Challenging question on mathematical economics

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

Consider the following system of two equations:

$$$$\frac{A_M \alpha_M (I_H) M}{I_H - I_L} = \frac{A_H \alpha_H (I_H) H}{1- I_H}$$$$

$$$$\frac{A_L \alpha_L (I_L) L}{I_L} = \frac{A_M \alpha_M (I_L) M}{I_H - I_L}$$$$

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

$$$$I_L = I_H - \frac{A_M}{A_H} \frac{\alpha_M(I_H)}{\alpha_H(I_H)} \frac{M}{H}(1-I_H)$$$$

$$$$I_H= I_L + \frac{A_M}{A_L} \frac{\alpha_M (I_L)}{\alpha_L (I_L)} \frac{M}{L} I_L$$$$

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

$$$$y= x - (1.25/1.6)(e^{-0.3x}) (60/50) (1-x)$$$$

$$$$x= y + 1.25 (e^{0.6y}) (60/40)y$$$$

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?

• 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... Commented Jun 8 at 19:49
• I did not pay much attention when I placed the bounty. Your answer will certainly be welcomed Commented Jun 8 at 20:29
• Ok, :-). Tomorrow I hope I have the time to review and post it. Commented Jun 8 at 20:36
• 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. Commented Jun 9 at 3:17
• @Maximilian Thank you for accepting my answer. Your question was excellent, congratulations. Commented Jun 9 at 23:57

1 Answer

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:

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:

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

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

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

$$$$I_L = I_H - \frac{A_M}{A_H} \frac{\alpha_M(I_H)}{\alpha_H(I_H)} \frac{M}{H}(1-I_H) \tag {1'}$$$$

$$$$I_H= I_L + \frac{A_M}{A_L} \frac{\alpha_L (I_L)}{\alpha_M (I_L)} \frac{M}{L} I_L \tag {2'}$$$$

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:

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

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.