Bayesian general equilibrium

Abstract

I introduce a general equilibrium model of non-optimizing agents that respond to aggregate variables (prices and the average demand profile of agent types) by putting a “prior” on their demand. An interim equilibrium is defined by the posterior demand distribution of agent types conditional on market clearing. A Bayesian general equilibrium (BGE) is an interim equilibrium such that aggregate variables are correctly anticipated. Under weak conditions, I prove the existence and the informational efficiency of BGE. I discuss the conditions under which the set of Bayesian and Walrasian equilibria coincide and show that the Walrasian equilibrium arises from a large class of non-optimizing behavior.

Appendices

Appendix 1: Bayesian inference and maximum entropy

Given a multinomial distribution $\mathbf {p}=(p_1,\dots ,p_K)$, where $p_k\ge 0$ and $\sum \nolimits _{k=1}^Kp_k=1$, Shannon (1948) defined its entropy by

$$\begin{aligned} H(\mathbf {p})=-\sum _{k=1}^Kp_k\log p_k. \end{aligned}$$

(7.1)

Jaynes (1957) proposed that when we want to assign probabilities $\mathbf {p}=(p_1,\dots ,p_K)$ given some background information (such as moment constraints), we should maximize the Shannon entropy (7.1) subject to the constraints imposed by the background information. This is the original maximum entropy principle (MaxEnt). A prototypical example of such an inference problem is the die problem in Jaynes (1978), which dates back to a 1962 lecture:

Suppose a die has been tossed $N$ times and we are told only that the average number of spots up was not 3.5 as one might expect for an “honest” die but 4.5. Given this information, and nothing else, what probability should we assign to $i$ spots in the next toss?

One drawback of the Shannon entropy is that it is not clear how to define it for distributions on a continuous space (say, Euclidean space). To circumvent this difficulty, Kullback and Leibler (1951) introduced the information measure

$$\begin{aligned} H(p;q)=\int p(x)\log \frac{p(x)}{q(x)}\mathrm {d}x, \end{aligned}$$

where $q(x)$ is the “prior” and $p(x)$ is the “posterior” density. Here we are implicitly using the Lebesgue measure $\mathrm {d}x$ and the density functions with respect to the Lebesgue measure, but it does not need to be so. A big advantage of the Kullback–Leibler information is that it is invariant to the choice of the reference measure: if measures $P,Q,\mu _1,\mu _2$ are mutually absolutely continuous and $p_i=\mathrm {d}P/\mathrm {d}\mu _i, q_i=\mathrm {d}Q/\mathrm {d}\mu _i$ are Radon–Nikodym derivatives (which correspond to density functions of probability distributions), then

$$\begin{aligned} H(p_1;q_1)=\int p_1\log \frac{p_1}{q_1}\mathrm {d}\mu _1=\int p_2\log \frac{p_2}{q_2}\mathrm {d}\mu _2=H(p_2,;q_2), \end{aligned}$$

so the choice of $\mu _1,\mu _2$ is irrelevant. Thus, given any “prior” measure $Q$ and “posterior” measure $P$, we can define the Kullback–Leibler information⁷ of $P$ with respect to $Q$ by

$$\begin{aligned} H(P;Q)=\int \frac{\mathrm {d}P}{\mathrm {d}Q}\log \frac{\mathrm {d}P}{\mathrm {d}Q}\mathrm {d}Q=\int p\log \frac{p}{q}\mathrm {d}\mu , \end{aligned}$$

(7.2)

where $\mu$ is any reference measure and $p=\mathrm {d}P/\mathrm {d}\mu$, $q=\mathrm {d}Q/\mathrm {d}\mu$ are Radon–Nikodym derivatives (density functions).

The Shannon entropy (7.1) corresponds to the Kullback–Leibler information (7.2) with respect to the uniform prior on a discrete set modulo the sign and an additive constant. Thus, the maximum entropy principle of Jaynes (1957) can be generalized to what I refer to as the minimum information principle, which prescribes to minimize the Kullback–Leibler information (7.2) subject to the given constraints. Axiomatizations of the minimum information principle have been obtained by Shore and Johnson (1980), Caticha and Giffin (2006), and Knuth and Skilling (2012).

Returning to Bayesian inference, Van Campenhout and Cover (1981) showed that Bayes’s theorem implies the minimum information principle in the following sense: the conditional distribution of a random variable $X_n$ given the empirical observation

$$\begin{aligned} \frac{1}{N}\sum _{n=1}^NT(X_n)=\bar{T}, \end{aligned}$$

where $X_n$’s are i.i.d. with prior density $g$, converges to $f_\lambda (x)=\mathrm {e}^{\lambda 'T(x)}g(x)$ (suitably normalized) as $N\rightarrow \infty$, where $\lambda$ is chosen to satisfy the population moment constraint

$$\begin{aligned} \int T(x)f_\lambda (x)\mathrm {d}x=\bar{T}. \end{aligned}$$

This $f_\lambda (x)$ turns out to be the solution to

$$\begin{aligned} \min _f H(f;g)~\text {subject to}~\int T(x)f(x)\mathrm {d}x=\bar{T}, \end{aligned}$$

i.e., the minimum information problem, and that $\lambda$ is the corresponding Lagrange multiplier.⁸ Although Van Campenhout and Cover (1981) proved the statement only for the case $T$ is a real function, Csiszár (1984) showed that the same conclusion holds even if $T$ is vector-valued and the sample moment constraints $\frac{1}{N}\sum \nolimits _{n=1}^NT(X_n)=\bar{T}$ are replaced by the condition that the sample moments belong to a specified convex set, in particular with inequality constraints. Thus, computing the posterior distribution (in the Bayesian sense) reduces to solving the minimum Kullback–Leibler information problem, at least in the large sample limit.

Appendix 2: Proof of equilibrium existence

Proof of Step 1

That $H^*(\xi ;p,x)$ is $C^1$ in $\xi$ and is differentiable under the integral sign follow by Assumption 1 and Lebesgue’s dominated convergence theorem. I only show that $H^*(\xi ;p,x)$ is continuous in $(p,x,\xi )$ since the case for its first derivatives is similar. Let $(p_n,x_n,\xi _n)\rightarrow (p,x,\xi )$ as $n\rightarrow \infty$, $f_n(y)=\mathrm {e}^{-\xi _n'y}$, and $f(y)=\mathrm {e}^{-\xi 'y}$ for $y\in \mathbb {R}_+^C$. Then for any sequence such that $y_n\rightarrow y$, we have $f_n(y_n)\rightarrow f(y)$. (This property is referred to as “$f_n$ continuously converges to $f$.”) Since $f_n\le 1, \left\{ f_n\right\}$ are uniformly $\mu _i(p_n,x_n)$-integrable, i.e.,

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\sup _n\int _{f_n>\alpha }f_n(y)\mu _i(\mathrm {d}y;p_n,x_n)=0. \end{aligned}$$

(Just take $\alpha \ge 1$.) Therefore, by Theorem 9.2, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\int f_n(y)\mu _i(\mathrm {d}y;p_n,x_n)=\int f(y)\mu _i(\mathrm {d}y;p,x). \end{aligned}$$

Hence, $\int \mathrm {e}^{-\xi 'y}\mu _i(\mathrm {d}y;p,x)$ is continuous in $(p,x,\xi )$, and so is $H^*(\xi ;p,x)$. $\square$

Proof of Proposition 3.2

$\fancyscript{E}$ has a Bayesian general equilibrium because (3.14) is stronger than Assumption 2. Suppose that $\fancyscript{E}$ has a non-degenerate equilibrium $(p,x,(f_i))$. Then

$$\begin{aligned} x_i=\int yf_i(y)\mu _i(\mathrm {d}y;p,x)\in \mathrm{cl }\mathrm{co }X_i(p,x). \end{aligned}$$

By market clearing and the nature of Lagrange multipliers, we have

$$\begin{aligned} \sum _{i=1}^In_i(x_i-e_i)=\bar{x}[f;\mu (p,x)]-\bar{e}\le 0 \end{aligned}$$

and $p\cdot \sum _{i=1}^In_i(x_i-e_i)=0$. Then for any $y=(y_i)$ with $y_i\in X_i(p,x)$, by (3.14) we get

$$\begin{aligned} p\cdot \sum _{i=1}^In_i(y_i-e_i)\ge 0=p\cdot \sum _{i=1}^In_i(x_i-e_i), \end{aligned}$$

so $p\cdot y\ge p\cdot x_i$ for all $i$ and $y\in X_i(p,x)$. Hence, by Definition 2.4, $(p,x)$ is a degenerate equilibrium.

Let $\fancyscript{E}^n=\left\{ I,\left\{ n_i\right\} ,\left\{ e_i\right\} ,\left\{ \mu _i^n\right\} \right\}$ be an economy such that

$$\begin{aligned} \mu _i^n(B;p,x)=\mu _i\left( (1-1/n)^{-1}B;p,x\right) \end{aligned}$$

for any Borel set $B$, that is, the prior $\mu _i^n(p,x)$ is obtained by shrinking the domain of $\mu _i(p,x)$ by $1-1/n$ about the origin. Then the offer set is $X_i^n(p,x)=\mathrm{supp }\mu _i^n(p,x)=(1-1/n)X_i(p,x)$, so

$$\begin{aligned} \sum _{i=1}^I n_i\inf \left\{ p\cdot y|y\in X_i^n(p,x)\right\}&=\left( 1-\frac{1}{n}\right) \sum _{i=1}^I n_i\inf \left\{ p\cdot y|y\in X_i(p,x)\right\} \\&<\sum _{i=1}^I n_i\inf \left\{ p\cdot y|y\in X_i(p,x)\right\} . \end{aligned}$$

Hence, by (3.14) we get

$$\begin{aligned} \sum _{i=1}^I n_i\inf \left\{ p\cdot (y-e_i)|y\in X_i^n(p,x)\right\} <\sum _{i=1}^I n_i\inf \left\{ p\cdot (y-e_i)|y\in X_i(p,x)\right\} =0. \end{aligned}$$

By Theorem 3.2, the economy $\fancyscript{E}^n$ has a non-degenerate equilibrium $(p^n,(x_i^n),(f_i^n))$, where

$$\begin{aligned} x_i^n=\int yf_i^n(y)\mathrm {d}\mu _i(p^n,x^n)\in \mathrm{cl }\mathrm{co }X_i(p^n,x^n). \end{aligned}$$

Then by market clearing and the nature of Lagrange multipliers, after some algebra we obtain

$$\begin{aligned} \sum _{i=1}^In_i(x_i^n-e_i)\le 0~\text {and}~ p^n\cdot \sum _{i=1}^In_i(x_i^n-e_i)=0, \end{aligned}$$

(8.1)

where $\bar{e}=\sum _{i=1}^I n_ie_i$ is the aggregate endowment. Since $x_i^n\in \mathrm{cl }\mathrm{co }X_i(p^n,x^n)\subset X\subset \mathbb {R}_+^C$ and $\left\{ x_i^n\right\}$ is bounded ($\because$ market clearing), by taking a subsequence if necessary, we may assume $x_i^n$ converges to some $x_i$ as $n\rightarrow \infty$. Since $\varDelta ^{C-1}$ is compact, we may also assume $p^n\rightarrow p$. Letting $n\rightarrow \infty$ in (8.1), we get $\sum _{i=1}^In_i(x_i-e_i)\le 0$ and $p\cdot \sum _{i=1}^In_i(x_i-e_i)=0$. By Assumption 4, we have $x_i\in \mathrm{cl }\mathrm{co }X_i(p,x)$, where $x=(x_i)$. By the same argument as above, $(p,x)$ is a degenerate equilibrium. $\square$

Appendix 3: Mathematical results

Lemma 9.1

(Chebyshev’s inequality) If $f,g:\mathbb {R}\rightarrow \mathbb {R}$ are increasing (decreasing) functions and $X$ is a random variable, then

$$\begin{aligned} \mathrm{E }[f(X)g(X)]\ge \mathrm{E }[f(X)]\mathrm{E }[g(X)]. \end{aligned}$$

Proof

Let $X'$ be an i.i.d. copy of $X$. Since $f,g$ are monotone, we have

$$\begin{aligned} (f(X)-f(X'))(g(X)-g(X'))\ge 0. \end{aligned}$$

Taking expectations of both sides, noting that $X,X'$ are i.i.d., and rearranging terms, we obtain

$$\begin{aligned} \mathrm{E }[f(X)g(X)]\ge \mathrm{E }[f(X)]\mathrm{E }[g(X)]. \end{aligned}$$

$\square$

Clearly if one of $f,g$ is increasing and the other is decreasing, the reverse inequality holds.

Theorem 9.1

(Generalized Kuhn–Tucker) Let $X$ be a linear vector space, $Z_1,Z_2$ normed spaces, $\Omega$ a convex subset of $X$, and $P$ the positive cone in $Z_1$. Assume that $P$ contains an interior point.

Let $f$ be a real-valued convex functional on $\Omega$, $G_1:\Omega \rightarrow Z_1$ a convex mapping, and $G_2:X\rightarrow Z_2$ an affine mapping. Assume the existence of a point $x_1\in \Omega$ for which $G_1(x_1)<0$ (i.e., $G_1(x_1 )$ is an interior point of $N=-P$) and $G_2(x_1)=0$, and that 0 is an interior point of $G_2(\Omega )$. Let

$$\begin{aligned} \mu _0=\inf f(x)~\text {subject to}~x\in \Omega ,~G_1(x)\le 0,~G_2(x)=0 \end{aligned}$$

(9.1)

and assume $\mu _0$ is finite. Then there exist $z_1^*\ge 0$ in $Z_1^*$ and $z_2^*\in Z_2^*$ such that

$$\begin{aligned} \mu _0=\inf _{x\in \Omega }\left[ f(x)+\left\langle G_1(x),z_1^*\right\rangle +\left\langle G_2(x),z_2^*\right\rangle \right] . \end{aligned}$$

(9.2)

Furthermore, if the infimum is achieved in (9.1) by $x_0\in \Omega$, it is achieved by $x_0$ in (9.2) and $\left\langle G_2(x_0),z_2^*\right\rangle =0$.

Proof

Similar to Luenberger (1969), Theorem 1, p. 217. $\square$

Proposition 9.1

Let $(X,\fancyscript{B},\mu )$ be a measure space, where $X$ is a topological space, $\fancyscript{B}$ is the Borel $\sigma$-algebra, and $\mu (X)>0$. Let $T:X\rightarrow \mathbb {R}^C$ be measurable. Then,

$$\begin{aligned} f(\xi ):=\log \left( \int \mathrm {e}^{-\xi 'T(x)}\mu (\mathrm {d}x)\right) \end{aligned}$$

is convex and lower semi-continuous on $\mathrm{dom }f$.⁹ Furthermore, $f$ is strictly convex if $\dim T(\mathrm{supp }\mu )=C$.¹⁰

Proof

See Proposition B.4 in Toda (2010). $\square$

Proposition 9.2

Let $(X,\fancyscript{B},\mu )$ be as in Proposition 9.1 and $\phi :X\rightarrow \mathbb {R}$ be measurable. If $\int \mathrm {e}^{t\phi (x)}\mu (\mathrm {d}x)<\infty$ for some $t>0$, then

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{t}\log \left( \int \mathrm {e}^{t\phi }\mathrm {d}\mu \right) =\mathrm{ess }\sup \phi . \end{aligned}$$

(9.3)

If $\phi$ is upper semi-continuous, then (9.3) is equal to $\sup \left\{ \phi (x)|x\in \mathrm{supp }\mu \right\}$.

Proof

Let $E_n=\left\{ x\in X|\phi (x)\ge -n\right\}$. If $\mu (E_n)=0$ for all $n$, then

$$\begin{aligned} \mu (X)=\mu \left( \textstyle \bigcup _{n=1}^\infty E_n\right) =\lim _{n\rightarrow \infty }\mu (E_n)=0, \end{aligned}$$

which contradicts $\mathrm{supp }\mu \ne \emptyset$. Therefore, $\mu (E_n)>0$ for some $n$. Letting

$$\begin{aligned} v=\mathrm{ess }\sup \phi =\sup \left\{ c|\mu (\left\{ x\in X|\phi (x)\ge c\right\} )>0\right\} , \end{aligned}$$

we have $v>-\infty$.

Let us first prove (9.3) when $v<\infty$. Define

$$\begin{aligned} X_+&=\left\{ x\in X|\phi (x)>v\right\} ,~X_-=\left\{ x\in X|\phi (x)\le v\right\} ,~\text {and}\\ X_n&=\left\{ x\in X\big |\phi (x)\ge v+\frac{1}{n}\right\} . \end{aligned}$$

Since $X_+=\bigcup _{n=1}^\infty X_n$ and $\mu (X_n)=0$ by the definition of $v$, we have $\mu (X_+)=0$. Obviously, $X_\pm$ are disjoint and $X_+\cup X_-=X$, so $\mu (X_-)=\mu (X)>0$. Fix $t_0>0$ such that $\int \mathrm {e}^{t_0\phi (x)}\mathrm {d}\mu <\infty$. Then, for all $t>0$ we obtain

$$\begin{aligned} \int \mathrm {e}^{t\phi (x)}\mathrm {d}\mu =\mathrm {e}^{tv}\int _{X_-}\mathrm {e}^{t(\phi (x)-v)}\mathrm {d}\mu . \end{aligned}$$

(9.4)

Denote the integral over $X_-$ in (9.4) by $I(t)$. Since $\phi (x)\le v$ for $x\in X_-$, for each $x\in X_-$ the integrand $\mathrm {e}^{t(\phi (x)-v)}$ is decreasing in $t$, so $I(t)$ is decreasing in $t$. (In particular, $0<I(t)<\infty$ for $t\ge t_0$.) Hence, for $t\ge t_0$ we obtain

$$\begin{aligned} \frac{1}{t}\log \left( \int \mathrm {e}^{t\phi (x)}\mathrm {d}\mu \right) =v+\frac{1}{t}\log I(t)\le v+\frac{1}{t}\log I(t_0). \end{aligned}$$

(9.5)

Letting $t\rightarrow \infty$ in (9.5), we obtain

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{1}{t}\log \left( \int \mathrm {e}^{t\phi (x)}\mathrm {d}\mu \right) \le v. \end{aligned}$$

(9.6)

To show the reverse inequality, take any $\epsilon >0$ and let

$$\begin{aligned} A=\left\{ x\in X_-|\phi (x)\ge v-\epsilon \right\} . \end{aligned}$$

By assumption and the definition of $X_\pm$, we have

$$\begin{aligned} \mu (A)=\mu (\left\{ x\in X|\phi (x)\ge v-\epsilon \right\} )>0. \end{aligned}$$

By taking a compact subset of $A$ if necessary, we may assume $0<\mu (A)<\infty$ since $\mu$ is regular. Therefore, we obtain

$$\begin{aligned} \frac{1}{t}\log \left( \int \mathrm {e}^{t\phi (x)}\mathrm {d}\mu \right)&\ge \frac{1}{t}\log \left( \mathrm {e}^{t(v-\epsilon )}\int _A\mathrm {e}^{t(\phi (x)-v+\epsilon )}\mathrm {d}\mu \right) \nonumber \\&\ge v-\epsilon +\frac{1}{t}\log \mu (A). \end{aligned}$$

(9.7)

Letting $t\rightarrow \infty$ in (9.7) and then $\epsilon \rightarrow 0$, we obtain

$$\begin{aligned} \liminf _{t\rightarrow \infty }\frac{1}{t}\log \left( \int \mathrm {e}^{t\phi (x)}\mathrm {d}\mu \right) \ge v. \end{aligned}$$

(9.8)

(9.3) follows by (9.6) and (9.8).

If $v=\infty$, let $F_n=\left\{ x\in X|\phi (x)\ge n\right\}$. By the definition of $v$, we have $\mu (F_n)>0$. Then we obtain the same result as (9.7) with $A$ replaced by $F_n$ and $v-\epsilon$ replaced by $n$. Letting $n\rightarrow \infty$ we get (9.3).

Finally let us show $\mathrm{ess }\sup \phi =\sup \left\{ \phi (x)|x\in \mathrm{supp }\mu \right\}$ if $\phi$ is upper semi-continuous. Let $u=\sup \left\{ \phi (x)|x\in \mathrm{supp }\mu \right\}$. If $u<\infty$, for all $\epsilon >0$ there exists an $x_0\in X$ such that $u-\epsilon <\phi (x_0)$. Since $\phi$ is upper semi-continuous, there exists an open neighborhood $U$ of $x_0$ such that $x\in U$ implies $\phi (x)>u-\epsilon$. Since $\mu (U\cap X)>0$ by assumption, it follows that $v\ge u-\epsilon$. Since $\epsilon >0$ is arbitrary, we obtain $v\ge u$. A similar reasoning holds for the case $u=\infty$.

To show the reverse inequality, take any $\epsilon >0$. By the definition of $v$, we have $\mu (\left\{ x\in X|\phi (x)\ge v-\epsilon \right\} )>0$. In particular, there exists an $x_0\in \mathrm{supp }\mu$ such that $\phi (x_0)\ge v-\epsilon$. Therefore,

$$\begin{aligned} u=\sup \left\{ \phi (x)|x\in \mathrm{supp }\mu \right\} \ge \phi (x_0)\ge v-\epsilon . \end{aligned}$$

Since $\epsilon >0$ is arbitrary, we obtain $u\ge v$. Therefore, $u=v$. $\square$

Finally, I need a convergence theorem of Lebesgue integrals with varying measures. Let $X$ be a locally compact second countable Hausdorff space (e.g., Euclidean space) with Borel $\sigma$-algebra $\fancyscript{B}$. We say that $f_n$ continuously converges to $f$, denoted by $f_n\rightarrow _c f$, if $\lim f_n(x_n)\rightarrow f(x)$ for any $x_n\rightarrow x$.¹¹

Theorem 9.2

Let $\mu , \left\{ \mu _n\right\}$ be finite Borel measures on $X$. Suppose that $f_n\ge 0, f_n\rightarrow _c f, \mu _n\rightarrow \mu$ weakly, and $\int f_n\mathrm {d}\mu _n<\infty$ for all $n$. Then

$$\begin{aligned} \lim _{n\rightarrow \infty }\int f_n\mathrm {d}\mu _n=\int f\mathrm {d}\mu \end{aligned}$$

if and only if $\left\{ f_n\right\}$ is uniformly $\left\{ \mu _n\right\}$-integrable, i.e.,

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\sup _n\int _{f_n>\alpha }f_n\mathrm {d}\mu _n=0. \end{aligned}$$

Proof

See Serfozo (1982), Theorem 3.5. $\square$