Time consistent Markov policies in dynamic economies with quasi-hyperbolic consumers

Abstract

We study the question of existence and computation of time-consistent Markov policies of quasi-hyperbolic consumers under a stochastic transition technology in a general class of economies with multidimensional action spaces and uncountable state spaces. Under standard complementarity assumptions on preferences, as well as a mild geometric condition on transition probabilities, we prove existence of time-consistent solutions in Markovian policies, and provide conditions for the existence of continuous and monotone equilibria. We present applications of our methods to habit formation models, environmental policies, and models of consumption under borrowing constraints, and hence show how our methods extend the results obtained by Harris and Laibson (Econometrica 69:935–957, 2001) to a broad class of dynamic economies. We also present a simple successive approximation scheme for computing extremal equilibrium, and provide some results on the existence of monotone equilibrium comparative statics in the model’s deep parameters.

Appendix: Proof of technical result

We begin with some useful definitions not provided earlier in the paper, but used in proofs of the proposition. An arbitrary set ($X,\ge )$ is partially ordered set (or poset) if $X$ is equipped with an order relation $\ge :X\times X\rightarrow X$ that is reflexive, antisymmetric and transitive. If every element of a poset $X$ is comparable, then $X$ is chain. If $X$ is a chain and countable, $X$ is a countable chain. An upper (respectively, lower) bound for a set $B\subset X$ is an element $x^{u}($respectively, $x^{l})\in X$ such that for any other element $x\in B,$ $x\le x^{u}$ (respectively, $x^{l}\le x)$. If there is a point $x^{u}$ (respectively, x$^{l})$ such that $x^{u}$ is the least element in the subset of upper bounds of $B\subset X$ (respectively, the greatest element in the subset of lower bounds of $B\subset X$), we say $x^{u}$ (respectively, $x^{l})$ is the supremum (respectively, infimum) of $B.$ Clearly if the supremum or infimum of a set $X$ exists, it must be unique.

We say a set $L\subset X$ is a lattice if for any two elements, say $x$ and $x^{\prime }$ in $L, L$ is closed under the operation of infimum (denoted by $x\wedge x^{\prime }),$ and supremum (denoted $x\vee x^{\prime }).$ The former is referred to as “the meet” of the two points, while the latter is “the join” . A subset $L_{1}$ of $L$ is a sublattice of $L$ if it contains the sup and the inf (with respect to $L$) of any pair of points in $L_{1}.$ A lattice is complete if any $L_{1}\subset L$, the least upper bound (denoted $\vee L_{1})$ and the greatest lower bound (denoted $\wedge L_{1})$ are both in $L$ . If this completeness property only holds for countable subsets $L_{c}$, the lattice is $\sigma -$ complete. In a poset $X$, if every subchain $C\subset$ $X$ is complete, then $X\$is referred to as a chain complete poset (or equivalent, a complete partially ordered set or CPO). A set $C$ is countable if it is either finite or there is a bijection from the natural numbers onto $C.$ If every countable chain $C\subset X$ is complete, then $X$ is referred to as a countably chain complete poset.

Let $(X_{1},\ge _{X_{1}})$ and $(X_{2},\ge _{X_{2}})$ be posets. A function (or, equivalently, operator) $f:X_{1}\rightarrow X_{2}$ is monotone (or order-preserving or isotone) if $f(x^{\prime })\ge _{X_{2}}f(x),$ when $x^{\prime }\ge _{X_{1}}x,$ for $x,x^{\prime }\in X_{1}$. A sequence $\{h_{n}\}$ in $H$ is order convergent if there exists two monotonic sequences of elements from $H$, one decreasing $\{h_{\downarrow n}\},$ and one increasing $\{h_{\uparrow n}\}$, such that $h=\inf h_{\downarrow n}=\sup h_{\uparrow n}$ and $h_{\uparrow n}\le h_{n}\le h_{\downarrow n}.$ A necessary and sufficient condition for an increasing sequence $h_{n}\rightarrow h$ to be order convergent is $h=\sup h_{n}.$

Let $(X,\ge )$ be a countably chain complete poset, i.e. where each increasing sequence has supremum, and each decreasing sequence has infimum. Assume that $X$ has the greatest element $\overline{\theta }$ and the least element $\underline{\theta }$. For a monotone sequence $\{x_{n}\}_{n=0}^{ \infty }$, let

$$\begin{aligned} \bigvee x_{n}:=\sup \limits _{n\in \mathbf {N}}x_{n}, \end{aligned}$$

and

$$\begin{aligned} \bigwedge x_{n}:=\inf \limits _{n\in \mathbf {N}}x_{n}. \end{aligned}$$

Denote by $F^{n}(x)$ the $n$-th orbit (or iteration) of $x$ under the function $F$, i.e. $F^{n}(x)=F\circ F\circ \ldots \circ F(x)$. We have the following two theorems, with the first pertaining to fixed point existence, and the second pertaining to fixed point comparative statics. The auxiliary fixed point theorem found in Sect. 2 is a corollary of both theorems.

Theorem 7

Let $X$ be a countably chain complete poset with the greatest element $\overline{\theta }$ and the least element $\underline{\theta }$, and $F:X\rightarrow X$ an increasing function, that is monotonically sup-inf-preserving i.e.

• if $x_n$ is increasing, then $F\left( \bigvee x_n\right) =\bigvee F(x_n)$ and

• if $x_n$ is decreasing, then $F\left( \bigwedge x_n\right) =\bigwedge F(x_n)$.

Then:

• (i) $\overline{\Phi }:=\bigwedge F^{n}(\overline{\theta })$ is the greatest fixed point and $\underline{\Phi }:=\bigvee F^{n}(\underline{\theta } )$ is the least fixed point.

• (ii) the set of fixed points is a nonempty countably chain complete poset with

  $$\begin{aligned} \overline{\Phi }=\bigvee \left\{ x:F(x)\ge x\right\} , \end{aligned}$$

  (10)

  and

  $$\begin{aligned} \underline{\Phi }=\bigwedge \left\{ x:F(x)\le x\right\} , \end{aligned}$$

  (11)

Proof

Proof of (i): Clearly $F(\overline{\theta })\le \overline{\theta }$. If for some $n,F^{n}(\overline{\theta })\ge F^{n+1}(\overline{\theta })$, then $F^{n+1}(\overline{\theta })=F(F^{n}(\overline{\theta }))\ge F(F^{n+1}( \overline{\theta }))=F^{n+2}(\overline{\theta })$. Hence, $F^{n}(\overline{ \theta })$ is decreasing, and $\overline{\Phi }$ is well defined. Since $F$ is monotonically inf-preserving, we have

$$\begin{aligned} F(\overline{\Phi })&= F\left( \bigwedge F^{n}(\theta )\right) , \\&= \bigwedge F^{n+1}(\theta ), \\&= \overline{\Phi }. \end{aligned}$$

Therefore, $\overline{\Phi }$ is fixed point of $F$. Let us take arbitrary fixed point $e=F(e)$. Clearly, $e\le \overline{\theta }$, and $e=F(e)\le F( \overline{\theta })$. If $e=\le F^{n}(\overline{\theta }),$ then $e=F(e)=\le F^{n+1}(\overline{\theta })$. Therefore, $e\le F^{n}(\overline{ \theta })$ for all $n$, which implies $e\le \overline{\Phi }$. Similarly, we prove that $\underline{\Phi }$ is well defined and it is the least fixed point of $F$.

Proof of (ii): Let $e_{n}$ be an increasing set of fixed points. Let $\bar{e} =\bigvee e_{n}$. Then,

$$\begin{aligned} F(\bar{e})&= F\left( \bigvee e_{n}\right) , \\&= \bigvee F(e_{n}), \\&= \bigvee e_{n}=\bar{e}. \end{aligned}$$

Similarly, we prove the thesis for decreasing sequences. Now, we finally prove equality (10). Let $x$ be arbitrary point such that $x\le F(x)$. Clearly $x\le \overline{\theta }$. Assume $x\le F^{n}( \overline{\theta })$. Then, $x\le F(x)\le F(F^{n}(\overline{\theta } ))=F^{n+1}(\overline{\theta })$. Hence, $x\le \overline{\Phi }$. Since $\overline{\Phi }\in \{x:F(x)\ge x\},$ equality (10) is proven. We prove (11) analogously. $\square$

We finally prove a theorem (and a corollary) on increasing selections for parameterized problems that we use in the paper to obtain our results on equilibrium monotone comparatives.

Theorem 8

Let $X$ be a countably chain complete poset with the greatest and least elements and $T$ a poset. If $F:X\times T\rightarrow X$ is increasing, and monotonically-sup-inf preserving on $X$ then $t\rightarrow \overline{\Phi }(t)$ and $t\rightarrow \underline{\Phi }(t)$ are isotone.

Proof

Let $t_{1}\le t_{2}$. From Theorem 7 we know that $m_{i}:= \overline{\Phi }(t_{i})=\vee \Gamma _{i}:=\vee \{x:F(x,t_{i})\le x\}$. Note that by isotonicity of $F(x,\cdot )$ we obtain $m_{1}=F(m_{1},t_{1})\le F(m_{1},t_{2})$. Hence $m_{1}\in \Gamma _{2}$. Since $m_{2}$ is the greatest element of $\Gamma _{2}$, hence $m_{1}\le m_{2}$. $\square$