Uniqueness of equilibrium in a Bewley–Aiyagari model

Abstract

I establish the uniqueness of a stationary equilibrium in a Bewley–Aiyagari model when the agents’ utility function exhibits constant relative risk aversion bounded above by 1 and the production function exhibits a certain gross substitute property.

Appendix

I first introduce the notations and preliminary results that are needed to prove Theorem 1.

Let $A =[0 ,{\overline{a}}]$, $Z =A \times S$.

Denote by b the agent’s savings in the next period. For every $(a ,s_{i} ,R) \in Z \times I$ and $b \in C (a ,s_{i} ,R)$, define the following function:

$$\begin{aligned} h (a ,s_{i} ,b ,R ,V) =u (Ra +ws_{i} -b) +\beta \sum \limits _{j =1}^{n}P_{i j} V (b ,s_{j} ,R) . \end{aligned}$$

(2)

From Lemma 1, the value function is bounded on $Z \times I$. Thus, I can use the standard Bellman’s equation approach to solve the agent’s problem.

Let $B (Z \times I)$ be the space of all bounded real-valued functions defined on $Z \times I$. Define the operator $T :B (Z \times I) \rightarrow B (Z \times I)$ by

$$\begin{aligned}T f (a ,s ,R) =\max _{b \in C (a ,s ,R)}h (a ,s ,b ,R ,f) . \end{aligned}$$

Standard dynamic programming arguments (Blackwell 1965) show that the value function V is the unique fixed point of T, i.e., there is a unique function $V \in B (Z \times I)$ such that $T V =V$. The equation $TV =V$ is called the Bellman equation.

From Theorem 9.8 in Stokey and Lucas (1989), the value function V is continuous and concave in the first argument, and thus h(a, s, b, R, V) is strictly concave in b. The unique savings level that attains the maximum of h(a, s, b, R, V) is the savings policy function:

$$\begin{aligned} g\left( a ,s ,R\right) ={\text {*}}{argmax}_{b \in C\left( a ,s ,R\right) }h\left( a ,s ,b ,R ,V\right) . \end{aligned}$$

Note that the savings policy function induces a consumption policy function $\sigma (a ,s ,R)$ by the equation $\sigma (a ,s ,R) =R a +ws -g (a ,s ,R)$.

Let $g_{f}\left( a ,s ,R\right) ={\text {*}}{argmax}_{b \in C\left( a ,s ,R\right) }h\left( a ,s ,b ,R ,f\right)$ and $\sigma _{f}\left( a ,s\right) =Ra +ws -g_{f}\left( a ,s ,R\right)$ for all $\left( a ,s ,R\right) \in Z \times I$.

For $f \in B\left( Z \times I\right)$ define

$$\begin{aligned} f^{ \prime }(a ,s ,R) : =\frac{ \partial f\left( a ,s ,R\right) }{ \partial a} . \end{aligned}$$

Denote by $u^{ \prime }$ the derivative of u and by $h^{ \prime }\left( a ,s ,b ,R ,f\right)$ the derivative of h with respect to b.

The envelope theorem (Benveniste and Scheinkman 1979) implies that Tf is differentiable.

In addition, if $\sigma _{f}\left( a ,s ,R\right) >0$ (which will always be the case in this paper, since $u^{ \prime }(0) =\infty$), then

$$\begin{aligned} (Tf)^{ \prime }\left( a ,s ,R\right) =Ru^{ \prime }(\sigma _{f}(a ,s ,R)) . \end{aligned}$$

The last equation is called the envelope condition. Berge’s maximum theorem [see Theorem 17.31 in Aliprantis and Border (2006)] implies that $\sigma _{f}$ is continuous. Thus, $(Tf)^{ \prime }$ is continuous as a composition of continuous functions. Since $TV =V$, the value function is continuously differentiable and $V^{ \prime }(a ,s ,R) =Ru^{ \prime }\left( \sigma (a ,s ,R)\right)$.

To prove Theorem 1 I need the following three Lemmas:

Lemma 4

Let $z :[0 ,\infty ) \rightarrow (0 ,\infty )$ be a concave function. Then $\frac{\alpha }{z(\alpha x)}$ is increasing in $\alpha$ on $(0 ,\infty )$ for all $x >0$.

Proof

Let $\alpha _{2} \ge \alpha _{1} >0$ and $x >0$. Since z is a concave function we have

$$\begin{aligned} \frac{z(\alpha _{2}x) -z(\alpha _{1}x)}{\alpha _{2}x -\alpha _{1}x} \le \frac{z(\alpha _{2}x) -z(0)}{\alpha _{2}x -0} , \end{aligned}$$

which implies

$$\begin{aligned} \alpha _{1}xz(\alpha _{2}x) \le \alpha _{2}xz(\alpha _{1}x) -z(0)(\alpha _{2}x -\alpha _{1}x) . \end{aligned}$$

Since $z(0) >0$ the last inequality implies $\alpha _{1}xz(\alpha _{2}x) \le \alpha _{2}xz(\alpha _{1}x)$ or

$$\begin{aligned}\frac{\alpha _{1}}{z(\alpha _{1}x)} \le \frac{\alpha _{2}}{z(\alpha _{2}x)} , \end{aligned}$$

which proves the Lemma. $\square$

Lemma 5

$\sigma (a ,s ,R)$ is concave in a for all $(s ,R) \in S \times I$.

Proof

See Jensen (2017). $\square$

The proof of the following simple lemma is given in Lehrer and Light (2018).

Lemma 6

Let $z :\mathbb {R} \rightarrow \mathbb {R}$ and $f :\mathbb {R} \rightarrow \mathbb {R}$ be strictly concave, continuously differentiable functions. Let $\phi >0$. Denote $x_{f} ={\text {*}}{argmax}_{x \in [0 ,\phi ]}f \left( x\right)$ and $x_{z} ={\text {*}}{argmax}_{x \in [0 ,\phi ]}z \left( x\right)$. If for all $x \in [0 ,\phi ]$ we have $f^{ \prime } \left( x\right) \ge z^{ \prime } (x)$, then $x_{f} \ge x_{z}$. Furthermore, if $x_{z} \in (0 ,\phi )$ and $f^{ \prime } \left( x\right) >z^{ \prime } \left( x\right)$ for all $x \in [0 ,\phi ]$, then $x_{f} >x_{z}$.

Theorem 1

Assume that $\gamma \le 1$. Then $R_{2} \ge R_{1}$ implies $g(a ,s ,R_{2}) \ge g(a ,s ,R_{1})$ for all $(a ,s) \in A \times S$ .

Proof

Assume $\frac{1}{\beta }>R_{2} \ge R_{1} >0$ and $w >0$. Let $f :A \times S \times I \rightarrow \mathbb {R}$ be differentiable and concave in the first argument. Assume that $f^{ \prime }(a ,s ,R_{2}) \ge f^{ \prime }(a ,s ,R_{1})$ for all $(a ,s) \in Z$. Let $(a ,s_{i}) \in Z$ and $b \in C(a ,s_{i} ,R_{1})$. We have

$$\begin{aligned} h^{ \prime }(a ,s_{i} ,b ,R_{1} ,f)&= -u^{ \prime }(R_{1}a +ws_{i} -b) +\beta \sum \limits _{j =1}^{n}P_{ij}f^{ \prime }(b ,s_{i} ,R_{1}) \nonumber \\&\le -u^{ \prime }(\frac{R_{1}}{R_{2}}R_{2}a +ws_{i} -b) +\beta \sum \limits _{j =1}^{n}P_{ij}f^{ \prime }(b ,s_{i} ,R_{2}) \nonumber \\&=h^{ \prime }(\frac{R_{1}}{R_{2}}a ,s_{i} ,b ,R_{2} ,f) . \end{aligned}$$

(3)

Note that $C(a ,s_{i} ,R_{1}) =[0 ,R_{1}a +s_{i}] =C(\frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2})$. Thus, Lemma 6 implies that $g_{f}(\frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}) \ge g_{f}(a ,s_{i} ,R_{1})$. That is, $\sigma _{f}(a ,s_{i} ,R_{1}) \ge \sigma _{f}(\frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2})$. We have

$$\begin{aligned} (Tf)^{ \prime }(a ,s_{i} ,R_{1})&=R_{1}u^{ \prime }(\sigma _{f}(a ,s_{i} ,R_{1})) \\&\le R_{1}u^{ \prime }\left( \sigma _{f}\left( \frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}\right) \right) \\&=\frac{R_{1}}{\sigma _{f}\left( \frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}\right) }u^{ \prime }\left( \sigma _{f}\left( \frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}\right) \right) \sigma _{f}(\frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}) \\&\le R_{2}u^{ \prime }(\sigma _{f}(a ,s_{i} ,R_{2})) =(Tf)^{ \prime }(a ,s_{i} ,R_{2}) . \end{aligned}$$

The first and last equalities follow from the envelope condition. The first inequality follows from the fact that $u^{ \prime }$ is decreasing. To see why the second inequality holds, first note that by defining $c_{2} =\sigma _{f}(\frac{R_{2}}{R_{2}}a ,s_{i} ,R_{2})$ and $c_{1} =\sigma _{f}(\frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2})$ and using¹⁹$\gamma \le 1$ we have $u^{ \prime }(\sigma _{f}(\frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}))\sigma _{f}(\frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}) \le u^{ \prime }(\sigma _{f}(a ,s_{i} ,R_{2}))\sigma _{f}(a ,s_{i} ,R_{2})$. Using Lemma 4 and the fact that $\sigma _{f}$ is concave in a yields

$$\begin{aligned} \frac{R_{1}}{\sigma _{f}\left( \frac{R_{1}}{R_{2}}a ,s_{i} ,R_{2}\right) } \le \frac{R_{2}}{\sigma _{f}\left( \frac{R_{2}}{R_{2}}a ,s_{i} ,R_{2}\right) } . \end{aligned}$$

Define $f_{n} =T^{n}f : =T(T^{n -1}f)$ for $n =1 ,2 , \ldots$ where $T^{0}f : =f$. I conclude that $f_{n}^{ \prime }(a ,s ,R)$ is increasing in R for every n. From the Banach fixed-point theorem $f_{n} \rightarrow V$. The envelope condition implies that $f_{n}^{ \prime } =u^{ \prime }\left( \sigma _{f_{n}}\right)$ for every n. Theorem 3.8 and Theorem 9.9 in Stokey and Lucas (1989) show that $\sigma _{f_{n}} \rightarrow \sigma$. Thus, $f_{n}^{ \prime }(a ,s ,R) =Ru^{ \prime }\left( \sigma _{f_{n}}\left( a ,s ,R\right) \right) \rightarrow Ru^{ \prime }\left( \sigma \left( a ,s ,R\right) \right) =V^{ \prime }(a ,s ,R)$ which implies that $V^{ \prime }(a ,s ,R)$ is increasing in R. Inequality (3), Lemma 6, and the fact that g is increasing in a imply that $g(a ,s ,R_{2}) \ge g(\frac{R_{1}}{R_{2}}a ,s ,R_{2}) \ge g(a ,s ,R_{1})$ - which proves the Theorem. $\square$

Theorem 2

Assume that $\gamma \le 1$. Then the aggregate savings increase with the interest rate for every fixed $w >0$. That is, $R_{2} \ge R_{1}$ implies $AS(R_{2} ,w) \ge AS(R_{1} ,w)$.

Proof

For two distributions $\lambda _{2} ,\lambda _{1} \in \mathcal {P}(Z)$ write $\lambda _{2} \succeq _{I_{a}}\lambda _{1}$ if for any bounded and continuous real-valued function f(a, s) that is increasing in a we have²⁰

$$\begin{aligned} \int _{Z}f(a ,s)\lambda _{2}^{\,}(d(a ,s)) \ge \int _{Z}f(a ,s)\lambda _{1}^{\,}(d(a ,s)). \end{aligned}$$

Let $R_{2} \ge R_{1} >0$ and fix $w >0$. I claim that if $\lambda ( \cdot ;R_{2}) \succeq _{I_{a}}\lambda ( \cdot ;R_{1})$ then $M\lambda ( \cdot ;R_{2}) \succeq _{I_{a}}M\lambda ( \cdot ;R_{1})$.

Assume that $\lambda ( \cdot ;R_{1}) ,\lambda ( \cdot ;R_{2}) \in \mathcal {P}(Z)$ and $\lambda ( \cdot ;R_{2}) \succeq _{I_{a}}\lambda ( \cdot ;R_{1})$. Let $f :Z \rightarrow \mathbb {R}$ be a bounded and continuous function. First note that

$$\begin{aligned} \int _{Z}f(a ,s)M\lambda (d(a ,s);R) =\int _{Z}\left[ \sum \limits _{j =1}^{n}P_{sj}f(g(a ,s ,R)) ,s_{j})\right] \lambda (d(a ,s);R). \end{aligned}$$

(4)

To see that the last equality holds, first let $f =1_{D \times B}$ for $D \times B \in \mathcal {B}(A) \times 2^{S}$ where $1_{D \times B}$ is the indicator function of $D \times B$. We have

$$\begin{aligned} \int _{Z}1_{D \times B}M\lambda (d(a ,s);R)&=M\lambda (D \times B;R) \\&=\int _{Z}P(s ,B)1_{D}(g(a ,s ,R))\lambda (d(a ,s);R) \\&=\int _{Z}[1_{D}(g(a ,s ,R))\sum \limits _{j =1}^{n}P_{sj}1_{B}(s_{j})]\lambda (d(a ,s);R) \\&=\int _{Z}\left[ \sum _{j =1}^{n}P_{sj}1_{D \times B}(g(a ,s ,R)) ,s_{j})\right] \lambda (d(a ,s);R) . \end{aligned}$$

A standard argument shows that (4) holds for a general f. Now assume that f is increasing in a. Then

$$\begin{aligned} \int _{Z}f(a ,s)M\lambda (d(a ,s);R_{2})&=\int _{Z}\left[ \sum \limits _{j =1}^{n}P_{sj}f(g(a ,s ,R_{2}) ,s_{j})\right] \lambda (d(a ,s);R_{2}) \\&\ge \int _{Z}\left[ \sum \limits _{j =1}^{n}P_{sj}f(g(a ,s ,R_{2}) ,s_{j})\right] \lambda (d(a ,s);R_{1}) \\&\ge \int _{Z}\left[ \sum \limits _{j =1}^{n}P_{sj}f(g(a ,s ,R_{1}) ,s_{j})\right] \lambda (d(a ,s);R_{1}) \\&=\int _{Z}f(a ,s)M\lambda (d(a ,s);R_{1}) . \end{aligned}$$

The first inequality follows from the facts that $\sum \limits _{j =1}^{n}P_{sj}f(g(a ,s ,R_{2})) ,s_{j})$ is increasing in a and $\lambda ( \cdot ;R_{2}) \succeq _{I_{a}}\lambda ( \cdot ;R_{1})$. The second inequality follows from the facts that $g(a ,s ,R_{2}) \ge g(a ,s ,R_{1})$ for all $(a ,s) \in Z$ and f is increasing in the first argument.

I conclude that $M^{t}\lambda ( \cdot ;R_{2}) \succeq _{I_{a}}M^{t}\lambda ( \cdot ;R_{1})$ for all $t =1 ,2 ,3 ,\cdots$. Since $\succeq _{I_{a}}$ is a closed order and $M^{t}\lambda ( \cdot ;R)$ converges to $\mu ( \cdot ;R)$ we have $\mu ( \cdot ;R_{2}) \succeq _{I_{a}}\mu ( \cdot ;R_{1})$. Thus,

$$\begin{aligned} AS(R_{2} ,w) =\int _{Z}a\mu (d(a ,s);R_{2}) \ge \int _{Z}a\mu (d(a ,s);R_{1}) =AS(R_{1} ,w) , \end{aligned}$$

which proves that the aggregate savings increase in the interest rate. $\square$

Lemma 3

Suppose that the elasticity of substitution ES(K, N) is bounded below by 1. Then the function K(R) / w(R) is strictly decreasing.

Proof

Step 1. If $F_{KN}(K ,1)K <F_{N}(K ,1)$ for all $K \ge 0$, where $F_{KN}$ is the mixed partial derivative, then K(R) / w(R) is strictly decreasing.

Let $M(R) =F_{K}^{ -1}(R +\delta -1 ,1)$. We have

$$\begin{aligned}\frac{ \partial }{ \partial R}\frac{K(R)}{w(R)} =\frac{ \partial }{ \partial R}\frac{M(R)}{F_{N}(M(R) ,1)} =\frac{M^{ \prime }(R)F_{N}(M(R) ,1) -M^{ \prime }(R)F_{KN}(M(R) ,1)M(R)}{(F_{N}(M(R),1)^{2}} . \end{aligned}$$

Since F is twice differentiable and strictly concave in K, M(R) is strictly decreasing and differentiable. Thus, $M^{\prime }(R)<0$. I conclude that $F_{N}(M(R) ,1) >F_{KN}(M(R) ,1)M(R)$ implies $\frac{ \partial }{ \partial R}\frac{K(R)}{w(R)} <0$ which proves step 1.

Step 2. If $ES(K ,1) \ge 1$ then $F_{KN}(K ,1)K <F_{N}(K ,1)$. Note that

$$\begin{aligned} \frac{1}{ES(K ,1)} = -\frac{ \partial \ln (F_{K}/_{}F_{N})}{ \partial \ln (K)} = -K\left( \frac{F_{KK}(K ,1)}{F_{K}(K ,1)} -\frac{F_{KN}(K ,1)}{F_{N}(K ,1)}\right) . \end{aligned}$$

The last equality and the facts that $F_{K}(K ,1) >0$, $F_{KK}(K ,1) <0$, and $ES(K ,1) \ge 1$ imply

$$\begin{aligned} \frac{F_{KN}(K ,1)}{F_{N}(K ,1)} =\frac{F_{KK}(K ,1)}{F_{K}(K ,1)} +\frac{1}{ES(K ,1)K} <\frac{1}{K}. \end{aligned}$$

That is, $F_{KN}(K ,1)K <F_{N}(K ,1)$ which proves step 2. I conclude that the function K(R) / w(R) is strictly decreasing. $\square$

Theorem 4

Assume that $\gamma \le 1$ and $ES(K ,N) \ge 1$. (i) If the discount factor increases then the unique equilibrium interest rate $R^{ *}$ decreases. (ii) If the risk of the invariant distribution of labor productivities increases (in the sense of second-order stochastic dominance) then the unique equilibrium interest rate $R^{ *}$ decreases. (iii) Suppose that the production function is given by the Cobb–Douglas production function $F(K ,N) ={\overline{T}}K^{\alpha }N^{1 -\alpha }$ for some $0<\alpha <1$ and ${\overline{T}} >0$. If the depreciation rate $\delta$ increases or the output elasticity $\alpha$ decreases then the unique equilibrium interest rate $R^{ *}$ decreases.

Proof

Consider the function $v(R;e) =AS(R ,1;e) -\frac{K(R;e)}{w(R;e)}$ where e is some parameter (the discount factor, the invariant distribution of labor productivities, the depreciation rate, or the output elasticity). $R^{ *}$ is an equilibrium price if $v(R^{ *};e) =0$. Theorem 3 implies that v is strictly increasing in R. It is easy to see that if v is increasing in e then the unique equilibrium $R^{ *}$ is decreasing in e. From Propositions 1 and 2 in Acemoglu and Jensen (2015), the aggregate supply of savings is increasing in the discount factor and in the risk of the invariant distribution of labor productivities, so parts (i) and (ii) follow. Furthermore, in the case of the Cobb–Douglas production function, the first-order conditions yield

$$\begin{aligned}K(R) =\genfrac[]{}{}{{\overline{T}}\alpha }{R +\delta -1}^{\frac{1}{1 -\alpha }} ,\ \ \ w(R) =(1 -\alpha )T\genfrac[]{}{}{{\overline{T}}\alpha }{R +\delta -1}^{\frac{\alpha }{1 -\alpha }} . \end{aligned}$$

It is easy to check that k(R) / w(R) is decreasing in the depreciation rate and increasing in the output elasticity, so part (iii) follows. $\square$