Machina Math Handout

n x1 xn g n x1 xn 0 l

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , the method of Lagrangians is nothing more than a roundabout way of generating our condition that the normal vector of ƒ(x1,..., xn) be λ times the normal vector of g(x1,..., xn), and the constraint g(x1,..., xn) = c be satisfied. We will not do second order conditions for constrained optimization (they are a royal pain). Econ 100A 11 Mathematical Handout F. OPTIMIZATION #2: COMPARATIVE STATICS OF SOLUTION FUNCTIONS Having obtained the first order conditions for a constrained or unconstrained optimization problem, we can now ask how the optimal values of the control variables change when the parameters change (for example, how the optimal quantity of a commodity will be affected by a price change or an income change). Consider a simple maximization problem with a single control variable x and single parameter α max ƒ( x ;α ) x For a given value of α, recall that the solution x* is the value that satisfies the first order condition ∂ ƒ( x*;α ) =0 ∂x Since the values of economic parameters can (and do) change, we have defined the solution function x*(α) as the formula that specifies the optimal value x* for each value of α. Thus, for each value of α, the value of x*(α) satisfies the first order condition for that value of α. So we can basically plug the solution function x*(α) into the first order condition to obtain the identity ∂ ƒ( x*(α );α ) ≡0 α ∂x We refer to this as the identity version of the first order condition. Comparative statics is the study of how changes in a parameter affect the optimal value of a control variable. For example, is x*(α) an increasing or decreasing function of α ? How sensitive is x*(α) to changes in α? To learn this about x*(α), we need to derive its derivative d x*(α)/dα. The easiest way to get d x*(α)/dα would be to solve the first order condition to get the formula for x*(α) itself, then differentiate it with respect to α to get the formula for d x*(α)/dα. But sometimes first order conditions are too complicated to solve. Are we up a creek? No: there is another approach, implicit differentiation, which always gets the formula for the derivative d x*(α)/dα. In fact, it can get the formula for d x*(α)/dα even when we can’t get the formula for the solution function x*(α) itself ! Implicit differentiation is straightforward. Since the solution function x*(α) satisfies the identity ∂ ƒ( x*(α );α ) ∂x ≡0 α we can just totally differentiate this identity with respect to α, to get ∂ 2 ƒ( x*(α );α ) d x*(α ) ⋅ dα ∂x 2 and solve to get Econ 100A d x*(α ) dα ≡ α + ∂ 2 ƒ( x*(α );α ) ∂x ∂α ∂ 2 ƒ( x*(α );α ) − ∂x ∂α 12 ≡0 α ∂ 2 ƒ( x*(α );α ) ∂x 2 Mathematical Handout For example, let’s go back to that troublesome problem max α⋅x2 – ex, with first order condition 2⋅α⋅x* – ex* = 0. Its solution function x* = x*(α) satisfies the first order condition identity 2 ⋅α ⋅ x*(α ) − e x*(α ) ≡ α 0 So to get th...
View Full Document

Ask a homework question - tutors are online