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

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?


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:


(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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