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 optimized
1
2
2
function depends on some parameters
how does optimum value change with these parameters?
Context: First order conditions of an optimization problem
Examples:
1
2
optimized utility as a function of income
maximized profit as a function of price
Two kinds of optimization:
1
2
3
unconstrained optimization (intuition is fairly straightforward)
constrained optimization (intuition is much more elusive)
Our analysis of the lagrangian was a special case of the envelope theorem
1
invoked the (constrained) envelope thm to prove λj = shadow value of bj
2
()
use same approach to compute shadow value” of any parameter
October 7, 2015
1 / 27
Comparative Statics Overview (pt 2):
2
Envelope theorem (bonus): in very special cases you can short-circuit
the implicit function theorem, and use the envelope theorem to get the
same information that the implicit function theorem would give you, but
much more easily
Examples:
1
2
input demand functions (Hotelling’s Lemma)
conditional input demand functions (Shephard’s Lemma)
These results are very special cases:
1
2
not what the envelope theorem was “meant for”
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 meaning
of “envelope” to be: “A curve or surface that is tangent to every one of a family
of 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).
b
f
x∗ (b)
(x∗ (b3 ); b3 )
(x∗ (b2 ); b2 )
(x∗ (b1 ); b1 )
x
db
df
()
df =
∂f (x∗ (b);b)
db
∂b
dx
∂f (x∗ (b); b)
+
dx
db db
October 7, 2015
4 / 27
Comparative Statics Overview (pt 3):
3
Implicit function theorem: used to compute relationship between
endogenous and exogenous variables.
1
Context: First order conditions of an optimization problem.
Examples
e.g., demands as a function of income
e.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 endowments
Cournot quantities as a function of demand parameter, # of firms
()
October 7, 2015
5 / 27
Two envelope Theorems: (summary)
(UET) Unconstrained envelope theorem:
α) solves maxx f (x;α
α) and x∗ (·) is diffable at α , then
if x∗ (α
df (x∗ (α );α )
d αk
=
∂f (x∗ (α );α )
.
∂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∗ (α );α )
d αk
(CET) Constrained envelope theorem:
α),λ
λ∗ (α
α)) solves maxx f (x;α
α) s.t. hj (x;α
α) ≥ 0, j = 1, ...m,
If (x∗ (α
∗
and x (·) is differentiable at α , then
α);α
α)
df (x∗ (α
d αk
=
α);α
α)
∂f (x∗ (α
∂αk
m
+
∑ λ∗j (αα)
j =1
α);α
α)
∂hj (x∗ (α
∂αk
α) = α0j − g j (x;α
α).
For NPP, we wrote g j (x) ≤ bj or bj − g j (x) ≥ 0; now have hj (x;α
In words: total deriv of f w.r.t. αk equals partial deriv of f w.r.t. αk plus
λ ∗ -weighted sum of partial derivs of hj ’s w.r.t. αk
only difference between UET and CET is the addition of constraints
again, reoptimization in x direction has no (first-order) impact on
()
df (x∗ (α );α )
d αk
October 7, 2015
6 / 27
Envelope Theorem: unconstrained version
back to slide 15
Assume f is C2 :
α) solves maxx f (x;α
α) and x∗ (·) is diffable at α , then
if x∗ (α
df (x∗ (α );α )
d αk
=
∂f (x∗ (α );α )
.
∂αk
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.)
α);α
α)
df (x∗ (α
d αk
=
α);α
α)
∂f (x∗ (α
∂αk
α);α
α) = 0,
which, since necessarily, 5x f (x∗ (α
=
()
α);α
α)
∂f (x∗ (α
∂αk
α);α
α) dxi (α
α)
∂f (x∗ (α
∂
x
d
α
k
i =1 |
{zi
}
n
+
∑
back to slide 15
=0
October 7, 2015
7 / 27
Envelope Theorem: graphical intuition
back to slide 16
back to slide 17
()
October 7, 2015
8 / 27
0.2
0.2
0.15
0.15
df/db(x*(3.3),3.3)
df/db(x*(0.9),0.9)
Envelope Theorem: tangent planes
0.1
0.05
0.1
0.05
0
0
1
1
1
0.5
1
0.5
0.5
db
0
0
dx
0.5
db
0
0
dx
Height of differential determined exclusively by magnitude of db
()
October 7, 2015
9 / 27
Envelope Theorem: tangent planes (alt view)
0.2
0.18
0.16
0.16
0.14
0.14
df/db(x*(3.3),3.3)
0.2
0.18
0.12
0.1
0.08
0.12
0.1
0.08
0.06
0.06
0.04
0.04
0.02
back to slide 15
0.02
1
1
0
1
0.5
0.5
db
()
0
0
1
0
dx
0.5
0.5
db
0
0
dx
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 Lemma
Hotelling’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 xi∗ (p, w) = −
∂π(x∗ (p,w);p,w)
∂w i
Again math is straightforward: by the unconstrained envelope theorem,
d π(x∗ (p, w); p, w)
dwi
=
∂π(x∗ (p, w); p, w)
∂wi
=
− xi∗ (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 xi∗ (p, w) in terms of primitives
()
October 7, 2015
12 / 27
Numerical example of Hotelling’s lemma
Recall π(x; p, w) = pf (x) − w · x
Let 5f = (1/x1 , 1/x2 ) and 5x π = (p/x1 − w1 , p/x2 − w2 )
The unconstrained optimum is x∗ (p, w) = (p/w1 , p/w2 ).
∂π(x∗ (p,w);p,w)
d π(x∗ (p,w);p,w)
,
dwi
Now compare the partial deriv,
, to the total deriv,
∂wi
in order to illustrate that the envelope theorem actually works
∂π(x∗ (p, w); p, w)
∂wi
d π(x∗ (p, w); p, w)
dwi
=
− xi∗ (p, w)
=
∂π(x∗ (p, w); p, w)
∂wi
=
−
p
from FOC
wi
∂π(x∗j (p, w); p, w) dxj∗ (p, w)
∂xj
dwi
j =1
2
+
∑
which, since for j 6= i, xj∗ (p, w) doesn’t depend on wi
=
=
=
∂π(x∗ (p, w); p, w)
∂wi
− xi∗ (p, w)
− xi∗ (p, w)
+
∂π(x∗i (p, w); p, w) dxi∗ (p, w)
∂xi
dwi
+ (p/xi∗ (p,w) − wi ) × −p/wi2
p/p/w − wi × −p/w 2
= − xi∗ (p, w)
i
i
| {z }
+
0
Hotellings Lemma applied to f (x) = ∑2i =1 log xi :
p
Unconstrained input demand: xi∗ (p, w) = w
i
()
October 7, 2015
13 / 27
Two envelope Theorems: (redux)
(UET) Unconstrained envelope theorem:
α) solves maxx f (x;α
α) and x∗ (·) is diffable at α , then
if x∗ (α
df (x∗ (α );α )
d αk
=
∂f (x∗ (α );α )
.
∂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∗ (α );α )
d αk
(CET) Constrained envelope theorem:
α),λ
λ∗ (α
α)) solves maxx f (x;α
α) s.t. hj (x;α
α) ≥ 0, j = 1, ...m,
If (x∗ (α
∗
and x (·) is differentiable at α , then
α);α
α)
df (x∗ (α
d αk
=
α);α
α)
∂f (x∗ (α
∂αk
m
+
∑ λ∗j (αα)
j =1
α);α
α)
∂hj (x∗ (α
∂αk
α) = α0j − g j (x;α
α).
For NPP, we wrote g j (x) ≤ bj or bj − g j (x) ≥ 0; now have hj (x;α
In words: total deriv of f w.r.t. αk equals partial deriv of f w.r.t. αk plus
λ ∗ -weighted sum of partial derivs of hj ’s w.r.t. αk
only difference between UET and CET is the addition of constraints
again, reoptimization in x direction has no (first-order) impact on
()
df (x∗ (α );α )
d αk
October 7, 2015
14 / 27
Constrained vs unconstrained envelope theorems
Major intuition deficit for the constrained th’m relative to unconstrained one
for the unconstrained theorem, we have our intuitive picture
x term doesn’t enter expr for
α)
df (x,α
d αj
α) = 0.
because 5x f (x,α
go to slide 7
go to slide 10
differential is flat in the each of the xi directions
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 direction
α) 6= 0.
5x f (x,α
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 Lagrangian
α).
view the Lagrangian as a function of (x,α
α) = 0.
necessarily 5x L(x,α
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:
√
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):
maxx ≥0 f (x ) =
λ∗ (b)
=
f 0 (x ∗ (b))/g 0 (x ∗ (b))
=
f 0 (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.
slide #8
Graph of L (x ; b) looks like graph of f (x , b) in
For each b, slope of L (·; b) is flat in x direction at x ∗ (b) = b.
specifically:
So
()
∂L (x ∗ (b);b)
∂x
d L (x ∗ (b);b)
db
=
1
√
2 x
− λ∗ (b) =
1
√
2 b
− 2√1 b = 0
∗
= ∂L (x∂b(b);b) = λ∗ (b).
October 7, 2015
16 / 27
Lagrange for max
√
x s.t. x ≤ 1
√
L(x, b) = x+λ(2)(2−x)
5
4
3
2
1
0
5
0
x
0
Lagrange for max
√
2
1
0
5
x
x s.t. x ≤ 3
0
b
Lagrange for max
2
1
6
4
2
0
√
3
3
x s.t. x ≤ 2
3
b
4
√
4
6
4
2
go to slide 8
Lagrange for max
√
L(x, b) = x+λ(4)(4−x)
√
L(x, b) = x+λ(3)(3−x)
√
L(x, b) = x+λ(1)(1−x)
Graph of Lagrange in (x , b) space
x s.t. x ≤ 4
2.5
2
1.5
1
0.5
0
0
5
0
x
()
2
b
4
6
5
x
0
2
b
4
6
October 7, 2015
17 / 27
The 2-variable version: the consumer’s optim problem
max u (x)
x
=
(x1 x2 )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)
dM (b)
dbj
=
=
f (x̄(b))
¯(b), b)
d L (x̄(b), λ
dbj
=
L (x̄(b), λ¯(b), b) = f (x̄) + λ¯(b − g (x̄)).
=
∑
n
i =1
n
m
fi (x̄(b)) −
+
m
o ∂x
∑ ¯λj (b)gij (x̄(b))
j =1
i
∂bj
∂λj
∑ ∂bj (bj − g j (x̄(b))) + ¯λj (b)
j =1
∂λ
The ∂∂bxi and ∂bj terms all disappear: multiplied by zeros
j
j
The only term remaining is ¯λj (b).
Conclude that
dM (b)
dbj
=
¯λj (b)
∂L (x̄(b),λ¯(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 α that enters nonlinearly into L
in general, α will be an argument of both f and g j ’s
α)
λ∗ ;α
α)
∂L (x∗ ,λ
dM (α
but αk won’t disappear
we’ll see, in general, d α =
∂α
k
k
()
October 7, 2015
21 / 27
Envelope Theorem: constrained version (general)
α),λ
λ∗ (α
α)) solves maxx f (x;α
α) s.t. hj (x;α
α) ≥ 0, j = 1, ...m,
If (x∗ (α
∗
and x (·) is differentiable at α , then
α);α
α)
df (x∗ (α
d αk
α);α
α)
∂f (x∗ (α
∂αk
=
j
j
j
m
+
∑ λ∗j (αα)
j =1
α);α
α)
∂hj (x∗ (α
∂αk
j
α) = bj − g j (x;α
α).
For NPP, we wrote g (x) ≤ b or b − g (x) ≥ 0; now can have hj (x;α
In words: total deriv of f w.r.t. αk equals partial deriv of f w.r.t. αk plus
λ ∗ -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;
m
α) = f (x;α
α) +
L (x,λλ;α
∑ λj hj (x;αα)
j =1
α) ≡ f (x∗ ,α
α)
L (x∗ ,λλ∗ ;α
∗
∗
∗
λ ;α
α)
α)
d L (x ,λ
df (x ,α
d αk
≡
=
()
|
d αk
=
{z
λ∗ ;α
α)
∂L (x∗ ,λ
∂αk
}
by the unconstrained envelope theorem
α, x∗ (α
α))
∂f (α
∂αk
+
m
∑ λ∗j (αα)
j =1
α, x∗ (α
α))
∂hj (α
∂αk
October 7, 2015
22 / 27
Constrained Envelope Theorem Example
E.g. of constrained envelope theorem (Shephard’s lemma):
Let ĉ (w, q̄ ) = w · x̂(w, q̄ ) be the minimized level of costs given input prices w
and output q̄ (i.e., x̂(w, q̄ ) is optimal input mix given params). Then
∂ĉ (·,·)
The i’th conditional input demand function is x̂i (·) = ∂w .
i
Again, not asking: how does minimized cost change when wi changes.
min ĉ (x; w, q̄ )
s.t .
q̄ − f (x) = 0
max −ĉ (x; w, q̄ )
s.t .
q̄ − f (x) = 0
Proof:
x
x
L (x, λ; w, q̄ ) =
L (x̂, λ∗ ; w, q̄ ) ≡
d L (x̂, λ∗ ; w, q̄ )
dwi
=
−w·x
+
λ(q̄ − f (x))
− w · x̂ which we define to be − ĉ (x̂; w, q̄ )
∂L (x̂, λ∗ ; w, q̄ )
∂ĉ (x̂; w, q̄ )
=−
= − x̂i (w)
∂wi
∂wi
Given a parameterized prodn function (see notes for CES),
env thm delivers an expression for x̂i (w) in terms of primitives
()
October 7, 2015
23 / 27
The mother of all envelope theorems
Average Cost
The LRATC curve (all inputs variable) is the
outer envelope of the SRATC curves (some
inputs fixed); when fixed inputs are at their
optimal levels, the slopes of LRATC and
SRATC curves are equal.
LRATC(·)
SRATC(·, k)
q
()
q̄
Output
October 7, 2015
24 / 27
∆ Cost moving from q to q + dq: (k is optimal for q)
ℓ
∆SRAT C
Isocost line at q + dq (new optimal input mix)
∆LRAT C
Isocost line at q + dq (increase only ℓ)
Isoquant for q + dq
increase only ℓ
optimal input adjustment
(k, ℓ)
Isoquant for q
Isocost line at q
k
k
()
October 7, 2015
25 / 27
∆ Cost moving from q̄ to q̄ + dq: (k is suboptimal for q̄)
ℓ
∆SRAT C
Isocost line at q̄ + dq (new optimal input mix)
Isocost line at q̄ + dq (increase only ℓ)
∆LRAT C
increase only ℓ
Isoquant for q̄ + dq
(k, ℓ′ )
optimal input adjustment
(k̄, ℓ̄)
Isoquant for q̄
Isocost line at q̄
k
k
()
October 7, 2015
26 / 27
Minimizing LR and SR average total cost: the math
q = f (`, k ). Given q, `(k ; q ) is determined: (k , `(k ; q )) must be on q-isoquant
acLR (k ; q̄ )
The long-run problem:
w `(k ;q̄ )+rk
q̄
=
max −acLR (k ; q̄ )
Unconstrained optimization:
k
dacLR (k LR (q̄ ); q̄ )
By envelope theorem:
dq
=
acSR (k ; q̄ , k̄ )
The short-run problem:
L (k ; q̄ , k̄ ) =
By envelope theorem:
=
When k̄ = k LR (q̄ ):
()
w `(k ;q )+r k̄
q̄
=
w `(k ; q̄ ) + r k̄
q̄
+
dacSR (k SR (q̄ ); q̄ )
k ≤ k̄
,
λ(k̄ − k )
∂acSR (k SR (q̄ ); q̄ )
∂q
+ w `(k SR (q̄ ); q̄ ) + r k̄
∂k
+ λ
q̄ 2
∂q
dq
−qw ∂∂`q
q̄ 2
=
dL
=
dq
λ = 0.
dacSR (k SR (q̄ ); q̄ )
dq
−
−qw ∂∂`q
∂acLR (k LR (q̄ ); q̄ )
∂q
+ w `(k LR (q̄ ); q̄ ) + rk LR (q̄ )
=
=
−qw ∂∂`q + w `(k LR (q̄ ); q̄ ) + rk LR (q̄ )
q̄ 2
=
dacLR (k LR (q̄ ); q̄ )
dq
October 7, 2015
27 / 27