<p>Hi: Using the lag operator, L, for <span class="math-container">$y_t$</span>, you can write <span class="math-container">$y_{t+1}(1- 5L) = 2 \longrightarrow y_{t+1} = \frac{2}{(1-5L)}$</span>. 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 <span class="math-container">$y_t$</span>. In the discrete time series literature, it's referred to as an explosive process. Think of the model for <span class="math-container">$y_{t}$</span> as an AR(1) where the lagged coefficient, 5, implies the non-stationarity of <span class="math-container">$y_{t}$</span>.</p>
<p>Also, if you are not familiar with discrete time series, check out Box-Jenkins or Hamilton. Box Jenkins is better for an introduction.</p>
<p>Since, <span class="math-container">$y_t$</span> does not have an equilibrium, this implies that <span class="math-container">$x_t$</span> does not have one either.</p>
<p>#=================================================================
ADDENDUM: 12-18-2023
#==================================================================</p>
<p>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.</p>

CC BY-SA 4.0
2023-12-17T13:26:45.750Z
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" ?
mark leeds
CC BY-SA 4.0
2023-12-17T14:30:29.287Z
Hi! Is $y=-1/2$ not an equilibrium for $y$?
Giskard
CC BY-SA 4.0
2023-12-18T06:00:13.023Z
@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.
mark leeds
CC BY-SA 4.0
2023-12-18T06:01:28.810Z
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.
mark leeds
CC BY-SA 4.0
2023-12-18T06:02:38.860Z
I'd be interested in seeing other responses because it would be good for me to see the economic interpretation of $y = -\frac{1}{2}$. Thanks.
mark leeds
CC BY-SA 4.0
2023-12-18T06:04:46.377Z
Hi @markleeds!  >> *equilibrium obviously means "does it always get to that point from any starting point"* << You seem to be describing a [global asymptotically stable](https://math.stackexchange.com/questions/3789696/global-stability-vs-global-asymptotic-stability) equilibrium. You are correct in saying that the dynamics system does not have that, it has an unstable equilibrium instead.
Giskard
CC BY-SA 4.0
2023-12-18T06:07:26.270Z
okay. so unstable equilibrium means, "if it gets there, it will stay there but it might not ever get there ?". Thanks for your great help. I thought that asymptotitically stable was what all equilibriums were so I've already learned something.
mark leeds
CC BY-SA 4.0
2023-12-18T06:09:22.333Z
"*if it gets there, it will stay there*" this is the definition of equilibrium. Unstable means that in any open environment of the equilibrium there is a state which puts the system on a track that will diverge from the equilibrium.
Giskard
CC BY-SA 4.0
2023-12-18T06:09:50.937Z
great stuff. thank you.
mark leeds
CC BY-SA 4.0
2023-12-17T13:22:10.377Z
2023-12-18T06:17:11.403Z
2023-12-18T06:17:11.403Z
mark leeds
<p><strong>Preamble</strong> <br></p>
<ol>
<li><p>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 <em>generally</em> you can solve the one with the constants as well.</p>
</li>
<li><p>I will use the <span class="math-container">$\triangleq$</span> symbol to define things. I will also write the discrete time dynamic equations in their 'difference' form, meaning instead of
<span class="math-container">$$
x_{t+1} = f(x_t)
$$</span>
I will write
<span class="math-container">$$
\Delta x_t = g(x_t),
$$</span>
where
<span class="math-container">$$
\Delta x_t \triangleq x_{t+1} - x_t.
$$</span></p>
</li>
<li><p>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.</p>
</li>
</ol>
<hr />
<p><strong>Matrix form</strong> <br></p>
<p>Suppose we have a dynamic system that is linear but non-homogeneous (the origin point is not an equilibrium):
<span class="math-container">\begin{align*}
    \Delta x_t &amp; = a_{11} x_t + a_{12} y_t + b_1  \\
    \Delta y_t &amp; = a_{21} x_t + a_{22} y_t + b_2.
\end{align*}</span>
If we define the (vector) variable <span class="math-container">$\underline{x}_t$</span> for all <span class="math-container">$t$</span> time periods as
<span class="math-container">\begin{align*}
    \underline{x}_t &amp; \triangleq \begin{bmatrix}{x_t \\ y_t} \end{bmatrix}
\end{align*}</span>
we can write the above dynamic system in matrix form:
<span class="math-container">\begin{align*}
    \Delta \underline{x}_t &amp; = A \underline{x}_t + \underline{b},
\end{align*}</span></p>
<hr />
<p><strong>Shifting/homogenizing the system</strong> <br></p>
<p>It is possible to 'homogenize' the above linear equation system, to redefine the variables in such a way that <span class="math-container">$\underline{b}$</span> disappears.</p>
<p>For simplicity's sake let us assume that <span class="math-container">$A$</span> is invertable. Let us define a  new (vector) variable <span class="math-container">$\underline{z}_t$</span> for all <span class="math-container">$t$</span> time periods as
<span class="math-container">\begin{align*}
    \underline{z}_t &amp; \triangleq \underline{x}_t + A^{-1}\underline{b}.
\end{align*}</span>
For <span class="math-container">$t+1$</span> this is
<span class="math-container">\begin{align*}
    \underline{z}_{t+1} &amp; = \underline{x}_{t+1} + A^{-1}\underline{b}.
\end{align*}</span>
Taking the difference of the two equations above:
<span class="math-container">\begin{align*}
    \underline{z}_{t+1} - \underline{z}_t &amp; = (\underline{x}_{t+1} + A^{-1}\underline{b}) - (\underline{x}_t + A^{-1}\underline{b}) \\
\\
\Delta \underline{z}_t &amp; = \Delta \underline{x}_t
\end{align*}</span>
The change in <span class="math-container">$\underline{z}_t$</span> is the same as the change in <span class="math-container">$\underline{x}_t$</span>; by creating <span class="math-container">$\underline{z}$</span> we merely a shifted <span class="math-container">$\underline{x}$</span>. Thus if we can find equilibrium values for <span class="math-container">$\underline{z}$</span> we will also find equilibrium values for <span class="math-container">$\underline{x}$</span>. Let us first get the dynamic system for <span class="math-container">$\underline{z}$</span>:<br />
<span class="math-container">\begin{align*}
\Delta \underline{z}_t &amp; = A \underline{x}_t + \underline{b} \\
\\
\Delta \underline{z}_t &amp; = A(\underline{x}_t + A^{-1}\underline{b}) = A \underline{z}_t. 
\end{align*}</span>
This is a homogeneous linear dynamic system. In equilibrium there is no change in the varible, so
<span class="math-container">\begin{align*}
\Delta \underline{z}^* &amp; = \underline{0} \\
\\
A \underline{z}^* &amp; = \underline{0} \\
\\
\underline{z}^* &amp; = A^{-1}\underline{0} = \underline{0}.
\end{align*}</span>
We have found the only equilibrium. (We relied heavily on <span class="math-container">$A$</span> being invertible.) The corresponding <span class="math-container">$\underline{x}_t$</span> value is
<span class="math-container">\begin{align*}
\underline{x}^* &amp; = \underline{z}^* - A^{-1}\underline{b} \\
&amp; = 0 - A^{-1}\underline{b}.
\end{align*}</span></p>
<hr />
<p><strong>Stability</strong> <br></p>
<p>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 <span class="math-container">$\underline{x}$</span> states and their corresponding <span class="math-container">$\underline{z}$</span> states, the distance is the same between the two state pairs. Specifically:
<span class="math-container">$$
d(\underline{x}^*,\underline{x}_t) =
d(\underline{z}^*,\underline{z}_t). 
$$</span>
A 2D illustration of the concept:</p>
<p><a href="https://i.sstatic.net/1vTb9.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1vTb9.png" alt="enter image description here" /></a></p>
<p>Since the definitions of stability (global, local, <a href="https://www.sciencedirect.com/topics/engineering/asymptotic-global-stability" rel="nofollow noreferrer">asymptotic</a>, 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.</p>

CC BY-SA 4.0
2023-12-18T14:23:59.870Z
Thank you so much; it is very helpful. I really appreciate it.
Hypatia
CC BY-SA 4.0
2023-12-18T11:14:16.797Z
2023-12-18T13:04:00.997Z
2023-12-18T13:04:00.997Z
Giskard
<p>I have the following system of equations:</p>
<p><span class="math-container">\begin{align*}
x_{t+1} &amp; = 3x_t + y_t \\
\\ 
y_{t+1} &amp; = 2 + 5y_t.
\end{align*}</span></p>
<p>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.</p>

<dynamic-optimization><difference-equations>
Nonhomogeneous linear dynamic system exercise
Hypatia
# Nonhomogeneous linear dynamic system exercise

(Asked by: username_0 on 2023-12-17)

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.

**Tags:** dynamic-optimization, difference-equations


---

## 2 answers

### Answer by username_2 on 2023-12-18. Score: 2 | Accepted answer

**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](https://i.sstatic.net/1vTb9.png)](https://i.sstatic.net/1vTb9.png)

Since the definitions of stability (global, local, [asymptotic](https://www.sciencedirect.com/topics/engineering/asymptotic-global-stability), 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.

> **Comments:**
> **username_0**: Thank you so much; it is very helpful. I really appreciate it.  

---

### Answer by username_1 on 2023-12-17. Score: 2

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
#==================================================================

username_2 pointed out that there is an equilibrium point so the behavior really depends on how one defines the equilibrium. See username_2's comments below the answer for more details. Otherwise, the answer above is incomplete.

> **Comments:**
> **username_1**: 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" ?  
> **username_2**: Hi! Is $y=-1/2$ not an equilibrium for $y$?  
> **username_1**: @username_2: 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.  
> **username_1**: 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.  
> **username_1**: I'd be interested in seeing other responses because it would be good for me to see the economic interpretation of $y = -\frac{1}{2}$. Thanks.  
> **username_2**: Hi @username_1!  >> *equilibrium obviously means "does it always get to that point from any starting point"* << You seem to be describing a [global asymptotically stable](https://math.stackexchange.com/questions/3789696/global-stability-vs-global-asymptotic-stability) equilibrium. You are correct in saying that the dynamics system does not have that, it has an unstable equilibrium instead.  
> **username_1**: okay. so unstable equilibrium means, "if it gets there, it will stay there but it might not ever get there ?". Thanks for your great help. I thought that asymptotitically stable was what all equilibriums were so I've already learned something.  
> **username_2**: "*if it gets there, it will stay there*" this is the definition of equilibrium. Unstable means that in any open environment of the equilibrium there is a state which puts the system on a track that will diverge from the equilibrium.  
> **username_1**: great stuff. thank you.  

---

