3.1 Duality and Hicksian Demand

Intro

A major theme in modern mathematics over the last two centuries has been the value of looking at problems from multiple angles. Often, taking a “mirror image” of some problem can reveal new angles for us to attack in solving it.

The theory of optimization has certainly benefited from this insight. Certain optimization problems, such as consumer utility optimization, have twin problems that mathematicians have taken to calling dual problems. In utility maximization, the consumer maximizes utility against a spending constraint. In today’s lesson, we’re going to study the dual of utility maximization: minimizing spending subject to a utility constraint. The solutions to these problems will be the same, but when solving one is difficult, solving the other might be easier.

Expenditure Minimization and Hicksian Demand

Here’s a table to compare and contrast the two problems, along with a third example you guys will be familiar with:

As summarized in the table above, the new problem we’re looking at is another one in which we optimally select an amount of $x_1$ and $x_2$ in order to optimize an objective. The objective is now to minimize spending $p_1 x_1 + p_2 x_2$. The rightmost column is just a fun example to emphasize that what we’re doing here is definitely not just about consumer theory, but rather optimization in general.

Now, the easiest way to pick $x_1$ and $x_2$ to make spending low is to buy nothing: that is, always set $x_1=x_2 = 0$. But this is sort of boring, and not a very good model of human behavior. To model a consumer’s desire to actually buy stuff, we’re going to impose a constraint that they reach a minimum level of utility, which we will call $\bar u$.

Constraints play an important role in both problems. The analog of buying nothing in the expenditure minimization problem is buying an infinite amount in the utility maximization problem: to maximize utility, you’d get an infinite amount of each good, but this isn’t realistic, so we add a budget constraint. Constraints discipline our models to better reflect reality.

The parameters (that is, the variables we don’t get to choose ourselves) of this new problem are similar to utility maximization. The main parameters are still the prices of the two goods. Instead of having an exogenous maximum number of dollars to spend, $m$, however, we now have an exogenous minimum utility level $\bar u \in \mathbb{R}$ we must attain.

In the main reading for today, you learned about Marshallian Demand, the solution function for the utility maximization problem, as well as indirect utility, the value function for the problem. You should already be relatively familiar with solutions, but value functions will probably be new for most of you.

Definition (Value Function): Let $u(\vec{x})$ be an objective function and $\vec{x}^\star(\vec{p})$ the solution given the parameters $\vec{p}$ and constraints. A Value Function $V(\vec{p})$ for the problem is defined by $V(\vec{p})=u(\vec{x}^\star(\vec{p}))$. That is, the Value Function gives the optimal level of the objective given parameters and constraints.

[Insert graph showing objective, value, and solution, with sliders to control parameter values]

A Value Function is like a “high score” in a video game. It’s the very highest we can make utility, or the very lowest we can make spending. We can solve for a value function by plugging the solution function back into the objective function, replacing the choice variables with their optimal values. The value function for utility maximization is called indirect utility, and the value function for spending is called expenditure, and is denoted $e(p_1,p_2,\bar u)$. We’ll walk through the process of mechanically solving for a value function at the end of this reading.

While Marshallian Demand $x_1^\star(\vec p, m), x_2^\star(\vec p, m)$ should be familiar to you at this point, it might still be confusing to see the difference between the solutions for the two problems. The solution for utility maximization (Marshallian Demand $\vec x^\star(\vec p, m)$) and the solution for expenditure minimization (Hicksian Demand $\vec h^\star(\vec p, \bar u)$) are both bundles of goods, but they differ in their parameters and interpretations. $x^\star$ gives how many apples to buy under a budget $m$ to maximize utility, while $h^\star$ gives how many apples to buy to minimize spending when one must reach minimum utility $\bar u$. Both are in units of “number of apples purchased,” but they come from slightly different places.

As a review, for the cost minimization problem we have

• Objective: minimize $p_1 x_1 + p_2 x_2$
• Constraint: $u(x_1, x_2) \geq \bar u$, where $u(x_1, x_2)$ is some given function like $u(x_1, x_2) = x_1 x_2$

With these, we can write a Lagrangian for the problem: \(\mathcal{L} = p_1 x_1 + p_2 x_2 + \lambda (\bar u - u(x_1, x_2))\)

This is a problem you can solve like any other Lagrangian, and in the next reading we show explicitly how you might do so. For the rest of this reading, we’re going to look at some properties of the expenditure minimization problem, in conjunction with utility maximization, that can let us understand them both better.

Duality

Expenditure minimization and utility maximization look a lot alike, and really do have deep connections to each other. How do we think about the duality of this relationship? Well, the tangency condition is an optimality condition, in the sense that along that line, there is no overlap between the set of bundles preferred to a point and the set of bundles cheaper than the point. Hence it doesn’t really matter if we think of each point along the tangency as the most effective use of a fixed budget, or the cheapest way to afford a fixed amount of utility. But if we look at a series of indifference curves and budget/iso-cost lines, we can’t tell which kind of problem we’re solving:

We could be maximizing utility subject to four budget constraints, or we could be minimizing cost subject to four utility constraints. Either way, the solution lies at the intersection of the tangency condition and the constraint.

The following theorem formalizes the connection between the problems. In particular, it connects the value functions of the two problems as inverses, and the solutions as identical.

Theorem 1 (Duality): Fix a price vector $\vec{p}$. (1) The indirect utility function and the expenditure function are inverse functions: $V(\vec{p}, e(\vec{p}, \bar u)) = \bar u$ and $e(\vec{p}, V(\vec{p}, m)) = m$. (2) Marshallian and Hicksian demand coincide at optimality: $\vec{x}(\vec{p}, e(\vec{p}, \bar u)) = \vec{h}(\vec{p}, \bar u)$ and $\vec{h}(\vec{p}, V(\vec{p}, m)) = \vec{x}(\vec{p}, m)$.

The upshot is that solving either problem also solves the other. So in practice, you can tackle whichever one is easier, and convert to the other when necessary. You’ll practice doing so on this week’s homework.

Practically speaking, the procedure for such problems will often look something like this:

• Solve for either Marshallian or Hicksian demand, maybe using a Lagrangian, KKT, or the tangency condition.
• Plug the optimal bundle (that is, the solution function) back into the objective function. If you found $x_1^\star = m / (2 p_1)$ and $x_2^\star = m / (2 p_2)$ when $u(x_1, x_2)=x_1 x_2$, then $V(p_1, p_2, m)=x_1^\star x_2^\star = m^2 / (4p_1 p_2)$.
• Invert whichever value function you solved for to get the other. For instance, if $V(p_1, p_2, m)=x_1^\star x_2^\star = m^2 / (4p_1 p_2)$ and you needed the expenditure function for whatever reason, you could find it by using Theorem 1: \(\begin{aligned} V(p_1, p_2, m) &= m^2 / (4p_1 p_2) \\ \bar u &= [e(p_1, p_2, \bar u)]^2 / (4p_1 p_2) \\ (4p_1 p_2)\, \bar u &= [e(p_1, p_2, \bar u)]^2 \\ e(p_1, p_2, \bar u) &= 2 \sqrt{p_1 p_2 \bar u} \end{aligned}\)
• Do something with the new value function. In the example above, if you were asked for the lowest required spending to reach 16 utils when $p_1=3$ and $p_2=3$, then you could solve by plugging in $e(3, 3, 16)=2 \sqrt{9*16} = 24$.

Useful Properties of Expenditure

The expenditure function $e(p_1, p_2, \bar u)$, the lowest spending required at prices $\vec p$ to achieve utility $\bar u$, itself also has a few useful properties worth knowing. For the following properties, you can assume the utility is “well behaved:” we’re dealing with continuous convex utility.

Shephard’s Lemma. $\frac{\partial e}{\partial p_i} = h_i(\vec{p}, \bar u)$, for $i=1,2$.

Shephard’s Lemma will be particularly useful for us, and we will be discussing it at greater length next time. You may find it useful on this week’s homework.

Homogeneous of degree 1 in prices. $e(\alpha \vec{p}, \bar u) = \alpha e(\vec{p}, \bar u)$

This one should make some intuitive sense. Say $\alpha=2$, so I am doubling the price of every good at the store by multiplying the $\vec p$ vector by the scalar $\alpha \in \mathbb{R}_+$. In that case, the cost of buying your optimal bundle would, indeed, go up by a factor of 2.

The remaining three are more technical, but no less useful:

Concave in prices. For all $\alpha \in [0,1]$ and any price vectors $\vec{p}_1$ and $\vec{p}_2$, \(e(\alpha \vec{p}_1 + (1-\alpha)\vec{p}_2, \bar u) \geq \alpha e(\vec{p}_1, \bar u) + (1-\alpha)e(\vec{p}_2, \bar u).\)

Non-decreasing in prices and utility. If $\vec{p}$ or $\bar u$ increase, then $e(\vec{p}, \bar u)$ weakly increases.

Continuous. The expenditure function is continuous.