Transcript Ch. 8 Comparative-Static Analysis of General

Ch. 8 Comparative-Static Analysis of
General-Function Models
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•
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•
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8.1
8.2
8.3
8.4
8.5
8.6
Models
•
8.7
Differentials
Total Differentials
Rules of Differentials (I-VII)
Total Derivatives
Derivatives of Implicit Functions
Comparative Statics of General-Function
Limitations of Comparative Statics
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7.4 P artialDerivatives, p.167
1)
y  f( x1,x 2 , ,x n )
variablesx i are all independent of one another
each x can varyby itself w/o affect the othersuch that
2)
Δy Δx1  f( x1  Δx,x 2 , ,x n ) Δx1
3)
lim Δy Δx1 
Δx 0

y  f1
x1
4)
Y  C  I 0  G0
5)
6)
C  a  b( Y-T);
T  d  tY;
a-bd  I  G
Y* 
1-b  tb
7)
8)
C* 
b(1-t)( I  G)  a-bd
1-b  tb
9)
T* 
t( I  G)  ta  d(1-b)
1-b  tb
partialderivativewith operator x1
( a  0; 0  b  1 )
( d  0; 0  t  1 )
 *
1
Y 
0
Go
1  b  bt

b(1  t)
C* 
0
Go
1  b  bt
 *
t
T 
0
Go
1  b  bt
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8.1 Differentials
8.1.1 Differentials and derivatives
8.1.2 Differentials and point elasticity
3
8.1.1 Differentials and derivatives
Problem: What if no explicit reduced-form solution exists
because of the general form of the model? Example: What
is Y / T when
Y = C(Y, T0) + I0 + G0
T0 can affect C direct and indirectly thru Y, violating the
partial derivative assumption
Solution:
• Find the derivatives directly from the original equations in
the model.
• Take the total differential
• The partial derivatives become the parameters
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If
1)
Y  C  I 0  G0
2)
3)
C  a  b( Y-T)
Y  a  b( Y-T)  I 0  G0
4)
Y  bY  a  bT  I 0  G0
Y  a  bT  I 0  G0 1  b
parameter data
parameters and exog. vars mutuallyindependent
Then
4)
If
1)
2)
3)

Y   b 1 b
T
Y  C  I 0  G0
C  C Y , T0 
Y  C Y , T0   I 0  G0
4)
Y *  Y * T0 , I 0 , G0 
5)
Y *  C Y * , T0  I 0  G0
6)
Y*

 C Y
no parameter data
assumingequil.exists

*
T0 , I 0 , G0 , T0 
I 0  G0
exog. var (C argumentsand T0 ) mutuallydependent
Then
7)
 *
Y ?
T0

 
problem: C Y * , T0  C Y * T0 , I 0 , G0 , T0

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8.1 Differentials
d
1)
y  f x   lim y x
derivative
x 0
dx
2)
y x  f x   
as x  0 then   0
3)
y x  f x   
4)
y  f x x  x
5)
y  f x x
as finite x  0
6)
dy  f x dx
differential
where f x  is the factor of proportionality
A first - order Taylorseriesapproximation of y1 is :
7)
f x1   f x0   f x0 x1  x0 
8)
y  f x0 x
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Differential: dy & dx as finite changes (p. 180)
fi·nite
Mathematics.
a. Being neither infinite nor
infinitesimal.
b. Having a positive or negative
numerical value; not zero.
c. Possible to reach or exceed by
counting. Used of a number.
d. Having a limited number of
elements. Used of a set.
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Difference Quotient, Derivative &
Differential
f(x)
y=f(x)
f(x0+x)
B
x
y
f(x0)
D
A
x0
x
f’(x)
f’(x0)x
C
x
x0+x
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Overview of Taxonomy Equations: forms and functions
Form
Primitive
Function
Specific
(parameters)
General
(no parameters)
Explicit
(causation)
y = a+bx
y = f(x)
Implicit
(no causation)
y3+x3-2xy = 0
F(y, x) = 0
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Overview of Taxonomy –
1st Derivatives & Total Differentials
Form
Differentiation
Function
Specific
(parameters)
Explicit
(causation)
d
y b
dx
Implicit
(no causation)
3 y
2
 
General
(no parameters)
dy  f ( x) dx
dy
 f ( x)
dx

 2 x dy  2 x 2  2 y dx  0


dy
2x 2  2 y

dx
3y 2  2x


dy  
Fx
dx
Fy
F
dy
 x
dx
Fy
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8.1.1 Differentials and derivatives
• From partial differentiation to total differentiation
• From partial derivative to total derivative using total
differentials
• Total derivatives measure the total change in y from
the direct and indirect affects of a change in xi
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8.1.1 Differentials and derivatives
• The symbols dy and dx are called the differentials of
y and x respectively
• A differential describes the change in y that results
for a specific and not necessarily small change in x
from any starting value of x in the domain of the
function y = f(x).
• The derivative (dy/dx) is the quotient of two
differentials (dy) and (dx)
• f '(x)dx is a first-order approximation of dy
y  f ( x)
dy  f ' ( x)dx
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8.1.1 Differentials and derivatives
• “differentiation”
– The process of finding the differential (dy)
• (dy/dx) is the converter of (dx) into (dy) as dx 0
– The process of finding the derivative (dy/dx) or
• Differentiation with respect to x
Differential
Derivative
 dy 
dy   dx
 dx 
dy   dy 
 
dx  dx 
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8.1.2 Differentials and point elasticity
• Let Qd = f(P)
(explicit-function general-form demand equation)
• Find the elasticity of demand with respect to price
dQd 
 dQd


dP  m arg inal function
%Qd
Qd 
d 



dP
Qd
%P
average function
P
P
elastic if  d  1, inelastic if  d  1
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8.2 Total Differentials
• Extending the concept of differential to smooth
continuous functions w/ two or more variables
• Let y = f (x1, x2) Find total differential dy
y
y
dy 
dx1 
dx2
x1
x2
dy  f1dx1  f 2 dx2
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8.2 Total Differentials (revisited)
• Differentiation of U wrt x1
• U/ x1 is the marginal utility of the good x1
• dx1 is the change in consumption of good x1
dU U U dx2
U dxn


 ... 
dx1 x1 x2 dx1
xn dx1
dU    U 
dx1   x1  x ...x cons tan t
2
n
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8.2 Total Differentials (revisited)
Total Differentiation:
Let Utility function U = U (x1, x2, …, xn)
U
U
U
dU 
dx1 
dx2   
dxn
x1
x2
xn
To find total derivative
divide through by the differential dx1 ( partial total derivative)
dU U U dx2
U dxn


 ... 
dx1 x1 x2 dx1
xn dx1
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8.2 Total Differentials
•
•
•
•
Let Utility function U = U (x1, x2, …, xn)
Differentiation of U wrt x1..n
U/ xi is the marginal utility of the good xi
dxi is the change in consumption of good xi
U
U
U
dU 
dx1 
dx2   
dxn
x1
x2
xn
• dU equals the sum of the marginal changes in the
consumption of each good and service in the
consumption function
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8.3 Rules of differentials,
the straightforward way
Find dy given function y=f(x1,x2)
1. Find partial derivatives f1 and f2 of x1 and x2
2. Substitute f1 and f2 into the equation
dy = f1dx1 + f2dx2
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8.3 Rules of Differentials
(same as rules of derivatives)
Let k is a constant function; u = u(x1); v = v(x2)
• 1. dk = 0
(constant-function rule)
• 2. d(cun) = cnun-1du
(power-function rule)
• 3. d(u  v) = du  dv (sum-difference rule)
• 4. d(uv) = vdu + udv (product rule)
• 5.  u  vdu  udv (quotient rule)
d  
v
v2
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8.3 Rules of Differentials (I-VII)
6. d u  v  w  du  dv  dw
7. d(uvw) = vwdu + uwdv + uvdw
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Rules of Derivatives & Differentials
for a Function of One Variable
d
1) c  0
dx
d n
2) x  nxn 1
dx
1' ) dc  0dx  0
2' ) dxn  nxn 1dx
d
3)  f x   g x   f x   g x 
dx
3' ) d  f x   g x   f x dx  g x dx
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Rules of Derivatives & Differentials
for a Function of One Variable
d
4)  f x g x   f x g x   f x g x 
dx
4' ) d  f x g x   f x g x dx  f x g x dx
d
5a ) cx  c
dx
d n
5b) cx  cnxn 1
dx
5a ' ) dcx  cdx
5b' ) dcxn  cnxn 1dx
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Rules of Derivatives & Differentials
for a Function of One Variable
d  f x  f x g x   f x g x 
6) 


2
dx  g x  
g x 
 f x  f x g x dx  f x g x dx
6' ) d 


2
g x 
 g x  
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8.3 Example 3, p. 188:
Find the total differential (dz) of the function
1)
z
2)
z
3)
4)
x y
2x2
x

2x 2 2x 2
z
z
dz 
dx  dy
x
y
z   x
y  2 x 2  4 x 2  4 xy
 



2
2 2
x x  2 x 2 2 x 2 
2x
2x 2
 

5)
6)
y
2 x 2  4 x 2  4 xy

 
x  2x  2 y
4x4
z
  x
y    y
  2  2  2
y y  2 x
2 x  y  2 x
 x  2 y 
1
dz 
dx

dy
3
2
2x
2x
2x3
1

 2
 2x
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8.3 Example 3 (revisited using the quotient rule for
total differentiation)

1
 x y
2
2
d

2
x
d
(
x

y
)

(
x

y
)
d
(
2
x
)

2
2
2
 2x  2x
1
 4 2 x 2 (dx  dy)  ( x  y )4 xdx
4x
1
 4 2 x 2 dx  2 x 2 dy  4 x 2 dx  4 yxdx
4x
1
 4 2 x 2 dy  2 x 2 dx  4 yxdx
4x
2x2
2 x 2  4 yx
 4 dy 
dx
4
4x
4x
1
x  2y
 2 dy 
dx
3
2x
2x
 







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8.4 Total Derivatives
•
•
•
•
8.4.1 Finding the total derivative
8.4.2 A variation on the theme
8.4.3 Another variation on the theme
8.4.4 Some general remarks
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8.4.1 Finding the total derivative from the
differential
8.4.1 pp.184- 5, Given
1)
y  f x1,x 2 , ,x n 
Total differential dy is equalto the sum of the partialchangesin y :
y
y
y
2)
dy 
dx1 
dx2   
dxn
x1
x2
xn
3)
dy  f1dx1  f 2 dx2  ...  f n dxn
The partialtotalderivativeof y wrt x1 , for example,
is found by dividingbothsidesby dx1
4)
dx
dx
dy
 f1  f 2 2    f n n
dx1
dx1
dx1
28
8.4.3 Another variation on the theme
8.4.3
1)
2)
y  f x1 , x2 , u, v 
x1  g u, v 
3)
x2  hu, v 
4)
dy  f x1 dx1  f x2 dx2  f u du  f v dv totaldifferential
5)
6)
dx1
dx2
dy
dv
 f x1
 f x2
 fu  fv
du
du
du
du
§y
 f x1 g u  f x2 hu  f u
partialtotalderivative
§u v c
29
8.4.3 Another variation on the theme
y  f ( x1 , x2 )
x1  5u 2  3v
x2  u  4v 3
dy  f1dx1  f 2 dx2
dx1  10udu  3dv
dx2  du  12v 2 dv


dy  f1 10udu  3dv  f 2 du  12v 2 dv
dy  f110udu  f1 3dv  f 2 du  f 212v 2 dv


dy   f110u  f 2 du  f1 3  f 212v 2 dv
§y
 f110u  f 2 ,
§ u c v
§y
 f1 3  f 212v 2
§ v c u
30
8.5 Derivatives of Implicit Functions
•
•
•
8.5.1 Implicit functions
8.5.2 Derivatives of implicit functions
8.5.3 Extension to the simultaneousequation case
31
8.5.1 Implicit functions
• Explicit function: y = f(x)  F(y, x)=0 but reverse may not
be true, a relation?
• Definition of a function: each x  unique y (p. 16)
• Transform a relation into a function by restricting the range
of y0, F(y,x)=y2+x2 -9 =0
32
8.5.1 Implicit functions
• Implicit function theorem: given F(y, x1 …, xm) = 0
a) if F has continuous partial derivatives
Fy, F1, …, Fm and Fy  0 and
b) if at point (y0, x10, …, xm0), we can construct a
neighborhood (N) of (x1 …, xm), e.g., by limiting the
range of y, y = f(x1 …, xm), i.e., each vector of x’s 
unique y
then i) y is an implicitly defined function y = f(x1 …, xm)
and ii) still satisfies F(y, x1 … xm) for every m-tuple in
the N such that F  0 (p. 195)
dfn: use  when two side of an equation are equal for any values of x and y
dfn: use = when two side of an equation are equal for certain values of x and y (p.197)
33
8.5.1 Implicit functions
• If the function F(y, x1, x2, . . ., xn) = k is an
implicit function of y = f(x1, x2, . . ., xn), then
Fy dy  Fx1 dx1  Fx2 dx2  ... Fxn d xn  0
where Fy = F/y;
Fx1 = F/x1
• Implicit function rule
• F(y, x) = 0; F(y, x1, x2 … xn) = 0, set dx2 to n = 0
34
8.5.1 Implicit functions
• Implicit function rule
Fy dy  Fx1 dx1  Fx2 dx2  ...  Fxn d xn  0
Fy dy   Fx1 dx1  Fx2 dx2  ...  Fxn d xn
Let dx 2  dxn  0 such that
Fx1
dy
y
|dx x .dxn 0 

dx1
x1
Fy
35
8.5.1 Deriving the implicit function rule (p. 197)
1) F ( y, x1 , x2 )  0
2) y  f ( x1 , x2 )
3) Fy dy  F1dx1  F2 dx2  0
4) dy  f1dx1  f 2 dx2
5) Fy  f1dx1  f 2 dx2   F1dx1  F2 dx2  0
36
8.5.1 Deriving the implicit function rule (p. 197)
6) Fy f1dx1  Fy f 2 dx2  F1dx1  F2 dx2  0
7) Fy f1  F1 dx1  Fy f 2  F2 dx2  0
8) Fy f1  F1 dx1  0
9) Fy f 2  F2 dx2  0
y
F1
10) f1 

x1
Fy
y
F2
11) f 2 
 ,
x2
Fy
37
Implicit function problem:
Exercise 8.5-5a, p. 198
• Given the equation F(y, x) = 0 shown below, is it an implicit
function y = f(x) defined around the point
(y = 3, x = 1)? (see Exercise 8.5-5a on p. 198)
• x3 – 2x2y + 3xy2 - 22 = 0
• If the function F has continuous partial derivatives Fy, F1, …,
Fm
• ∂F/∂y =-2x2+6xy
∂F/∂x =3x2-4xy+3y2
38
Implicit function problem
Exercise 8.5-5a, p. 198
• If at a point (y0, x10, …, xm0) satisfying the equation F (y, x1
…, xm) = 0, Fy is nonzero (y = 3, x = 1)
• This implicit function defines a continuous function f with
continuous partial derivatives
• If your answer is affirmative, find dy/dx by the implicitfunction rule, and evaluate it at point (y = 3, x = 1)
• ∂F/∂y =-2x2+6xy
∂F/∂x =3x2-4xy+3y2
• dy/dx = - Fx/Fy =- (3x2-4xy+3y2 )/-2x2+6xy
• dy/dx = -(3*12-4*1*3+3*32 )/(-2*12+6*1*3)=-18/16=-9/8
39
8.5.2 Derivatives of implicit functions
• Example
If F(z, x, y) = x2z2 + xy2 - z3 + 4yz = 0, then
Fy
z
2 xy  4 z

 2
2
y
Fz
2 x z  3z  4 y
40
8.5 Implicit production function
•
•
•
•
F (Q, K, L)
K/L = -(FL/FK)
Q/L = -(FL/FQ)
Q/K = -(FK/FQ)
Implicit production function
MRTS: Slope of the isoquant
MPPL
MPPK (pp. 198-99)
41
Overview of the Problem –
8.6.1 Market model
• Assume the demand and supply functions for a
commodity are general form explicit functions
Qd = D(P, Y0)
(Dp < 0; DY0 > 0)
Qs = S(P, T0)
(Sp > 0; ST0 < 0)
• where
Q is quantity, P is price, (endogenous variables)
Y0 is income, T0 is the tax (exogenous variables)
no parameters, all derivatives are continuous
• Find
P/Y0,
Q/Y0,
P/T0
Q/T0
42
Overview of the Procedure 8.6.1 Market model
• Given
Qd = D(P, Y0)
(Dp < 0; DY0 > 0)
Qs = S(P, T0)
(Sp > 0; ST0 < 0)
• Find
P/Y0,
P/T0,
Q/Y0,
Q/T0
Solution:
• Either take total differential or apply implicit function rule
• Use the partial derivatives as parameters
• Set up structural form equations as Ax = d,
• Invert A matrix or use Cramer’s rule to solve for x/d
43
8.5.3 Extension to the simultaneousequation case
• Find total differential of each implicit function
• Let all the differentials dxi = 0 except dx1
and divide each term by dx1 (note: dx1 is a choice )
• Rewrite the system of partial total derivatives of the
implicit functions in matrix notation
44
8.5.3 Extension to the simultaneous-equation case
1) F1 ( y1 , y2 , x1 )  0
2) F2 ( y1 , y2 , x2 )  0
3)
F1
F
F
dy1  1 dy2  1 dx1  0
y1
y2
x1
4)
F2
F
F
dy1  2 dy2  2 dx2  0
y1
y2
x2
5)
F1
F
F
dy1  1 dy2   1 dx1  0dx2
y1
y2
x1
6)
F2
F
dy1  2 dy2 
y1
y2
0dx1 
F2
dx2
x2
45
8.5.3 Extension to the simultaneous-equation case
• Rewrite the system of partial total derivatives of the
implicit functions in matrix notation (Ax=d)
7)
F1 dy1 F1 dy2
F

 1
y1 dx1 y2 dx1
x1
10)
F1 dy1 F1 dy2


y1 dx2 y2 dx2
8)
F2 dy1 F2 dy2


y1 dx1 y2 dx1
11)
F2 dy1 F2 dy2
F

 2
y1 dx2 y2 dx2
x2
 F1
 y
9)  1
 F2
 y
 1
0
F1   dy1 
F
y2   dx1   1 

   x1 

F2   dy2  
0





dx
y2   1 
 F1
 y
12)  1
 F2
 y
 1
0
F1   dy1 
y2   dx2   0 

   F2 
F2   dy2   x 
2 




dx
y2   2 
46
7.6 Note on Jacobian Determinants
• Use Jacobian determinants to test the existence of
functional dependence between the functions /J/
• Not limited to linear functions as /A/ (special case
of /J/
• If /J/ = 0 then the non-linear or linear functions
are dependent and a solution does not exist.
y1 x1
J 
y 2 x1
F 1
y1 x2
y1

F 2
y 2 x2
y1
F 1
y 2
2  0
F
y 2
47
8.5.3 Extension to the simultaneousequation case
• Solve the comparative statics of endogenous variables
in terms of exogenous variables using Cramer’s rule
 F

dy1 1  x1

dx1 J  F 2

 x1
1
F 

y2 
2
F

y2 
1
48
8.6 Comparative Statics of GeneralFunction Models
•
•
8.6.1 Market model
8.6.2 Simultaneous-equation
approach
•
8.6.3 Use of total derivatives
•
8.6.4 National income model
•
8.6.5 Summary of the procedure
49
Overview of the Problem –
8.6.1 Market model
• Assume the demand and supply functions for a
commodity are general form explicit functions
Qd = D(P, Y0)
(Dp < 0; DY0 > 0)
Qs = S(P, T0)
(Sp > 0; ST0 < 0)
• where
Q is quantity, P is price, (endogenous variables)
Y0 is income, T0 is the tax (exogenous variables)
no parameters, all derivatives are continuous
• Find
P/Y0,
Q/Y0,
P/T0
Q/T0
50
Overview of the Procedure 8.6.1 Market model
• Given
Qd = D(P, Y0)
(Dp < 0; DY0 > 0)
Qs = S(P, T0)
(Sp > 0; ST0 < 0)
• Find
P/Y0,
P/T0,
Q/Y0,
Q/T0
Solution:
• Either take total differential or apply implicit function rule
• Use the partial derivatives as parameters
• Set up structural form equations as Ax = d,
• Invert A matrix or use Cramer’s rule to solve for x/d
51
General Function Comparative Statics:
A Market Model (8.6.1)
Let thedemandand supply functionsfor a commoditybe :
1) Qd  DP, Y0 
2) Qs  S P, T0 
( DP/  0; DY/0  0)
( S P/  0; ST/0  0)
WhereY0 is incomeand T0 is the tax on thecommodity.
All derivativecontinuous.
Endogenous: Q, P
Exogenous:Y0 , T0
P arameters: D, S are functionsof P , Y0 , and T0
3) F 1(P, Q; Y0 , T0 )  D(P , Y0 ) – Q  0
4) F 2(P, Q; Y0 , T0 ) S(P , T0 ) – Q  0
52
General Function Comparative Statics: A
Market Model
3) D( P , Y0 )  Q  0
4) S ( P , T0 )  Q  0
Find dQ * dY0 , dQ * dT0 , dP * dY0 , dP * dT 0
T ake the totaldifferential of equations (3) & (4) ;
5) DP/ dP  DY/0 dY0  dQ  0
6) S P/ dP  ST/0 dT0  dQ  0
P ut only endog. vars on left
7) DP/ dP  dQ   DY/0 dY0
8) S P/ dP  dQ   ST/0 dT0
53
General Function Comparative Statics: A
Market Model
7) DP/ dP  dQ   DY/0 dY0
8) S P/ dP  dQ   ST/0 dT0
Put equations(7) & (8) in m atrix form at ( Ax  d );
 DP/  1  dP   DY/0
0  dY0 
9)  /
   

/ 
dT
0

S
d
Q
S

1
0
T 0 
 P
  
Calculate the sign of the Jacobiandet .
DP/
10) J  /
SP
1
 S P/  DP/  0
1
54
General Function Comparative Statics: A Market Model
 DP/  1  dP   DY/0
0  dY0 
11)  /
   

/ 
dT
0

S
d
Q
S

1
0
T 0 
 P
  
Take the partial  total derivatives of equation(11)
wrt exogenousvar s dY0 and dT0
 DP/
12)  /
 SP
 dP 
 1  dY0   DY/0 
  dQ   

 1 
  0 
 dY 
 0
 DP/
13)  /
 SP
 dP 
 1  dT0   0 
  dQ     S / 
 1 
  T0 
 dT 
 0
55
General Function Comparative Statics: A Market Model
 DP/  1  dP   DY/0
0  dY0 
11)  /
   

/ 
dT
0

S
d
Q
S

1
0
T 0 
 P
  
Take the partial  total derivatives of equation(11)
wrt exogenousvar s dY0 and dT0
 dP 
 DP/  1  dY0   DY/0 
12)  /
  dQ   

S

1
0




 P

 dY 
 0
Solve for the vectors of endogenousderivatives by
inverting the A m atrices of equations12
1
13) /
S P  DP/
 1
 S /
 P
 dP 
1   DY/0   dY0 
   dQ  ;
/ 
DP   0  

 dY 
 0
56
General Function Comparative Statics: A Market Model
 dP 
1   DY/0   dY0 
 1
1
13) /
   dQ 
/
/ 
/ 
S P  DP  S P DP   0  

 dY 
 0
Solve equations13 for the endogenousderivatives;
Calculate the sign of the derivatives
( DP/  0; DY/0  0)
J 
DP/
1
S P/
1
( S P/  0; ST/0  0)
 S P/  DP/  0
DY/0
dP
14)
 /
 0;
/
dY0 S P  DP
S P/ DY/0
dQ
15)
 /
 0; ;
/
dY0 S P  DP
57
General Function Comparative Statics: A Market Model
 dP 
1   0   dT0 
 1
1


16) /
/  
/
/ 
/ 
S P  DP  S P DP   ST0   dQ 
 dT 
 0
Solve equations16 for the endogenousderivatives;
Calculate the sign of the derivatives
( DP/  0; DY/0  0)
J 
DP/
1
S P/
1
( S P/  0; ST/0  0)
 S P/  DP/  0
 ST0
dP
17)
 /
 0;
/
dT0 S P  DP
/
/ /
dQ  DP ST0
18)
 /
0
/
dT0 S P  DP
58
Market model comparative static solutions by
Cramer’s rule

13)
dP

dY0
D
1
D
Y0
0
 1 Y0

 0;
J
J
D
D

P
Y0
S D
S
0
P Y0
dQ
14)
 P

0
dY0
J
J
13' ) An increase in incom e(Y0 ) causes an increase in equilibrium price paid.
14' ) An increase in incom ecauses an increase in equilibrium quantity consum ed.
0
1
S
S

1 
T0
T0
dP
dQ
15)


 0; 16)

dT0
J
J
dT0
D
P
S
P
0

J
S
T0


D S
P T0
0
J
15' ) An increase in taxes (T0 ) causes an increase in equilibrium prices paid
16' ) An increase in taxes causes a decrease in equilibrium quantity consum ed
59
Market model comparative static solutions by
matrix inversion
1
13)
J
 1
 S
 P
 dP 
D
1   D   dY 
dP Y0
D   Y    0 ; 14)

 0;
0
d
Q


dY
J

0
P   0  
 dY 
 0
S D
dQ P Y0
15)

0
dY0
J
14' ) An increase in incom e(Y0 ) will cause an increase in equilibrium price paid.
15' ) An increase in incom e will cause an increase in equilibrium quantity consum ed.
 dP 
S
D S


0
 

1 
T0
P T0
1  1
dP
dQ
dT
0

S

; 17)
16)  S D  

 0; 18)

0

d
Q
J 
dT0
J
dT0
J

 P P   T0  

 dT0 
17' ) An increase in taxes (T0 ) will cause an increase in equilibrium prices paid
18' ) An increase in taxes will cause a decrease in equilibrium quantity consum ed
60
8.7 Limitations of Comparative Statics
• Comparative statics answers the question:
how does the equilibrium change w/ a
change in a parameter.
• The adjustment process is ignored
• New equilibrium may be unstable
• Before dynamic, optimization
61