5. Local Stability of Optimal Paths and Speed of Convergence

In this section, we will consider the stability and speed of convergence in the small neighborhood around the steady state. We are interested about this because in the long run, we are at the steady state. But when there are some shocks happening, we might be slightly moved off the steady state. So this topic is very useful in determining the behavior of the economy in that situation. To grasp the essence of this problem in a simpler way, we will focus on one dimension cases. Since we only care about a small neighborhood, so the linear approximation using Taylor's expansion will be frequently used in this section.

Note that the optimal path is not the same as the saddle path. The optimal path is a sequence of state variable over time that generates maximum overall returns, while the saddle path is a line in the \((k,c)\) space or \((k,\lambda)\) space that involves not only the state variable, but also the control or the co-state variable. First, we will focus on the optimal path for only the state variable and discuss the local stability and speed of convergence in discrete time and then in continuous time. The slope of the saddle path will be discussed in a specific set-up of the Neoclassical Growth Model in section 9.4.

5.1 Discrete Time Local Stability of Optimal Paths and Speed of Convergence

5.1.1 Linearization of the optimal path around the steady state

Denote \(x^\star\) as the steady state of state variable. Suppose we have obtained \(x_{t+1}=g(x_t)\) as the optimal decision rule, which gives us the optimal path of the state variable. To consider the small neighborhood around the steady state, we can do the first-order Taylor approximation around \(x^\star\): $$x_{t+1}=g(x_t)\approx g(x^\star)+g'(x^\star)(x_t-x^\star)$$ By the definition of steady state, \(g(x^\star)=x^\star\), so $$x_{t+1}-x^\star\approx g'(x^\star)(x_t-x^\star)$$

5.1.2 Stability and speed of convergence

Consider the following two cases: - If \(|g'(x^\star)|<1\), then \(x_{t+1}-x^\starstable around its steady state. - The smaller the absolute value of \(g'(x^\star)\), the faster it will converge. - If \(|g'(x^\star)|\ge1\), then \(x_{t+1}-x^\star\ge x_t-x^\star\). - In this case, once the system diverges from the steady state, then it either stays there forever or goes away gradually. - The system is not stable around its steady state.

5.1.3 Construct a quadratic function of \(g'(x^\star)\) using EE

Substitute the optimal decision rule \(x_{t+1}=g(x_t)\) into the EE to obtain $$F_y(x,g(x))+\beta F_x(g(x),g(g(x)))=0$$ Our goal is to write an equation for \(g'(x^\star)\), so let's take derivative of the above EE w.r.t. \(x\) and evaluate it at \(x=x^\star\): $$\begin{aligned}0=&F_{yx}(x,g(x))+F_{yy}(x,g(x))g'(x)\\&+\beta\left[F_{xx}(g(x),g(g(x)))g'(x)+F_{xy}(g(x),g(g(x)))g'(g(x))g'(x)\right]\end{aligned}$$ Since we evaluate at \(x=x^\star\), we can use the fact that \(g(x^\star)=x^\star\) to rewrite: $$\begin{aligned}0=&F_{yx}(x^\star,x^\star)+F_{yy}(x^\star,x^\star)g'(x^\star)\\&+\beta\left[F_{xx}(x^\star,x^\star)g'(x^\star)\right]+\beta\left[F_{xy}(x^\star,x^\star)(g'(x^\star))^2\right]\\\Rightarrow&\beta F_{xy}(x^\star,x^\star)(g'(x^\star))^2+(F_{yy}(x^\star,x^\star)+\beta F_{xx}(x^\star,x^\star))g'(x^\star)+F_{yx}(x^\star,x^\star)=0\end{aligned}\tag{5.1}$$ which is a quadratic function w.r.t. \(g'(x^\star)\). To simplify the notation, denote \(\lambda\) as \(g'(x^\star)\) and rewrite: $$Q(\lambda)\equiv\beta F_{xy}\lambda^2+(F_{yy}+\beta F_{xx})\lambda+F_{yx}=0\tag{5.2}$$ Since \(F_{xy}=F_{yx}\), by Vieta's formulas, \(\lambda_1\lambda_2=\frac1\beta\). Since \(\beta\in(0,1)\), if one root \(|\lambda_1|<1\), then the other root \(\lambda_2\) must be larger than one in absolute value, i.e. $$|\lambda_2|=\frac{1}{\beta|\lambda_1|}>1$$ which means there are at most one solution of \(g'(x^\star)\) with an absolute value smaller than 1. Note that it is also possible to have two roots both larger than 1.

5.1.4 Use \(g'(x^\star)\) to construct a linear path around steady state

In a small neighborhood around the steady state, we can suppose the system stays on the below linear path, which uses the linear approximation of \(g(x_t)\) by Taylor expansion and the \(g'(x^\star)\) obtained from \(Q(\lambda)=0\): $$x_{t+1}=g(x_t)=x^\star+g'(x^\star)(x_t-x^\star),\quad\text{for }\forall t\ge0\tag{5.3}$$ By construction, this sequence satisfies the EE approximately around the steady state. So, if \(|g'(x^\star)|<1\), the system will converge back to the steady state, which makes it satisfy TC and thus the path is optimal. But if \(|g'(x^\star)|\ge1\), then the sequence will not converge back to the previous steady state, and we won't know if it satisfies TC or not and thus won't know if it's optimal.

The steady state \(x^\star\) with \(|g'(x^\star)|<1\) is called stable steady state; while the steady state \(x^\star\) with \(|g'(x^\star)|\ge1\) is called unstable steady state.

5.1.5 Solve for \(g'(x^\star)\)

Recall the quadratic equation (5.2): $$Q(\lambda)\equiv\beta F_{xy}\lambda^2+(F_{yy}+\beta F_{xx})\lambda+F_{yx}=0$$ Trivially, we can solve this quadratic equation to obtain the two roots: $$\lambda_1=\frac{-\frac{F_{yy}+\beta F_{xx}}{F_{xy}}-\sqrt{\left(\frac{F_{yy}+\beta F_{xx}}{F_{xy}}\right)^2-4\beta}}{2\beta}$$ $$\lambda_2=\frac{-\frac{F_{yy}+\beta F_{xx}}{F_{xy}}+\sqrt{\left(\frac{F_{yy}+\beta F_{xx}}{F_{xy}}\right)^2-4\beta}}{2\beta}$$ If one root satisfies \(|\lambda|<1\), then there is one stable steady state. If both roots don't satisfy this condition, there is no stable steady state. Note that this conclusion holds without dependence on linearization. The optimal path in an infinitesimally small neighborhood around the steady state is precisely linear, and we are interested in that infinitesimally small neighborhood to theoretically distinguish between stable and unstable steady states. The linearized sequence used in the argument is just for avoiding mentioning infinitesimal and make it seem like a normal sequence.

5.2.1 From time-domain to state domain

Recall that in the continuous state sequence set-up, we start with \(k(0)=k_0\), then we want to find a sequence of \((x(t),\dot x(t))\) that satisfies the EE: for \(\forall t\ge0\), $$F_x(x(t),\dot x(t))+\rho F_{\dot x}(x(t),\dot x(t))=F_{\dot x x}(x(t),\dot x(t))\dot x(t)+F_{\dot x\dot x}(x(t),\dot x(t))\ddot x(t)\tag{5.4}$$ and converges to steady state and thus satisfies the TC: $$0=\lim_{T\to\infty}e^{-\rho T}F_{\dot x}(x(T),\dot x(T))x(T)$$ In this way, we solve the problem in the time domain. We can also solve this problem in the state domain by finding a function \(\dot x=g(x)\). The function \(g(x)\) is the optimal decision rule in the continuous problem. In order to have the function \(g(x)\) solve the problem, consider the following relationship between \(g(x)\) and \(\ddot x\): $$\dot x=g(x),\qquad\ddot x=\frac{d\dot x}{dt}=\frac{dg(x)}{dt}=g'(x)\frac{dx}{dt}=g'(x)\dot x=g'(x)g(x)$$ So, if \(g(x)\) solves the problem in the state domain, then when we replace \(\ddot x\) with \(g'(x)g(x)\) and replace \(\dot x\) with \(g(x)\) in the EE, the EE will also hold, which means \(g(x)\), after transformation, also solves the problem in the time domain. We have already seen the equivalence of solving the problem (finding an optimal sequence \((x(t),\dot x(t))\)) in the time domain and in the state domain (finding an optimal decision \(\dot x=g(x)\)). Let's focus on the state domain solution to discuss the local stability and speed of convergence around the steady state.

5.2.2 Linearization of the optimal decision rule around the steady state

Same as for the discrete time version, we can also do the Taylor expansion on the optimal decision rule \(\dot x(t)=g(x(t))\) around \(x=x^\star\): $$\dot x(t)=g(x(t))\approx g(x^\star)+g'(x^\star)(x(t)-x^\star)$$ Since \(g(x^\star)=0\), we can obtain the approximated optimal decision rule in the neighborhood of \(k=\bar k\): $$\dot x(t)=g'(x^\star)(x(t)-x^\star)$$

5.2.3 Stability and speed of convergence

Consider the following two cases: - If \(g'(x^\star)<0\), then \(\dot x(t)<0\) when \(x(t)>x^\star\) and \(\dot x(t)>0\) when \(x(t)stable around its steady state. - The more negative \(g'(x^\star)\) is, the faster it will converge. - If \(g'(x^\star)>0\), then \(\dot x(t)<0\) when \(x(t)0\) when \(x(t)>x^\star\). - The system is not stable around its steady state.

5.2.4 Construct a quadratic function of \(g'(x^\star)\)

First note that we have derived \(\dot x=g(x)\) and \(\ddot x=g'(x)\dot x=g'(x)g(x)\). Recall the EE: $$F_x(x,\dot x)+\rho F_{\dot x}(x,\dot x)-F_{\dot x x}(x,\dot x)\dot x-F_{\dot x\dot x}(x,\dot x)\ddot x=0$$ As in the discrete time model, we can substitute the optimal decision rule into the EE: $$0=F_x(x,g(x))+\rho F_{\dot x}(x,g(x))-F_{\dot x x}(x,g(x))g(x)-F_{\dot x\dot x}(x,g(x))g'(x)g(x)\tag{5.5}$$ Then, differentiate equation (5.5) w.r.t. \(x\) evaluated at \(x=x^\star\) to obtain (using \(g(x^\star)=0\)): $$\begin{aligned}0=&F_{xx}(x^\star,0)+F_{x\dot x}(x^\star,0)g'(x^\star)+\rho F_{\dot x x}(x^\star,0)+\rho F_{\dot x\dot x}(x^\star,0)g'(x^\star)\\&-F_{\dot x x}(x^\star,0)g'(x^\star)-F_{\dot x\dot x}(x^\star,0)(g'(x^\star))^2\\=&-F_{\dot x\dot x}\cdot(g'(x^\star))^2+(\rho F_{\dot x\dot x}+F_{x\dot x}-F_{\dot x x})g'(x^\star)+\rho F_{\dot x x}+F_{xx}\\=&-F_{\dot x\dot x}\cdot(g'(x^\star))^2+\rho F_{\dot x\dot x}\cdot g'(x^\star)+\rho F_{\dot x x}+F_{xx}\end{aligned}$$ (the last line uses \(F_{x\dot x}=F_{\dot x x}\) to cancel). So we have obtained the quadratic equation. Denote \(\lambda=g'(x^\star)\) and rewrite: $$Q(\lambda)\equiv-F_{\dot x\dot x}\lambda^2+\rho F_{\dot x\dot x}\lambda+\rho F_{\dot x x}+F_{xx}=0\tag{5.6}$$ Solve this quadratic equation: $$\lambda_1=\frac{\rho-\sqrt{\rho^2+4\frac{\rho F_{\dot x x}+F_{xx}}{F_{\dot x\dot x}}}}{2},\qquad\lambda_2=\frac{\rho+\sqrt{\rho^2+4\frac{\rho F_{\dot x x}+F_{xx}}{F_{\dot x\dot x}}}}{2}$$ Again, by Vieta's theorem, \(\lambda_1+\lambda_2=\rho\). So, if \(\frac{\rho F_{\dot x x}+F_{xx}}{F_{\dot x\dot x}}>0\), then \(\lambda_2>\rho\), and thus \(\lambda_1=g'(k^\star)<0\) will be a solution of a stable steady state.