MICROECONOMIC THEORY I PROBLEM SET 1 MARCIN PĘSKI Properties of rational preferences. MWG 1.B1 and 1.B.2. Solutions: Tutorial Utility and preferences. MWG 1.B.4. Solutions: Tutorial Choice structure. MWG 1.D.1 and MWG 1.D.4. Solutions: Tutorial Lexicographic prefences. MWG 3.C.1. Solutions: Let us formally define lexicographic ordering on R+: L (x 1,..., x L ) (y 1,..., y L ) if and only if there exists k = 1,..., L, +1 such that (a) for each l < k, x l = y l, and (b) x k < y k (the last part does not apply if k = L + 1). Completeness: For each (x 1,..., x L ), (y 1,..., y L ) R+, L either two bundles are equal (and both of them are equivalent with respect to the lexic. preferences), or the exists the first coordinate such that x k > y k or vice versa. In the former case, (x 1,..., x L ) (y 1,..., y L ), and in the latter (x 1,..., x L ) (y 1,..., y L ). Transitivity:Suppose that (x 1,..., x L ) (y 1,..., y L ) and (y 1,..., y L ) (z 1,..., z L ). Let k and k be the coordinates from the definition for, respectively, the first and the second relation. Let k = min (k, k ). Then, x l = z l for all l < k, and x k < z k (again, the last inequality does not apply if k = L+1). In particular, (x 1,..., x L ) (z 1,..., z L ). Strong monotonicity. It follows directly from definition. Strict convexity. It follows from the fact that each indifference set contains exactly one element (so that the definition applies trivially). 1
2 MARCIN PĘSKI Leontieff preferences. Suppose that L = 2. Suppose that preferences on X = {(a, b) : a, b 0} are represented by the following utility functions: min (a, b), max (a, b). Check convexity, monotonicity, homotheticity, and local non-satiation of these preferences. Solutions: Both preferences are homothetic, locally non-satiated and min (a, b) (but not max (a, b)) is also convex. Quasi-linear preferences. (Based on MWG 3.C.5b) Let X = {(x 1,..., x L ) : l L 1 x l 0}. To shorten the notation, for each x X, we will write x = (y, x L ), where y R L 1 +. A preference relation on X is quasi-linear with respect to the last good L if for any two bundles x = (y, x L ) and x = (y, x L), for any constant α > 0, adding α amount of good L to both consumption bundles does not change the ranking of bundles. (y, x L ) (y, x L) if and only if (y, x L + α) (y, x L + α), good L is desirable: for each α > 0, (y, x L ) (y, x L + α). Suppose that a consumer has a preference relation on X that is quasi-linear with respect to the last good L, continuous, and strictly monotone. In this exercise, we are going to show that is quasi-linear with respect to the last good L if and only if there exists function ψ : R L 1 + R such that is represented by utility function u (y, x L ) = ψ (y) + x L. (1) Show that if is represented by utility function u (y, x L ) = ψ (y) + x L for some function ψ : R L 1 + R, then is quasi-linear with respect to the last good L. (2) In the rest of the exercise, we are going to show the implication from part (a) holds also in the other direction, i.e. that is quasi-linear with respect to the last good L only if there exists function ψ : R L 1 + R such that is represented by utility function u (y, x L ) = ψ (y) + x L. As the first step, explain, that there exists a continuous utility function v : R L 1 + R R that represents.
ADVANCED MICRO THEORY II 3 (3) Let Y R L 1 + be the set of tuples y for which there exists at least one x R, such that v (y, x) = 0. Show that for each y Y, for any x L, x L R, if v (y, x L ) = v (y, x L) = 0, then x L = x L. (4) Assume for now on that Y = R L 1 + (we will come back to proving that later). By the previous point, for each y R L 1 +, there exists exactly one x L (y) R such that v (y, x L (y)) = 0. Let ψ (y) = x L (y). Show that utility function u : R L 1 + R R defined by represents preferences. u (y, x L ) = ψ (y) + x L (5) (Difficult.) We need to go back to our missing step. Without using part (d) of the excercise, show that Y = R L 1 +. Solutions: Part (2). It follows from Proposition in the class and the fact that we assume that is continuous and strictly monotone. Part (3). It follows from the fact that good Lis always desirable for quasilinear preferences (or, from strong monotonicity of the preferences). Part (4). Take any two bundles two bundles x = (y, x L ) and x = (y, x L). Observe that x = (y, x L ) = (y, ψ (y) + ψ (y) + x L ) = (y, ψ (y) + u (x)), and, similarly, x = (y, ψ (y ) + u (x )). Notice that by the choice of function ψ (.), we have (y, ψ (y)) (y, ψ (y )). By the first property of quasi-linear preferences, (y, ψ (y) + u (x)) (y, ψ (y ) + u (x)). By the second property of quasi-linear preferences, u (x) u (x ) if and only if x = (y, ψ (y) + u (x)) (y, ψ (y ) + u (x)) (y, ψ (y ) + u (x) + u (x ) u (x)) = x. Part (5) (Thanks to David Walker-Jones for providing a solution.) We will show that Y = R L 1 + by assuming y R L 1 + \Y and reaching a subsequent contradiction
4 MARCIN PĘSKI which shows R L 1 + \ Y =, and thus Y = R L 1 + as desired. Notice through construction of v that Y can be easily shown to be non-empty, and further, that wlog we can assume v(y, x L ) < 0 x L R. 1 Consider the line S in R L 1 + with gradient(1,1,1,...1) that passes though y and continues on to y, where y > y 0. Because v is strictly monotone v(y, x 0 L) > 0, and thus since v is continuous there is a boundary point z between Y and its complement Y c on S. There are two cases that must be considered: when z Y, and when z Y c. Each case will result in the desired contradiction. Case 1: Assume z Y. This means x z L st v(z, x z L) = 0. Construct an infinite sequence {y n } n=1 of points on the line S all contained in Y c that converges to z. Now choose a δ > 0 and note that since v is strictly monotone v(z, x z L + δ) = ɛ > 0. Since v is continuous v(y n, x z L + δ) converges to v(z, x z L + δ) = ɛ. Thus k st if n k then v(y n, x z L + δ) > 0. This is impossible since v is continuous, n y n Y c, and v(y n, x z L) < 0. Case 2: Assume z Y c. Construct an infinite sequence {y n } n=1 of points on the line S all contained in Y that converges to z. Let v(z, x 0 L) = c < 0. Pick δ > 0 and let v(z, x 0 L + δ) = c + ɛ < 0 where ɛ > 0 since v monotone. Further, since v is continuous, v(y n, x 0 L) converges to v(z, x 0 L). Thus k st if n k then v(y n, x 0 L) < c + ɛ. Choose such an n k and observe that v(z, x 0 L + δ) > v(y n, x 0 L). Let x n L be the value st v(y n, x n L) = 0. Since v is quasi-linear in x L we have that 1 Y can be ensured to be non-empty by the construction of v: from the proposition in lecture, and as is discussed in part (b), we know that since is continuous and monotone it can be represented by a continuous utility function v 0. We can further pick a point (y 0, x 0 L ) in RL 1 + R and create a new continuous representation of defined by v(y, x L ) = v 0 (y, x L ) v 0 (y 0, x 0 L ). Thus v(y0, x 0 L ) = 0, and Y is non-empty by construction. Further because v is continuous, v(y, x L ) must either be negative x L R, or positive x L R. Since v(y, x L ) can be made negative x L R if it is not already (by simply letting v(y, x L ) = v 0 (y, x L ) + v 0 (y 0, x 0 L )), we can assume v(y, x L ) < 0 x L R. Given what we eventually show, these kinds of alterations are not concerning in the least.
ADVANCED MICRO THEORY II 5 v(z, x 0 L + δ + (x n L x 0 L)) > v(y n, x 0 L + (x n L x 0 L)) = 0. 2 This is of course impossible since v is continuous, z Y c, and v(z, x 0 L) < 0. 2 We know x n L x0 L > 0 since v is monotone and v(yn, x 0 L ) < 0 while v(yn, x 0 L + (xn L x0 L )) = v(y n, x n L ) = 0