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I'm going through the paper of Acemoglu and Autor (2011, Handbook of Labor 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:

enter image description here

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_M (I_L)}{\alpha_L (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 curve cross at $I_H=0.541$ and $I_L=0.176$. 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? enter image description here

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  • $\begingroup$ What do you mean by 'an answer from a reputable source'? I can elaborate an answer, but it is a mathematical answer, the 'source' can be just some mathematics theorem... $\endgroup$ Commented Jun 8 at 19:49
  • $\begingroup$ I did not pay much attention when I placed the bounty. Your answer will certainly be welcomed $\endgroup$
    – Maximilian
    Commented Jun 8 at 20:29
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    $\begingroup$ Ok, :-). Tomorrow I hope I have the time to review and post it. $\endgroup$ Commented Jun 8 at 20:36
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    $\begingroup$ I am also curious, especially about the values at the corner. In equation (1), if x=0, then y < 0 (unless $\alpha_M(IH) < 0$). Similarly, in equation (2), if y=0, x=0 too. However, in the diagram shown, both corner values of (1) and (2) attain strictly positive values. Therefore, I think there must be some tricks added to your $\alpha$ functions. $\endgroup$
    – teddi
    Commented Jun 9 at 3:17
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    $\begingroup$ @Maximilian Thank you for accepting my answer. Your question was excellent, congratulations. $\endgroup$ Commented Jun 9 at 23:57

1 Answer 1

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What I say below in my answer is what I can say without having read Acemoglu and Autor's article (unfortunately I haven’t it), an answer based on what you report in your question, of course.

I think your graph is correct.

$0 < I_L < I_H < 1$; this condition always holds.

This condition implies that you have to consider only the part of the graph where $ 0 < I_L < 1$ and $0 < I_H < 1$, that is the part in the blue dotted square in the graph below, the $(0,1)\times (0,1)$ square:

enter image description here

Moreover, I have added the 45° green line, as the condition quoted above says that it is always $I_H>I_L$, so that the graph should be comprised in the part of the plane below the 45° green line.

This condition is respected in your graph, in the $(0,1)\times (0,1)$ square.

Instead, this condition is not respected in the graph of Acemoglu and Autor's paper you posted.

As I said, I didn't read the paper of Acemoglu and Autor but, on the basis of what you refer, I suppose there is an inaccuracy in the graph.

According to the condition $I_L < I_H $, the values of the functions for $I_H=0$ can't be strictly positive.

Indeed, in your graph you have a $0$ value for the blue function and a negative value for the red function. As I show also below, it seems to me that using functions like those in your graph, everything fits. $$***$$

How can Acemoglu and Autor's graph be qualitatively obtained?

We have, in addition, the assumption:

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 $i \in [0,1]$.

How can we obtain a qualitative graph of the two functions, without assuming a particular function for the $\alpha_j(i)$?

The graph of the two functions can be qualitatively depicted evaluating $I_L$ when $L_H=0$, and calculating the derivatives and evaluating their signs.

I re-write below the two equations of Acemoglu and Autor in implicit form, to use them in the following:

\begin{equation} f_1(I_L, I_H)=\frac{A_M \alpha_M (I_H) M}{I_H - I_L} - \frac{A_H \alpha_H (I_H) H}{1- I_H}=0 \tag{1}\end{equation}

\begin{equation} f_2 (I_L, I_H)=\frac{A_L \alpha_L (I_L) L}{I_L} - \frac{A_M \alpha_M (I_L) M}{I_H - I_L}=0 \tag{2}\end{equation}

Moreover, it could be useful to write the equations in terms of $I_L$ and $I_H$ as you did in your question:

\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) \tag {1'}\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 \tag {2'}\end{equation}

I suppose, from your question, that the variables are all positive, so we can see from $(1')$ that for $I_H=0$ $I_L$ assumes a negative value, as in your graph (the red fuction).

From $(2')$ it can be easily seen that the function starts from the origin (the blue function) (of course, then, we must exclude the origin as by assumption $I_L\neq0$ and $I_H\neq 0$ and of course equation $(2')$ is equivalent to $(2)$ for $I_L\neq 0$ only, and for $I_L=0$ function $(2)$ is not defined).

Now, let's calculate the derivatives, and evaluate their signs.

For equation $(2)$ I use the implicit function theorem. I re-write it as:

\begin{equation} f_2 (I_L, I_H)= A_L \frac {\alpha_L(I_L)}{a_M(I_L)} L(I_H-I_L) - A_M M I_L=0 \tag{2''}\end{equation}

From $(2'')$ we can calculate the derivative ($I_{L2}$ is the function implicitly defined by $(2)$ or $(2'')$)

$$\frac{dI_{L2}}{dI_H}=- \frac { \frac {\partial f_2}{\partial I_H}} {\frac{\partial f_2}{\partial I_L}}= \frac {- A_L \frac {\alpha_L(I_L)}{a_M(I_L)}L}{-A_L \frac {\alpha_L(I_L)}{_M(I_L)}L +(I_H-I_L)A_L L\left (\frac{\alpha_L(I_L)}{\alpha_M(I_L)}\right)'-A_M M}>0 $$

where $\left(\frac{\alpha_L(I_L)}{\alpha_M(I_L)}\right)'$ is the derivative of the ratio with respect to $I_L$, which is negative by assumption.

Analogously, we can calculate the derivative for equation $(1)$ using the implicit function theorem, or simply differentiating $(1')$, obtaining that its sign is positive.

In conclusion, we can draw a graph similar to your graph.

$$***$$

The point left to understand is whether the two functions cross in the $(0,1)\times (0,1)$ square, that is an equilibrium exists (in your particular example, yes).

We can resort to the intermediate value theorem, applied to the difference function $g(I_H)$ between the two functions defined by $(1)$ and $(2)$ above, on the interval $[0,1]$. That is, define the function:

$$g(I_H)= I_{L1}(I_H)- I_{L2}(I_H)$$

where $I_{L1}$ and $I_{L2}$ are the functions implicitly defined by $(1)$ and $(2)$, (and explicitated in $(2')$ and $(1')$), that is respectively the red and blue functions in the graph.

For $I_H=0$ we have already seen that $g(I_H) <0$.

For $I_H=1$ we can show that $g(I_H)>0$.

Indeed, from $(1')$ it is easily seen that for $I_H=1$ the function $I_{L1}=1$ (the red function), as also in your graph.

From equation $(2')$, we can see that, for $I_H=1$, $I_{L2}<1$ (the blue function) (I omit the details otherwise this post becomes cumbersome).

Therefore, $g(I_H)$ is positive for $I_H=1$ .

As $g(I_H)$ is a continuous function, as difference of two continuous functions, we can conclude, from the intermediate value theorem, that there must be a point in $(0,1)$ where $g(I_H)=0$, that is where the two red and blue functions in the graph cross.

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