I have read the statement below and don't quite understand what it means. This is probably because I don't have full understanding of duality with support function in math, but just to foster understanding... The statement is:

"The utility maximization problem(UMP) has a non-linear utility objective and a linear price constraint. The expenditure minimization problem is exactly the other way around: a linear price objective and a non-linear utility constraint. This is at the heart of the idea of the meaning of dual - to recast the problem with the objective as a constraint and the constraint as the objective. An important observation to make is linearity in pricing in them. This linearity, either in objective or constraint, is the cornerstone in enabling us to use the duality theorem."


3 Answers 3


Consider the standard utility maximization problem:

$$\begin{align}\max_{\vec x} & \quad U(\vec x) \\ & \quad \vec p \cdot \vec x \leq Y \end{align}$$

Here utility is your objective function, and the price inequality is the constraint. You solve for the optimal bundle via a Lagrangian.

Now consider its dual problem.

$$\begin{align}\min_{\vec x} & \quad p \cdot \vec x \\ & \quad U(\vec x) \geq \bar U \end{align}$$

Now the objective function is expenditure, and a desired minimum utility is the constraint.

The utility that you gain from the maximization solution, if that is set to be $\bar U$, then the solution bundle to the minimization problem will be the same. This holds if you solve the bundle for the minimization problem first.

From this result (which you can verify for yourself) eventually comes the ever useful Roy's Identity.

$${x^m_i}^* = - \frac{\frac{\partial V}{\partial p_i}}{\frac{\partial V}{\partial Y}}$$

Where $V(\vec p,Y)$ is the indirect utility function.

If you are confused about the meaning of support function, recall its definition. (I will use the one from Mas-Colell)

For any nonempty closed set $K \subset \mathbb{R}^L$, the support function of $K$ is defined for any $\vec p \in \mathbb{R}^L$ to be $$\mu_k(\vec p) = \inf \{\vec p \cdot \vec x: \quad \vec x \in K \}$$

which is just a description of a set's minimum value. What does the (familiar) set here look like?...

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    $\begingroup$ Kitsune, thanks for the response. will have to take a closer look, but have a good July 4th weekend. Kitsune, the name reminds me of an apparel brand I like. Japanese word for fox... cool. $\endgroup$ Jul 1, 2016 at 13:02

Denote a bundle of goods $\mathbf{x}=(x_1,...,x_n)$ and a price vector $\mathbf{p}=(p_1,...,p_n)$. The cost of any bundle is the scalar product $\mathbf{p}\cdot\mathbf{x}$.

Consider first the UMP, which consists in finding the bundle $\mathbf{x}$ that maximizes utility $u(\mathbf{x})$ subject to the budget constraint that the bundle of good costs less than an income $R$: $$\underset{\mathbf{x}}{\max}u(\mathbf{x}) \text{ s.t. } \mathbf{p}\cdot\mathbf{x}\leq R.$$

Consider now the EMP, which consists in finding the bundle $\mathbf{x}$ that minimizes expenditures $\mathbf{p}\cdot\mathbf{x}$ subject to achieving a minimal level of utility $\underline{u}$: $$\underset{\mathbf{x}}{\min}\mathbf{p}\cdot\mathbf{x} \text{ s.t. } u(\mathbf{x})\geq \underline{u}.$$

These problems are dual in the sense that you can relate the solution of the first one to the solution of the second one. The solution of the UMP gives you the Marshallian demands, $\mathbf{x}^M(\mathbf{p},R)$, as the function of the prices and the income. The solution of the EMP gives you the Hicksian demands, $\mathbf{x}^H(\mathbf{p},\underline{u})$, as the function of the prices and the minimal level of utility. The link between these two functions is heavily studied in microeconomics, it is rich in many economic intuitions and useful mathematical relationships.

Duality enables you to find one solution once you know the other. For instance suppose we know the function $\mathbf{x}^M(\mathbf{p},R)$ but we do not know the function $\mathbf{x}^H(\mathbf{p},\underline{u})$, how can we find $\mathbf{x}^H(\mathbf{p},\underline{u})$? First, denote the indirect utility of the first problem by $v(\mathbf{p},R)=u(\mathbf{x}^M(\mathbf{p},R))$. Second define the income $\underline{R}$ as the solution of $v(\mathbf{p},\underline{R})=\underline{u}$ (the existence of $\underline{R}$ comes from the linearity property that is emphasized in the last sention of the citation). $\underline{R}$ is the minimum amount that is required to achieve utility $\underline{u}$. I will not show, but you can understand intuitively that we have: $$\mathbf{x}^H(\mathbf{p},\underline{u})=\mathbf{x}^M(\mathbf{p},\underline{R})$$

Lastly, note that we should write $\underline{R}$ as a function of $\mathbf{p}$ and $\underline{u}$ to be rigorous.

  • $\begingroup$ Thank you. Busy with July 4th weekend but will re-digest your response. $\endgroup$ Jul 1, 2016 at 13:01

The idea Is quite simple, you can either

  1. maximize your utility, given a budget constraint
  2. you want to reach a certain target of utility and you want to minimize your expenditure

Now solve the maximization problem in 1, you will obtain the maximum utility that you can obtain. Now that you have obtained this number, you can use it as a target in the minimization problem and you will find out (not surprisingly) that the minimum expenditure needed is the same as the income you have in the maximization problem.

In more simply terms, this two problems are equivalent/dual.


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