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
∂ ƒ( x*;α )
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 )
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 )
α we can just totally differentiate this identity with respect to α, to get
∂ 2 ƒ( x*(α );α ) d x*(α )
and solve to get Econ 100A d x*(α )
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...
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