EC400 Part II, Math for Micro: Lecture 5 Leonardo Felli NAB.SZT 15 - - PowerPoint PPT Presentation

EC400 Part II, Math for Micro: Lecture 5

Leonardo Felli

NAB.SZT

15 September 2010

One Inequality Constraint:

Let f and g be continuous functions defined on U ∈ Rn. Assume that x∗ is the solution of the problem: max

x

f (x) s.t. g(x) ≤ b. Assume also that x∗ is not a critical point of g(x) if g(x∗) = b. Then given the Lagrangian function: L(x, λ) = f (x) − λ (g(x) − b)

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 2 / 24

There exists a real number λ∗ such that: ∇xL(x∗, λ∗) = ∇f (x∗) − λ∗∇g(x∗) = 0. λ∗ (g(x∗) − b) = 0 λ∗ ≥ 0 g(x∗) ≤ b

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 3 / 24

Example:

ABC is a perfectly competitive and profit maximizing firm. It produces y from input x according to the production function f (x) = x1/2. The price of output y is 2, and of input x is 1. Negative levels of x are impossible. Also, the firm cannot buy more than a > 0 units of input x.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 4 / 24

The firm’s maximization problem is therefore max

x

f (x) = 2x1/2 − x s.t. g(x) = x ≤ a x ≥ 0 We ignore for now the last constraint x ≥ 0. The Lagrangian is: L(x, λ) = 2 x1/2 − x − λ[x − a]

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 5 / 24

The first order condition with respect to x is: (x∗)−1/2 − 1 − λ∗ = 0 Therefore: (x∗)−1/2 − 1 − λ∗ = λ∗(x∗ − a) = λ∗ ≥ x∗ ≤ a

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 6 / 24

We can now solve the system of equations. It is the easiest to consider the two separate cases: λ∗ > 0 and λ∗ = 0. Assume λ∗ > 0. The constraint is binding. Then x∗ = a. The full solution is then (constraint x∗ ≥ 0 is satisfied): x∗ = a, λ∗ = 1 √a − 1 When is this solution viable?

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 7 / 24

We need to keep consistency so if we assume that λ∗ > 0 it needs to be the case that: 1 √a − 1 > 0 ⇔ a < 1. What if λ∗ = 0? this means that the constraint is not binding. From the first order condition: (x∗)−1/2 − 1 = 0 ⇔ x∗ = 1 The solution is therefore (constraint x∗ ≥ 0 is satisfied): x∗ = 1, λ∗ = 0 and this solution holds for all a ≥ 1.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 8 / 24

Several Inequality constraints

The generalization is easy. Key difference: some constraints may be binding while some may not. An example: consider the problem: max

x,y,z

x y z s.t. x + y + z ≤ 1 x ≥ 0, y ≥ 0, z ≥ 0.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 9 / 24

The Lagrangian is L(x, y, z, λ1, λ2, λ3, λ4) = = x y z − λ1(x + y + z − 1) + λ2x + λ3y + λ4z Solving the Lagrange problem yields a set of critical points. The optimal solution will be a subset of this. However, we can already restrict this set of critical points because clearly λ2 = λ3 = λ4 = 0. Indeed, if, say, λ2 > 0, then by complementary slackness x = 0. But then the value of (x y z) is 0, and obviously we can do better than that (for example x = y = z = 3/10).

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 10 / 24

Thus, the non-negativity conditions cannot bind. This leaves us with a problem with one constraint x + y + z − 1 ≤ 0. We have to decide whether λ1 > 0 or λ1 = 0. Obviously, the constraint must bind. If x + y + z < 1 we can increase one of the variables, satisfy the constraint, and increase the value of the function.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 11 / 24

From the first order conditions: xy − λ1 = zy − λ1 = xz − λ1 = We then find that xy = yz = zx and hence it follows, from the binding constraint, that the optimal solution is: x = y = z = 1 3

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 12 / 24

We have looked at: max

x,y

f (x y) s.t. g(x, y) ≤ b. We have characterized necessary conditions for a maximum. So that if x∗ is a solution to a constrained optimization problem (it maximizes f subject to some constraints), it is also a critical point of the Lagrangian. We then find the critical points of the Lagrangian.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 13 / 24

Can we then say that these are the solutions to the constrained

• ptimization problem?

In other words, can we say that if these are maximizers of the Lagrangian these are also maximizers of f (subject to the constraint)? To determine the answer, let (x′, y′, λ) satisfy all necessary conditions for a maximum. We can now show that if x′, y′ is a maximizer of the Lagrangian, it also maximizes f . To see this notice that λ [g(x′, y′) − b] = 0.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 14 / 24

Thus, f (x′, y′) = f (x′, y′) − λ[g(x′, y′) − b]. By λ ≥ 0 and g(x, y) ≤ b for all other (x, y), then f (x, y) − λ[g(x, y) − b] ≥ f (x, y). Since x′, y′ maximizes the Lagrangian, then for all other x, y : f (x′, y′) − λ[g(x′, y′) − b] ≥ f (x, y) − λ[g(x, y) − b] Which implies that f (x′, y′) ≥ f (x, y) So that if (x′, y′) maximizes the Lagrangian, it also maximizes f (x, y) subject to g(x, y) ≤ b.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 15 / 24

Recall the main results from unconstrained optimization: If f is a concave function defined on a convex subset U ⊂ Rn, x0 is a point in the interior of U such that Df (x0) = 0, then x0 maximizes f (x) in U : f (x) ≤ f (x0) for all x. You have seen in class that in the constrained optimization problem, if f is concave and g is convex, then the Lagrangian function is also concave. This means that first order conditions are necessary and sufficient.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 16 / 24

Main Theorem:

Theorem (The Kuhn-Tucker Theorem)

Consider the problem of maximizing f (x) subject to the constraint that g(x) ≤ b. Assume that f and g are differentiable, f is concave, g is convex, and that the constraint qualification holds. Then x∗ solves this problem if and only if there exists a scalar λ∗ such that ∂L(x∗, λ∗) ∂xi = ∂ ∂xi f (x∗) − λ ∂ ∂xi g(x∗) = 0, ∀i λ∗ ≥ g(x∗) ≤ b λ∗[b − g(x∗)] =

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 17 / 24

Several Inequality Constraints

Consider the problem: max

x

f (x) s.t. g1(x) ≤ b1 . . . gm(x) ≤ bm. The Lagrangian is then L(x, λ) = f (x) − λ(g(x) − b). where λ = (λ1, . . . , λm) and (g(x) − b) =    g1(x) − b1 . . . gm(x) − bm   

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 18 / 24

Assume that there exist λ∗ such that: ∂L(x∗, λ∗) ∂xi = 0, ∀i = 1, . . . , n λ∗

j

≥ 0, ∀j = 1, . . . , m λ∗

j (gj(x) − b)

= 0, ∀j = 1, . . . , m Assume that g1 to ge are binding and that ge+1 to gm are not: λe+1 = 0, . . . , λm = 0. Denote gE =    g1 . . . ge    λE = (λ1, . . . , λe) .

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 19 / 24

Suppose that the Hessian of L with respect to x at (x∗, λ∗) is negative definite on the linear constraint set {v : DgE(x∗)v = 0}, that is: v = 0, DgE(x∗)v = 0 → vT(D2

xL(x∗, λ∗))v < 0,

Then x∗ is a strict local constrained maximizer of f on the constraint set. The question is then how to establish that whether D2

xL(x∗, λ∗) is

negative definite on a constraint set.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 20 / 24

To establish this we make use of the bordered Hessian: Q =

• DgE(x∗)

DgE(x∗)T D2

xL(x∗, λ∗)

• If the last (n − e) leading principal minors of Q alternate in sign with

the sign of the determinant of the largest matrix the same as the sign

• f (−1)n, then sufficient second order conditions hold for a candidate

point x∗ to be a solution to a constrained maximization problem.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 21 / 24

Intuition:

What is D2L(λ, x)?             D2

λL(λ, x) = 0 ∂2L ∂λ1∂x1 = − ∂g1 ∂x1

· · ·

∂2L ∂λ1∂xn = − ∂g1 ∂xn

. . . ... . . .

∂2L ∂λe∂x1 = − ∂ge ∂x1

· · ·

∂2L ∂λe∂xn = − ∂ge ∂xn

−DgE(x∗)T

∂2L ∂2x1

· · ·

∂2L ∂x1∂xn

. . . ... . . .

∂2L ∂x1∂xn

· · ·

∂2L ∂2xn

            Hence, the bordered Hessian is similar to the Hessian of L.

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 22 / 24

SOC with Two Variables and One Constraint

Consider the problem: max

x1,x2

f (x1, x2) s.t. g1(x1, x2) ≤ b. The Lagrangian is L(x1, x2, λ) = f (x1, x2) − λ(g(x1, x2) − b). Assume that there exist λ∗ such that: ∂L(x∗

1, x∗ 2, λ∗)

∂xi = 0, ∀i = 1, 2 λ∗ ≥ λ∗(g(x∗

1, x∗ 2) − b)

= g1(x∗

1, x∗ 2)

≤ b

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 23 / 24

Assume that g is binding: λ∗ > 0. The bordered Hessian matrix is: B =          ∂g ∂x1 ∂g ∂x2 ∂g ∂x1 ∂2f ∂x2

1

− λ∗ ∂2g ∂x2

1

∂2f ∂x1 ∂x2 − λ∗ ∂2g ∂x1 ∂x2 ∂g ∂x2 ∂2f ∂x1 ∂x2 − λ∗ ∂2g ∂x1 ∂x2 ∂2f ∂x2

2

− λ∗ ∂2f ∂x2

2

         The sufficient SOC are |B| > 0 at (x∗

1, x∗ 2).

Leonardo Felli (LSE, NAB.SZT) EC400 Part II, Math for Micro: Lecture 5 15 September 2010 24 / 24