Question About Implicit Function Theorem and Comparative Statics - Mathematics for Economists by Simon and Blume Chapter 15 Exercise 32

Question

I am studying Implicit Function Theorem and its application on comparative statics using Mathematics for Economists by Simon and Blume. Here is the question:

Consider a pure exchange economy with two consumers, numbered $1$ and $2$, as well as two consumption goods, $x$ and $y$. Suppose that consumer $1$ has initial endowment $(e_1, 0)$, and that consumer $2$ has initial endowment $(0, e_2)$. To describe the consumers' preferences, let $u_1$ and $u_2$ be $C^2$, strictly concave ($u_i^{''} < 0$) functions of a single variable and let $\alpha$ be a scalar between $0$ and $1$. For $i = 1, 2$, we assume that consumer $i$'s preferences over consumption bundles $(x, y)$ are described by the utility function \begin{equation} U_i(x_i, y_i) = \alpha u_i(x_i) + (1-\alpha)u_i(y_i) \end{equation} Let $p$ and $q$ denote the price of a unit of good $1$ and $2$, respectively. Compute and interpret the comparative statics that results from an increase in $\alpha$.

I tried this question myself. Could someone please help me check whether my answer is correct or not? In addition, I have difficulties interpreting the economic meaning or intuition of the comparative statics that results from a change in $\alpha$. I would really appreciate it if someone could help me with the interpretation!

(I know there is a an Answer Pamphlet, but there is no solution to this exercise.)

Here is my attempt:

The constrained optimization problem is: \begin{equation} Max \space\space\space\space U_i(x_i, y_i) = \alpha u_i(x_i) + (1-\alpha)u_i(y_i) \\ s.t. \space\space\space\space px_i + qy_i = value \space of \space initial \space endowment. \end{equation} Setting the MRS equal to the price ratio, we have \begin{equation} \frac{\frac{\partial U_i}{\partial x_i}(x_i, y_i)}{\frac{\partial U_i}{\partial y_i}(x_i, y_i)} = \frac{\alpha u_i^{'}(x_i)}{(1-\alpha)u_i^{'}(y_i)} = \frac{p}{q}. \end{equation} Setting good $2$ as the numeraire ($q = 1$), we have the following system of equations describing the optimal choices for consumers $1$ and $2$: \begin{equation} F_1(x_1, x_2, y_1, y_2, p, e_1, e_2, \alpha) = \frac{\alpha}{1 - \alpha}u_1^{'}(x_1) - pu_1^{'}(y_1) = 0 \\ F_2(x_1, x_2, y_1, y_2, p, e_1, e_2, \alpha) = px_1 + y_1 - pe_1 = 0 \space\space\space\space\space\space\space\space\space\space\space\space\space \\ F_3(x_1, x_2, y_1, y_2, p, e_1, e_2, \alpha) = \frac{\alpha}{1 - \alpha}u_2^{'}(x_2) - pu_2^{'}(y_2) = 0 \\ F_4(x_1, x_2, y_1, y_2, p, e_1, e_2, \alpha) = x_1 + x_2 - e_1 = 0 \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \\ F_5(x_1, x_2, y_1, y_2, p, e_1, e_2, \alpha) = y_1 + y_2 - e_2 = 0 \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \end{equation} Denote it system $(1)$. Since the question does not say which specific solution of system (1) we should consider, I just choose to begin by setting $e_1 = e_2 = 1$ and $\alpha = \frac{1}{2}$. In this case, the unique solution of system $(1)$ is: \begin{equation} x_1 = y_1 = x_2 = y_2 = \frac{1}{2} \\ p = 1 \end{equation} Denote it system $(2)$. We ask how a change in $\alpha$ affects the equilibrium consumption bundles a prices, keeping $e_1$ and $e_2$ fixed. Since, at the point $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, 1, 1, 1, \frac{1}{2})$, \begin{equation} det \begin{pmatrix} \frac{\partial F_1}{\partial x_1} & \frac{\partial F_1}{\partial x_2} & \frac{\partial F_1}{\partial y_1} & \frac{\partial F_1}{\partial y_2} & \frac{\partial F_1}{\partial p} \\ \frac{\partial F_2}{\partial x_1} & \frac{\partial F_2}{\partial x_2} & \frac{\partial F_2}{\partial y_1} & \frac{\partial F_2}{\partial y_2} & \frac{\partial F_2}{\partial p} \\ \frac{\partial F_3}{\partial x_1} & \frac{\partial F_3}{\partial x_2} & \frac{\partial F_3}{\partial y_1} & \frac{\partial F_3}{\partial y_2} & \frac{\partial F_3}{\partial p} \\ \frac{\partial F_4}{\partial x_1} & \frac{\partial F_4}{\partial x_2} & \frac{\partial F_4}{\partial y_1} & \frac{\partial F_4}{\partial y_2} & \frac{\partial F_4}{\partial p} \\ \frac{\partial F_5}{\partial x_1} & \frac{\partial F_5}{\partial x_2} & \frac{\partial F_5}{\partial y_1} & \frac{\partial F_5}{\partial y_2} & \frac{\partial F_5}{\partial p} \end{pmatrix} = \begin{pmatrix} u_1^{''}(\frac{1}{2}) & 0 & -u_1^{''}(\frac{1}{2}) & 0 & -u_1^{'}(\frac{1}{2}) \\ 1 & 0 & 1 & 0 & \frac{1}{2} \\ 0 & u_2^{''}(\frac{1}{2}) & 0 & -u_2^{''}(\frac{1}{2}) & -u_2^{'}(\frac{1}{2}) \\ 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \end{pmatrix} \neq 0, \end{equation} system $(1)$ can be solved for $x_1$, $x_2$, $y_1$, $y_2$, $p$ as a function of $e_1$, $e_2$, $\alpha$ near $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, 1, 1, 1, \frac{1}{2})$. The linearization of system $(1)$ is \begin{equation} \frac{\alpha}{1 - \alpha}u_1^{''}(x_1)dx_1 + 0dx_2 - pu_1^{''}(y_1)dy_1 + 0dy_2 - u_1^{'}(y_1)dp = 0de_1 + 0de_2 - \frac{1}{(1 - \alpha)^2}u_1^{'}(x_1)d\alpha \\ pdx_1 + 0dx_2 + 1dy_1 + 0dy_2 - (e_1 - \alpha)dp = pde_1 + 0de_2 + 0d\alpha \\ 0dx_1 + \frac{\alpha}{1 - \alpha}u_2^{''}(x_2)dx_2 + 0dy_1 - pu_2^{''}(y_2)dy_2 - u_2^{'}(y_2)dp = 0de_1 + 0de_2 - \frac{1}{(1 - \alpha)^2}u_2^{'}(x_2)d\alpha \\ 1dx_1 + 1dx_2 + 0dy_1 + 0dy_2 + 0dp = 1de_1 + 0de_2 + 0d\alpha \\ 0dx_1 + 0dx_2 + 1dy_1 + 1dy_2 + 0dp = 0de_1 + 1de_2 + 0d\alpha \end{equation} Denote it system $(3)$. Set $de_1 = de_2 = 0$. We first solve the last two equations in system $(3)$ for $dx_1$ and $dy_1$: \begin{equation} dx_1 = -dx_2 \\ dy_1 = -dy_2 \end{equation} Substitute these expressions for $dx_1$ and $dy_1$ into the first three equations of system $(3)$ at $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, 1, 1, 1, \frac{1}{2})$, we have: \begin{equation} -u_1^{''}(\frac{1}{2})dx_2 + u_1^{''}(\frac{1}{2})dy_2 - u_1^{'}(\frac{1}{2})dp = -4u_1^{'}(\frac{1}{2})d\alpha \\ -1dx_2 - 1dy_2 - \frac{1}{2}dp = 0d\alpha \\ u_2^{''}(\frac{1}{2})dx_2 - u_2^{''}(\frac{1}{2})dy_2 - u_2^{'}(\frac{1}{2})dp = -4u_2^{'}(\frac{1}{2})d\alpha \end{equation} Multiply the first equation through by $\frac{\frac{1}{2}}{u_1^{'}(\frac{1}{2})}$ and the third equation through by $\frac{\frac{1}{2}}{u_2^{'}(\frac{1}{2})}$, we have: \begin{equation} \begin{pmatrix} -\frac{\frac{1}{2}u_1^{''}(\frac{1}{2})}{u_1^{'}(\frac{1}{2})} & \frac{\frac{1}{2}u_1^{''}(\frac{1}{2})}{u_1^{'}(\frac{1}{2})} & -\frac{1}{2} \\ -1 & -1 & -\frac{1}{2} \\ \frac{\frac{1}{2}u_2^{''}(\frac{1}{2})}{u_2^{'}(\frac{1}{2})} & -\frac{\frac{1}{2}u_2^{''}(\frac{1}{2})}{u_2^{'}(\frac{1}{2})} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} dx_2 \\ dy_2 \\ dp \end{pmatrix} = \begin{pmatrix} -2d\alpha \\ 0 \\ -2d\alpha \end{pmatrix} \end{equation} Denote it system $(4)$. Let $r_i(z) = -\frac{zu_i^{''}(z)}{u_i^{'}(z)}$ be the Arrow-Pratt measure of relative risk aversion. We have that $r_i(z)$ is strictly positive for $i = 1, 2$. Rewrite system $(4)$ as: \begin{equation} \begin{pmatrix} r_1(\frac{1}{2}) & -r_1(\frac{1}{2}) & -\frac{1}{2} \\ -1 & -1 & -\frac{1}{2} \\ -r_2(\frac{1}{2}) & r_2(\frac{1}{2}) & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} dx_2 \\ dy_2 \\ dp \end{pmatrix} = \begin{pmatrix} -2d\alpha & 0 & -2d\alpha \end{pmatrix} \end{equation} Let $R_1 = r_1(\frac{1}{2}) > 0$, $R_2 = r_2(\frac{1}{2}) > 0$, and \begin{equation} D = det\begin{pmatrix} R_1 & -R_1 & -\frac{1}{2} \\ -1 & -1 & -\frac{1}{2} \\ -R_2 & R_2 & -\frac{1}{2} \end{pmatrix} = R_1 + R_2 > 0 \end{equation} Then by Cramer's Rule, we have \begin{equation} dx_2 = \frac{det\begin{pmatrix} -2d\alpha & -R_1 & \frac{1}{2} \\ 0 & -1 & -\frac{1}{2} \\ -2d\alpha & R_2 & -\frac{1}{2} \end{pmatrix}}{D} = -\frac{R_1 + R_2}{D}d\alpha \\ dy_2 = \frac{det\begin{pmatrix} R_1 & -2d\alpha & -\frac{1}{2} \\ -1 & 0 & -\frac{1}{2} \\ -R_2 & -2d\alpha & -\frac{1}{2} \end{pmatrix}}{D} = -\frac{R_1 + R_2}{D}d\alpha \\ dp = \frac{det\begin{pmatrix} R_1 & -R_1 & -2d\alpha \\ -1 & -1 & 0 \\ -R_2 & R_2 & -2d\alpha \end{pmatrix}}{D} = \frac{4(R_1 + R_2)}{D}d\alpha \end{equation} Therefore, \begin{equation} \frac{\partial x_1}{\partial \alpha} = \frac{R_1 + R_2}{D} \\ \frac{\partial x_2}{\partial \alpha} = -\frac{R_1 + R_2}{D} \\ \frac{\partial y_1}{\partial \alpha} = \frac{R_1 + R_2}{D} \\ \frac{\partial y_2}{\partial \alpha} = -\frac{R_1 + R_2}{D} \\ \frac{\partial p}{\partial \alpha} = \frac{4(R_1 + R_2)}{D} \end{equation}

Again, I would like to know if my answer is correct; especially the step where I write "since the question does not say which specific solution of system (1) we should consider, I just choose to begin by setting $e_1 = e_2 = 1$ and $\alpha = \frac{1}{2}$."

Moreover, I would really appreciate it if someone could help me with the interpretation of the comparative statics of that results from an increase in $\alpha$.

Answer 1

If you assume $e_1=e_2=1$ as you have done (but not that $\alpha=1/2$), then the solution (as given in equation (48) in Simon and Blume) is:

$$ \begin{align*} x_1=y_1&=\alpha\\ x_2=y_2&=1-\alpha\\ p&=\frac{\alpha}{1-\alpha} \end{align*} $$

Since in this case there are explicit expressions for the solution, it is not necessary to use the implicit function theorem to find the comparative statics with respect to $\alpha$. You can just differentiate the expressions above:

$$ \begin{align*} \frac{\partial x_1}{\partial \alpha}=\frac{\partial y_1}{\partial \alpha}&=1\\ \frac{\partial x_2}{\partial \alpha}=\frac{\partial y_2}{\partial \alpha}&=-1\\ \frac{\partial p}{\partial \alpha}&=\frac{1}{(1-\alpha)^2} \end{align*} $$

When $\alpha=1/2$, this is the same as what you have obtained (though your expressions have a superfluous $d\alpha$ on the right).

---

Since an increase in $\alpha$ increases the weight on the utility from good $x$, the signs of the comparative statics results should not be surprising. An increase in the desirability of good $x$ raises its price. Consumer $1$ has all of the endowment of good $x$ and so is now wealthier, while consumer $2$ is less wealthy. Hence consumer $1$'s consumption of both goods rises while consumer $2$'s consumption falls.

---

It is not clear whether Simon and Blume want you to solve the problem for $e_1=e_2=1$ or more generally. However, given the question appears in a chapter on the implicit function theorem, probably they want you to solve it when $e_1$ and $e_2$ are not specified. (It is the equality of $e_1$ and $e_2$ that makes it possible to find an explicit solution and make the comparative statics with respect to $\alpha$ easy to calculate.)

Answer 2

I just want to add an observation, too long for a comment, to the excellent explanations of smcc.

You wrote

[...]if my answer is correct; especially the step where I write "since the question does not say which specific solution of system (1) we should consider, I just choose to begin by setting $e_1 = e_2 = 1$ and $\alpha = \frac{1}{2}$.

Setting $\alpha=1/2$ is correct from a mathematical point of view: you find the derivatives just around that point.

However, not only this assumption is not necessary, but makes your answer too restrictive, and without much sense from an economic point of view: this way you will know how the model works just at one point.

In comparative static we want to know the signs of the derivatives more in general, and to have chiefly qualitative results, in order to have results significant from an economic point of view.

If you set $\alpha=1/2$, you know, for example, that the derivative of $x_1$ with respect to $\alpha$ is positive for $\alpha=1/2$, and then? For other values of $\alpha$ how is the derivative? It can have in principle any sign, different from point to point.

This is not very theoretical informative, and not very useful from an economic point of view. $^1$

Therefore, I suppose, more general comparative static results are requested in your exercise.

So, avoid restrictions, and in your calculations just take $e_1$ and $e_2$ as constant (the required derivative is only with respect to $\alpha$) and let $\alpha$ be a variable.

Of course, the condition that the determinant of the Jacobian must be different from zero still holds, but simply your static comparative results will hold only at the points where this determinant is different from zero.

This way the comparative results you find would hold on all the set under consideration, $\mathbb{R^n_+}$ I suppose, except possible points where the Jacobian determinant is zero.

---

$^1$ ^{Think of, for instance, a classic result of comparative static, the multipliers in macroeconomics. And imagine you know that the multiplier of government expenditure is positive only for a specific value, say if the government expenditure is $1000$. For other values you don't know. How can this be useful for an economic policy?}