I'm a master student in Economics, and I'm going through a paper which involves some mathematics. I'm gonna try to make the argument as far as possible from economics.
Consider the following system of two equations:
\begin{equation}
\frac{A_M \alpha_M (I_H) M}{I_H – I_L} = \frac{A_H \alpha_H (I_H) H}{1- I_H}
\end{equation}
\begin{equation}
\frac{A_L \alpha_L (I_L) L}{I_L} = \frac{A_M \alpha_M (I_L) M}{I_H – I_L}
\end{equation}
Everything is exogenous (basically, the exogenous variables are calibrated to some arbitrary values. In other words, take them as constants), except for the two endogenous variables of the system, $I_L$ and $I_H$.
What you shall know now is that $0 < I_L < I_H < 1$; this condition always holds.
It would be simple to plot the related curves in the $I_H$ and $I_L$ space if we did not have $\alpha()$, which is a function depending on the endogenous.
The paper does not exactly define $\alpha$. However, what it says is that $\alpha_L (i) / \alpha_M (i)$ and $\alpha_M (i) / \alpha_H (i)$ are continuously differentiable and strictly decreasing, with $i \in [0,1]$.
The paper says that from the two equations above, it is possible to get the following graph:
Could you please demonstrate the mathematical steps needed to produce this graph?
That's my attempt. I expressed the first and second equation in terms of $I_L$ and $I_H$, respectively, as follows
\begin{equation}
I_L = I_H – \frac{A_M}{A_H} \frac{\alpha_M(I_H)}{\alpha_H(I_H)} \frac{M}{H}(1-I_H)
\end{equation}
\begin{equation}
I_H= I_L + \frac{A_M}{A_L} \frac{\alpha_L (I_L)}{\alpha_M (I_L)} \frac{M}{L} I_L
\end{equation}
Then, for graphing this two functions, I assumed $\alpha_H \equiv e^{-i}$, $\alpha_M \equiv e^{-1.3i}$ and $\alpha_L \equiv e^{-1.9i}$, such that $\alpha_M/\alpha_H$ and $\alpha_L/\alpha_M$ are both strictly decreasing in $i$.
For the sake of the graphical representation, I set $I_H = x$, $I_L = y$, and the following parametrization: $A_L= 1$, $A_M=1.25$, $A_H=1.6$, $L=40$, $M=60$, $H=50$, such that
\begin{equation}
y= x – (1.25/1.6)(e^{-0.3x}) (60/50) (1-x)
\end{equation}
\begin{equation}
x= y + 1.25 (e^{-0.6y}) (60/40)y
\end{equation}
and when I graph them, I get a reasonable result since the two curves crosses at $I_H=0.56$ and $I_L=0.211$. However, as you can see below (the red curve refers to the first function), my graph intersects the x and y axes differently compared to the paper. What I am doing wrong?
Best Answer
I think the following is wrong.
It should be \begin{equation} I_H= I_L + \frac{A_M}{A_L} \color{red}{\frac{\alpha_M (I_L)}{\alpha_L (I_L)}} \frac{M}{L} I_L \end{equation}
The two equations can be simplified as $$y=x-C_1(1-x)F_1(x)\tag1$$ $$x=y+\frac{C_2y}{F_2(y)}\tag2$$ where $$F_1(x):=\frac{\alpha_M(x)}{\alpha_H(x)},F_2(y):=\frac{\alpha_L(y)}{\alpha_M(y)}$$ are continuously differentiable and strictly decreasing with $$C_1:=\frac{A_MM}{A_HH},C_2:=\frac{A_M M}{A_LL},0 < y < x < 1$$
For $(2)$, we have $x\to 0$ as $y\to 0^+$ while the two curves of the paper do not pass through the origin.
So, I think the graph of the paper might not be very rigorous (about how they intersect axes).
Added :
I'm going to write what I've observed.
In the following, let us assume that $C_1,C_2,F_1(x),F_2(y)$ are positive.
About $(1)$ :
$\frac{dy}{dx}$ is positive since $$\frac{dy}{dx}=1+C_1F_1(x)+C_1\underbrace{(x-1)F_1'(x)}_{\text{positive}}\gt 0$$
We have $y\to 1$ as $x\to 1^-$.
It passes through $(0, -C_1F_1(0))$ where $-C_1F_1(0)\lt 0$.
About $(2)$ :
$\frac{dy}{dx}$ is positive since $$\frac{dy}{dx}=\frac{(F_2(y))^2}{(F_2(y))^2+C_2\underbrace{(F_2(y)-yF_2'(y))}_{\text{positive}}}\gt 0$$
We have $x\to 0$ as $y\to 0^+$.
It passes through $(1,\alpha)$ where $0\lt \alpha\lt 1$ since for $y=1$, we have $x=1+\frac{C_2}{F_2(1)}\gt 1$.
These imply that the two curves have at least one intersection point in $0\lt x\lt 1$.
However, since we know almost nothing about $F_1(x)$ and $F_2(y)$, I think it should be impossible to know how the two curves look like.