Exponential discounting bias

Abstract

We address intertemporal utility maximization under a general discount function that nests the exponential discounting and the quasi-hyperbolic discounting cases as particular specifications. Under the suggested framework, the representative agent adopts, at some initial date, an optimal behavior that shapes her consumption trajectory over time. This agent desires to take a constant discount rate to approach the optimization problem, but bounded rationality, under the form of a present bias, deviates the individual from the intended goal. As a result, decreasing impatience will end up dominating the agent’s behavior. The individual will not be aware of her own time inconsistency and, therefore, she will not revise her plans as time elapses, what makes the problem relatively simple to address from a computational point of view. The general discounting framework is used to approach a standard optimal growth model in discrete time. Transitional dynamics and stability properties of the corresponding dynamic setup are studied. An extension of the standard utility maximization model to the case of habit persistence is also considered.

Appendices

Appendix A: Computation of equation (6)

To derive Eq. (6), observe that (4) and (5) might be presented, respectively, as

$$\begin{aligned} U_{s}(c)=u(c_{s})+D(s+1,s)u(c_{s+1})+D(s+2,s)u(c_{s+2})+\cdots \end{aligned}$$

(18)

and

$$\begin{aligned} U_{s+1}(c)=u(c_{s+1})+D(s+2,s+1)u(c_{s+2})+D(s+3,s+1)u(c_{s+3})+\cdots \nonumber \\ \end{aligned}$$

(19)

Under the assumption of exponential discounting bias, (18) and (19) come

$$\begin{aligned} U_{s}(c)=u(c_{s})+\beta ^{1-\theta _{0}+\theta _{1}}u(c_{s+1})+\beta ^{2(1-\theta _{0})+\theta _{1}}u(c_{s+2})+\cdots \end{aligned}$$

(20)

and

$$\begin{aligned} U_{s+1}(c)=u(c_{s+1})+\beta ^{1-\theta _{0}+\theta _{1}}u(c_{s+2})+\beta ^{2(1-\theta _{0})+\theta _{1}}u(c_{s+3})+\cdots \end{aligned}$$

(21)

Now, we rearrange the terms in Eq. (20) in order to display an equivalent expression,

$$\begin{aligned} U_{s}(c)&= u(c_{s}) \nonumber \\&+\beta ^{1-\theta _{0}}\left[ u(c_{s+1})+\beta ^{1-\theta _{0}+\theta _{1}}u(c_{s+2})+\beta ^{2(1-\theta _{0})+\theta _{1}}u(c_{s+3})\right. \nonumber \\&\left. +\cdots -(1-\beta ^{\theta _{1}})u(c_{s+1})\right] \end{aligned}$$

(22)

Replacing $U_{s+1}(c)$ as defined in (21) into (22), we finally arrive to

$$\begin{aligned} U_{s}(c)=u(c_{s})+\beta ^{1-\theta _{0}}\left[ U_{s+1}(c)-(1-\beta ^{\theta _{1}})u(c_{s+1})\right] \end{aligned}$$

(23)

Equation (23) is identical to Eq. (6). In the main text, we replace $s$ by $t$ in order to make clear that the function holds for any two consecutive periods. After displaying this equation, all the subsequent analysis uses $t$ as the time subscript.

Appendix B: Solution of the optimization problem

In order to maximize intertemporal utility subject to the budget constraint, we resort to a conventional dynamic programming procedure.⁷ This requires recovering Eq. (6) and defining function $V(a_{t})$, such that,

$$\begin{aligned} V(a_{t})=\underset{c}{\max }\left\{ u(c_{t})+\beta ^{1-\theta _{0}}\left[ V(a_{t+1})-(1-\beta ^{\theta _{1}})u(c_{t+1})\right] \right\} \end{aligned}$$

(24)

Note that, in (24), $V(a_{t+1})$ is a function of $a_{t}$ and $c_{t}$, given the budget constraint.

The corresponding first order conditions are

$$\begin{aligned}&\frac{\partial V(a_{t})}{\partial c_{t}}=0\Rightarrow \nonumber \\&\quad u^{\prime }(c_{t})+\beta ^{1-\theta _{0}}\left[ V^{\prime }(a_{t+1})\frac{ \partial a_{t+1}}{\partial c_{t}}-(1-\beta ^{\theta _{1}})u^{\prime }(c_{t+1})\frac{dc_{t+1}}{dc_{t}}\right] =0 \end{aligned}$$

(25)

and

$$\begin{aligned} \frac{\partial V(a_{t})}{\partial a_{t}}&= \beta ^{1-\theta _{0}}\frac{ \partial V(a_{t+1})}{\partial a_{t}}\Rightarrow \nonumber \\ V^{\prime }(a_{t})&= \beta ^{1-\theta _{0}}V^{\prime }(a_{t+1})\frac{ \partial a_{t+1}}{\partial a_{t}} \end{aligned}$$

(26)

The transversality condition of the problem is given by

$$\begin{aligned} \underset{t\rightarrow \infty }{\lim }a_{t}D_{EB}(t,s)V(a_{t})=0 \end{aligned}$$

(27)

To proceed with the analysis, one must compute the derivatives that are present in the expressions of the first-order conditions. Note that $\frac{\partial a_{t+1}}{\partial c_{t}}=-1$ and $\frac{\partial a_{t+1}}{\partial a_{t}}=1+r$; these values are straightforward to obtain from the resource constraint (7). Relatively to the derivative $\frac{dc_{t+1}}{dc_{t}}$, that appears in condition (25), it indicates the expected change on the value of consumption from date $t$ to date $t+1$. If we recall that our central assumption is that the representative agent intends to discount the future at a constant rate, although she is unable to follow her plan, the expected change in consumption will be the effective change that would be obtained under constant discounting. Constant discounting implies a well known rule of motion for consumption, $c_{t+1}=\beta (1+r)c_{t}$, which holds for every value of $t$. Thus, the expected change in consumption to be used in (25) is $\frac{dc_{t+1}}{dc_{t}}=$ $\beta (1+r)$.

The previous reasoning allows to rewrite the first-order conditions as follows,

$$\begin{aligned} \beta ^{1-\theta _{0}}V^{\prime }(a_{t+1})=u^{\prime }(c_{t})-\beta ^{2-\theta _{0}}(1-\beta ^{\theta _{1}})(1+r)u^{\prime }(c_{t+1}) \end{aligned}$$

(28)

and

$$\begin{aligned} V^{\prime }(a_{t})=\beta ^{1-\theta _{0}}(1+r)V^{\prime }(a_{t+1}) \end{aligned}$$

(29)

Combining the two optimality conditions, (28) and (29), one obtains an equation of motion for consumption. Observe that (28) is equivalent to

$$\begin{aligned} V^{\prime }(a_{t+1})=\frac{1}{\beta ^{1-\theta _{0}}}u^{\prime }(c_{t})-\beta (1-\beta ^{\theta _{1}})(1+r)u^{\prime }(c_{t+1}) \end{aligned}$$

(30)

or, one period earlier,

$$\begin{aligned} V^{\prime }(a_{t})=\frac{1}{\beta ^{1-\theta _{0}}}u^{\prime }(c_{t-1})-\beta (1-\beta ^{\theta _{1}})(1+r)u^{\prime }(c_{t}) \end{aligned}$$

(31)

Replacing (30) and (31) into (29) and taking a logarithmic utility function, one easily arrives to Eq. (8) in the main text.

An alternative way to approach the optimization problem, that is similar to the previous one, consists in solving it through the Pontryagin’s principle. This requires writing the corresponding current-value Hamiltonian function, which takes the form

$$\begin{aligned} H(a_{t},c_{t},p_{t})=u(c_{t})+\beta ^{1-\theta _{0}}\left[ p_{t+1}(w+ra_{t}-c_{t})-(1-\beta ^{\theta _{1}})u(c_{t+1})\right] \end{aligned}$$

(32)

In function (32), $p_{t}$ represents the shadow-price of the state variable $a_{t}$, also known as the problem’s costate variable. First-order conditions are, in this case,

$$\begin{aligned} \frac{\partial H}{\partial c_{t}}&= 0\Rightarrow \nonumber \\ \beta ^{1-\theta _{0}}p_{t+1}&= u^{\prime }(c_{t})-\beta ^{2-\theta _{0}}(1-\beta ^{\theta _{1}})(1+r)u^{\prime }(c_{t+1})\end{aligned}$$

(33)

$$\begin{aligned} \beta ^{1-\theta _{0}}p_{t+1}&-&p_{t}=-\frac{\partial H}{\partial a_{t}} \Rightarrow p_{t}=\beta ^{1-\theta _{0}}(1+r)p_{t+1} \end{aligned}$$

(34)

The transversality condition holds, as presented in (27). Combining ( 33) and (34) and, again, taking the logarithmic utility function, we arrive to the same difference equation for consumption, (8). The two methods had to deliver the same outcome, what becomes clear once we note that there is a correspondence between $V^{\prime }(a_{t})$ in the dynamic programming problem and the shadow-price $p_{t}$ used to write the Hamiltonian function.

Appendix C: Proofs of Propositions

1.1 Proof of Proposition 1

To prove this proposition, one just needs to solve (9), for $\psi ^{*}:=\psi _{t+1}=\psi _{t}$. A quadratic equation is generated and, as a result, two solutions exist. The equation is

$$\begin{aligned} \left( \psi ^{*}\right) ^{2}-\beta ^{1-\theta _{0}}\left[ 1+\beta (1-\beta ^{\theta _{1}})\right] (1+r)\psi ^{*}+\beta ^{3-2\theta _{0}}(1-\beta ^{\theta _{1}})(1+r)^{2}=0\nonumber \\ \end{aligned}$$

(35)

Simple algebra allows to arrive to the pair of solutions of (35) that are given in the proposition.

1.2 Proof of Proposition 2

The steady-state is defined as the pair of values $(k^{*},c^{*})$ such that $k_{t+1}=k_{t}$ and $c_{t+1}=c_{t}=c_{t-1}$. Applying these conditions to the equations in system (12), one calculates the values in the proposition.

The evaluation of the consumption’s equation in the steady-state yields the following quadratic equation that can be solved with respect to the term $\eta A\left( k^{*}\right) ^{-(1-\eta )}$,

$$\begin{aligned} \beta ^{3-2\theta _{0}}(1\!-\!\beta ^{\theta _{1}})\left[ \eta A\left( k^{*}\right) ^{\!-\!(1\!-\!\eta )}\right] ^{2}\!-\!\beta ^{1\!-\!\theta _{0}}\left[ 1\!+\!\beta (1-\beta ^{\theta _{1}})\right] \eta A\left( k^{*}\right) ^{-(1-\eta )}+1=0\nonumber \\ \end{aligned}$$

(36)

The solution of (36) is

$$\begin{aligned} \eta A\left( k^{*}\right) ^{-(1-\eta )}=\frac{1}{\beta ^{1-\theta _{0}}} \vee \eta A\left( k^{*}\right) ^{-(1-\eta )}=\frac{1}{\beta ^{2-\theta _{0}}(1-\beta ^{\theta _{1}})} \end{aligned}$$

(37)

From the above expressions, one withdraws the steady-state values of capital, $k_{1}^{*}$ and $k_{2}^{*}$ that are presented in the proposition. To obtain the steady-state values of consumption, one just needs to replace $k_{1}^{*}$ and $k_{2}^{*}$ in expression $c^{*}=A\left( k^{*}\right) ^{\eta }-k^{*}$; this expression follows directly from constraint (10).

1.3 Proof of Proposition 3

The existence of one stable dimension implies that one of the eigenvalues of the Jacobian matrix locates inside the unit circle, while the other two fall outside the unit circle. The eigenvalues are straightforward to obtain once we solve the corresponding characteristic equation; they are $\lambda _{1}=\eta$, $\lambda _{2}=\frac{1}{\beta (1-\beta ^{\theta _{1}})}$, $\lambda _{3}=\frac{1}{\beta ^{1-\theta _{0}}\eta }$. Observing that $\lambda _{1}\in (0,1)$, $\lambda _{2},\lambda _{3}>1$, one confirms the statement in the proposition.

1.4 Proof of Proposition 4

The saddle-path stable trajectory can be obtained by computing the eigenvector associated to the eigenvalue inside the unit circle. This requires solving the system

$$\begin{aligned} \left[ \begin{array}{c@{\quad }c@{\quad }c} \chi _{1}-\lambda _{1} &{} -1 &{} 0 \\ (\chi _{2}-\chi _{1})\chi _{3} &{} 1+\chi _{2}+\chi _{3}-\lambda _{1} &{} -\chi _{2} \\ 0 &{} 1 &{} -\lambda _{1} \end{array} \right] \cdot \left[ \begin{array}{l} p_{1} \\ p_{2} \\ p_{3} \end{array} \right] =\left[ \begin{array}{l} 0 \\ 0 \\ 0 \end{array} \right] \end{aligned}$$

(38)

Letting $p_{1}=1$, the eigenvector might be written as

$$\begin{aligned} P=\left[ \begin{array}{c} 1 \\ \chi _{1}-\lambda _{1} \\ \chi _{1}/\lambda _{1}-1 \end{array} \right] \end{aligned}$$

(39)

or, equivalently,

$$\begin{aligned} P=\left[ \begin{array}{c} 1 \\ 1/\beta ^{1-\theta _{0}}-\eta \\ 1/(\eta \beta ^{1-\theta _{0}})-1 \end{array} \right] \end{aligned}$$

(40)

The slope of the contemporaneous relation between consumption and capital is given by the ratio $p_{2}/p_{1}$, i.e., $c_{t}-c_{1}^{*}=\left( p_{2}/p_{1}\right) (k_{t}-k_{1}^{*})$, which corresponds to expression

$$\begin{aligned} c_{t}-c_{1}^{*}=\left( 1/\beta ^{1-\theta _{0}}-\eta \right) (k_{t}-k_{1}^{*}) \end{aligned}$$

(41)

Taking into consideration the steady-state for which the analysis is being undertaken, (41) is equivalent to the equation in the proposition.

1.5 Proof of Proposition 5

By taking $\psi ^{*}:=\psi _{t+1}=\psi _{t}=\psi _{t-1}$, one transforms (16) into the quadratic equation $\left( \psi ^{*}\right) ^{2}-\ell _{2}\psi ^{*}+\ell _{1}=0$, with $\ell _{1}:=\beta ^{2(1-\theta _{0})}(1-\beta ^{\theta _{1}})(1+r)\left[ b+\beta (1+r)\right]$ and $\ell _{2}:=\beta ^{1-\theta _{0}}\left\{ (1+r)+(1-\beta ^{\theta _{1}}) \left[ b+\beta (1+r)\right] \right\}$. The solutions of this equation are the steady-state values $\psi _{1}^{*}=\beta ^{1-\theta _{0}}(1+r)$ $\vee$ $\psi _{2}^{*}=\beta ^{1-\theta _{0}}(1-\beta ^{\theta _{1}})[b+\beta (1+r)]$.

Appendix D: Computation of equation (11)

Begin by defining function

$$\begin{aligned} V(k_{t})=\underset{c}{\max }\left\{ u(c_{t})+\beta ^{1-\theta _{0}}\left[ V(k_{t+1})-(1-\beta ^{\theta _{1}})u(c_{t+1})\right] \right\} \end{aligned}$$

(42)

First-order conditions are,

$$\begin{aligned}&\frac{\partial V(k_{t})}{\partial c_{t}}=0\Rightarrow \nonumber \\&\quad u^{\prime }(c_{t})+\beta ^{1-\theta _{0}}\left[ V^{\prime }(k_{t+1})\frac{ \partial k_{t+1}}{\partial c_{t}}-(1-\beta ^{\theta _{1}})u^{\prime }(c_{t+1})\frac{dc_{t+1}}{dc_{t}}\right] =0 \end{aligned}$$

(43)

and

$$\begin{aligned} \frac{\partial V(k_{t})}{\partial k_{t}}=\beta ^{1-\theta _{0}}\frac{ \partial V(k_{t+1})}{\partial k_{t}}\Rightarrow V^{\prime }(k_{t})=\beta ^{1-\theta _{0}}V^{\prime }(k_{t+1})\frac{\partial k_{t+1}}{\partial k_{t}} \end{aligned}$$

(44)

The transversality condition also applies. As in section 4, the derivative $\frac{dc_{t+1}}{dc_{t}}$ represents the expected change on consumption, under constant discounting. For the constraint assumed in the current case,

$$\begin{aligned} \frac{dc_{t+1}}{dc_{t}}=\beta \eta Ak_{t+1}^{-(1-\eta )} \end{aligned}$$

(45)

Replacing (45) into (44), one obtains relation

$$\begin{aligned} V^{\prime }(k_{t+1})=\frac{1}{\beta ^{1-\theta _{0}}}u^{\prime }(c_{t})-\beta (1-\beta ^{\theta _{1}})\eta Ak_{t+1}^{-(1-\eta )}u^{\prime }(c_{t+1}) \end{aligned}$$

(46)

or, one period earlier,

$$\begin{aligned} V^{\prime }(k_{t})=\frac{1}{\beta ^{1-\theta _{0}}}u^{\prime }(c_{t-1})-\beta (1-\beta ^{\theta _{1}})\eta Ak_{t}^{-(1-\eta )}u^{\prime }(c_{t}) \end{aligned}$$

(47)

The second optimal condition is, in the current model’s specification,

$$\begin{aligned} V^{\prime }(k_{t})=\beta ^{1-\theta _{0}}\eta Ak_{t}^{-(1-\eta )}V^{\prime }(k_{t+1}) \end{aligned}$$

(48)

Now, we replace (46) and (47) into (48); as a result, the equation of motion for consumption takes the form

$$\begin{aligned} u^{\prime }(c_{t+1})=\frac{\beta ^{1-\theta _{0}}\left[ 1+\beta (1-\beta ^{\theta _{1}})\right] \eta Ak_{t}^{-(1-\eta )}u^{\prime }(c_{t})-u^{\prime }(c_{t-1})}{\beta ^{3-2\theta _{0}}(1-\beta ^{\theta _{1}})(\eta A)^{2}k_{t}^{-(1-\eta )}k_{t+1}^{-(1-\eta )}} \end{aligned}$$

(49)

Taking utility function $u(c)=\ln c$, it is straightforward to realize that Eq. (49) is equivalent to Eq. (11) in the main text.

Appendix E: Computation of equation (15)

Let us start by analyzing the problem under constant discounting. Define $V(a_{t})$ in the current circumstance as

$$\begin{aligned} V(a_{t})=\underset{c}{\max }\left[ u(c_{t},c_{t-1})+\beta V(a_{t+1})\right] \end{aligned}$$

(50)

First-order conditions are:

$$\begin{aligned} u^{\prime }(c_{t},c_{t-1})+\beta V^{\prime }(a_{t+1})\frac{\partial a_{t+1}}{ \partial c_{t}}\Rightarrow \frac{1}{c_{t}-bc_{t-1}}=\beta V^{\prime }(a_{t+1}) \end{aligned}$$

(51)

and

$$\begin{aligned} V^{\prime }(a_{t})=\beta V^{\prime }(a_{t+1})\frac{\partial a_{t+1}}{ \partial a_{t}}\Rightarrow V^{\prime }(a_{t})=\beta (1+r)V^{\prime }(a_{t+1}) \end{aligned}$$

(52)

The transversality condition, as presented in previous occasions, holds as well. Combining (51) and (52), and advancing one time period, the difference equation for consumption comes

$$\begin{aligned} c_{t+1}=\left[ b+\beta (1+r)\right] c_{t}-b\beta (1+r)c_{t-1} \end{aligned}$$

(53)

Under the exponential discounting bias,

$$\begin{aligned} V(a_{t})=\underset{c}{\max }\left\{ u(c_{t},c_{t-1})+\beta ^{1-\theta _{0}} \left[ V(a_{t+1})-(1-\beta ^{\theta _{1}})u(c_{t+1},c_{t})\right] \right\} \end{aligned}$$

(54)

The optimality conditions required for finding the consumption dynamic equation are, in the current case,

$$\begin{aligned}&u^{\prime }(c_{t},c_{t-1})+\beta ^{1-\theta _{0}}\left[ V^{\prime }(a_{t+1}) \frac{\partial a_{t+1}}{\partial c_{t}}-(1-\beta ^{\theta _{1}})u^{\prime }(c_{t+1},c_{t})\frac{dc_{t+1}}{dc_{t}}\right] =0\Rightarrow \nonumber \\&V^{\prime }(a_{t+1})=\frac{1}{\beta ^{1-\theta _{0}}}\frac{1}{c_{t}-bc_{t-1}} -(1-\beta ^{\theta _{1}})\left[ b+\beta (1+r)\right] \frac{1}{c_{t+1}-bc_{t}} \end{aligned}$$

(55)

and

$$\begin{aligned} V^{\prime }(a_{t})=\beta V^{\prime }(a_{t+1})\frac{\partial a_{t+1}}{\partial a_{t}}\Rightarrow V^{\prime }(a_{t})=\beta ^{1-\theta _{0}}(1+r)V^{\prime }(a_{t+1}) \end{aligned}$$

(56)

Note that, in (55), $\frac{dc_{t+1}}{dc_{t}}=b+\beta (1+r)$ given the expression for the evolution of consumption in (53). Taking together (55) and (56) and proceeding as in the previous settings, one obtains the consumption difference equation in the proposition.

Appendix F: stability of the steady-state points in the Model with habit persistence

The study of stability requires, in this case, addressing the local properties of system

$$\begin{aligned} \left\{ \begin{array}{l} \psi _{t+1}=b+\frac{\psi _{1}^{*}\psi _{2}^{*}}{\frac{\psi _{1}^{*}+\psi _{2}^{*}}{1-b/\psi _{t}}-\frac{\psi _{t}}{1-b/\phi _{t} }} \\ \phi _{t+1}=\psi _{t} \end{array} \right. \end{aligned}$$

(57)

Values $\psi _{1}^{*}$ and $\psi _{2}^{*}$ are the steady-state values, as presented in section 5; variable $\phi _{t}$ is the lag variable associated with $\psi _{t}$. The linearization of the equations in (57), with respect to each steady-state value yields:

• For $\psi _{1}^{*}$:

  $$\begin{aligned} \left[ \begin{array}{l} \psi _{t+1}-\psi _{1}^{*} \\ \phi _{t+1}-\psi _{1}^{*} \end{array}\right] =\left[ \begin{array}{ll} \frac{b}{\psi _{1}^{*}}+\frac{\psi _{1}^{*}}{\psi _{2}^{*}} &{} - \frac{b}{\psi _{2}^{*}} \\ 1 &{} 0 \end{array} \right] .\left[ \begin{array}{l} \psi _{t}-\psi _{1}^{*} \\ \phi _{t}-\psi _{1}^{*} \end{array} \right] \end{aligned}$$

  (58)

• For $\psi _{2}^{*}$:

  $$\begin{aligned} \left[ \begin{array}{l} \psi _{t+1}-\psi _{2}^{*} \\ \phi _{t+1}-\psi _{2}^{*} \end{array} \right] =\left[ \begin{array}{ll} \frac{b}{\psi _{2}^{*}}+\frac{\psi _{2}^{*}}{\psi _{1}^{*}} &{} - \frac{b}{\psi _{1}^{*}} \\ 1 &{} 0 \end{array} \right] .\left[ \begin{array}{l} \psi _{t}-\psi _{2}^{*} \\ \phi _{t}-\psi _{2}^{*} \end{array} \right] \end{aligned}$$

  (59)

The eigenvalues of the Jacobian matrices for the cases $\psi _{1}^{*}$ and $\psi _{2}^{*}$ are, respectively, $\left. \lambda _{1}\right| _{\psi _{1}^{*}}=\frac{b}{\psi _{1}^{*}}$; $\left. \lambda _{2}\right| _{\psi _{1}^{*}}=\frac{\psi _{1}^{*}}{\psi _{2}^{*}}$; $\left. \lambda _{1}\right| _{\psi _{2}^{*}}=\frac{b }{\psi _{2}^{*}}$; $\left. \lambda _{2}\right| _{\psi _{2}^{*}}= \frac{\psi _{2}^{*}}{\psi _{1}^{*}}$. The presented eigenvalues are all positive.

Equilibrium point $\psi _{1}^{*}$ is stable if:

• $\left. \lambda _{1}\right| _{\psi _{1}^{*}}<1\Rightarrow b<\beta ^{1-\theta _{0}}(1+r)$ and

• $\left. \lambda _{2}\right| _{\psi _{1}^{*}}<1\Rightarrow b> \frac{1-\beta (1-\beta ^{\theta _{1}})}{1-\beta ^{\theta _{1}}}(1+r)$.

Equilibrium point $\psi _{2}^{*}$ is stable if:

• $\left. \lambda _{1}\right| _{\psi _{2}^{*}}<1\Rightarrow b< \frac{\beta (1+r)}{1-\beta ^{1-\theta _{0}}(1-\beta ^{\theta _{1}})}$ and

• $\left. \lambda _{2}\right| _{\psi _{2}^{*}}<1\Rightarrow b< \frac{1-\beta (1-\beta ^{\theta _{1}})}{1-\beta ^{\theta _{1}}}(1+r)$.

Saddle-path stability holds when only one of the eigenvalues, for each of the above pairs, falls inside the unit circle and instability prevails when both eigenvalues of the Jacobian matrix fall outside the unit circle. Results point for a crucial role of the habit persistence parameter $b$; it is the value of this parameter that will determine the stability profile.

Note that $\left. \lambda _{2}\right| _{\psi _{1}^{*}}=1/\left. \lambda _{2}\right| _{\psi _{2}^{*}}$. Observe, as well, that the condition allowing for $\left. \lambda _{2}\right| _{\psi _{2}^{*}}<1$ is identical to (17), i.e., it is the one that will determine which of the two points $\psi _{1}^{*}$ and $\psi _{2}^{*}$ is preferable from an utility maximization point of view. As a result, the desired steady-state will have one of its dimensions necessarily unstable, while the other steady-state is such that the same dimension is stable. We reemphasize, though, that this instability of the steady-state the agent wants to achieve is not an obstacle for the success of the agent’s optimal strategy because consumption is a control variable.

Recover the example that has taken the following array of parameters: $\beta =0.97$, $r=0.05$, $\theta _{0}=0.95$ and $\theta _{1}=50$. In this case, independently of the value of $b\in [0,1),\,\left. \lambda _{1}\right| _{\psi _{1}^{*}}<1$ and $\left. \lambda _{1}\right| _{\psi _{2}^{*}}<1$. Relatively to the other eigenvalue, it is true that:

• If $b<0.3243$, then $\left. \lambda _{1}\right| _{\psi _{1}^{*}}>1$ and $\left. \lambda _{1}\right| _{\psi _{2}^{*}}<1$;

• If $b>0.3243$, then $\left. \lambda _{1}\right| _{\psi _{1}^{*}}<1$ and $\left. \lambda _{1}\right| _{\psi _{2}^{*}}>1$;

In this case, a low level of habit persistence implies that $\psi _{1}^{*}$ is saddle-path stable and $\psi _{2}^{*}$ is a stable equilibrium point. This result is reversed once the habit persistence parameter acquires a value larger than $0.3243$. Condition $b<0.3243$ also implies $\psi _{1}^{*}>\psi _{2}^{*}$, meaning that the unstable dimension is associated with the steady-state point that the household prefers to achieve.