Econ 50Q Section 4: Hicksian Demand and Duality

Duality and Hicksian Demand

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.

Utility Maximization and Cost Minimization

For utility maximization, the objective function has been the “utility function” whose output is measured in utils, and the constraint has been the budget constraint: that is, \(\begin{aligned} \text{Objective function: }f(x_1,x_2) &= u(x_1,x_2)\\ \text{Constraint (when set equal to zero): }g(x_1,x_2) &= m - p_1x_1 - p_2x_2 \end{aligned}\) Visually, we can picture the budget constraint and a number of indifference curves (i.e., the level sets of the objective function), and see that our objective is to get to the highest indifference curve (i.e., the highest level set of the objective function) while staying on the constraint:

We wrote the optimal bundle for this problem as $(x_1^\star, x_2^\star)$. This bundle was a function of the market prices $p_1$ and $p_2$, as well as the exogenously given income of $m$. We called the solution functions for this problem demand functions, though technically we should have called them the ordinary or Marshallian demand functions: \(\text{Ordinary demand functions: }\begin{cases}x_1^\star(p_1,p_2,m)\\x_2^\star(p_1,p_2,m)\end{cases}\)

What if we were to flip this script, though? What if we were to say that we wanted to find the cheapest way of affording some utility level $U$? Then our problem would become \(\begin{aligned} \min_{x_1,x_2}\ \ \ & p_1x_1 + p_2x_2 \\ \text{s.t.}\ \ \ & u(x_1,x_2) = U \end{aligned}\) Now, using the same $f()$ and $g()$ framing as above, we have \(\begin{aligned} \text{Objective function: }f(x_1,x_2) &= p_1x_1 + p_2x_2\\ \text{Constraint (when set equal to zero): }g(x_1,x_2) &= U - u(x_1,x_2) \end{aligned}\) Visually, we now think of the target indifference curve as the constraint, with the level curves of the objective function being “iso-cost” lines with slope $-p_1/p_2$; now the objective is to find the least-cost way of affording the level of utility shown by the indifference curve:

Let’s write the optimal bundle for this problem as $(x_1^c, x_2^c)$. This bundle is also a function of the market prices $p_1$ and $p_2$, but instead of the income $m$ being the exogenous factor that determines the constraint, now the constraint is determined by the exogenously given utility of $U$. The solution functions for this problem are called the compensated or Hicksian demand functions: \(\text{Compensated demand functions: }\begin{cases}x_1^c(p_1,p_2,U)\\x_2^c(p_1,p_2,U)\end{cases}\)

Worked example

Mathematically, when we solved the utility maximization subject to a budget constraint problem using a Lagrangian, we characterized the optimal solution by: \(\begin{aligned} \text{Tangency condition: } & MRS(x_1^\star,x_2^\star) = {p_1 \over p_2}\\ \text{Constraint condition: } & p_1x_1^\star + p_2x_2^\star = m \end{aligned}\) If we were to solve the cost minimization problem by the same method, we would end up with the optimal bundle $(x_1^c,x_2^c)$ being characterized by two similar conditions: \(\begin{aligned} \text{Tangency condition: } & MRS(x_1^\star,x_2^\star) = {p_1 \over p_2}\\ \text{Constraint condition: } & u(x_1,x_2) = U \end{aligned}\) Because the constraint is not a budget constraint, but rather a utility constraint – that is, we’re constraining ourselves to be along the indifference curve representing bundles, the optimal bundle is the one along the utility constraint where the tangency condition holds.

Let’s look at the example of a Cobb-Douglas utility function, $u(x_1,x_2) = x_1x_2$. The MRS for this utility function is \(MRS = {x_2 \over x_1}\) so the tangency condition is \(x_2 = {p_1 \over p_2} x_1\) For the utility maximization problem, we plug this into the budget constraint $p_1x_1 + p_2x_2 = m$, and get the optimal bundle \(x_1^\star(p_1,p_2,m) = {m \over 2p_1}\) \(x_2^\star(p_1,p_2,m) = {m \over 2p_2}\) For the cost minimization problem, we plug this in to the utility constraint to find the “compensated” bundle $(x_1^c,x_2^c)$: \(\begin{aligned} x_1x_2 &= U\\ x_1\left[{p_1 \over p_2} x_1\right] &= U\\ x_1^2 &= {p_2 \over p_1}U\\ x_1^c &= \sqrt{\frac{p_2}{p_1}U} \end{aligned}\) Plugging this back into the tangency condition gives us \(x_2^c = {p_1 \over p_2} x_1^c = \sqrt{\frac{p_1}{p_2}U}\) Visually, we can compare these two methods:

The left-hand graph shows the utility maximization problem with a budget constraint for income $m$; the right-hand graph shows the cost minimization problem with a utility constraint for utility $U$. In each case, the optimum is the intersection of the relevant constraint with the tangency condition.

Solution Functions and Value Functions

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 $U$.

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 $U \in \mathbb{R}$ we must attain.

In the reading for lecture 10, 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.

Value Function is like a “high score” in a video game. For example, think about a simple maximization problem: maximizing the rectangular area enclosed by a fence of length $F$ by choosing a length $L$ and width $W$. If $F = 40$, this means that the $2L + 2W = 40$:

Here, our objective function is the area $A(L,W) = LW$ and the constraint is that the perimeter be 40 feet: $2L + 2W = F$. More formally, we can write this problem as \(\max_{(L,W)} A(L,W) = LW\) \(\text{s.t. } 2L + 2W = F\) The Lagrangian for this problem is \(\mathcal{L}(L,W,\lambda) = LW + \lambda(F - 2L - 2W)\) and the first-order conditions are \(\begin{aligned} \frac{\partial \mathcal{L}(L,W,\lambda)}{\partial L} &= W - 2\lambda = 0 &\Rightarrow& W = 2\lambda \\ \frac{\partial \mathcal{L}(L,W,\lambda)}{\partial W} &= L - 2\lambda = 0 &\Rightarrow& L = 2\lambda \\ \frac{\partial \mathcal{L}(L,W,\lambda)}{\partial \lambda} &= F - 2L - 2W =0 &\Rightarrow& 2L + 2W = F \end{aligned}\) Solving this gives us our solution functions \(\begin{aligned} L^\star(F) &= {F \over 4}\\ W^\star(F) &= {F \over 4}\\ \lambda(F) &= {F \over 8} \end{aligned}\) To get our value function, we take our optimal values of $L^\star$ and $W^\star$ and plug them back into the objective function $A(L,W) = LW$. This gives us the maximum possible area for any length of fence $F$: \(V(F) = A(L^\star(F), W^\star(F)) = L^\star(F) \times W^\star(F) = {F \over 4} \times {F \over 4} = {F^2 \over 16}\) (Intuitively, all this is saying is that for any amount of perimeter $F$, the best we can do is to create a square with side lengths $F/4$, which has an area of $(F/4)^2 = F^2/16$.)

With this interpretation, note that $\lambda$ represents the derivative of this $A^\star(F)$ function: \({dA^\star(F) \over dF} = {d \over dF}\left({F^2 \over 16}\right) = {F \over 8}\) Visually, we can see this as the slope of $A^\star(F)$:

In other words, $\lambda$ tells us the amount by which the objective function rises due to a one-unit relaxation of the constraint. (We can also see that if we take the derivative of the Lagrangian with respect to $F$, we get $\lambda$.)

In consumer theory, value functions represent 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 $E(p_1,p_2,U)$.

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: $V(\vec{p}, E(\vec{p}, U)) = U$ and $E(\vec{p}, V(\vec{p}, m)) = m$. (2) Marshallian and Hicksian demand coincide at optimality: $\vec{x}(\vec{p}, E(\vec{p}, U)) = \vec{x}^c(\vec{p}, U)$ and $\vec{x}^c(\vec{p}, V(\vec{p}, m)) = \vec{x}(\vec{p}, m)$.

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.
• Invert whichever value function you solved for to get the other.
• 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 $E(3, 3, 16)=2 \sqrt{9*16} = 24$.

Worked example, continued

Let’s see how this plays out with the example from above, with the utility function $u(x_1,x_2) = x_1x_2$.

The Marshallian demand functions were \(\begin{aligned} x_1^\star(p_1,p_2,m) &= {m \over 2p_1}\\ x_2^\star(p_1,p_2,m) &= {m \over 2p_2} \end{aligned}\) Plugging these back into the utility function gives us the indirect utility function \(V(p_1,p_2,m) = u(x_1^\star(p_1,p_2,m),x_2^\star(p_1,p_2,m)) = {m \over 2p_1} \times {m \over 2p_2} = {m^2 \over 4p_1p_2}\) We can find the expenditure function by inverting this (i.e., setting it equal to $U$ and solving for $m$): \(\begin{aligned} U &= {m^2 \over 4p_1 p_2} \\ 4p_1 p_2U &= m^2 \\ m &= 2 \sqrt{p_1 p_2 U} \equiv E(p_1,p_2,U) \end{aligned}\) It works the other way, too! We found that the Hicksian demand functions were \(\begin{aligned} x_1^c(p_1,p_2,U) &= \sqrt{ {p_2 \over p_1}U}\\ x_2^c(p_1,p_2,U) &= \sqrt{ {p_1 \over p_2}U} \end{aligned}\) Plugging these back into the expenditure function gives us \(E(p_1,p_2,U) = p_1x_1^c(p_1,p_2,U) + p_2x_2^c(p_1,p_2,U) = p_1\sqrt{ {p_2 \over p_1}U} + p_2\sqrt{ {p_1 \over p_2}U} = 2\sqrt{p_1p_2U}\) Inverting this (i.e., setting it equal to $m$ and solving for $U$) gives us back the indirect utility function: \(\begin{aligned} m &= 2 \sqrt{p_1 p_2 U}\\ m^2 &= 4p_1p_2U\\ U &= {m^2 \over 4p_1 p_2} \equiv V(p_1, p_2, m) \end{aligned}\)

Useful Properties of Expenditure

The expenditure function $E(p_1, p_2, U)$, the lowest spending required at prices $\vec p$ to achieve utility $U$, 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. $\frac{\partial E}{\partial p_i} = x_i^c(\vec{p}, U)$, 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.

To see that it works, note that in the worked example above we have \(E(p_1,p_2,U) = 2\sqrt{p_1p_2U}\) and we can see that the partial derivatives of this with respect to the prices give us back our Hicksian demand! \(\begin{aligned} \frac{\partial E}{\partial p_1} &= \sqrt{ {p_2 \over p_1} U} \equiv x_1^c(p_1,p_2,U)\\ \frac{\partial E}{\partial p_2} &= \sqrt{ {p_1 \over p_2} U} \equiv x_2^c(p_1,p_2,U)\\ \end{aligned}\)

Homogeneous of degree 1 in prices. $E(\alpha \vec{p}, U) = \alpha E(\vec{p}, U)$

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 $\vec{p}_2$, \(E(\alpha \vec{p}_1 + (1-\alpha)\vec{p}_2, U) \geq \alpha E(\vec{p}_1, U) + (1-\alpha)E(\vec{p}_2, U).\)

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

Continuous. The expenditure function is continuous.

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