<p>Not without further assumptions. Let <span class="math-container">$F(X,Y)=H(X,Y)=|X-Y|$</span>. This satisfies your assumptions, but every pair with <span class="math-container">$X=Y$</span> is an equilibrium.</p>

CC BY-SA 4.0
2023-05-22T13:04:42.553Z
Thanks! But can I leverage the uniqueness when only one variable is endogenous? I guess it is not enough, but maybe with some other element I can do it. What do you think?
Karl Smith
CC BY-SA 4.0
2023-05-22T13:20:38.477Z
Yes, in that case you have at most one equilibrium.
Michael Greinecker
CC BY-SA 4.0
2023-05-22T13:39:24.467Z
So just to clarify. If I can show that $X^{*}(Y,\omega)$ is unique and that $Y^{*}(X,\omega)$ is also unique. How can I then argue/prove that $(X(\omega),Y(\omega))$ is unique?
Karl Smith
CC BY-SA 4.0
2023-05-22T13:56:06.110Z
In that case, you have two endogenous variables, exactly as in the example. That is not enough for uniqueness.
Michael Greinecker
CC BY-SA 4.0
2023-05-22T13:59:56.427Z
OK thanks. So what else would I need?
Karl Smith
CC BY-SA 4.0
2023-05-22T14:11:42.103Z
What kind of assumptions make sense for your model and guarantee uniqueness is hard to say without knowing the model.
Michael Greinecker
CC BY-SA 4.0
2023-05-22T12:38:25.343Z
2023-05-22T12:38:25.343Z
Michael Greinecker
<p>Nothing ensures that will be a unique solution.</p>
<p>Indeed, on the contrary, saying</p>
<blockquote>
<p>I can prove that, if X is exogenously given, <span class="math-container">$Y^{*}(\omega, X)$</span> is
uniquely determined. Similarly, I can prove that, if Y is exogenously
given, <span class="math-container">$X^{*}(\omega, Y)$</span> is uniquely determined</p>
</blockquote>
<p>you are stating that the system has  <strong>infinite solutions.</strong><span class="math-container">$^1$</span></p>
<p>Consider your system, assuming a given value of <span class="math-container">$X$</span>, say <span class="math-container">$X_1$</span>.</p>
<p><span class="math-container">\begin{align*}
F(X_1,Y;\omega)=0 \\
H(X_1,Y;\omega)=0
\end{align*}</span></p>
<p>According to your statement above, there is a unique <span class="math-container">$Y_1$</span> such that <span class="math-container">$(X_1,Y_1)$</span> is a solution.</p>
<p>That is, the pair <span class="math-container">$(X_1,Y_1)$</span> is a solution.</p>
<p>Now, re-consider the system with a different, given value of <span class="math-container">$X$</span>, say <span class="math-container">$X_2$</span>.</p>
<p><span class="math-container">\begin{align*}
F(X_2,Y;\omega)=0 \\
H(X_2,Y;\omega)=0
\end{align*}</span></p>
<p>There will be a <span class="math-container">$Y_2$</span> such that <span class="math-container">$(X_2,Y_2)$</span> is a solution.</p>
<p>That is, the pair <span class="math-container">$(X_2,Y_2)$</span> is also a solution.</p>
<p>And so on.</p>
<p><strong>Assuming infinite given value of <span class="math-container">$X$</span>, <span class="math-container">$X_i$</span>, <span class="math-container">$i=1,2,...$</span> you have infinite solutions.</strong><span class="math-container">$^2$</span></p>
<p><span class="math-container">$\;$</span></p>
<p>The second part of your statement</p>
<blockquote>
<p>Similarly, I can prove that, if Y is exogenously given, <span class="math-container">$X^{*}(\omega,
&gt; Y)$</span> is uniquely determined</p>
</blockquote>
<p>just says that, for each given <span class="math-container">$Y_i$</span>, there is a unique <span class="math-container">$X_i$</span>, that is that the <strong><span class="math-container">$Y_i$</span> will be different from each other: <span class="math-container">$Y_i\neq Y_j$</span> for <span class="math-container">$i\neq j$</span></strong> (actually, you are just stating that there is a bijection between the <span class="math-container">$X_i$</span> and the <span class="math-container">$Y_i$</span>).</p>
<p><span class="math-container">$\;$</span></p>
<p>For an example, if the two equations where linear, they would be linearly dependent, that is, they will  be two coincident lines.</p>
<p><span class="math-container">$$***$$</span></p>
<p>Outside linear equations systems, <strong>there isn't a general method</strong> to state existence and unicity of the solutions of a system of equations.</p>
<p>A possible example where  the existence and unicity of the solution can be stated, for one single equation,  is the case of a <strong>continuous strictly increasing function <span class="math-container">$f(x)$</span></strong> on an interval <span class="math-container">$(a,b)$</span> of <span class="math-container">$\mathbb{R}$</span> (it is a consequence of the Intermediate Value Theorem): the equation <span class="math-container">$f(x)=C$</span>, <span class="math-container">$C\in\mathbb{R}$</span>, has a unique solution in <span class="math-container">$(a,b)$</span>.</p>
<p>An important theorem of mathematical analysis can be used, the <a href="https://en.wikipedia.org/wiki/Banach_fixed-point_theorem" rel="nofollow noreferrer"><strong>Contraction mapping theorem or Banach-Caccioppoli fixed point theorem</strong></a>, because many problems of finding the solutions of a system of equations can be viewed as a fixed-point  problem of a contraction mapping.</p>
<p>This theorem is used in numerical analysis to find solutions of equations,  but its  use in applications requires  sufficient information about the functions.</p>
<hr />
<p><span class="math-container">$^1$</span> <sup> Obviously, I understand that you mean that a solution exists, for any exogenously given <span class="math-container">$X$</span> or <span class="math-container">$Y$</span>. </sup></p>
<p><span class="math-container">$^2$</span> <sup> Actually, there will be  a continuum of solutions, as <span class="math-container">$X$</span> varies in <span class="math-container">$\mathbb{R}$</span>.</sup></p>

CC BY-SA 4.0
2023-05-22T15:00:28.387Z
2023-05-22T19:57:43.383Z
2023-05-22T19:57:43.383Z
BakerStreet
<p>I have a model with two endogenous variables X and Y, and a set of parameters that I denote by <span class="math-container">$\omega$</span>. Equilibrium <span class="math-container">$X^{*}(\omega)$</span> and <span class="math-container">$Y^{*}(\omega)$</span> in this model is the solution to:</p>
<p><span class="math-container">\begin{align*}
F(X,Y;\omega)=0 \\
H(X,Y;\omega)=0
\end{align*}</span></p>
<p>I would like to know whether this equilibrium is unique (given parameters). I can prove that, if X is exogenously given, <span class="math-container">$Y^{*}(\omega, X)$</span> is uniquely determined. Similarly, I can prove that, if Y is exogenously given, <span class="math-container">$X^{*}(\omega, Y)$</span> is uniquely determined. Can I leverage these two results to show that the equilibrium with two endogenous variables is unique? I was thinking about proving it by contradiction, but not sure if it makes sense.</p>
<p>Thanks!</p>

<equilibrium>
Uniqueness of equilibrium
Karl Smith
# Uniqueness of equilibrium

(Asked by: username_0 on 2023-05-22)

I have a model with two endogenous variables X and Y, and a set of parameters that I denote by $\omega$. Equilibrium $X^{\*}(\omega)$ and $Y^{\*}(\omega)$ in this model is the solution to:

\begin{align\*}
F(X,Y;\omega)=0 \\
H(X,Y;\omega)=0
\end{align\*}

I would like to know whether this equilibrium is unique (given parameters). I can prove that, if X is exogenously given, $Y^{\*}(\omega, X)$ is uniquely determined. Similarly, I can prove that, if Y is exogenously given, $X^{\*}(\omega, Y)$ is uniquely determined. Can I leverage these two results to show that the equilibrium with two endogenous variables is unique? I was thinking about proving it by contradiction, but not sure if it makes sense.

Thanks!

**Tags:** equilibrium

---

## 2 answers

### Answer by username_1 on 2023-05-22. Score: 2

Not without further assumptions. Let $F(X,Y)=H(X,Y)=|X-Y|$. This satisfies your assumptions, but every pair with $X=Y$ is an equilibrium.

> **Comments:**
> **username_0**: Thanks! But can I leverage the uniqueness when only one variable is endogenous? I guess it is not enough, but maybe with some other element I can do it. What do you think?  
> **username_1**: Yes, in that case you have at most one equilibrium.  
> **username_0**: So just to clarify. If I can show that $X^{*}(Y,\omega)$ is unique and that $Y^{*}(X,\omega)$ is also unique. How can I then argue/prove that $(X(\omega),Y(\omega))$ is unique?  
> **username_1**: In that case, you have two endogenous variables, exactly as in the example. That is not enough for uniqueness.  
> **username_0**: OK thanks. So what else would I need?  
> **username_1**: What kind of assumptions make sense for your model and guarantee uniqueness is hard to say without knowing the model.  

---

### Answer by username_2 on 2023-05-22. Score: 1

Nothing ensures that will be a unique solution.

Indeed, on the contrary, saying

> I can prove that, if X is exogenously given, $Y^{\*}(\omega, X)$ is
> uniquely determined. Similarly, I can prove that, if Y is exogenously
> given, $X^{\*}(\omega, Y)$ is uniquely determined

you are stating that the system has **infinite solutions.**$^1$

Consider your system, assuming a given value of $X$, say $X\_1$.

\begin{align\*}
F(X\_1,Y;\omega)=0 \\
H(X\_1,Y;\omega)=0
\end{align\*}

According to your statement above, there is a unique $Y\_1$ such that $(X\_1,Y\_1)$ is a solution.

That is, the pair $(X\_1,Y\_1)$ is a solution.

Now, re-consider the system with a different, given value of $X$, say $X\_2$.

\begin{align\*}
F(X\_2,Y;\omega)=0 \\
H(X\_2,Y;\omega)=0
\end{align\*}

There will be a $Y\_2$ such that $(X\_2,Y\_2)$ is a solution.

That is, the pair $(X\_2,Y\_2)$ is also a solution.

And so on.

**Assuming infinite given value of $X$, $X\_i$, $i=1,2,...$ you have infinite solutions.**$^2$

$\;$

The second part of your statement

> Similarly, I can prove that, if Y is exogenously given, $X^{\*}(\omega,
> > Y)$ is uniquely determined

just says that, for each given $Y\_i$, there is a unique $X\_i$, that is that the **$Y\_i$ will be different from each other: $Y\_i\neq Y\_j$ for $i\neq j$** (actually, you are just stating that there is a bijection between the $X\_i$ and the $Y\_i$).

$\;$

For an example, if the two equations where linear, they would be linearly dependent, that is, they will be two coincident lines.

$$\*\*\*$$

Outside linear equations systems, **there isn't a general method** to state existence and unicity of the solutions of a system of equations.

A possible example where the existence and unicity of the solution can be stated, for one single equation, is the case of a **continuous strictly increasing function $f(x)$** on an interval $(a,b)$ of $\mathbb{R}$ (it is a consequence of the Intermediate Value Theorem): the equation $f(x)=C$, $C\in\mathbb{R}$, has a unique solution in $(a,b)$.

An important theorem of mathematical analysis can be used, the [**Contraction mapping theorem or Banach-Caccioppoli fixed point theorem**](https://en.wikipedia.org/wiki/Banach_fixed-point_theorem), because many problems of finding the solutions of a system of equations can be viewed as a fixed-point problem of a contraction mapping.

This theorem is used in numerical analysis to find solutions of equations, but its use in applications requires sufficient information about the functions.

---

$^1$  Obviously, I understand that you mean that a solution exists, for any exogenously given $X$ or $Y$.

$^2$  Actually, there will be a continuum of solutions, as $X$ varies in $\mathbb{R}$.


---

