<p>I just want to add an observation, too long for a comment, to the excellent explanations of smcc.</p>
<p>You wrote</p>
<blockquote>
<p>[...]if my answer is correct; especially the step where I write &quot;<em>since the question does not say
which specific solution of system (1) we should consider, I just
choose to begin by setting <span class="math-container">$e_1 = e_2 = 1$</span> and <span class="math-container">$\alpha =
 \frac{1}{2}$</span>.</em></p>
</blockquote>
<p><strong>Setting <span class="math-container">$\alpha=1/2$</span> is correct from a mathematical point of view:</strong> you find the derivatives just around that point.</p>
<p>However, not only this assumption is not necessary, but <strong>makes your answer too restrictive</strong>, and without much sense from an economic point of view: this way you will know how the model works just at <strong>one</strong> point.</p>
<p>In <strong>comparative static</strong> we want to know the signs of the derivatives more in general, and to have chiefly <strong>qualitative results</strong>, in order to have  results significant from an economic point of view.</p>
<p>If you set <span class="math-container">$\alpha=1/2$</span>, you  know, for example,  that the derivative of <span class="math-container">$x_1$</span> with respect to <span class="math-container">$\alpha$</span> is  positive for <span class="math-container">$\alpha=1/2$</span>, and then? For other values of <span class="math-container">$\alpha$</span>  how is the derivative? It can have in principle any sign, different from point to point.</p>
<p><strong>This is not very theoretical informative, and not very useful from an economic point of view. <span class="math-container">$^1$</span></strong></p>
<p>Therefore, I suppose,  more general comparative static results are requested in your exercise.</p>
<p>So, avoid restrictions, and in your calculations just take <span class="math-container">$e_1$</span> and
<span class="math-container">$e_2$</span> as constant (the required derivative is only with respect to <span class="math-container">$\alpha$</span>)  and let <span class="math-container">$\alpha$</span> be a variable.</p>
<p>Of course, the condition that the determinant of the Jacobian must be different from zero still holds, but simply <strong>your static comparative results will hold only at the points where this determinant is different from zero.</strong></p>
<p>This way <strong>the comparative results you find would hold on all the set under  consideration, <span class="math-container">$\mathbb{R^n_+}$</span> I suppose</strong>, except possible points where the Jacobian determinant is  zero.</p>
<hr />
<p><span class="math-container">$^1$</span> <sup> 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  <span class="math-container">$1000$</span>. For other values you don't know. How can this  be useful for an economic policy?</sup></p>

CC BY-SA 4.0
2023-08-03T20:40:38.617Z
Thank you so much! That makes a lot of sense!
Beerus
CC BY-SA 4.0
2023-08-03T20:41:09.383Z
You are welcome!
BakerStreet
CC BY-SA 4.0
2023-08-03T19:57:52.050Z
2023-08-05T00:07:20.070Z
2023-08-05T00:07:20.070Z
BakerStreet
<p>If you assume <span class="math-container">$e_1=e_2=1$</span> as you have done (but not that <span class="math-container">$\alpha=1/2$</span>), then the solution (as given in equation (48) in Simon and Blume) is:</p>
<p><span class="math-container">$$
\begin{align*}
x_1=y_1&amp;=\alpha\\
x_2=y_2&amp;=1-\alpha\\
p&amp;=\frac{\alpha}{1-\alpha}
\end{align*}
$$</span></p>
<p>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 <span class="math-container">$\alpha$</span>. You can just differentiate the expressions above:</p>
<p><span class="math-container">$$
\begin{align*}
\frac{\partial x_1}{\partial \alpha}=\frac{\partial y_1}{\partial \alpha}&amp;=1\\
\frac{\partial x_2}{\partial \alpha}=\frac{\partial y_2}{\partial \alpha}&amp;=-1\\
\frac{\partial p}{\partial \alpha}&amp;=\frac{1}{(1-\alpha)^2}
\end{align*}
$$</span></p>
<p>When <span class="math-container">$\alpha=1/2$</span>, this is the same as what you have obtained (though your expressions have a superfluous <span class="math-container">$d\alpha$</span> on the right).</p>
<hr />
<p>Since an increase in <span class="math-container">$\alpha$</span> increases the weight on the utility from good <span class="math-container">$x$</span>, the signs of the comparative statics results should not be surprising. An increase in the desirability of good <span class="math-container">$x$</span> raises its price. Consumer <span class="math-container">$1$</span> has all of the endowment of good <span class="math-container">$x$</span> and so is now wealthier, while consumer <span class="math-container">$2$</span> is less wealthy. Hence consumer <span class="math-container">$1$</span>'s consumption of both goods rises while consumer <span class="math-container">$2$</span>'s consumption falls.</p>
<hr />
<p>It is not clear whether Simon and Blume want you to solve the problem for <span class="math-container">$e_1=e_2=1$</span> or more generally. However, given the question appears in a chapter on the implicit function theorem, probably they want you to solve it when <span class="math-container">$e_1$</span> and <span class="math-container">$e_2$</span> are not specified. (It is the equality of <span class="math-container">$e_1$</span> and <span class="math-container">$e_2$</span> that makes it possible to find an explicit solution and make the comparative statics with respect to <span class="math-container">$\alpha$</span> easy to calculate.)</p>

CC BY-SA 4.0
2023-08-03T17:08:52.263Z
Thank you so much for your explanation! I think your economic interpretation of the comparative statics makes sense. As for the computation, the reason I specified all $e_1 = 1$, $e_2 = 1$, and $\alpha = \frac{1}{2}$ is that the Implicit Function Theorem only allows us to compute $\frac{\partial F_k}{\partial x_h}$ within an open ball around a specific solution to the system of equations. \\ In addition, I think the textbook example treated $\alpha$ as a constant (instead of a variable). However, the exercise asked us to treat $\alpha$ as a variable. That is why I put a $d\alpha$ there.
Beerus
CC BY-SA 4.0
2023-08-03T17:15:08.863Z
It is indeed not clear whether Simon and Blume want me to solve it for $e_1 = 1$, $e_2 = 1$, and $\alpha = \frac{1}{2}$. So I just follow their example, $e_1  = e_2 = 1$ , and randomly assign $\alpha = \frac{1}{2}$ because of the requirement of the Implicit Function Theorem.
Beerus
CC BY-SA 4.0
2023-08-03T17:36:00.793Z
It is not necessary to specify the values of $e_1$, $e_2$ and $\alpha$. As long as they are such that there is a solution and the required condition on the Jacobian holds, then you can apply the implicit function theorem to find the derivatives of the demands and price. 

You are correct that the $d\alpha$ should be there in your second-to-last set of equations (the one with the differentials), but the last set of equations (the one with the partial derivatives) should not have the $d\alpha$ as you have $\partial \cdot/\partial \alpha$ on the lefthand side.
smcc
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2023-08-03T20:26:44.283Z
Ohhhh, you are right! I have amended it. Thanks a lot!
Beerus
CC BY-SA 4.0
2023-08-03T12:59:25.897Z
2023-08-03T19:08:45.350Z
2023-08-03T19:08:45.350Z
smcc
<p>I am studying Implicit Function Theorem and its application on comparative statics using <em>Mathematics for Economists</em> by Simon and Blume. Here is the question:</p>
<blockquote>
<p>Consider a pure exchange economy with two consumers, numbered <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, as well as two consumption goods, <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Suppose that consumer <span class="math-container">$1$</span> has initial endowment <span class="math-container">$(e_1, 0)$</span>, and that consumer <span class="math-container">$2$</span> has initial endowment <span class="math-container">$(0, e_2)$</span>. To describe the consumers' preferences, let <span class="math-container">$u_1$</span> and <span class="math-container">$u_2$</span> be <span class="math-container">$C^2$</span>, strictly concave (<span class="math-container">$u_i^{''} &lt; 0$</span>) functions of a single variable and let <span class="math-container">$\alpha$</span> be a scalar between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>. For <span class="math-container">$i = 1, 2$</span>, we assume that consumer <span class="math-container">$i$</span>'s preferences over consumption bundles <span class="math-container">$(x, y)$</span> are described by the utility function
<span class="math-container">\begin{equation}
U_i(x_i, y_i) = \alpha u_i(x_i) + (1-\alpha)u_i(y_i)
\end{equation}</span>
Let <span class="math-container">$p$</span> and <span class="math-container">$q$</span> denote the price of a unit of good <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, respectively. <strong>Compute and interpret the comparative statics that results from an increase in <span class="math-container">$\alpha$</span>.</strong></p>
</blockquote>
<p>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 <span class="math-container">$\alpha$</span>. I would really appreciate it if someone could help me with the interpretation!</p>
<p>(I know there is a an Answer Pamphlet, but there is no solution to this exercise.)</p>
<p>Here is my attempt:</p>
<blockquote>
<p>The constrained optimization problem is:
<span class="math-container">\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}</span>
Setting the MRS equal to the price ratio, we have
<span class="math-container">\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}</span>
Setting good <span class="math-container">$2$</span> as the numeraire (<span class="math-container">$q = 1$</span>), we have the following system of equations describing the optimal choices for consumers <span class="math-container">$1$</span> and <span class="math-container">$2$</span>:
<span class="math-container">\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}</span>
<em><strong>Denote it system <span class="math-container">$(1)$</span></strong></em>. <em>Since the question does not say which specific solution of system (1) we should consider, I just choose to begin by setting <span class="math-container">$e_1 = e_2 = 1$</span> and <span class="math-container">$\alpha = \frac{1}{2}$</span>.</em> In this case, the unique solution of system <span class="math-container">$(1)$</span> is:
<span class="math-container">\begin{equation}
x_1 = y_1 = x_2 = y_2 = \frac{1}{2} \\
p = 1
\end{equation}</span>
<em><strong>Denote it system <span class="math-container">$(2)$</span></strong></em>. We ask how a change in <span class="math-container">$\alpha$</span> affects the equilibrium consumption bundles a prices, keeping <span class="math-container">$e_1$</span> and <span class="math-container">$e_2$</span> fixed. Since, at the point <span class="math-container">$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, 1, 1, 1, \frac{1}{2})$</span>,
<span class="math-container">\begin{equation}
det \begin{pmatrix} \frac{\partial F_1}{\partial x_1} &amp; \frac{\partial F_1}{\partial x_2} &amp; \frac{\partial F_1}{\partial y_1} &amp; \frac{\partial F_1}{\partial y_2} &amp; \frac{\partial F_1}{\partial p} \\ 
\frac{\partial F_2}{\partial x_1} &amp; \frac{\partial F_2}{\partial x_2} &amp; \frac{\partial F_2}{\partial y_1} &amp; \frac{\partial F_2}{\partial y_2} &amp; \frac{\partial F_2}{\partial p} \\ 
\frac{\partial F_3}{\partial x_1} &amp; \frac{\partial F_3}{\partial x_2} &amp; \frac{\partial F_3}{\partial y_1} &amp; \frac{\partial F_3}{\partial y_2} &amp; \frac{\partial F_3}{\partial p} \\
\frac{\partial F_4}{\partial x_1} &amp; \frac{\partial F_4}{\partial x_2} &amp; \frac{\partial F_4}{\partial y_1} &amp; \frac{\partial F_4}{\partial y_2} &amp; \frac{\partial F_4}{\partial p} \\
\frac{\partial F_5}{\partial x_1} &amp; \frac{\partial F_5}{\partial x_2} &amp; \frac{\partial F_5}{\partial y_1} &amp; \frac{\partial F_5}{\partial y_2} &amp; \frac{\partial F_5}{\partial p} \end{pmatrix} = \begin{pmatrix} u_1^{''}(\frac{1}{2}) &amp; 0 &amp; -u_1^{''}(\frac{1}{2}) &amp; 0 &amp; -u_1^{'}(\frac{1}{2}) \\
1 &amp; 0 &amp; 1 &amp; 0 &amp; \frac{1}{2} \\
0 &amp; u_2^{''}(\frac{1}{2}) &amp; 0 &amp; -u_2^{''}(\frac{1}{2}) &amp; -u_2^{'}(\frac{1}{2}) \\
1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 1 &amp; 1 &amp; 0 \end{pmatrix} \neq 0,
\end{equation}</span>
system <span class="math-container">$(1)$</span> can be solved for <span class="math-container">$x_1$</span>, <span class="math-container">$x_2$</span>, <span class="math-container">$y_1$</span>, <span class="math-container">$y_2$</span>, <span class="math-container">$p$</span> as a function of <span class="math-container">$e_1$</span>, <span class="math-container">$e_2$</span>, <span class="math-container">$\alpha$</span> near <span class="math-container">$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, 1, 1, 1, \frac{1}{2})$</span>. The linearization of system <span class="math-container">$(1)$</span> is
<span class="math-container">\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}</span>
<em><strong>Denote it system <span class="math-container">$(3)$</span></strong></em>. Set <span class="math-container">$de_1 = de_2 = 0$</span>. We first solve the last two equations in system <span class="math-container">$(3)$</span> for <span class="math-container">$dx_1$</span> and <span class="math-container">$dy_1$</span>:
<span class="math-container">\begin{equation}
dx_1 = -dx_2 \\
dy_1 = -dy_2
\end{equation}</span>
Substitute these expressions for <span class="math-container">$dx_1$</span> and <span class="math-container">$dy_1$</span> into the first three equations of system <span class="math-container">$(3)$</span> at <span class="math-container">$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, 1, 1, 1, \frac{1}{2})$</span>, we have:
<span class="math-container">\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}</span>
Multiply the first equation through by <span class="math-container">$\frac{\frac{1}{2}}{u_1^{'}(\frac{1}{2})}$</span> and the third equation through by <span class="math-container">$\frac{\frac{1}{2}}{u_2^{'}(\frac{1}{2})}$</span>, we have:
<span class="math-container">\begin{equation}
\begin{pmatrix} -\frac{\frac{1}{2}u_1^{''}(\frac{1}{2})}{u_1^{'}(\frac{1}{2})} &amp; \frac{\frac{1}{2}u_1^{''}(\frac{1}{2})}{u_1^{'}(\frac{1}{2})} &amp; -\frac{1}{2} \\
-1 &amp; -1 &amp; -\frac{1}{2} \\
\frac{\frac{1}{2}u_2^{''}(\frac{1}{2})}{u_2^{'}(\frac{1}{2})} &amp; -\frac{\frac{1}{2}u_2^{''}(\frac{1}{2})}{u_2^{'}(\frac{1}{2})} &amp; -\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}</span>
<em><strong>Denote it system <span class="math-container">$(4)$</span></strong></em>. Let <span class="math-container">$r_i(z) = -\frac{zu_i^{''}(z)}{u_i^{'}(z)}$</span> be the Arrow-Pratt measure of relative risk aversion. We have that <span class="math-container">$r_i(z)$</span> is strictly positive for <span class="math-container">$i = 1, 2$</span>. Rewrite system <span class="math-container">$(4)$</span> as:
<span class="math-container">\begin{equation}
\begin{pmatrix} r_1(\frac{1}{2}) &amp; -r_1(\frac{1}{2}) &amp; -\frac{1}{2} \\ -1 &amp; -1 &amp; -\frac{1}{2} \\ -r_2(\frac{1}{2}) &amp; r_2(\frac{1}{2}) &amp; -\frac{1}{2} \end{pmatrix}
\begin{pmatrix} dx_2 \\ dy_2 \\ dp \end{pmatrix} = 
\begin{pmatrix} -2d\alpha &amp; 0 &amp; -2d\alpha \end{pmatrix}
\end{equation}</span>
Let <span class="math-container">$R_1 = r_1(\frac{1}{2}) &gt; 0$</span>, <span class="math-container">$R_2 = r_2(\frac{1}{2}) &gt; 0$</span>, and
<span class="math-container">\begin{equation}
D = det\begin{pmatrix} R_1 &amp; -R_1 &amp; -\frac{1}{2} \\ -1 &amp; -1 &amp; -\frac{1}{2} \\ -R_2 &amp; R_2 &amp; -\frac{1}{2} \end{pmatrix} = R_1 + R_2 &gt; 0
\end{equation}</span>
Then by Cramer's Rule, we have
<span class="math-container">\begin{equation}
dx_2 = \frac{det\begin{pmatrix} -2d\alpha &amp; -R_1 &amp; \frac{1}{2} \\ 0 &amp; -1 &amp; -\frac{1}{2} \\ -2d\alpha &amp; R_2 &amp; -\frac{1}{2} \end{pmatrix}}{D} = -\frac{R_1 + R_2}{D}d\alpha \\
dy_2 = \frac{det\begin{pmatrix} R_1 &amp; -2d\alpha &amp; -\frac{1}{2} \\ -1 &amp; 0 &amp; -\frac{1}{2} \\ -R_2 &amp; -2d\alpha &amp; -\frac{1}{2} \end{pmatrix}}{D} = -\frac{R_1 + R_2}{D}d\alpha \\
dp = \frac{det\begin{pmatrix} R_1 &amp; -R_1 &amp; -2d\alpha \\ -1 &amp; -1 &amp; 0 \\ -R_2 &amp; R_2 &amp; -2d\alpha \end{pmatrix}}{D} = \frac{4(R_1 + R_2)}{D}d\alpha
\end{equation}</span>
Therefore,
<span class="math-container">\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}</span></p>
</blockquote>
<p>Again, I would like to know if my answer is correct; especially the step where I write &quot;<em>since the question does not say which specific solution of system (1) we should consider, I just choose to begin by setting <span class="math-container">$e_1 = e_2 = 1$</span> and <span class="math-container">$\alpha = \frac{1}{2}$</span>.</em>&quot;</p>
<p>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 <span class="math-container">$\alpha$</span>.</p>

<microeconomics><mathematical-economics><optimization><linear-algebra><comparative-statics>
Question About Implicit Function Theorem and Comparative Statics - Mathematics for Economists by Simon and Blume Chapter 15 Exercise 32
Beerus
# Question About Implicit Function Theorem and Comparative Statics - Mathematics for Economists by Simon and Blume Chapter 15 Exercise 32

(Asked by: username_0 on 2023-08-02)

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 $\alpha$.

**Tags:** microeconomics, mathematical-economics, optimization, linear-algebra, comparative-statics


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## 2 answers

### Answer by username_2 on 2023-08-03. Score: 3 | Accepted answer

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.)

> **Comments:**
> **username_0**: Thank you so much for your explanation! I think your economic interpretation of the comparative statics makes sense. As for the computation, the reason I specified all $e_1 = 1$, $e_2 = 1$, and $\alpha = \frac{1}{2}$ is that the Implicit Function Theorem only allows us to compute $\frac{\partial F_k}{\partial x_h}$ within an open ball around a specific solution to the system of equations. \\ In addition, I think the textbook example treated $\alpha$ as a constant (instead of a variable). However, the exercise asked us to treat $\alpha$ as a variable. That is why I put a $d\alpha$ there.  
> **username_0**: It is indeed not clear whether Simon and Blume want me to solve it for $e_1 = 1$, $e_2 = 1$, and $\alpha = \frac{1}{2}$. So I just follow their example, $e_1  = e_2 = 1$ , and randomly assign $\alpha = \frac{1}{2}$ because of the requirement of the Implicit Function Theorem.  
> **username_2**: It is not necessary to specify the values of $e_1$, $e_2$ and $\alpha$. As long as they are such that there is a solution and the required condition on the Jacobian holds, then you can apply the implicit function theorem to find the derivatives of the demands and price.   You are correct that the $d\alpha$ should be there in your second-to-last set of equations (the one with the differentials), but the last set of equations (the one with the partial derivatives) should not have the $d\alpha$ as you have $\partial \cdot/\partial \alpha$ on the lefthand side.  
> **username_0**: Ohhhh, you are right! I have amended it. Thanks a lot!  

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### Answer by username_1 on 2023-08-03. Score: 1

I just want to add an observation, too long for a comment, to the excellent explanations of username_2.

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?

> **Comments:**
> **username_0**: Thank you so much! That makes a lot of sense!  
> **username_1**: You are welcome!  

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