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A Note on
Perfect Complements
1 Utility when Goods are Perfect
Complements
At some point, we have been considering the case in which two goods, say (x 1 , x 2 ), can only be consumed in a fixed proportion to each other. In such a case, we say that x 1 and x 2 are perfect complements. In this case, there is no way to substitute one good for the other one. In other words, the elasticity of substitution between x 1 and x 2 equals zero. How do we handle such a situation mathematically? Well, by, what we called a min function (Leon- tief function, limitational function). But what is a min function, and how do we maximize utility when the utility function is a min function (Leontief utility)? This note is addressing these very questions.
2 Digression: What is a Function?
Let us define the set of natural (counting) numbers by N = { 1 , 2 , 3 , ...}. Furthermore, let the n-fold Cartesian product of R, be given by Rn^ ≡ R × R × ... × R (where the product is taken n times), and n ∈ N
Generally, a (real-valued) function is a rule that assigns a unique real number to each element of its domain. Let us denote the domain by X ⊂ Rn, with n ∈ N. Each element of Rn^ is a vector of dimension n (an ordered n-tuple). So, the function may be a function of a single variable (in which case, n = 1) or of several variables (in which case n > 1).
We typically denote functions the following way. Let u denote our function of interest. Then we write:
u(x) : X → R. (1)
This is understood the following way. The function u assigns to each element of its domain x ∈ X a unique element from the set denoted at the right hand side of the right-arrow (R in our case — i.e., a real number).
In (1), it is understood that (i) X ⊂ Rn; (ii) x = (x 1 , x 2 , ..., xn) ∈ X. Let us work through three simple examples, next.
Example 1. Let n = 2 and X = R^2 +. That is x = (x 1 , x 2 ) ≥ 0, and u(x) = u(x 1 , x 2 ) = ax 1 + bx 2. Function u(x) assigns a real value to each member of its domain x = (x 1 , x 2 ) ∈ X. For example, consider x = (1, 1). Then,u(1, 1) = a + b. Or, u(0, 0) = 0.
Example 2. Let n = 2, X = R^2 +, and u(x) = u(x) = x^11 / 2 x^12 / 2. Function u(x) assigns a real value to each member of its domain x = (x 1 , x 2 ) ∈ X. For example, consider x = (1, 1). Then u(1, 1) = 1. Or u(4, 4) = 4.
Example 3. Let n = 2, X = R^2 +, and u(x) =
a xδ/δ + b yδ/δ
) 1 /δ , δ < 1.
As you recall, this is a CES utility function. Then u(1, 1) = (a/δ + b/δ)^1 /δ^ = (1/δ)^1 /δ^ (a + b)^1 /δ.
Notice that utility is negative in case δ < 0. This does not concern us in any way, as discussed in class.
We will come back to this CES utility function below. Specifically, we will argue that the min function is obtained as the limit of the CES utility function where the elasticity of substitution between x 1 and x 2 approaches zero.
3 The min Function
In order to keep things simple, we (1) interpret our function u as a utility function, and we (2) restrict ourselves to the case with two goods: n = 2; X = R^2 +. That is, we focus on the case
u(x 1 , x 2 ) : R^2 + → R. (2)
To deal with perfect complements, we introduce the min function here:
u(x 1 , x 2 ) = min{ax 1 , bx 2 } , a, b ∈ R++.
So, for any given parameter values of a and b, this function assigns to each (x 1 , x 2 ) ∈ X the smaller value, either a x 1 or b x 2. Formally,
u(x 1 , x 2 ) =
a x 1 , if a x 1 < b x 2 b x 2 , if a x 1 ≥ b x 2
Example 4. Let n = 2 and X = R^2 +. Suppose a < b. Then u(1, 1) = a. However, if a > b, then u(1, 1) = b. And it is very simple to calculate u(x 1 , x 2 ) for any other values of (x 1 , x 2 ). E.g., u(2, 1) = 2a if 2a < b, other- wise, u(2, 1) = b.
4 Marginal Utility and Indifference Curves
Given our utility function (3), marginal utilities are easily derived.
u 1 (x 1 , x 2 ) ≡
∂u(x 1 , x 2 ) ∂x 1
a , if a x 1 < b x 2 0 , if a x 1 ≥ b x 2
u 2 (x 1 , x 2 ) ≡
∂u(x 1 , x 2 ) ∂x 2
0 , if a x 1 < b x 2 b , if a x 1 ≥ b x 2
Consequently, the marginal rate of substitution of x 1 for x 2 is given by u 1 (x 1 , x 2 )/u 2 (x 1 , x 2 ) = 0 for ax 1 ≥ bx 2. So, the slope of the indifference curve is zero for ax 1 ≥ bx 2 and is represented as a horizontal line in (x 1 , x 2 ) space. Likewise, the marginal rate of substitution of x 2 for x 1 is given by u 2 (x 1 , x 2 )/u 1 (x 1 , x 2 ) = 0 for ax 1 < bx 2. That is, it is represented as a verti- cal line in (x 1 , x 2 ) space. Figure 2 illustrates Indifference curves for different values of utility ¯u.
A
B
C
D
0.5 1.0 1.5 2.
x 1
x 2
Figure 2: Indifference curves. Note: a = 1, b = 2.
Figure 2, displays three indifference curves for our min function. Clearly, u(A) = u(B) = u(C) > u(D). The M RSx 1 ,x 2 (A) = 0, and the M RSx 2 ,x 1 (B) = 0 = M RSx 2 ,x 1 (D). There is one question remaining. What about the M RSx 1 ,x 2 (C)? The answer is rather simple: the M RSx 1 ,x 2 (C) is not defined, as division by zero is against the rules. In other words, the slope of the indifference curve at C
(or, more generally, along the ray from the origin along which ax 1 = bx 2 ) is not defined. This has significant implications for utility maximization, as discussed in the proceeding section of this note.
5 The min Function and Utility Maximization
Let (p 1 , p 2 ) denote the prices of (x 1 , x 2 ), and I some given income. Obvi- ously, with a min utility function, we cannot apply the necessary first order conditions for an interior solution
u 1 (x 1 , x 2 ) u 2 (x 1 , x 2 )
p 1 p 2
as the left hand side (marginal rate of substitution of x 1 for x 2 ) is not defined for all (x 1 , x 2 ). Also, our Kuhn-Tucker conditions are not of help in this case. In order to make some progress, consider the following Figure 3.
C
B
D
A
0.5 1.0 1.5 2.
x 1
x 2
Figure 3: Optimal choice. Note: a = 1, b = 2.
In Figure 3, the budget set (all consumption bundles (x 1 , x 2 ) that cost less than the income I) is depicted as the shaded area (in yellow). This set corresponds to the opportunity set, as we discussed in class. All consumption bundles in this area can be afforded. Consumption bundles outside this set are either not affordable or not available (e.g., negative quantities of either x 1 or x 2 ). What is the best — that is, utility-maximizing – choice?
where the last line follows from Shephard’s lemma. Furthermore, the substi- tution effects are zero: ∂xc 1 /(∂pi) = ∂xc 2 /(∂pi) = 0, i = 1, 2.
7 The min Function as a Special Case of the
CES Function
In our last section, we derive the min Function as the limit of the CES Function when the elasticity of substitution approaches zero. As a starting point, consider a CES utility function. In a simple version, we may consider
u(x 1 , x 2 ) = [(a x 1 )−ρ^ + (b x 2 )−ρ]−^1 /ρ^ , a, b > 0 , ρ > − 1 , (13)
with (x 1 , x 2 ) ∈ R^2 +. In class, we denoted the exponent by δ = −ρ. For this section, it is slightly easier to use the ρ (as it will turn out to be a positive exponent, rather than a negative one). With this notation at hand, we can define the elasticity of substitution of x 1 for x 2 by
σ =
1 + ρ
For perfect complements, the elasticity of substitution equals zero. That is, we aim to show that in the limit, as ρ approaches plus infinity, CES function (13) becomes u(x 1 , x 2 ) = min{a x 1 , b x 2 }. (15) Assume, without loss of generality, that ax 1 ≥ bx 2.^1 That is,
min{ax 1 , bx 2 } = bx 2. (16)
We consider the limit as ρ approaches plus infinity. That is, we do not care about non-positive of ρ and, without loss of generality, assume that ρ > 0. As we assume ax 1 ≥ bx 2 , and noting that ρ > 0,
(ax 1 )−ρ^ ≤ (bx 2 )−ρ^. (17)
Moreover, let us write the utility function as:
u−^1 = [(a x 1 )−ρ^ + (b x 2 )−ρ]^1 /ρ^. (18) (^1) You may assume ax 1 ≤ bx 2. The analysis presented is the same, just then,
min{ax 1 , bx 2 } = ax 1.
Now, we construct two inequalities. First, we replace (ax 1 ) with the weakly smaller (bx 2 ). Observing (17),
u−^1 = [(a x 1 )−ρ^ + (b x 2 )−ρ]^1 /ρ ≤ [(b x 2 )−ρ^ + (b x 2 )−ρ]^1 /ρ^ = [2(bx 2 )−ρ]^1 /ρ^ = 2^1 /ρ(bx 2 )−^1.
So, we know that u−^1 ≤ 21 /ρ(bx 2 )−^1. (19)
Second, obviously
u−^1 = [(a x 1 )−ρ^ + (b x 2 )−ρ]^1 /ρ^ ≥ [(b x 2 )−ρ]^1 /ρ^ = (bx 2 )−^1 ,
so we know that u−^1 ≥ (bx 2 )−^1. (20)
Putting inequalities (19) and (20) together, we know that for all ρ > 0:
21 /ρ(bx 2 )−^1 ≥ u−^1 ≥ (bx 2 )−^1. (21)
As a final step, consider the limit as ρ goes to plus infinity:
lim ρ→∞ 21 /ρ(bx 2 )−^1 = (bx 2 )−^1 , (22)
as 2^1 /ρ^ approaches unity in the limit. Now, we “sandwiched” u−^1 in the limit:
lim ρ→∞ u−^1 = (bx 2 )−^1. (23)
It follows that limρ→∞ u = b x 2 :
lim ρ→∞ u(x 1 , x 2 ) = lim ρ→∞ [(a x 1 )−ρ^ + (b x 2 )−ρ]−^1 /ρ^ = b x 2 = min{ax 1 , bx 2 } , (24)
as was to be shown. As mentioned in the footnote, a parallel argument can be given for the case of ax 1 ≤ bx 2.
References
[1] Arrow, K.J., H.B. Chenery, B.S. Minhas, R.M. Solow (1961), Capital- labor substitution and economic efficiency, The Review of Economics and Statistics 43 , 225-250.
