Exercise set K

Please, see the general comment on the tutorial exercises

Question K.1

Consider an unconstrained optimization problem

\[ V(\theta) = \max_{x} f(x,\theta) \]

where:

• \(f(x,\theta) \colon \mathbb{R}^N \times \mathbb{R}^K \to \mathbb{R}\) is an objective function

• \(x \in \mathbb{R}^N\) are decision/choice variables

• \(\theta \in \mathbb{R}^K\) are parameters

• \(V(\theta) \colon \mathbb{R}^K \to \mathbb{R}\) is a value function

Prove the following statement

Statement

Let \(f(x,\theta) \colon \mathbb{R}^N \times \mathbb{R}^K \to \mathbb{R}\) be a differentiable function, and \(x^\star(\theta)\) be the maximizer of \(f(x,\theta)\) for every \(\theta\). Suppose that \(x^\star(\theta)\) is differentiable function itself. Then the value function of the problem \(V(\theta) = f\big(x^\star(\theta),\theta)\) is differentiable w.r.t. \(\theta\) and

\[ \frac{\partial V}{\partial \theta_j} = \frac{\partial f}{\partial \theta_j} \big(x^\star(\theta),\theta\big), \forall j \]

Hint

Total derivative + first order conditions

Question K.2

Consider an equality constrained optimization problem

\[\begin{split} V(\theta) = \max_{x} f(x,\theta) \\ \text {subject to} \\ g_i(x,\theta) = 0, \; i\in\{1,\dots,I\}\\ \end{split}\]

where:

• \(f(x,\theta) \colon \mathbb{R}^N \times \mathbb{R}^K \to \mathbb{R}\) is an objective function

• \(x \in \mathbb{R}^N\) are decision/choice variables

• \(\theta \in \mathbb{R}^K\) are parameters

• \(g_i(x,\theta) = 0, \; i\in\{1,\dots,I\}\) where \(g_i \colon \mathbb{R}^N \times \mathbb{R}^K \to \mathbb{R}\), are equality constraints

• \(V(\theta) \colon \mathbb{R}^K \to \mathbb{R}\) is a value function

Prove the following statement

Statement

Assume that the maximizer correspondence in the problem above is single-valued and can be represented by the function \(x^\star(\theta) \colon \mathbb{R}^K \to \mathbb{R}^N\), with the corresponding Lagrange multipliers \(\lambda^\star(\theta) \colon \mathbb{R}^K \to \mathbb{R}^K\).

Assume that both \(x^\star(\theta)\) and \(\lambda^\star(\theta)\) are differentiable, and that the constraint qualification assumption holds. Then

\[ \frac{\partial V}{\partial \theta_j} = \frac{\partial \mathcal{L}}{\partial \theta_j} \big(x^\star(\theta),\lambda^\star(\theta),\theta\big), \forall j \]

where \(\mathcal{L}(x,\lambda,\theta)\) is the Lagrangian of the problem.

Question K.3

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

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

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

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

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

Hint

Envelope theorem should be useful

Solutions

Question K.1

See Theorem 19.4 in Simon and Blume (1994), pp. 453-454

Question K.2

Here is a version of proof. The Lagrangian is

\[ \mathcal{L}(x^\star(\theta),\lambda^\star(\theta), \theta) = f(x^\star(\theta),\theta)- \sum_{i=1}^{I}\lambda_{i}^\star(\theta) g_{i}(x^\star(\theta), \theta ) \]

For any \(j=1,\dots,K\) the partial derivative with respect to \(\theta_{j}\) is

\[\begin{split} \begin{array}{rl} \frac{\partial\mathcal{L}(x^\star(\theta), \lambda^\star(\theta), \theta)}{\partial{\theta_{j}}} &= \frac{\partial f(x^\star(\theta), \theta)}{\partial{\theta_{j}}} - \sum_{i=1}^{I} \left[ \frac{\partial\lambda_{i}^\star(\theta)}{\partial \theta_{j}}g_{i}(x^\star(\theta), \theta) + \lambda_{i}^\star(\theta) \frac{\partial g_{i}(x^\star(\theta), \theta) }{\partial \theta_{j}}\right] \\ &= \frac{\partial f(x^\star(\theta), \theta)}{\partial{\theta_{j}}} - \sum_{i=1}^{I} \lambda_{i}^\star(\theta) \frac{\partial g_{i}(x^\star(\theta), \theta) }{\partial \theta_{j}} \end{array}\end{split}\]

where the last equality follows from the equality constraints \(g_{i}(x^\star(\theta), \theta)=0\) for all \(i\). The partial derivative of value function is

\[\begin{split} \begin{array}{rl} \frac{\partial V(\theta)}{\partial{\theta_{j}}} &= \frac{d f(x^\star(\theta),\theta)}{d \theta_{j}} \\ &= \sum_{n=1}^{N} \frac{\partial f(x^\star(\theta), \theta)}{\partial x_{n}} \frac{\partial x_n(\theta)}{\partial \theta_{j}} +\frac{\partial f(x^\star(\theta), \theta) }{\partial \theta_{j}} \\ &= \sum_{n=1}^{N} \left(\sum_{i=1}^{I}\lambda_{i} \frac{\partial g_i(x^\star(\theta), \theta)}{\partial x_{n}} \right)\frac{\partial x_{n}(\theta)}{ \partial \theta_{j}} +\frac{\partial f(x^\star(\theta), \theta) }{\partial \theta_{j}} \\ &= \sum_{i=1}^{I}\lambda_{i} \sum_{n=1}^{N} \frac{\partial g_i(x^\star(\theta), \theta)}{\partial x_{n}}\frac{\partial x_{n}(\theta)}{\partial\theta_{j}} +\frac{\partial f(x^\star(\theta), \theta) }{\partial \theta_{j}} \\ &= - \sum_{i=1}^{I}\lambda_{i} \frac{\partial g_i(x^\star(\theta), \theta)}{\partial \theta_j} +\frac{\partial f(x^\star(\theta), \theta) }{\partial \theta_{j}}\\ &= \frac{\partial\mathcal{L}(x^\star(\theta), \lambda^\star(\theta), \theta)}{\partial{\theta_{j}}} \end{array}\end{split}\]

where the third equality follows from the first-order conditions,

\[\begin{split} \begin{array}{rl} 0 = \frac{\partial\mathcal{L}(x^\star(\theta), \lambda^\star(\theta), \theta)}{\partial{x_n}} &= \frac{\partial f(x^\star(\theta), \theta)}{\partial{x_n}} - \sum_{i=1}^{I} \lambda_{i}^\star(\theta) \frac{\partial g_{i}(x^\star(\theta), \theta) }{\partial x_{n}}, \\ \end{array} \end{split}\]

and the fifth equality follows from the constraint and its total derivative

\[\begin{split} g_{i}(x^\star(\theta), \theta)=0\\ \Longrightarrow \sum_{n=1}^{N} \frac{\partial g_i(x^\star(\theta), \theta)}{\partial x_{n}}\frac{\partial x_{n}(\theta)}{ \partial\theta_{j}} + \frac{\partial g_i(x^\star(\theta), \theta)}{\partial \theta_{j}}= 0 \end{split}\]

Question K.3

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

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

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

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

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

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

Then, we have

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