<p>Totally differentiate each equation with respect to <span class="math-container">$(k,x,y,z)$</span>:</p>
<p><span class="math-container">\begin{align}
V(a)[p_{xx}dx+p_{xy}dy]&amp;=dk \tag{1}\\
V'(z)[p_xdx+p_ydy]+pV''(z)dz&amp;=0 \tag{2}\\
V(z)[p_{xy}dx+p_{yy}dy]+p_yV'(z)dz&amp;=0 \tag{3}
\end{align}</span></p>
<p>This system can be written in matrix form as</p>
<p><span class="math-container">$$\begin{pmatrix}
V(a)p_{xx} &amp; V(a)p_{xy} &amp; 0\\
V'(z)p_x &amp; V'(z)p_y &amp; pV''(z)\\
V(z)p_{xy} &amp; V(z)p_y &amp; p_yV'(z)
\end{pmatrix}\begin{pmatrix}dx\\ dy \\ dz\end{pmatrix}=\begin{pmatrix}dk\\ 0 \\ 0\end{pmatrix}\tag{*}$$</span></p>
<p>and solved by inverting the <span class="math-container">$3\times3$</span> matrix (assuming it is invertible):</p>
<p><span class="math-container">$$\begin{pmatrix}dx\\ dy \\ dz\end{pmatrix}=\begin{pmatrix}
V(a)p_{xx} &amp; V(a)p_{xy} &amp; 0\\
V'(z)p_x &amp; V'(z)p_y &amp; pV''(z)\\
V(z)p_{xy} &amp; V(z)p_y &amp; p_yV'(z)
\end{pmatrix}^{-1}\begin{pmatrix}dk\\ 0 \\ 0\end{pmatrix}=\begin{pmatrix}\frac{\partial x}{\partial k}\\ \frac{\partial y}{\partial k}\\\frac{\partial z}{\partial k}\end{pmatrix}dk.$$</span></p>
<p>However, if you are only interested in <span class="math-container">$\partial x/\partial k$</span>, and are doing this by hand, then this entails more working than necessary, so an alternative would be as follows:</p>
<p>Solving <span class="math-container">$(1)$</span> for <span class="math-container">$dy$</span> gives</p>
<p><span class="math-container">$$dy=\frac{1}{V(a)p_{xy}}dk-\frac{p_{xx}}{p_{xy}}dx$$</span></p>
<p>Substituting into <span class="math-container">$\frac{p_{xy}}{V'(z)}\times (2)$</span> and <span class="math-container">$\frac{p_{xy}}{V(z)}\times(3)$</span> gives</p>
<p><span class="math-container">\begin{align}
\left(p_xp_{xy}-p_yp_{xx}\right)dx+\frac{p_{xy}pV''(z)}{V'(z)}dz&amp;=-\frac{p_y}{V(a)}dk \tag{4}\\
\left(p_{xy}^2 -p_{xx}p_{yy}\right) dx+\frac{p_{xy}p_yV'(z)}{V(z)}dz&amp;=-\frac{p_{yy}}{V(a)}dk \tag{5}
\end{align}</span></p>
<p>Etc!</p>
<p>Just eliminate <span class="math-container">$dz$</span> from the equations to get an equation in the form <span class="math-container">$dx=Mdk$</span>. The expression for the partial derivative you want is <span class="math-container">$M$</span>.</p>
<hr />
<p>The more generally applicable method that will allow you to find the derivatives of <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, and <span class="math-container">$z$</span> with respect to <span class="math-container">$a$</span> and <span class="math-container">$k$</span> goes as follows:</p>
<p>Let <span class="math-container">$F:\mathbb{R}^5\to\mathbb{R}^3$</span> be given by</p>
<p><span class="math-container">\begin{align}
F_1(a,k,x,y,z)&amp;=p_x(x,y)V(a)-k\\
F_2(a,k,x,y,z)&amp;=p(x,y)V'(z)-1\\
F_3(a,k,x,y,z)&amp;=p_y(x,y)V(z)-1
\end{align}</span></p>
<p>With this notation, <span class="math-container">$F(a,k,x,y,z)=0$</span> is the equation that potentially defines <span class="math-container">$(x,y,z)$</span> as an implicit function of <span class="math-container">$(a,k)$</span>.</p>
<p>The Jacobian of <span class="math-container">$F$</span> with respect to <span class="math-container">$(x,y,z)$</span> is (the same <span class="math-container">$3\times 3$</span> matrix as in equation <span class="math-container">$(*)$</span> above):</p>
<p><span class="math-container">$$J_{F,(x,y,z)}\begin{pmatrix}
\frac{\partial F_1}{\partial x} &amp; \frac{\partial F_1}{\partial y}&amp; \frac{\partial F_1}{\partial z}\\
\frac{\partial F_2}{\partial x} &amp; \frac{\partial F_2}{\partial y}&amp; \frac{\partial F_2}{\partial z}\\
\frac{\partial F_3}{\partial x} &amp; \frac{\partial F_3}{\partial y}&amp; \frac{\partial F_3}{\partial z}
\end{pmatrix}=\begin{pmatrix}
p_{xx}V(a) &amp; p_{xy}V(a)&amp; 0\\
p_xV''(x) &amp; p_yV'(x)&amp; pV''(z)\\
p_{xy}V(z)&amp; p_{yy}V(z)&amp; p_yV'(z)
\end{pmatrix}$$</span></p>
<p>and the Jacobian of <span class="math-container">$F$</span> with respect to <span class="math-container">$(a,k)$</span> is</p>
<p><span class="math-container">$$J_{F,(a,k)}\begin{pmatrix}
\frac{\partial F_1}{\partial a} &amp; \frac{\partial F_1}{\partial k}\\
\frac{\partial F_2}{\partial a} &amp; \frac{\partial F_2}{\partial k}\\
\frac{\partial F_3}{\partial a} &amp; \frac{\partial F_3}{\partial k}
\end{pmatrix}=\begin{pmatrix}
p_{x}(x,y)V'(a) &amp; -1\\
0 &amp; 0\\
0 &amp; 0
\end{pmatrix}$$</span></p>
<p>If <span class="math-container">$J_{F,(x,y,z)}$</span> is invertible, then the Jacobian matrix of partial derivatives is given by</p>
<p><span class="math-container">$$\begin{pmatrix}
\frac{\partial x}{\partial a} &amp; \frac{\partial x}{\partial k}\\
\frac{\partial y}{\partial a} &amp; \frac{\partial y}{\partial k}\\
\frac{\partial z}{\partial a} &amp; \frac{\partial z}{\partial k}
\end{pmatrix}=-[J_{F,(x,y,z)}]^{-1}J_{F,(a,k)}$$</span></p>

CC BY-SA 4.0
2024-11-23T09:43:07.043Z
2024-11-23T09:43:07.043Z
smcc
<p>I have solved an optimization problem which yields the following set of first order conditions:
<span class="math-container">\begin{align}
    p_x(x,y)V(a)=k  \\
    p(x,y)V'(z)=1 \\
    p_y(x,y)V(z)=1
  \end{align}</span>
Where x, y and z are the control variables whereas parameters <span class="math-container">$a$</span> and <span class="math-container">$k$</span> are positive real numbers. <span class="math-container">$p(\cdot)$</span> is a probability function, strictly increasing and strictly concave in both arguments. No assumption made on the cross-derivative (but can be made if needed). <span class="math-container">$V(\cdot)$</span> is also strictly increasing and strictly concave. x, y and z must be non-negative.</p>
<p>My objective is to find how <span class="math-container">$x^*$</span> responds in equilibrium to an increase in the parameter k , i.e. the sign of the derivative of <span class="math-container">$x^*$</span>  wrt k.</p>
<p>I'm am familiar with the implicit function theorem with one equation but I am not sure how to set it up when I have 3 equations and multiple variables affecting each other, like in this instance. Could anybody help me solve this problem, step by step?</p>

<microeconomics><mathematical-economics><comparative-statics>
Comparative statics and Implicit function theorem with 3 equations
Dario
# Comparative statics and Implicit function theorem with 3 equations

(Asked by: username_0 on 2024-11-22)

I have solved an optimization problem which yields the following set of first order conditions:
\begin{align}
p\_x(x,y)V(a)=k \\
p(x,y)V'(z)=1 \\
p\_y(x,y)V(z)=1
\end{align}
Where x, y and z are the control variables whereas parameters $a$ and $k$ are positive real numbers. $p(\cdot)$ is a probability function, strictly increasing and strictly concave in both arguments. No assumption made on the cross-derivative (but can be made if needed). $V(\cdot)$ is also strictly increasing and strictly concave. x, y and z must be non-negative.

My objective is to find how $x^\*$ responds in equilibrium to an increase in the parameter k , i.e. the sign of the derivative of $x^\*$ wrt k.

I'm am familiar with the implicit function theorem with one equation but I am not sure how to set it up when I have 3 equations and multiple variables affecting each other, like in this instance. Could anybody help me solve this problem, step by step?

**Tags:** microeconomics, mathematical-economics, comparative-statics


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## 1 answers

### Answer by username_1 on 2024-11-23. Score: 4 | Accepted answer

Totally differentiate each equation with respect to $(k,x,y,z)$:

\begin{align}
V(a)[p\_{xx}dx+p\_{xy}dy]&=dk \tag{1}\\
V'(z)[p\_xdx+p\_ydy]+pV''(z)dz&=0 \tag{2}\\
V(z)[p\_{xy}dx+p\_{yy}dy]+p\_yV'(z)dz&=0 \tag{3}
\end{align}

This system can be written in matrix form as

$$\begin{pmatrix}
V(a)p\_{xx} & V(a)p\_{xy} & 0\\
V'(z)p\_x & V'(z)p\_y & pV''(z)\\
V(z)p\_{xy} & V(z)p\_y & p\_yV'(z)
\end{pmatrix}\begin{pmatrix}dx\\ dy \\ dz\end{pmatrix}=\begin{pmatrix}dk\\ 0 \\ 0\end{pmatrix}\tag{\*}$$

and solved by inverting the $3\times3$ matrix (assuming it is invertible):

$$\begin{pmatrix}dx\\ dy \\ dz\end{pmatrix}=\begin{pmatrix}
V(a)p\_{xx} & V(a)p\_{xy} & 0\\
V'(z)p\_x & V'(z)p\_y & pV''(z)\\
V(z)p\_{xy} & V(z)p\_y & p\_yV'(z)
\end{pmatrix}^{-1}\begin{pmatrix}dk\\ 0 \\ 0\end{pmatrix}=\begin{pmatrix}\frac{\partial x}{\partial k}\\ \frac{\partial y}{\partial k}\\\frac{\partial z}{\partial k}\end{pmatrix}dk.$$

However, if you are only interested in $\partial x/\partial k$, and are doing this by hand, then this entails more working than necessary, so an alternative would be as follows:

Solving $(1)$ for $dy$ gives

$$dy=\frac{1}{V(a)p\_{xy}}dk-\frac{p\_{xx}}{p\_{xy}}dx$$

Substituting into $\frac{p\_{xy}}{V'(z)}\times (2)$ and $\frac{p\_{xy}}{V(z)}\times(3)$ gives

\begin{align}
\left(p\_xp\_{xy}-p\_yp\_{xx}\right)dx+\frac{p\_{xy}pV''(z)}{V'(z)}dz&=-\frac{p\_y}{V(a)}dk \tag{4}\\
\left(p\_{xy}^2 -p\_{xx}p\_{yy}\right) dx+\frac{p\_{xy}p\_yV'(z)}{V(z)}dz&=-\frac{p\_{yy}}{V(a)}dk \tag{5}
\end{align}

Etc!

Just eliminate $dz$ from the equations to get an equation in the form $dx=Mdk$. The expression for the partial derivative you want is $M$.

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The more generally applicable method that will allow you to find the derivatives of $x$, $y$, and $z$ with respect to $a$ and $k$ goes as follows:

Let $F:\mathbb{R}^5\to\mathbb{R}^3$ be given by

\begin{align}
F\_1(a,k,x,y,z)&=p\_x(x,y)V(a)-k\\
F\_2(a,k,x,y,z)&=p(x,y)V'(z)-1\\
F\_3(a,k,x,y,z)&=p\_y(x,y)V(z)-1
\end{align}

With this notation, $F(a,k,x,y,z)=0$ is the equation that potentially defines $(x,y,z)$ as an implicit function of $(a,k)$.

The Jacobian of $F$ with respect to $(x,y,z)$ is (the same $3\times 3$ matrix as in equation $(\*)$ above):

$$J\_{F,(x,y,z)}\begin{pmatrix}
\frac{\partial F\_1}{\partial x} & \frac{\partial F\_1}{\partial y}& \frac{\partial F\_1}{\partial z}\\
\frac{\partial F\_2}{\partial x} & \frac{\partial F\_2}{\partial y}& \frac{\partial F\_2}{\partial z}\\
\frac{\partial F\_3}{\partial x} & \frac{\partial F\_3}{\partial y}& \frac{\partial F\_3}{\partial z}
\end{pmatrix}=\begin{pmatrix}
p\_{xx}V(a) & p\_{xy}V(a)& 0\\
p\_xV''(x) & p\_yV'(x)& pV''(z)\\
p\_{xy}V(z)& p\_{yy}V(z)& p\_yV'(z)
\end{pmatrix}$$

and the Jacobian of $F$ with respect to $(a,k)$ is

$$J\_{F,(a,k)}\begin{pmatrix}
\frac{\partial F\_1}{\partial a} & \frac{\partial F\_1}{\partial k}\\
\frac{\partial F\_2}{\partial a} & \frac{\partial F\_2}{\partial k}\\
\frac{\partial F\_3}{\partial a} & \frac{\partial F\_3}{\partial k}
\end{pmatrix}=\begin{pmatrix}
p\_{x}(x,y)V'(a) & -1\\
0 & 0\\
0 & 0
\end{pmatrix}$$

If $J\_{F,(x,y,z)}$ is invertible, then the Jacobian matrix of partial derivatives is given by

$$\begin{pmatrix}
\frac{\partial x}{\partial a} & \frac{\partial x}{\partial k}\\
\frac{\partial y}{\partial a} & \frac{\partial y}{\partial k}\\
\frac{\partial z}{\partial a} & \frac{\partial z}{\partial k}
\end{pmatrix}=-[J\_{F,(x,y,z)}]^{-1}J\_{F,(a,k)}$$


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