The Implicit Function Theorem (IFT): key points
1
The solution to any economic model can be characterized as the
level set corresponding to zero of some function
1
Model: S = S (p; t ), D = D (p), S = D; p =price; t =tax;
2
Level Set: LS (p; t ) = S (p; t ) − D (p) = 0.
2
When you do comparative statics analysis of a problem, you are
studying the slope of the level set that characterizes the problem.
1
2
3
3
The implicit function theorem tells you
1
2
4
Intuitive comp. stat. question: change t, what happens to p?
Intuitive idea: ∆t; how much ∆p is needed to keep on LS (·; ·)?
Math answer is the slope of LS
when this slope is well defined
if it is well-defined, what are the derivatives of the implicit function
It’s an extremely powerful tool
1
explicit function p(t ) could be nasty; no closed form
E.g., : LS (p; t ) = tp15 + t 13 + p95 −
√
p = 0; what’s p(t )?
2
don’t need to know p(t ) in order to know
3
can compute
dp(t )
dt
dp(t )
.
dt
from partials of LS (·; ·).
Level Sets → locally well-defined functions diffable functions
x
g(α)
α
0
Case 1: x is a well-defined, differentiable function of α
x
α
g(α)
(ᾱ, x̄)
α
0
Case 2: x is a well-defined, but not differentiable function of α
α
g(α)
x
(ᾱ, x̄)
α
0
α
Implicit function theorem (single variable version) I
∂f (ᾱ,x̄ )
Theorem: Given f : R2 → R1 , f ∈ C1 and (ᾱ, x̄ ) ∈ R2 , if ∂x 6= 0,
∃ nbds U α of ᾱ, U x of x̄ & a unique g : U α → U x , g ∈ C1 s.t. ∀α ∈ U α ,
f (α, g (α))
= f (ᾱ, x̄ ) i.e., (α, g (α)) is on the level set of f through (ᾱ, x̄ )
∂f (α, g (α)) ∂f (α, g (α))
0
g (α) = −
∂α
∂x
Trivial Proof of the second line:
f (α, x )
:= f (α, g (α))
∂f (α, g (α)) ∂f (α, g (α)) dg (α)
+
∂α
∂x
dα
dg (α)
dα
=
0
=
0
=
−
∂f (α,g (α))
∂α
∂f (α,g (α))
∂x
Implicit function theorem (single variable version) II
Example:
1
f (p; t ) = tp15 + t 13 + p95 −
2
∂f
∂t
3
if ∂∂pf =
6 0,
p
= 15 tp14 + 95 p94 − 1/2 √1p
1
15
12
14
94
√
= − p + 13t
15 tp + 95 p − 1/2 p
= p15 + 13t 12 ;
dp
dt
√
∂f
∂p
∂f (ᾱ,x̄ )
Note that the following is not true: if ∂x 6= 0,∃ nbd U α of ᾱ, & a
unique g : U α → R, g ∈ C1 s.t. ∀α ∈ U α , f (α, g (α)) = f (ᾱ, x̄ ).
Also true (if you are a mathematician
but not an economist):
.
1
if ∂∂ft =
6 0,
dt
dp
= − 15 tp14 + 95 p94 − 1/2 √1p
p15 + 13t 12
Implicit function theorem (single variable version) III
Example: (slightly less dramatic)
1
LS(x ; α) = x 2 + α2 − 1
2
3
∂LS
∂α = 2α;
if ∂∂LS
6= 0, ddxα
x
∂LS
∂x
=−
= 2x
dLS dLS
dα
dx
= −α/x.
√
Suppose ᾱ = 0.9, x̄ = 1 − 0.81 ≈ 0.44.
Use IFT to estimate effect of a 100% increase in α to 1.8
= −0.9/0.44 ≈ −2;
x (1.8) ≈ 0.44 + (−2 × 0.9) ≈ −1.36
dx
dα
i.e., way off, true answer: no real x solves LS
hazardous to base policy decisions on comp stat analysis
Implicit function theorem (single variable version) IV
1
0.5
x
0
−0.5
−1
−1.5
0
0.2
0.4
0.6
0.8
α
1
1.2
1.4
1.6
1.8
Implicit function theorem (intermediate version)
∂f (ᾱ,x̄)
α, x̄ ) ∈ Rn × R1 , if ∂x 6= 0,
Theorem: Given f : Rn+1 → R1 , f ∈ C1 and (ᾱ
j
x
∀j, ∃ nbds U of ᾱj , U of x̄ & a unique g : U j → U x , g ∈ C1 s.t. ∀αj ∈ U j ,
α, g (α
α))
f (α
α, x̄ ) i.e., g puts us on the level set of f containing (ᾱ
α, x̄ )
= f (ᾱ
α)
α, g (α
α)) ∂f (α
α, g (α
α))
dg (α
∂f (α
= −
or in vector form
d αj
∂αj
∂x
α) = − 5α f (α
α, g (α
α))/ fx (α
α, g (α
α)).
5g (α
This is a straightforward extension of the first theorem. Example:
1
Model: S = S (p; t ), D = D (p; y ), S = D; t =tax, y =income;
2
Level Set: LS (p; t , y ) = S (p; t ) − D (p; y ) = 0.
3
Can use IFT to compute 5p = (pt , py )
4
Once you have gradient, can get any directional derivative
what part of the theorem guarantees this?
5
Use IFT to approx impact of any combination of param changes
Equilibrium price as a function of tax and income
S(p, t′ )
P
S(p, t)
p(t′ , y ′ )
p(t, y)
D(p, y ′ )
D(p, y)
Q
The zero level set of S (p, t ) − D (p, y )
Zero Level set of Supply−demand model
3
equilibrium price
2.5
2
1.5
1
0.5
0
1
0.8
1
0.6
0.8
0.6
0.4
0.4
0.2
income
0.2
0
0
tax
Implicit function theorem (multivariate version)
α, x̄) ∈ Rn × Rm , if
Theorem: Given f : Rn+m → Rm , f ∈ C1 & (ᾱ
α
x
α
α, U of x̄ & !g : U → U x , g ∈ C1 s.t. ∀α
α ∈ Uα ,
α, x̄)) 6= 0, ∃U of ᾱ
det (Jfx (ᾱ
α, g(α
α)) = f(ᾱ
α, x̄)
f(α







∂g 1 (α )
∂α1
∂g 2 (α )
∂α1
..
.
···
..
∂g m (α )
∂α1
···
.
···
∂g 1 (α )
∂αn
∂g 2 (α )
∂αn
..
.
∂g m (α )
∂αn

α, x̄ )
i.e., g puts us on level set of f containing (ᾱ







α, g(α
α))−1 
= − Jfx (α




∂f 1 (α ,g(α ))
∂α1
∂f 2 (α ,g(α ))
∂α1
..
.
∂f m (α ,g(α ))
∂α1
···
···
..
.
···
∂f 1 (α ,g(α ))
∂αn
∂f 2 (α ,g(α ))
∂αn
..
.
∂f m (α ,g(α ))
∂αn
α, g(α
α)) is the Jacobian of f treating α as parameters
where Jfx (α







Inverse function theorem
Theorem: (Inverse function theorem): Given η : Rm → Rm , η ∈ C1 and
α and det (Jη (x̄)) 6= 0, then ∃U x of x̄ and U α of ᾱ
α s.t. η is
x̄ ∈ Rn , if η (x̄) = ᾱ
a one-to-one and onto map from U x to U α . Moreover, there exists a C1
η(x)) = x,
inverse function η −1 : U α → U x , defined by, for η (x) ∈ U α , η −1 (η
η(x)) = (Jη (x))−1 .
with Jacobian defined by J η −1 (η
What some other people do
Many economists don’t apply IFT directly; they totally differentiate
The IFT gives a formula for the Jacobian of the implicit funct g
Total differentiators end up with an expression for differential of g
Relationship between Jacobian and differential is as it always is
Recall: for every n × n matrix, ∃ unique L : Rn → Rn .
α; x)d α
dx = Jg(α
Summary:
α; x)
IFT route ends up at:
Jg(α
α; x)d α
Total differentiation route ends up at: dx = Jg(α
Total Differentiation Example: Perloff I
max pLα K β − wL − rK
(1)
L,K
Requirement: stay on the zero level set of f , where
πL
πK
f ( w , r ; L, K )
=
pLα−1 K β − w
pLα K β−1 − r
=
=
0
(2)
Totally differentiate:
0
=
0
=
∂πL
∂πL
∂πL
dL +
dK +
dw
∂L
∂K
∂w
∂πK
∂πK
∂πK
dL +
dK +
dr
∂L
∂K
∂r
(3)
(4)
Move exog variables to right hand side; put in matrix form:
"
∂πL
∂L
∂πK
∂L
∂πL
∂K
∂πK
∂K
#
"
∂πL
dL
= − ∂w
dK
0
0
∂πK
∂r
#
dw
1
=
dr
0
0
1
dw
dr
(5)
Total Differentiation Example: Perloff II
Invert
"
∂πL
dL
∂L
= − ∂π
K
dK
∂L
∂πL
∂K
∂πK
∂K
#−1 "
∂πL
∂w
0
0
#
∂πK
∂r
dw
dr
(6)
General form:



α,g(α
α))
∂f 1 (α
dg 1
∂α
 ∂f 2 (αα,g1(αα))
 dg 2 



∂α1
α, g(α
α))−1 
 .. = − Jfx (α
..

 . 

.
α,g(α
α))
∂f m (α
dg m
∂α1
···
···
..
.
···
α,g(α
α))
∂f 1 (α
∂αn
2 α
α))
∂f (α ,g(α
∂αn


d α1
  2
 d α 
  .  (7)
..
 . 
 .
.
α,g(α
α))
∂f m (α
d αn
∂αn
or, noting that the two matrices on the r.h.s. of the last equality are just
Jg(α), we can rewrite the above more economically as
(8)
dg = Jg(α)d α
IFT goes directly from step (2) to (7); skips steps in between.
To actually solve system, typically use Cramer’s rule
Cramer’s Rule: technique
Cramer’s rule is a method to solve Ax = b.
Could invert A and computer x = A−1 b.
But inversion involves effort
Cramer’s rule involves much less.
Let Ai denote the matrix A with the i’th column replaced by b.
Then for i = 1, ...n, xi = det(Ai ) det(A).
much easier to compute a few determinants than an inverse.
how on earth can this work????
Some intuition for Cramer’s rule
Recall
For i = 1, ...n, xi = det(Ai ) det(A).
column interpretation of x when Ax = b.
xi ’s are the weights on columns of A so that
b can be written written as linear combination of these columns
volume interpretation of det(A)
det(A) is the volume of parallelepiped defined by columns of A.
In the figure below, A = [v1 , v2 ], and b is close to v1 ,
so x1 close to 1, x2 close to 0
now
A 1 = [ b , v2 ] ,
A2 = [v1 , b],
what can you say about A1 vs A?
what can you say ratio of their determinants?
what can you say about A2 vs A?
Also have explain why the signs work out properly
Cramer’s Rule: example
a1
a1
b
a2
The columns of A and vector b
a2
det(A) = area parallelogram defined by cols of A
a1
b
b
a2
Multivariate IFT: graphical representation
α
The two tangent planes combined
Level set of f 1 corresponding to 0
ᾱ
x̄1
x̄2
Level set of f 2 corresponding to 0
ᾱ
x̄1
x̄2
The tangent plane to the 0-level set of f 1
dα
The two tangent planes: enlarged
dx1
dα
The tangent plane to the 0-level set of f 2
dα
dx2
dx1
dx2
f : R1+2 → R2 , i.e., two endog variables x, one exog α
1
x is in the horizontal plane; α on vertical plane
2
(ᾱ, x̄) ∈ LS1 ∩ LS2 is a “soln”, i.e., belongs to both zero level sets
3
as α changes, x changes to keep vector in LS1 ∩ LS2
4
(α, x) slides along the groove in the bottom right panel
1