Transition Dynamics in Growth Model

EC417_Macroeconomics

It’s a very important concept in Growth Model.

In this page, we try to answer two questions:

1. Can we analyze the speed of Convergence?
2. How analyze stability if two- or $N$ -dimensional state $x$? Because now we can no longer draw the Phase Diagram.

Suppose we already arrive at the Steady State, let $y\in {\mathbb{R}}^n$ and the function $m:{\mathbb{R}}^n\to {\mathbb{R}}^n$ define a dynamic system:

$$\dot{y}(t)=m(y(t))$$

If it is at Steady State, it’s obvious that $m(y^{\ast })=0$.

Consider a first order approximation of $m$ around $y^{\ast }$: (see Taylor Expansion):

$$\dot{y}\approx m(y^{\ast })+m^{\prime }(y^{\ast })(y-y^{\ast })$$

where $m^{\prime }(y^{\ast })$ is the $n\times n$ Jacobian of $m$ evaluated at $y^{\ast }$, we then define the gap of $y$: $\hat{y}=y-y^{\ast }$, thus in the traditional Growth Model, we could have:

$$\hat{y}=\left[\begin{array}{c}c-c^{\ast }\\ k-k^{\ast }\end{array}\right]$$

The Jacobian of $m$ is $A$, in this example, $A$ is

$$A=m^{\prime }(y^{\ast })=\left[\begin{array}{cc}\frac{\partial \dot{c}}{\partial c}& \frac{\partial \dot{c}}{\partial k}\\ \frac{\partial \dot{k}}{\partial c}& \frac{\partial \dot{k}}{\partial k}\end{array}\right]$$

We have $\dot{\hat{y}}\approx A\hat{y}$, it simply means that how $y$ would evolve around the steady state if the economy is deviated from the Steady State.

To solve this system, we need to find the eigenvalues and eigenvectors of matrix $A$. From Acemoglu, with initial value $\hat{y}(0)$, and let $A$ be a $n\times n$ matrix. Suppose that $\ell \le n$ of the eigenvalues of $A$ have negative real parts. Then there exists an $\ell$ -dimensional subspace $L$ of ${\mathbb{R}}^n$ such that starting from any $\hat{y}(0)\in L$, we have $\hat{y}(t)\to 0$ as $t\to \infty$. If all eigenvalues of $A$ have negative real parts, then $L={\mathbb{R}}^n$.

Simply saying:

• if $\ell =m$, we have “saddle-path stable”, we have a unique optimal trajectory. And the negative eigenvalue would govern the speed of convergence.
• if $\ell <m$, we have “unstable”, there would be infinite trajectories diverging from the steady state. $y(t)$ does not converge to steady state.
• if $\ell >m$, multiple optimal trajectories.

We skip the proof of this theorem because it is unlikely to be in the exams.

Linearized Growth Model

We continue using the traditional Growth Model as an example. The dynamic system is:

$$A=\left[\begin{array}{cc}0& \frac{1}{\sigma }{f}^{\prime \prime }(k^{\ast })c^{\ast }\\ -1& \rho \end{array}\right]$$

We then introduce how to find the eigenvalues of matrix $A$.

Eigenvalues and Eigenvectors

The eigenvalue $\lambda$ should satisfy:

$$\det (A-\lambda I)=0$$

where $I$ is the identity matrix. Thus we have:

$$A-\lambda I=\left[\begin{array}{cc}-\lambda & \frac{1}{\sigma }{f}^{\prime \prime }(k^{\ast })c^{\ast }\\ -1& \rho -\lambda \end{array}\right]$$

Then calculate the determinant:

$$\det (A-\lambda I)=(-\lambda )(\rho -\lambda )-\left(-1\cdot \frac{1}{\sigma }{f}^{\prime \prime }(k^{\ast })c^{\ast }\right)={\lambda }^2-\rho \lambda +\frac{1}{\sigma }{f}^{\prime \prime }(k^{\ast })c^{\ast }$$

Thus we have the eigenvalue equation:

$${\lambda }^2-\rho \lambda +\frac{1}{\sigma }{f}^{\prime \prime }(k^{\ast })c^{\ast }=0$$

Simply calculation leads:

$${\lambda }_{1,2}==\frac{\rho \pm \sqrt{{\rho }^2-4\cdot \frac{1}{\sigma }{f}^{\prime \prime }(k^{\ast })c^{\ast }}}{2}$$

Since $f^{\prime \prime }(k^{\ast })<0$, thus the two eigenvalues have opposite signs, we let ${\lambda }_1<0<{\lambda }_2$, then we have $\ell =1$, it is saddle-path stable.

Now we introduce how to solve the matrix differential equation.

Solution to Matrix Differential Equation

We try to solve the following equation:

$$\dot{\hat{y}}=A\hat{y}\hat{y}(0)={\hat{y}}_0$$

The intuition is we diogonalize matrix $A$ to decouple the system. Thus we have $n$ independent differential equations.

Step 1: We dioagonalize matrix $A$.

Suppose $A$ has $n$ linearly independent eigenvectors, we denote the eigenvectors as $v_1,v_2,\dots ,v_n$ and construct matrix $P$ as:

• $P=(v_1,v_2,\dots ,v_n)$
• Diagonal matrix: $\Lambda =\text{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$

By definition, we have $AP=P\Lambda$ , thus we have

$$A=P\Lambda P^{-1}$$

For example, recall that $$ A = \begin{bmatrix} 0 & \frac{1}{\sigma} f”(k^)c^ \ -1 & \rho \end{bmatrix}

We suppose the eigenvalues are $\lambda_1$ and $\lambda_2$, eigenvectors are:

v_1 = \begin{bmatrix} v_{11} \ v_{12} \end{bmatrix} \quad v_2= \begin{bmatrix} v_{21} \ v_{22} \end{bmatrix}

$$Thus,wehave:$$

P = \begin{bmatrix} v_{11} & v_{21} \ v_{12} & v_{22} \end{bmatrix} \quad \Lambda = \begin{bmatrix} \lambda_1 & 0 \ 0 & \lambda_2 \end{bmatrix}