i:

$$\begin{aligned} P_{\mathbb {D}}\left[ v\right] _{i}=v_{i}-\frac{1}{\mathcal {\nu }}\sum _{j\in I}v_{j}. \end{aligned}$$

The particular instance $v_{i}=m_{i}-n_{i}\in M_{i}(x_{i})-N(X_{i},x_{i}),$ yields

$$\begin{aligned}&\frac{1}{\mathcal {\nu }}\sum _{i}\sum _{j}\left\langle x_{i}-\bar{x}_{i}, \left[ m_{i}-n_{i}\right] -\left[ m_{j}-n_{j}\right] \right\rangle \\&\quad =\sum _{i}\left\langle x_{i}-\bar{x}_{i},\left[ m_{i}-n_{i}\right] -\frac{1 }{\mathcal {\nu }}\sum _{j}\left[ m_{j}-n_{j}\right] \right\rangle =\left\langle x-\bar{x},P_{\mathbb {D}}\left[ m-n\right] \right\rangle \\&\quad \in \left\langle x-\bar{x},M(x)-N(x,X)-N(x,\mathbb {D})\right\rangle =\left\langle x-\bar{x},M(x)-N(X,x)\right\rangle \\&\quad \le \left\langle x-\bar{x},M(x)\right\rangle . \end{aligned}$$

Combined with (25) the last inequality implies

$$\begin{aligned} E\left\| x^{+1}-\bar{x}\right\| ^{2}\le \left\| x-\bar{x} \right\| ^{2}-\frac{4s}{(\mathcal {\nu }-1)}\left\langle M(x),\bar{x} -x\right\rangle +s^{2}\delta ^{2}. \end{aligned}$$

Reintroducing the allocation $x^{k}$ which prevails at stage k, it holds for $k=0,1,...$

$$\begin{aligned} E\left\| x^{k+1}-\bar{x}\right\| ^{2}\le \left\| x^{k}-\bar{x} \right\| ^{2}-\frac{4s_{k}}{(\mathcal {\nu }-1)}\left\langle M(x^{k}),\bar{ x}-x^{k}\right\rangle +s_{k}^{2}\delta ^{2}. \end{aligned}$$

(26)

Now invoke the following auxiliary result due to Robbins and Siegmund; see 5.2.1 in Benveniste et al. (1990):

Lemma 2

(On a “supermartingale”) Suppose $A_{k},B_{k},C_{k},D_{k}$ are finite, non-negative random variables, adapted to a $\sigma$-field $\mathcal {F}_{k}\subseteq \mathcal {F}_{k+1}$, which satisfy

$$\begin{aligned} E\left[ A_{k+1}\left| \mathcal {F}_{k}\right. \right] \le A_{k}(1+B_{k})-C_{k}+D_{k}. \end{aligned}$$

Then, on the event $\left\{ \sum _{k}B_{k}<\infty \& \sum _{k}D_{k}<\infty \right\} ,$ it holds almost surely that $\sum _{k}C_{k}<+\infty$ & $A_{k}\rightarrow$ some finite A. $\Diamond$

In the present setting, let $\mathcal {F}_{k}$ be generated by $\left\{ x^{0},\ldots ,x^{k}\right\} .$ In view of (26), posit

$$\begin{aligned} A_{k}:=\left\| x^{k}-\bar{x}\right\| ^{2}, \quad B_{k}:=0, \quad C_{k}:=\frac{4s_{k}}{(\mathcal {\nu }-1)}\left\langle M(x^{k}), \quad \bar{x} -x^{k}\right\rangle , and D_{k}:=s_{k}^{2}\delta ^{2}. \end{aligned}$$

For sure, $\sum _{k}B_{k}<+\infty$ and $\sum _{k}D_{k}<+\infty$. Now consider a scenario such that $A>0$—if any. Along the corresponding trajectory,

$$\begin{aligned} \lim \inf \left\{ \left\langle m^{k},\bar{x}-x^{k}\right\rangle \right\} >0, \end{aligned}$$

hence $\sum _{k}C_{k}=+\infty$. This contradiction tells that $A=0\$almost surely. $\square$