Evolution of Walrasian equilibrium in an exchange economy

Abstract

We study the stochastic stability of a dynamic trading process in an exchange economy. We use a simplified version of a trading model à la Shapley and Shubik (J Polit Econ 85:937–968, 1977). Two types of agents equipped with Leontief preferences trade goods in markets by offering endowments, and actual trades occur at market clearing prices. Better behavior tends to spread through the same type of agents by imitation, and agents also make mistakes occasionally. We provide a sufficient condition for the perturbed dynamic process to have a unique stochastically stable state that is a Walrasian equilibrium allocation. In this sense, we give a rationale for Walrasian behavior.

Appendices

Appendix I: Proof of Theorem 1 for A = 1

1.1 AI.1 Transition among states in $\bar Z$ when A = 1

1.1.1 AI.1.1 Moving states $z^{b_n,q_n}$

When A = 1, the two offer curves OC _I and OC _II coincides with the diagonal.

Consider any $n\in\{1,\cdots,\bar n\}$. We now show that:

$$ p^+_I=p^-_{II}>{b_n\over q_n}> p^-_{I}=p^-_{II}. $$

(56)

First, since A = 1, by Eqs. 26, 6, and 19 we have: $p^-_{I} = p^-_{II}$, and $p^+_{I} = p^+_{II}$. (This is slightly different from Eq. 27.) Second, from Eq. 26 we still have Eqs. 28 and 29. Third, as $n\leqq \bar n=\hat n-1=(N/2)-1$ (where N is given in Eq. 7), we have K > 2nδ, and so from Eq. 29 we have $p^-_I < b_n/q_n$.

Therefore, we need to consider the least number of mutations to achieve $p^-_I$ and $p^+_{II}$. We have the minimization problems (MP₁) and (MP₂) defined in Eqs. 34 and 35, respectively.¹⁹

Lemma 3′

Let A = 1, then:

• (a) The solution of (MP₁) is $(\alpha,\beta)=(0,\beta^-_{I})$, where:

  $$ \beta^-_I ={(K-b_n)b_n-q_nb_{n-1} \over (\bar q-q_n) b_{n-1 }}. $$

  (57)

• (b) The solution of (MP₂) is $(\alpha,\beta)=(0,\beta^+_{II})$, where:

  $$ \beta^+_{II}={(K-q_n)q_n-b_{n}q_{n+1} \over (K-q_n)(q_n-\underline{q})}. $$

  (58)

We compare the terms $\beta^-_I$ and $\beta^+_{II}$ in the following lemma:

Lemma 4′

Let A = 1. Then:

• (a) $\beta^-_I<\beta^+_{II}$ for all $n\in\{1,\cdots,\bar n\}$ ,

• (b) $\beta^-_I$ is strictly decreasing in n for all $n\in\{1,\cdots,\bar n\}$ .

We next focus on $\beta^-_I$.

Claim 1′

Suppose A = 1.

• (a) For any z ∈ Z, if $(1/2)\|z-z^{b_n,q_n}\|<\beta^-_I m$, then Eq. 38 holds.

• (b) Let z ∈ Z be such that z(b_n − 1) = 1, z(b _n ) = m − 1, $\beta^-_I m\leqq z(\bar q)<\beta^-_I m+1$, and $z(q_n)=m-z(\bar b)$. Then Eq. 39 holds.

Claim 1′ can be proved in the same manner as Claim 1, and so we omit the proof. Also, Claim 1′ gives similar results as discussed in Remark 3.

1.1.2 AI.1.2 Moving states $z^{\widetilde b_n,\widetilde q_n}$

We use the same $\widetilde p^-_I$, $\widetilde p^+_I$, $\widetilde p^-_{II}$ and $\widetilde p^+_{II}$ given in Section 3.5.4. We focus on ${\widetilde{p}}^-_{II}$ and ${\widetilde{p}}^+_I$, and have the minimization problems ($\widetilde{{\rm MP}_{1}}$) and ($\widetilde {\rm{MP}_2}$) defined in Eqs. 42 and 43. Then the solutions of ($\widetilde {\rm{MP}_1}$) and ($\widetilde {\rm{MP}_2}$) are $({\widetilde{\alpha}}^-_{II},0)$ and $({\widetilde{\alpha}}^+_I,0)$ respectively, where ${\widetilde{\alpha}}^-_{II}=\beta^-_I$, and ${\widetilde{\alpha}}^+_I=\beta^+_{II}$.

1.1.3 AI.1.3 Least number of mutation for moving state $z^{\hat b,\hat q}$

We consider the critical prices $p^L_I$, $p^R_I$, $p^L_{II}$, and $p^R_{II}$ given in Eq. 45. As A = 1, we have:

$$ p^L_{I}= p^L_{II}>{\hat b\over \hat q}>p^R_{I}= p^R_{II}. $$

(59)

We focus on $p^L_{II}$ and $p^R_I$, and have minimization problems (MP _L ) and (MP _R ) given in Eqs. 48 and 49. Their solutions are $(0,\beta^L_{II})$ and $( \alpha^R_{I},0)$, respectively, and $\beta^L_{II}=\alpha^R_I$.

Lemma 5′

Let A = 1. Then $\beta^L_{II}>\beta^-_I|_{n=1}$

As in Remark 3a, leaving the basin of attraction of $z^{\hat b,\hat q}$ requires the number of mutations to be at least $\beta^L_{II}m$ ($=\alpha^R_Im$).

1.2 AI.2 Proof of Theorem 1 when A = 1

The main strategy of the proof is almost identical to the one used for the case of A > 1. We will follow the same steps of the proof with some minor modifications.

(Step 1)

• (Stage A) When A = 1, the two offer curves OC _I and OC _II coincide with the diagonal, on which the point $(\hat b,\hat q)$ is in the middle, and the points (b _n ,q _n ) and $(\widetilde b_n,\widetilde q_n)$ are in the left and right side of it, respectively. Notice that, when A = 1, each (b _n ,q _n ) satisfies Eq. 23 and each $(\widetilde b_n,\widetilde q_n)$ satisfies Eq. 24 with equalities. Thus, the arguments in Cases A1 through A4 in the proof for the case of A > 1 are still valid.

• (Stage B) Now we will modify Cases B1 and B2 as follows:

  • (Case B1′) Consider the set $\{(b_n,q_n):n=1,\cdots,\bar n\}$. For each (b _n ,q _n ), we choose the state z ∈ Z where z(b _n − 1) = 1, $\beta^-_Im\leqq z({\underline{b}})<\beta^-_I m+1$, $z(b_n)=m-z({\underline{b}})-1$, and z(q ⁿ) = m.

    We then choose the path $(z^{b_n,q_n}\to z\to z^{b_{n-1},q_n})$. We have $r(z^{b_n,q_n}\to z\to z^{b_{n-1},q_n})< \beta^-_I+2$ by Claim 1′.

  • (Case B2′) Consider the set $\{(\widetilde b_n,\widetilde q_n):n=1,\cdots,\bar n\}$. For each $(\widetilde b_n,\widetilde q_n)$, we choose the state z ∈ Z where $z(\widetilde q_{n-1})=1$, $\widetilde\alpha^-_{II}m\leqq z({\underline{q}})<\widetilde\alpha^-_{II}m+1$, $z(\widetilde q_n)=m-z({\underline{q}})-1$, and $z(\widetilde b^n)=m$. We then choose the path $(z^{\widetilde b_n,\widetilde q_n}\to z\to z^{\widetilde b_{n },\widetilde q_{n-1}})$. We have $r(z^{\widetilde b_n,\widetilde q_n}\to z\to z^{\widetilde b_{n },\widetilde q_{n-1}})< \widetilde\alpha^-_{II}+2$ by Claim 1′.

  With these two stages, we complete the construction by choosing the edges for transient states as discussed at the end of Step 1 of the previous proof of Theorem 1. Thus, we have constructed a $\hat z$-tree, h.

(Step 2)  The resistance of h is

$$ r(h) < \sum\limits_{n=1}^{\bar n}[2(\beta^-_I|_n m+2)]+\#(S)\times\#(T)-2\bar n-1. $$

(60)

Now, we consider any $z'\not=z$ and any tree z′-tree, h′. We must leave all monomorphic states, or all except one, which cannot be $\hat z$. Therefore, we can at most save one of the terms $\beta^-_I|_{n}$ and $\widetilde \alpha^-_{II}|_n$. The maximum of these terms is $\beta^-_I|_{n=1}<\beta^L_{II}$ (by Lemma 5′). Thus, the resistance

$$ \begin{array}{rll} r(h') & \geqq & \sum\limits_{n=1}^{\bar n} [2 \beta^-_I|_n m]+(\beta^L_{II}-\beta^-_I|_{n=1})m +\#(S)\times\#(T)-2\bar n -1\\ &> & r(h) - 4 n^{**} + (\beta^L_{II}-\beta^-_I|_{n=1})m. \end{array} $$

(61)

Therefore, as $(\beta^L_{II}-\beta^-_I|_{n=1})>0$, for all large m, we have r(h′) > r(h).

Hence state $\hat z$ is the unique state in C ^* given in Fact 1. Therefore, we have μ ^*(z) = 1. This proves Theorem 1 when A = 1.²⁰ □

Appendix II: Proof of Lemmas

Proof of Lemma 1

By symmetry, it suffices to prove Part a.

At z ^b,q, the price p = b/q > b/A(K − b′) = P _I (b, b′). The price p′ of z ^b,q|b′ is p′ = (b′ + (m − 1)b)/q. Therefore, when m is sufficiently large, we have p′ > p _I (b, b′), so U(b′; z ^b,q|b′) > U(b, z ^b,q|b′).□

Proof of Lemma 2

By symmetry, it suffices to prove Part a.

Since $b'\geqq \hat b=AK/(A+1)$, we have $b'/[A(K-b')]\geqq 1/A$. Then critical price $P_I(b,b')=b'/[A(K-b)] >b'/[A(K-b')]\geqq 1/A$. Therefore, (b − b′)/P _I (b,b′) < A[b − b′]. We can write P _I (b, b′) = b/[A(K − b) + (b − b′)/P _I (b, b′)], so we have: $P_I(b,b')>b/[A(K-b')]\geqq b/q$. Hence, for all large m, we have P _I (b, b′) > p′ where p′ = (b′ + (m − 1)b)/q. Thus, U(b′; z ^{b, q|b′}) > U(b,z ^b,q|b′).□

Proof of Lemmas 3b and 3′b

We consider any $A\geqq 1$. By using the constraint equation in Eq. 35, we can express β as a function of α. Then checking the derivative of α + β with respect to α, we find out that (MP₂) has the solution where:

$$ \alpha =0, \,\rm{ if }\,p^+_{II}>{\bar b -b_n\over q_n-{\underline{q}}} \qquad \rm{ and }\qquad \beta = 0, \,\rm{ if }\,p^+_{II}<{\bar b -b_n\over q_n-{\underline{q}}}. $$

(62)

Using Eqs. 26, 19, 7 and 8, by calculation we have:

$$ p^+_{II}>{\bar b-b_n\over q_n-\underline{q}} \qquad\Leftrightarrow\qquad A\left[{K\over 1+A}+n\delta\right]>{K\over 1+A}-(n+1)\delta. $$

(63)

As the r.h.s of Eq. 63 holds, by Eq. 62 the solution of (MP₂) has α = 0. Hence, by the constraint (Eq. 35), we obtain the solution $\beta=\beta^+_{II}$ as given in Lemmas 3b and 3′b for the cases of A > 1 and A = 1 respectively. □

Proof of Lemmas 3a and 3′a

We consider any $A\geqq 1$. By using Eq. 34, it follows that the solution of (MP₁) is such that:

$$ \alpha =0, \,\rm{ if }\,p^-_{I}>{b_n-\underline{b}\over \bar b-q_n} \qquad\rm{ and }\qquad \beta = 0, \,\rm{ if }\, p^-_{I}<{b_n-\underline{b}\over \bar b-q_n}. $$

(64)

Using Eqs. 26, 19, 7 and 8, by calculation we have:

$$ \,p^-_{I}\, >\,(<)\, {b_n-\underline{b}\over \bar b-q_n} \qquad \Leftrightarrow \qquad A\left[{K\over 1+A}-n\delta\right]\, <\,(>)\, {K\over 1+A}+(n-1)\delta. $$

(65)

Suppose A = 1. As K/2 − nδ < K/2 + (n − 1)δ, we have α = 0, so by the constraint in Eq. 34 we have $\beta=\beta^-_I$ as claimed in Lemma 3′a. (For this proof of Lemma 3′a, we do not require δ to be small.)

Suppose A > 1, we have AK/(1 + A) > K/(1 + A). Therefore, for all small δ we have β = 0. So by the constraint in Eq. 34 we have $\alpha=\alpha^-_I$, as claimed in Lemma 3a. □

Proof of Lemma 4

We express

$$ \alpha^-_I={T^-_1-T^-_2\over T^-_3}, \qquad \rm{ and}\qquad \beta^+_{II}={T^+_1-T^+_2\over T^+_3}, $$

(66)

where

$$ \begin{array}{rll} T^-_1 &=& A[K-(1+A)n\delta][AK+(1+A)n\delta] , \\ T^-_2 &=& [AK+(1+A)(n-1)\delta][AK-(1+A)n\delta] ,\\ T^-_3 &=&A[K-(1+A)n\delta][AK+(1+A)(n-1)\delta] ,\\ T^+_1 &=& A [K+(1+A)n\delta][AK-(1+A)n\delta] , \\ T^+_2 &=& [AK+(1+A)n\delta][AK-(1+A)(n+1)\delta] ,\\ T^+_3 &=& A[K+(1+A)n\delta][AK-(1+A)(n+1)\delta]. \end{array} $$

(67)

(Proof of Part a)

Note that:

$$ \begin{array}{rll} {d[T^-_1-T^-_2]\over d\delta}|_{\delta=0}&=&AK(1+A)[(1-A)n+1] , \rm{and }\\ {d[T^+_1-T^+_2]\over d\delta}|_{\delta=0}&=&AK(1+A)[(A-1)n+1]. \end{array} $$

(68)

So $(d[T^-_1-T^-_2]/ d\delta)|_{\delta=0}<(d[T^+_1-T^+_2]/ d\delta)|_{\delta=0}$. Since $(T^-_1-T^-_2)|_{\delta=0}=0=(T^+_1-T^+_2)|_{\delta=0}$ and $T^-_3|_{\delta=0}=T^+_3|_{\delta=0}$, we have:

$$ {d \alpha^-_I\over d\delta}|_{\delta=0}<{d \beta^+_{II}\over d\delta}|_{\delta=0}. $$

(69)

Then $\alpha^-_I<\beta^+_{II}$ for all small δ.

(Proof of Part b)

Note that:

$$ \begin{array}{rll} {d T^-_1\over dn} &=&-A(1+A)(A-1)K\delta , \\ {d T^-_2\over dn} &=&-(1+A)(1+A)(2n-1)\delta\delta \rm{ and} \\ {d T^-_3\over dn} &=&-A(1+A)\delta[(A-1)K+(1+A)(2n-1)\delta] . \end{array} $$

(70)

When δ > 0 is sufficiently small, we disregard terms involving δ ². Then:

$$ \begin{array}{rll} {d T^-_1\over dn} - {d T^-_2\over dn} &\simeq& -A(1+A)(A-1)K\delta <0 \rm{ and}\\ {d T^-_3\over dn} &\simeq& -A(1+A)(A-1)K\delta<0 . \end{array} $$

(71)

Therefore,

$$ {d\alpha^-_I\over d n}\simeq -A(1+A)(A-1)K\delta\left[{T^-_3-(T_1-T_2) \over (T^-_3)^2}\right]. $$

(72)

The term $[T^-_3-(T_1-T_2)]/(T^-_3)^2$ converges to $1/T^-_3>0$ as δ converges to 0. Then we have $(d\alpha^-_I)/ (d n) <0$ for all small δ > 0. Thus, $\alpha^-_I$ is strictly decreasing in n. □

Proof of Lemma 4′

Since A = 1, we have:

$$\begin{array}{rll} \beta^-_I &=& { \left[\left({K\over 2}\right)-n\delta\right]\left[\left({K\over 2}\right)+ n\delta\right]-\left[\left({K\over 2}\right)-n\delta\right]\left[\left({K\over 2}\right) +\left(n-1\right)\delta\right]\over \left[\left({K\over 2}\right)+\left(n-1\right)\delta\right]\left[\left({K\over 2}\right) +\left(n-1\right)\delta\right]} ={T_1\over T_2} \rm{ and}\\ \beta^+_{II}&=&{ \left[\left({K\over 2}\right)-n\delta\right]\left[\left({K\over 2}\right)+n\delta\right] -\left[\left({K\over 2}\right)-\left(n+1\right)\delta\right]\left[\left({K\over 2}\right) +n\delta\right]\over \left[\left({K\over 2}\right)+n\delta\right]\left[\left({K\over 2}\right) -\left(n+1\right)\delta\right] }={T_1'\over T_2'}, \end{array} $$

(73)

where:

$$\begin{array}{rll} T_1 &=& \left({K\over 2}\right)\delta -n\delta\delta ,\\ T_2 &=& \left({K\over 2}\right)\left({K\over 2}\right)+\left({K\over 2}\right)2(n-1)\delta +(n-1)(n-1)\delta\delta ,\\ T_1'&=& \left({K\over 2}\right)\delta +n\delta\delta \rm{ and}\\ T_2' &=& \left({K\over 2}\right)\left({K\over 2}\right)-\left({K\over 2}\right)\delta - n (n+1)\delta\delta. \end{array} $$

(74)

(Proof of Part a)

Clearly, We have: T ₁ < T ₁′ and T ₂ > T ₂′, so $\beta^-_I<\beta^+_{II}$.

(Proof of Part b)

As d T ₁/d n < 0 and dT ₂/d n > 0, $\beta^-_I$ is strictly decreasing in n. □

Proof of Lemma 5

Note that by Eqs. 50, 66 and 67, we have:

$$ \begin{array}{rll} \beta^L_{II} & =& {(A+1)\delta \over AK-(A+1)\delta } \rm{ and} \\ \alpha^-_I|_{n=1}& =&{A[K-(1+A)\delta][AK+(1+A)\delta] - AK[AK-(1+A)\delta]\over A[K-(1+A)\delta]AK}. \end{array} $$

(75)

Therefore,

$$\begin{array}{rll} {d\beta^L_{II}\over d\delta}|_{\delta=0} & =& {(1+AK) (AK+1) \over (AK)(AK) } \rm{ and}\\ { d \alpha^-_I|_{n=1} \over d\delta}|_{\delta=0}& =&{ (1+A)(2-A) \over AK}. \end{array} $$

(76)

Since A > 1, we have (AK + 1)/(AK) > 1 > 2 − A, so $(d\beta^L_{II}/ d\delta )|_{\delta=0}> (d \alpha^-_I|_{n=1} / d\delta) |_{\delta=0}$. □

Proof of Lemma 5′

As A = 1, by Eqs. 50 and 73 we have:

$$ \beta^L_{II}={\delta\over K/2-\delta} \qquad\rm{ and}\qquad \beta^-_I|_{n=1}={\delta-2\delta^2/K\over K/2.} $$

(77)

Thus, $\beta^L_{II}> \beta^-_I|_{n=1}$.□