🔬 Tutorial problems kappa \(\kappa\)

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left out are for additional practice on your own.

\(\kappa\).1

Formulation

Roy’s identity

Consider the choice problem of a consumer endowed with strictly concave and differentiable utility function \(u \colon \mathbb{R}^N_{++} \to \mathbb{R}\) where \(\mathbb{R}^N_{++}\) denotes the set of vector in \(\mathbb{R}^N\) with strictly positive elements.

The budget constraint is given by \(p \cdot x \le m\) where \(p \in \mathbb{R}^N_{++}\) are prices and \(m>0\) is income.

Then the demand function \(x^\star(p,m)\) and the indirect utility function \(v(p,m)\) (value function of the problem) satisfy the equations

\[ x_i^\star(p,m) = -\frac{\partial v}{\partial p_i}(p,m) \Big/ \frac{\partial v}{\partial m}(p,m), \; \forall i \in \{1,\dots,N\} \]

1. Prove the statement

2. Verify the statement by direct calculation (i.e. by expressing the indirect utility and plugging its partials into the identity) using the following specification of utility

\[ u(x) = \prod_{i=1}^N x_i^{\alpha_i}, \; \alpha_i > 0 \]

Hint

Envelope theorem should be useful here.

Solution

We first show the Roy’s identity.

The value function is \(v(p, m)=\max\{u(x): p \cdot x \leq m\}\) where \(u\colon \mathbb{R}^{N}_{++}\to \mathbb{R}\). The Lagrangian of the maximization problem is \(\mathcal{L}(x, \lambda, p, m) = u(x) - \lambda (\sum_{i=1}^{N}p_{i}x_{i}-m)\). The Envelope Theorem implies

\[\begin{split} \frac{\partial v}{\partial p_{j}}=\frac{\partial\mathcal{L}}{\partial{p_{j}}}=-\lambda x_{j} \qquad (j=1,\dots, N)\\ \frac{\partial{v}}{\partial{m}}= \frac{\partial{\mathcal{L}}}{\partial{m}}=\lambda \end{split}\]

Note, by the way, that here \(\lambda\) is indeed the shadow price of the budget constraint: the slope of the indirect utility along the \(m\) dimension is \(\lambda\)

It follows from the previous equations that

\[ x_{j}=- \frac{1}{\lambda} \frac{\partial v}{\partial p_{j}} = - \frac{\partial v}{\partial p_{j}} \bigg/ \frac{\partial v}{\partial m}. \]

Next, let \(u(x)= \prod_{i=1}^{N}x_{i}^{\alpha_i}\) where \(\alpha_i>0\) for all \(i\). Since \(\log(\cdot)\) function is strictly monotone, the optimization problem is equivalent to maximize \(u(x)=\sum_{i=1}^{N}\alpha_{i}\log(x_i)\). The corresponding Lagrangian is

\[ \mathcal{L}(x, \lambda, p, m) = \sum_{i=1}^{N}\alpha_{i}\log(x_{i}) - \lambda (\sum_{i}^{N}p_{i}x_{i}- m) \]

The first-order conditions yield

\[\begin{split} \frac{\partial\mathcal{L}}{\partial{x_{j}}} = \frac{\alpha_{i}}{x_{j}}-\lambda p_{j=0} \qquad (j=1,\dots, N)\\ \frac{\partial\mathcal{L}}{\partial{\lambda}} = \sum_{i}^{N}p_{i}x_{i}- m = 0 \\ \Longrightarrow \lambda = \frac{\sum_{i=1}^{N}\alpha_{i}}{m}\\ x_{j} = \frac{\alpha_{j}}{\lambda p_{j}} =\frac{\alpha_{j}}{\sum_{i=1}^{N}\alpha_{i}} \frac{m}{p_{j}} \end{split}\]

Hence, the optimal value function is

\[ v(p,m)=u(x^{*}(p,m))=\prod_{i=1}^{N}\left(\frac{\alpha_{j}}{\sum_{i=1}^{N}\alpha_{i}} \frac{m}{p_{j}}\right)^{\alpha_{i}} \]

To verify Roy’s identity, observe that

\[\begin{split} \frac{\partial{v}}{\partial{p_{j}}} =\frac{\partial}{\partial{p_{j}}} e^{\log v} = e^{\log v} \frac{\partial}{\partial{p_{j}}}\log v = v \frac{\partial}{\partial{p_{j}}}\log v\\ = v\frac{\partial}{\partial{p_{j}}}\left\{\sum_{i=1}^{N}\log\left(\frac{\alpha_{i}}{\sum_{i}\alpha_{i}}\right)+\alpha_{i}\log \frac{m}{p_{i}} \right\}\\ = v \left(\alpha_{j} \frac{1}{\frac{m}{p_{j}}}\frac{-m}{p_{j}^{2}} \right) =\frac{-\alpha_j}{p_{j}}v \end{split}\]

\[\begin{split} \frac{\partial{v}}{\partial{m}} =v\frac{\partial}{\partial{p_{j}}} \log v \\ = v \sum_{i=1}^{N}\alpha_{i}\frac{1}{\frac{m}{p_{j}}} \frac{1}{p_{i}} = \frac{\sum_{i=1}^{N}\alpha_{i}}{m} v \end{split}\]

Then, we have

\[ -\frac{\partial{v}}{\partial{p_{j}}}\Bigg/ \frac{\partial{v}}{\partial{m}} =\frac{\frac{-\alpha_j}{p_{j}}v} {\frac{\sum_{i=1}^{N}\alpha_{i}}{m} v} =\frac{\alpha_{j}}{\sum_{i=1}^{N}\alpha_{i}} \frac{m}{p_{j}} =x_{j}. \]

\(\kappa\).2

Formulation

First, find the maximum and minimum square distances from the origin to the ellipse \(x^2+xy+y^2 = 3\).

Second, using the envelope theorem, approximate these square distances for the ellipse \(x^2+xy+\tfrac{9}{10}y^2 = 3\).

[Simon and Blume, 1994] Exercises 18.2 and 19.12

Hint

Use \(x^2+y^2\) as your objective function.

Solution

For the first part the problem is:

\[\begin{split} \max x^2 + y^2 \\ \text{subject to } x^2 + xy + y^2 = 3. \end{split}\]

The Lagrangian is: $\( L = x^2 + y^2 - \lambda(x^2 + xy + y^2 - 3). \)$

The first-order conditions are: $\( L_x = 2x - 2\lambda x - \lambda y = 0 \)\( \)\( L_y = 2y - \lambda x - 2\lambda y = 0 \)\( \)\( L_{\lambda} = -(x^2 + xy + y^2 - 3) = 0. \)$

There are four solutions: $\( (x,y,\lambda) = \begin{cases} (-\sqrt{3}, \sqrt{3}, 2) \\ (\sqrt{3}, -\sqrt{3}, 2) \\ (1, 1, 2/3) \\ (-1, -1, 2/3) \end{cases} \)$

The constraint qualification condition holds at all four solutions. Clearly, the first two solutions are maxima, and the second two are minima.

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Now rewrite the problem as:

\[\begin{split} \max x^2 + y^2 \\ \text{subject to } x^2 + xy + b y^2 = 3, \end{split}\]

where \(b = \frac{9}{10}\).

\[ L = x^2 + y^2 - \lambda(x^2 + xy + b y^2 - 3). \]

\[ \frac{\partial L}{\partial b} = -\lambda y^2 \]

For \(b = 1\), the max is at \((x,y,\lambda) = (\pm \sqrt{3},\mp \sqrt{3},2)\) with \(f^* = 6\); and the min is at \((x,y,\lambda) = (\pm 1,\pm 1,2/3)\) with \(f^* = 2\).

Based on the up to linear term of the Taylor expansion for \(b = 0.9\), \(f^* \approx 6 + (-2 \cdot 3)\cdot(-.1) = 6.6\) at the max and \(f^* \approx 2 + (-(2/3)\cdot 1)\cdot(-.1) = 2 + (2/30) = 2\frac{1}{15}\) at the min.

\(\kappa\).3

Formulation

Consider the problem of maximizing \(f(x,y)=x^2+x+4y^2\) subject to the inequality constraint

\[ 2x+2y \leq 1, \; x \geq 0, \; y \geq 0. \]

Using the envelope theorem, approximate the maximum value of \(x^2+x+4.1 y^2\) on the same constraint set.

[Simon and Blume, 1994] Exercises 19.13

Hint

As the envelope theorem requires knowing the optimizer and the corresponding Lagrange multiplier, the problem has to be first solved using KKT method.

Solution

The Jacobian of the constraint functions is $\( \begin{pmatrix} 2 & 2 \\ -1 & 0 \\ 0 & -1 \end{pmatrix}. \)$

At most two constraints can be binding at the same time, and any \(2 \times 2\) submatrix of this Jacobian has rank two. Therefore, the constraint qualification condition will hold at any solution candidate. Form the Lagrangian

\[ L(x,y,\lambda_1,\lambda_2,\lambda_3) = x^2 + x + 4y^2 - \lambda_1(2x + 2y - 1) + \lambda_2 x + \lambda_3 y, \]

where we convert the nonnegativity constraints to \(\leq 0\) form. Next, write the KKT conditions:

\[\begin{split} \begin{aligned} 1)\quad & \frac{\partial L}{\partial x} = 2x + 1 - 2\lambda_1 + \lambda_2 = 0, \\ 2)\quad & \frac{\partial L}{\partial y} = 8y - 2\lambda_1 + \lambda_3 = 0, \\ 3)\quad & \lambda_1(2x + 2y - 1) = 0, \\ 4)\quad & \lambda_2 x = 0, \\ 5)\quad & \lambda_3 y = 0, \\ 6)\quad & \lambda_1 \geq 0, \\ 7)\quad & \lambda_2 \geq 0, \\ 8)\quad & \lambda_3 \geq 0, \\ 9)\quad & 2x + 2y \leq 1, \\ 10)\quad & x \geq 0, \\ 11)\quad & y \geq 0. \end{aligned} \end{split}\]

Rewrite condition 1 without minus signs as \(2x + 1 + \lambda_2 = 2\lambda_1\). Along with conditions 7 and 10, this equation implies that \(2\lambda_1 \geq 1 > 0\), and this, by condition 3, implies that the first constraint is binding:

\[ 2x + 2y = 1. \]

Now we need to examine a few cases. Let’s first look at the case \(\lambda_2 > 0\). In this case, it follows from condition 4 that \(x = 0\), from equation above that \(y = 0.5\), and from condition 5 that \(\lambda_3 = 0\). Plugging \(y = 0.5\) and \(\lambda_3 = 0\) into condition 2 yields \(\lambda_1 = 2\). Plugging \(x = 0\) and \(\lambda_1 = 2\) into condition 1 yields \(\lambda_2 = 3\). So, the assumption that \(\lambda_2 > 0\) leads to the candidate

\[ (x,y,\lambda_1,\lambda_2,\lambda_3) = (0,0.5,2,3,0). \]

Now, let’s try the ooposite case \(\lambda_2 = 0\). Cobmining conditions 1 and 2 we find

\[ 2x+1 + \lambda_2 = 8y + \lambda_3 = 2\lambda_1. \]

Substituting \(\lambda_2 = 0\) and \(2x=1-2y\) from above, we find

\[ 1 - 2y + 1 = 8y + \lambda_3, \quad\text{or}\quad 2 = 10y + \lambda_3. \]

Combining the latter equation with condition 5 leads to the conclusion that either \(y = 0\) or \(\lambda_3 = 0\), but not both. If \(y = 0\), then \(\lambda_3 = 2\); by the earlier equation, \(x = 0.5\); and by the previous condition, \(\lambda_1 = 1\). If, instead, \(\lambda_3 = 0\), then \(y = 0.2\), \(x = 0.3\), and \(\lambda_1 = 0.8\). Consequently, the case \(\lambda_2 = 0\) leads to the two candidates

\[\begin{split} (x, y, \lambda_1, \lambda_2, \lambda_3) = \begin{cases} (0.5, 0, 1, 0, 2) &\text{or}\\[6pt] (0.3, 0.2, 0.8, 0, 0). \end{cases} \end{split}\]

By evaluating the objective function at each of these three candidates, we find that the constrained maximum occurs at the point \(x = 0\), \(y = 0.5\), where \(\lambda_1 = 2\), \(\lambda_2 = 3\), and \(\lambda_3 = 0\).

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Now, to the answer of the approximation question. Consider the following parameterized version of the problem:

\[ \max \quad x^2 + x + a y^2 \]

\[ \text{subject to} \quad 2x + 2y \leq 1,\; x \geq 0,\; y \geq 0. \]

The Lagrangian is

\[ L = x^2 + x + a y^2 - \lambda_1(2x + 2y - 1) + \lambda_2 x + \lambda_3 y. \]

For \(a = 4\), \(x^* = 0\), \(y^* = 0.5\), \(\lambda_1^* = 2\), \(\lambda_2^* = 3\), \(\lambda_3^* = 0\), and \(f^* = 1\). At these values,

\[ \frac{\partial L}{\partial a} = y^2 = 0.5^2 = 0.25 \]

so using linear Taylor approximation

\[ f^*(4.1) \approx f^*(4) + \frac{\partial L}{\partial a}\cdot\Delta a = 1 + 0.25(0.1) = 1.025. \]