<p>We have an equilibrium given by
$$h_1(x,y)=f(x)+z-y=0,$$
$$h_2(x,y)=x-g(y)=0$$.
The implicit function theorem then says that (omitting the arguments):
$$\frac{\partial x}{\partial z}=\frac{-\det\left(
\begin{matrix}
  \frac{\partial h_1}{\partial z} &amp; \frac{\partial h_1}{\partial y} \\
  \frac{\partial h_2}{\partial z} &amp; \frac{\partial h_2}{\partial y} 
 \end{matrix}\right)}{\det\left(
\begin{matrix}
  \frac{\partial h_1}{\partial x} &amp; \frac{\partial h_1}{\partial y} \\
  \frac{\partial h_2}{\partial x} &amp; \frac{\partial h_2}{\partial y} 
 \end{matrix}\right)},\quad \frac{\partial y}{\partial z}=\frac{-\det\left(
\begin{matrix}
  \frac{\partial h_1}{\partial x} &amp; \frac{\partial h_1}{\partial z} \\
  \frac{\partial h_2}{\partial x} &amp; \frac{\partial h_2}{\partial z} 
 \end{matrix}\right)}{\det\left(
\begin{matrix}
  \frac{\partial h_1}{\partial x} &amp; \frac{\partial h_1}{\partial y} \\
  \frac{\partial h_2}{\partial x} &amp; \frac{\partial h_2}{\partial y} 
 \end{matrix}\right)}.$$</p>

<p>My source for this is <em>Mathematical Methods and Models for Economists</em> by Angel de la Fuente, although I don't have the book to hand now and can't remind myself of the intuition.</p>

<p>This implies
$$\frac{\partial x}{\partial z}=\frac{g'(y)}{1-f'(x) g'(y)}$$
$$\frac{\partial y}{\partial z}=\frac{1}{1-f'(x) g'(y)}.$$</p>

<p>For the implicit function theorem to hold and this solution to be valid, we need $1-f'(x) g'(y)\neq0$.</p>

<hr>

<p>More generally, the way this works is as follows: you write down a system of equations whose roots characterize the equilibrium:</p>

<p>$$F_1(\mathbf{x};a)=0,F_2(\mathbf{x};a)=0,\ldots,F_n(\mathbf{x};a)=0$$</p>

<p>(where $a$ is the parameter of interest). From them, we construct the vector-valued function</p>

<p>$$\mathbf{F}(x)=[F_1(\mathbf{x};a),F_2(\mathbf{x};a),\ldots,F_n(\mathbf{x};a)]$$
which has the Jacobian matrix
$$\mathbf J = \frac{d\mathbf F}{d\mathbf x}
= \begin{bmatrix}
    \dfrac{\partial F_1}{\partial x_1} &amp; \cdots &amp; \dfrac{\partial F_1}{\partial x_m}\\
    \vdots &amp; \ddots &amp; \vdots\\
    \dfrac{\partial F_n}{\partial x_1} &amp; \cdots &amp; \dfrac{\partial F_n}{\partial x_m} \end{bmatrix}.$$</p>

<p>To calculate the derivative of $x_i$ with respect to $a$, we construct the modified Jacobian in which we replace the $i^{\text{th}}$ column with derivative WRT $a$ instead of $x_i$. So, for $x_1$ this would look like
$$\mathbf{J}_{x_1}=\begin{bmatrix}
    \dfrac{\partial F_1}{\partial a} &amp; \dfrac{\partial F_1}{\partial x_2} &amp; \cdots &amp; \dfrac{\partial F_1}{\partial x_m}\\
    \vdots &amp; \vdots&amp;\ddots &amp; \vdots\\
    \dfrac{\partial F_n}{\partial a}&amp;\dfrac{\partial F_n}{\partial x_2} &amp; \cdots &amp; \dfrac{\partial F_n}{\partial x_m} \end{bmatrix}.$$</p>

<p>The derivative of interest is then calculated as $$\frac{\partial x_i}{\partial a}=\frac{-\det\mathbf{J}_{x_i}}{\det\mathbf{J}}.$$</p>

<p>We need $\det\mathbf{J}\neq0$ for the implicit function theorem to be valid.</p>

CC BY-SA 3.0
2015-03-06T09:48:10.163Z
2015-03-06T09:48:10.163Z
Ubiquitous
<p>Consider</p>

<p>$$f(x) + z = y \\
x = g(y)$$</p>

<p>I have a feedback effect in mind:</p>

<ol>
<li>We Shock z by 1 percent</li>
<li>as a direct response, y increases</li>
<li>x increases with y through the second equation</li>
<li>Through f(x), y increases. Repeat 3.-4. until new equilibrium</li>
</ol>

<p>I'm interested in solving the elasticity of $x,y$ w.r.t. $z$. I guess I need some sort of concavity requirement in $f \circ g$. How do I approach this?</p>

<mathematical-economics>
Solving for Feedback Effects / Amplification
FooBar
# Solving for Feedback Effects / Amplification

(Asked by: username_0 on 2015-03-06)

Consider

$$f(x) + z = y \\
x = g(y)$$

I have a feedback effect in mind:

1. We Shock z by 1 percent
2. as a direct response, y increases
3. x increases with y through the second equation
4. Through f(x), y increases. Repeat 3.-4. until new equilibrium

I'm interested in solving the elasticity of $x,y$ w.r.t. $z$. I guess I need some sort of concavity requirement in $f \circ g$. How do I approach this?

**Tags:** mathematical-economics

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## 1 answers

### Answer by username_1 on 2015-03-06. Score: 2 | Accepted answer

We have an equilibrium given by
$$h\_1(x,y)=f(x)+z-y=0,$$
$$h\_2(x,y)=x-g(y)=0$$.
The implicit function theorem then says that (omitting the arguments):
$$\frac{\partial x}{\partial z}=\frac{-\det\left(
\begin{matrix}
\frac{\partial h\_1}{\partial z} & \frac{\partial h\_1}{\partial y} \\
\frac{\partial h\_2}{\partial z} & \frac{\partial h\_2}{\partial y}
\end{matrix}\right)}{\det\left(
\begin{matrix}
\frac{\partial h\_1}{\partial x} & \frac{\partial h\_1}{\partial y} \\
\frac{\partial h\_2}{\partial x} & \frac{\partial h\_2}{\partial y}
\end{matrix}\right)},\quad \frac{\partial y}{\partial z}=\frac{-\det\left(
\begin{matrix}
\frac{\partial h\_1}{\partial x} & \frac{\partial h\_1}{\partial z} \\
\frac{\partial h\_2}{\partial x} & \frac{\partial h\_2}{\partial z}
\end{matrix}\right)}{\det\left(
\begin{matrix}
\frac{\partial h\_1}{\partial x} & \frac{\partial h\_1}{\partial y} \\
\frac{\partial h\_2}{\partial x} & \frac{\partial h\_2}{\partial y}
\end{matrix}\right)}.$$

My source for this is *Mathematical Methods and Models for Economists* by Angel de la Fuente, although I don't have the book to hand now and can't remind myself of the intuition.

This implies
$$\frac{\partial x}{\partial z}=\frac{g'(y)}{1-f'(x) g'(y)}$$
$$\frac{\partial y}{\partial z}=\frac{1}{1-f'(x) g'(y)}.$$

For the implicit function theorem to hold and this solution to be valid, we need $1-f'(x) g'(y)\neq0$.

---

More generally, the way this works is as follows: you write down a system of equations whose roots characterize the equilibrium:

$$F\_1(\mathbf{x};a)=0,F\_2(\mathbf{x};a)=0,\ldots,F\_n(\mathbf{x};a)=0$$

(where $a$ is the parameter of interest). From them, we construct the vector-valued function

$$\mathbf{F}(x)=[F\_1(\mathbf{x};a),F\_2(\mathbf{x};a),\ldots,F\_n(\mathbf{x};a)]$$
which has the Jacobian matrix
$$\mathbf J = \frac{d\mathbf F}{d\mathbf x}
= \begin{bmatrix}
\dfrac{\partial F\_1}{\partial x\_1} & \cdots & \dfrac{\partial F\_1}{\partial x\_m}\\
\vdots & \ddots & \vdots\\
\dfrac{\partial F\_n}{\partial x\_1} & \cdots & \dfrac{\partial F\_n}{\partial x\_m} \end{bmatrix}.$$

To calculate the derivative of $x\_i$ with respect to $a$, we construct the modified Jacobian in which we replace the $i^{\text{th}}$ column with derivative WRT $a$ instead of $x\_i$. So, for $x\_1$ this would look like
$$\mathbf{J}\_{x\_1}=\begin{bmatrix}
\dfrac{\partial F\_1}{\partial a} & \dfrac{\partial F\_1}{\partial x\_2} & \cdots & \dfrac{\partial F\_1}{\partial x\_m}\\
\vdots & \vdots&\ddots & \vdots\\
\dfrac{\partial F\_n}{\partial a}&\dfrac{\partial F\_n}{\partial x\_2} & \cdots & \dfrac{\partial F\_n}{\partial x\_m} \end{bmatrix}.$$

The derivative of interest is then calculated as $$\frac{\partial x\_i}{\partial a}=\frac{-\det\mathbf{J}\_{x\_i}}{\det\mathbf{J}}.$$

We need $\det\mathbf{J}\neq0$ for the implicit function theorem to be valid.


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