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I have the following system of equations:

\begin{align*} x_{t+1} & = 3x_t + y_t \\ \\ y_{t+1} & = 2 + 5y_t. \end{align*}

I know how to solve it when I do not have a constant, but I couldn't write it in the matrical form. What is the method for this kind of system to find the long-term equilibrium of this system? And how can I comment on whether equilibrium is unique or there are multiple equilibria? Any help is appreciated in advance.

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2 Answers 2

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Preamble

  1. This answer does not provide a step-by-step solution to the above exercise. It shows how if you can solve a linear dynamic system without the constants (a homogeneous linear dynamic system) then generally you can solve the one with the constants as well.

  2. I will use the $\triangleq$ symbol to define things. I will also write the discrete time dynamic equations in their 'difference' form, meaning instead of $$ x_{t+1} = f(x_t) $$ I will write $$ \Delta x_t = g(x_t), $$ where $$ \Delta x_t \triangleq x_{t+1} - x_t. $$

  3. I will use a somewhat unconvential matrix notation, I will simply use capital letters for matrices, I will not bold and nonitalicize them. Similarly I will simply underline vector variables.


Matrix form

Suppose we have a dynamic system that is linear but non-homogeneous (the origin point is not an equilibrium): \begin{align*} \Delta x_t & = a_{11} x_t + a_{12} y_t + b_1 \\ \Delta y_t & = a_{21} x_t + a_{22} y_t + b_2. \end{align*} If we define the (vector) variable $\underline{x}_t$ for all $t$ time periods as \begin{align*} \underline{x}_t & \triangleq \begin{bmatrix}{x_t \\ y_t} \end{bmatrix} \end{align*} we can write the above dynamic system in matrix form: \begin{align*} \Delta \underline{x}_t & = A \underline{x}_t + \underline{b}, \end{align*}


Shifting/homogenizing the system

It is possible to 'homogenize' the above linear equation system, to redefine the variables in such a way that $\underline{b}$ disappears.

For simplicity's sake let us assume that $A$ is invertable. Let us define a new (vector) variable $\underline{z}_t$ for all $t$ time periods as \begin{align*} \underline{z}_t & \triangleq \underline{x}_t + A^{-1}\underline{b}. \end{align*} For $t+1$ this is \begin{align*} \underline{z}_{t+1} & = \underline{x}_{t+1} + A^{-1}\underline{b}. \end{align*} Taking the difference of the two equations above: \begin{align*} \underline{z}_{t+1} - \underline{z}_t & = (\underline{x}_{t+1} + A^{-1}\underline{b}) - (\underline{x}_t + A^{-1}\underline{b}) \\ \\ \Delta \underline{z}_t & = \Delta \underline{x}_t \end{align*} The change in $\underline{z}_t$ is the same as the change in $\underline{x}_t$; by creating $\underline{z}$ we merely a shifted $\underline{x}$. Thus if we can find equilibrium values for $\underline{z}$ we will also find equilibrium values for $\underline{x}$. Let us first get the dynamic system for $\underline{z}$:
\begin{align*} \Delta \underline{z}_t & = A \underline{x}_t + \underline{b} \\ \\ \Delta \underline{z}_t & = A(\underline{x}_t + A^{-1}\underline{b}) = A \underline{z}_t. \end{align*} This is a homogeneous linear dynamic system. In equilibrium there is no change in the varible, so \begin{align*} \Delta \underline{z}^* & = \underline{0} \\ \\ A \underline{z}^* & = \underline{0} \\ \\ \underline{z}^* & = A^{-1}\underline{0} = \underline{0}. \end{align*} We have found the only equilibrium. (We relied heavily on $A$ being invertible.) The corresponding $\underline{x}_t$ value is \begin{align*} \underline{x}^* & = \underline{z}^* - A^{-1}\underline{b} \\ & = 0 - A^{-1}\underline{b}. \end{align*}


Stability

We have not discussed the stability of the equilibrium in either system. The shift operation we performed above does not change distance, meaning for an any two $\underline{x}$ states and their corresponding $\underline{z}$ states, the distance is the same between the two state pairs. Specifically: $$ d(\underline{x}^*,\underline{x}_t) = d(\underline{z}^*,\underline{z}_t). $$ A 2D illustration of the concept:

enter image description here

Since the definitions of stability (global, local, asymptotic, etc.) are all dependent on how a trajectory's distance from equilibrium evolves, and these are the same in the two systems, the equilibria's stability will be the same as well.

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    $\begingroup$ Thank you so much; it is very helpful. I really appreciate it. $\endgroup$
    – Hypatia
    Dec 18, 2023 at 14:23
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Hi: Using the lag operator, L, for $y_t$, you can write $y_{t+1}(1- 5L) = 2 \longrightarrow y_{t+1} = \frac{2}{(1-5L)}$. But you cannot write that an an infinite series because the thing multiplying the L is greater than 1. So, this implies that there is no equilibrium for $y_t$. In the discrete time series literature, it's referred to as an explosive process. Think of the model for $y_{t}$ as an AR(1) where the lagged coefficient, 5, implies the non-stationarity of $y_{t}$.

Also, if you are not familiar with discrete time series, check out Box-Jenkins or Hamilton. Box Jenkins is better for an introduction.

Since, $y_t$ does not have an equilibrium, this implies that $x_t$ does not have one either.

#================================================================= ADDENDUM: 12-18-2023 #==================================================================

Giskard pointed out that there is an equilibrium point so the behavior really depends on how one defines the equilibrium. See Giskard's comments below the answer for more details. Otherwise, the answer above is incomplete.

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  • $\begingroup$ Since this is an economics list, the book by Sargent, "Macroeconomic Theory", (1979), Second edition, contains a lot of the Box-Jenkins material referred to in the answer. It's not an intro book so Box-Jenkins text might still be preferred by you ? Or maybe just google for "Introduction to ARIMA Models" ? $\endgroup$
    – mark leeds
    Dec 17, 2023 at 13:26
  • $\begingroup$ Thank you so much for your help. $\endgroup$
    – Hypatia
    Dec 17, 2023 at 14:21
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    $\begingroup$ Hi! Is $y=-1/2$ not an equilibrium for $y$? $\endgroup$
    – Giskard
    Dec 17, 2023 at 14:30
  • $\begingroup$ @Giskard: It's an interesting question. Your solution satisfies the second equation but equilibrium obviously means "does it always get to that point from any starting point". I come at it from a time-series perspective ( I'm not an economist AT ALL ) so, from an econ perspective, maybe that is an equilibrium ? I don't feel confident enough to say. In stat time-series world, one would definitely say that that model for $y_t$ is non-stationary because the long term mean doesn't exist because the process explodes to infinity. $\endgroup$
    – mark leeds
    Dec 18, 2023 at 6:00
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    $\begingroup$ So, time series statistician types ( that's my academic background ) claim that the process does not have a long term mean which means that there's no equilibrium for the process to converge to. If the interpretation is different in economics, then you could very well be correct. $\endgroup$
    – mark leeds
    Dec 18, 2023 at 6:01

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