Comparative Statics Overview (pt 1)

Comparative Statics Overview (pt 1):

1 Envelope theorem: simplifies computation of change in value function(optimized level of objective function) as parameters change

1 Some function is being optimized1 function depends on some parameters2 how does optimum value change with these parameters?

2 Context: First order conditions of an optimization problemExamples:

1 optimized utility as a function of income2 maximized profit as a function of price

Two kinds of optimization:1 unconstrained optimization (intuition is fairly straightforward)2 constrained optimization (intuition is much more elusive)

3 Our analysis of the lagrangian was a special case of the envelope theorem1 invoked the (constrained) envelope thm to prove λj = shadow value of bj2 use same approach to compute shadow value” of any parameter

() October 7, 2015 1 / 27


2 Envelope theorem (bonus): in very special cases you can short-circuitthe implicit function theorem, and use the envelope theorem to get thesame information that the implicit function theorem would give you, butmuch more easily

Examples:1 input demand functions (Hotelling’s Lemma)2 conditional input demand functions (Shephard’s Lemma)

These results are very special cases:1 not what the envelope theorem was “meant for”2 just happens to work cos some part of objective function is linear

Why is the envelope theorem called the envelope theorem?(From Wikipedia) The American Heritage Dictionary (3ed) gives one meaningof “envelope” to be: “A curve or surface that is tangent to every one of a familyof curves or surfaces”.

() October 7, 2015 2 / 27

LRATC as lower envelope of SRATC’s

() October 7, 2015 3 / 27

M(b) as the upper envelope of f (x ,b)

∀b, tangent line to M(b) = f (x∗(b),b) lies on tangent plane to f , eval at (x∗(b),b).

db

b

f

(x∗(b1); b1)

(x∗(b2); b2)

(x∗(b3); b3)

x∗(b)

df

dx

x

When you evaluate df , the dx term has no effect because it is multiplied by zero

df = ∂f(x∗(b);b)∂b db+

∂f(x∗(b); b)∂x︸︷︷︸=0

dxdb db

1

() October 7, 2015 4 / 27


3 Implicit function theorem: used to compute relationship betweenendogenous and exogenous variables.

1 Context: First order conditions of an optimization problem.Examples

e.g., demands as a function of incomee.g., input demands as a function of output

2 Context: Some kind of system of equilibrium equations.Examples:

market equilibrium prices as a function of economy endowmentsCournot quantities as a function of demand parameter, # of firms

() October 7, 2015 5 / 27

Two envelope Theorems: (summary)(UET) Unconstrained envelope theorem:

if x∗(ααα) solves maxx f (x;ααα) and x∗(·) is diffable at ααα, then df (x∗(ααα);ααα)dαk

= ∂f (x∗(ααα);ααα)∂αk

.

total derivative of f w.r.t. αk equals partial derivative of f w.r.t. αk .i.e., reoptimization in x direction has no (first-order) impact on df (x∗(ααα);ααα)

dαk

(CET) Constrained envelope theorem:

If (x∗(ααα),λλλ∗(ααα)) solves maxx f (x;ααα) s.t. hj (x;ααα)≥ 0, j = 1, ...m,and x∗(·) is differentiable at ααα, then

df (x∗(ααα);ααα)

dαk=

∂f (x∗(ααα);ααα)

∂αk+

m

∑j=1

λ∗j (ααα)

∂hj (x∗(ααα);ααα)

∂αk

For NPP, we wrote g j (x)≤ bj or bj −g j (x)≥ 0; now have hj (x;ααα) = α0j −g j (x;ααα).In words: total deriv of f w.r.t. αk equals partial deriv of f w.r.t. αk plusλλλ∗-weighted sum of partial derivs of hj ’s w.r.t. αk

only difference between UET and CET is the addition of constraintsagain, reoptimization in x direction has no (first-order) impact on df (x∗(ααα);ααα)

dαk() October 7, 2015 6 / 27

Envelope Theorem: unconstrained version back to slide 15

Assume f is C2:



.

i.e., total derivative of f w.r.t. αk equals partial derivative of f w.r.t. αk .

(a sufficient condition for x(·) to be diffable at x is that it’s the unique maximizer.)


dαk=


∂αk+

n

∑i=1


∂xi︸︷︷︸=0

dxi (ααα)

dαk

which, since necessarily,5xf (x∗(ααα);ααα) = 0, back to slide 15

=∂f (x∗(ααα);ααα)

∂αk

() October 7, 2015 7 / 27

Envelope Theorem: graphical intuition back to slide 16

back to slide 17

() October 7, 2015 8 / 27

Envelope Theorem: tangent planes

0

0.5

1

0

0.5

1

0

0.05

0.1

0.15

0.2

dxdb

df/db

(x*(0

.9),0.

9)

0

0.5

1

0

0.5

1

0

0.05

0.1

0.15

0.2

dxdb

df/db

(x*(3

.3),3.

3)

Height of differential determined exclusively by magnitude of db

() October 7, 2015 9 / 27

Envelope Theorem: tangent planes (alt view) back to slide 15

0

0.5

1

00.510

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

dxdb

df/db

(x*(0

.9),0.

9)

0

0.5

1

00.510

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

dxdb

df/db

(x*(3

.3),3.

3)

() October 7, 2015 10 / 27

Tangent plane when x not optimized given b

x b

In this case, movements in the x direction do affect height of differential

() October 7, 2015 11 / 27

Unconstrained Envelope Th’m Eg: Hotelling’s LemmaHotelling’s lemma:Let π(x;p,w) = pf (x)−w ·x, given output price p and input price vector w.

Let x∗(p,w) be the solution to the unconstrained max, given p,w.

The i ’th input demand function is x∗i (p,w) = − ∂π(x∗(p,w);p,w)∂wi

Again math is straightforward: by the unconstrained envelope theorem,

dπ(x∗(p,w);p,w)

dwi=

∂π(x∗(p,w);p,w)

∂wi= − x∗i (p,w)

Note that this is a “bonus” result:goal here is not: how does profit change when wi changes?

care about r.h.s. of equation not l.h.s

Why is this result so useful?given a functional form for f , get expr for x∗i (p,w) in terms of primitives

() October 7, 2015 12 / 27

Numerical example of Hotelling’s lemmaRecall π(x;p,w) = pf (x)−w ·xLet5f = (1/x1,1/x2) and5xπ = (p/x1−w1,p/x2−w2)

The unconstrained optimum is x∗(p,w) = (p/w1,p/w2).

Now compare the partial deriv, ∂π(x∗(p,w);p,w)∂wi

, to the total deriv, dπ(x∗(p,w);p,w)dwi

,� in order to illustrate that the envelope theorem actually works


∂wi= − x∗i (p,w) = − p

wifrom FOC

dπ(x∗(p,w);p,w)

dwi=


∂wi+

2

∑j=1

∂π(x∗j (p,w);p,w)

∂xj

dx∗j (p,w)

dwi

which, since for j 6= i , x∗j (p,w) doesn’t depend on wi

=∂π(x∗(p,w);p,w)

∂wi+

∂π(x∗i (p,w);p,w)

∂xi

dx∗i (p,w)

dwi

= − x∗i (p,w) + (p/x∗i (p,w)−wi )×−p/w2i

= − x∗i (p,w) +(

p/p/wi −wi︸︷︷︸0

)×−p/w2

i = − x∗i (p,w)

Hotellings Lemma applied to f (x) = ∑2i=1 logxi :

Unconstrained input demand: x∗i (p,w) = pwi

() October 7, 2015 13 / 27

Two envelope Theorems: (redux)(UET) Unconstrained envelope theorem:



.

total derivative of f w.r.t. αk equals partial derivative of f w.r.t. αk .i.e., reoptimization in x direction has no (first-order) impact on df (x∗(ααα);ααα)

dαk

(CET) Constrained envelope theorem:

If (x∗(ααα),λλλ∗(ααα)) solves maxx f (x;ααα) s.t. hj (x;ααα)≥ 0, j = 1, ...m,and x∗(·) is differentiable at ααα, then


dαk=


∂αk+

m

∑j=1

λ∗j (ααα)


∂αk

For NPP, we wrote g j (x)≤ bj or bj −g j (x)≥ 0; now have hj (x;ααα) = α0j −g j (x;ααα).In words: total deriv of f w.r.t. αk equals partial deriv of f w.r.t. αk plusλλλ∗-weighted sum of partial derivs of hj ’s w.r.t. αk

only difference between UET and CET is the addition of constraintsagain, reoptimization in x direction has no (first-order) impact on df (x∗(ααα);ααα)

dαk() October 7, 2015 14 / 27

Constrained vs unconstrained envelope theorems

Major intuition deficit for the constrained th’m relative to unconstrained onefor the unconstrained theorem, we have our intuitive picture

x term doesn’t enter expr for df (x,ααα)dαj

because5xf (x,ααα) = 0. go to slide 7

differential is flat in the each of the xi directions go to slide 10

to a first order approx, a movement in any x direction doesn’t matter

for the constrained theorem, tangent plane isn’t flat in any direction5xf (x,ααα) 6= 0.the flat differential property no longer holds

i.e., movement in x directions does make a first-order difference

how can it be that the theorem still works???

Answer: when the lagrangian λ is set to just the right value...,soln to constrained prob is unconstrained extremum of Lagrangianview the Lagrangian as a function of (x,ααα).

necessarily5xL(x,ααα) = 0.

apply the intuition that we have for the unconstrained theorem.

() October 7, 2015 15 / 27

Convert constrained problem to unconstrained one

Consider the following trivial constrained problem:

maxx≥0 f (x) =√

x s.t. g(x)≤ b, where g(x) = x and b ≥ 0.Form the lagrangian: L(x ,λ;b) =

√x + λ(b)(b− x).

Trivally, the soln is: x∗(b) = b, for all b ≥ 0.Solve for λ(b):

λ∗(b) = f ′(x∗(b))/g′(x∗(b)) = f ′(x∗(b)) = 0.5/

√x∗(b) = 1/2

√b

Now we’ll replace λ in Lagrangian with its soln value:L(x ;b) =

√x + λ(b− x) =

√x + (b−x)/2

√b.

Graph of L(x ;b) looks like graph of f (x ,b) in slide #8

For each b, slope of L(·;b) is flat in x direction at x∗(b) = b.

� specifically: ∂L(x∗(b);b)∂x = 1

2√

x−λ∗(b) = 1

2√

b− 1

2√

b= 0

� So dL(x∗(b);b)db = ∂L(x∗(b);b)

∂b = λ∗(b).

() October 7, 2015 16 / 27

Graph of Lagrange in (x ,b) space go to slide 8

0 2 4 60

50

1

2

3

4

5

0 2 4 60

50

1

2

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4

0 2 4 6050

1

2

3

4

0 2 4 605

0

0.5

1

1.5

2

2.5

3

xx

xx

bb

bb

L(x,b)=

√ x+λ(1)(1−x

)

L(x,b)=

√ x+λ(2)(2−x

)

L(x,b)=

√ x+λ(3)(3−x

)

L(x,b)=

√ x+λ(4)(4−x

)

Lagrange for max√x s.t. x ≤ 1 Lagrange for max

√x s.t. x ≤ 2

Lagrange for max√x s.t. x ≤ 3 Lagrange for max

√x s.t. x ≤ 4

1

() October 7, 2015 17 / 27

The 2-variable version: the consumer’s optim problem

maxx

u(x) = (x1x2)0.35 s.t. y−p ·x ≥ 0

The objective function u is not flat in either direction xi .

() October 7, 2015 18 / 27

Unconstrained max of L at constrained max of f

L(x∗(y),λ∗(y);y) = u(x∗(y)) + λ∗(y)

(y−p ·x∗(y)

)

But the Lagrangian function is flat w.r.t. both x1 and x2() October 7, 2015 19 / 27

Small dx leaves L unchanged (to 1st order approx)

L(

x∗(y),λ∗(y);y)

vs L(

x∗(y + ∆y

),λ∗(y + ∆y

);y + ∆y

)

First order increase in L due only to ∆y (adjustments in x barely matter)() October 7, 2015 20 / 27

Envelope Theorem: constrained version (NPP)RECALL: In the NPP, λj is the “shadow value” of the j ’th constraint.

M(b) = f (x(b)) = L(x(b),λλλ(b),b) = f (x) + λλλ(b−g(x)).

dM(b)

dbj=

dL(x(b),λλλ(b),b)

dbj=

n

∑i=1

{fi (x(b))−

m

∑j=1

λj (b)g ji (x(b))

}∂xi

∂bj

+m

∑j=1

∂λj

∂bj(bj −g j (x(b))) + λj (b)

The ∂xi∂bj

and ∂λj

∂bjterms all disappear: multiplied by zeros

The only term remaining is λj(b).

Conclude that dM(b)dbj

= λj(b) = ∂L(x(b),λλλ(b),b)∂bj

Note that L depends on bj in a particularly simple (linear) way

so bj ’s disappears in the expression for ∂M(b)∂bj

.

moreover, bj doesn’t enter into the objective function f

more generally, there’s a parameter vector ααα that enters nonlinearly into Lin general, ααα will be an argument of both f and g j ’s

we’ll see, in general, dM(ααα)dαk

= ∂L(x∗,λλλ∗;ααα)∂αk

but αk won’t disappear

() October 7, 2015 21 / 27

Envelope Theorem: constrained version (general)If (x∗(ααα),λλλ∗(ααα)) solves maxx f (x;ααα) s.t. hj (x;ααα)≥ 0, j = 1, ...m,

and x∗(·) is differentiable at ααα, then


dαk=


∂αk+

m

∑j=1

λ∗j (ααα)


∂αk

For NPP, we wrote g j (x)≤ bj or bj −g j (x)≥ 0; now can have hj (x;ααα) = bj −g j (x;ααα).In words: total deriv of f w.r.t. αk equals partial deriv of f w.r.t. αk plusλλλ∗-weighted sum of partial derivs of hj ’s w.r.t. αk (Prev slide a special case.) .

Proof: apply unconstrained Env thm to (generalized) Lagrangian;

L(x,λλλ;ααα) = f (x;ααα) +m

∑j=1

λjhj(x;ααα)

L(x∗,λλλ∗;ααα) ≡ f (x∗,ααα)

df (x∗,ααα)

dαk≡ dL(x∗,λλλ∗;ααα)

dαk=

∂L(x∗,λλλ∗;ααα)

∂αk︸︷︷︸by the unconstrained envelope theorem

=∂f (ααα,x∗(ααα))

∂αk+

m

∑j=1

λ∗j (ααα)

∂hj(ααα,x∗(ααα))

∂αk() October 7, 2015 22 / 27

Constrained Envelope Theorem ExampleE.g. of constrained envelope theorem (Shephard’s lemma):Let c(w, q) = w · x(w, q) be the minimized level of costs given input prices wand output q (i.e., x(w, q) is optimal input mix given params). Then

The i ’th conditional input demand function is xi (·) = ∂c(·,·)∂wi

.

Again, not asking: how does minimized cost change when wi changes.

Proof: minx

c(x;w, q) s.t. q− f (x) = 0

maxx−c(x;w, q) s.t. q− f (x) = 0

L(x,λ;w, q) = −w ·x + λ(q− f (x))

L(x,λ∗;w, q) ≡ −w · x which we define to be − c(x;w, q)

dL(x,λ∗;w, q)

dwi=

∂L(x,λ∗;w, q)

∂wi= − ∂c(x;w, q)

∂wi= − xi(w)

Given a parameterized prodn function (see notes for CES),� env thm delivers an expression for xi (w) in terms of primitives

() October 7, 2015 23 / 27

The mother of all envelope theorems

The LRATC curve (all inputs variable) is theouter envelope of the SRATC curves (someinputs fixed); when fixed inputs are at theiroptimal levels, the slopes of LRATC and

SRATC curves are equal.

q q

Average

Cost

Output

SRATC(·, k)

LRATC(·)

1

() October 7, 2015 24 / 27

∆ Cost moving from q to q + dq: (k is optimal for q)

k

ℓ

(k, ℓ)

k

optimal input adjustment

increase only ℓ

Isocost line at q

Isocost line at q + dq (new optimal input mix)

Isocost line at q + dq (increase only ℓ)

Isoquant for q

Isoquant for q + dq

∆LRATC

∆SRATC

1

() October 7, 2015 25 / 27

∆ Cost moving from q to q + dq: (k is suboptimal for q)

k

ℓ

(k, ℓ)

(k, ℓ′)

k

optimal input adjustment

increase only ℓ

Isocost line at q

Isocost line at q + dq (new optimal input mix)

Isocost line at q + dq (increase only ℓ)

Isoquant for q

Isoquant for q + dq

∆LRATC

∆SRATC

1

() October 7, 2015 26 / 27

Minimizing LR and SR average total cost: the mathq = f (`,k). Given q, `(k ;q) is determined: (k , `(k ;q)) must be on q-isoquant

The long-run problem: acLR(k ; q) = w`(k ;q)+rkq

Unconstrained optimization: maxk−acLR(k ; q)

By envelope theorem:dacLR(kLR(q); q)

dq=

∂acLR(kLR(q); q)

∂q

=−qw ∂`

∂q +(w`(kLR(q); q) + rkLR(q)

)

q2

The short-run problem: acSR(k ; q, k) = w`(k ;q)+r kq , k ≤ k

L(k ; q, k) = − w`(k ; q) + r kq

+ λ(k− k)

By envelope theorem:dacSR(kSR(q); q)

dq=

∂acSR(kSR(q); q)

∂q=

dLdq

=−qw ∂`

∂q +(w`(kSR(q); q) + r k

)

q2 + λ∂k∂q

When k = kLR(q): λ = 0.

dacSR(kSR(q); q)

dq=

−qw ∂`∂q +

(w`(kLR(q); q) + rkLR(q)

)

q2 =dacLR(kLR(q); q)

dq

() October 7, 2015 27 / 27

Comparative Statics Overview (pt 1)

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