### L04_Expenditure Min_Duality_2014

```Expenditure Minimization
Expenditure Minimization
• Set up optimization problem
E  p x x  p y y, U  U(x, y)
L  p x x  p y y  (U - U(x, y))
FOC
L x : p x  U x  0  p x  U x
L y : p y  U y  0  p y  U y
L : U  U(x, y)  0  U  U(x, y)
With the result that
px Ux

py Uy
Expenditure Minimization: SOC
• The FOC ensure that the optimal consumption
bundle is at a tangency.
• The SOC ensure that the tangency is a minimum, and
not a maximum by ensuring that away from the
tangency, along the indifference curve, expenditure
rises.
Y
E=E’
E*<E’
E=E*
X
Expenditure Minimization: SOC
• The second order condition for constrained minimization will hold
if the following bordered Hessian matrix is positive definite:

let L( x, y )  px x  p y y   U  U ( x, y )
 L

H   L x
 L y

H 
L x
Lxx
Lyx
L y 

Lxy 
Lyy 
0
U x
U x
 U xx

Will hold if the Hessian of the
Lagrangian is is Positive Definite
0
U x
U y
 0 and H  U x
 U xx
 U xy  0
U y
 U yx
 U yy
  U x   0, and  2 U xU yU xy  U x 2U yy  U y 2U xx  0
2
Note, -(-Ux )2 =-Ux 2 < 0 and (so long as μ > 0), 2UxUxy Uy -Uy2Uxx-Ux2Uyy > 0, so these
conditions are equivalent to checking that the utility function is strictly quasi-concave.
Expenditure Minimization
• Solve FOC to get:
px Ux

py Uy
U  U(x, y)
And get the compensated, Hicksian, demand curves
x c*  x c  p x , p y , U 
yc*  yc  p x , p y , U 
*   c  p x , p y , U 
Expenditure Minimization
• Back into the expenditure function determine
minimum expenditure:
E*  p x x c  p x , p y , U   p y y c  p x , p y , U 
• Solve for Ū to get the indirect utility function:
V*  V*  p x , p y , E   V*  p x , p y , M 
Interpreting μ: Envelope Result


L*  p x x c (p x , p y , U)  p y y c (p x , p y , U)  (p x , p y , U) U  U x c (p x , p y , U), y c (p x , p y , U)
Differentiate with respect to U
L*
U
L*
U
L*
U
L*
U


U
c*
c* c*
*
*
c*
c* 
 px x  py y  
  U U  * U c*



U
x
,
y
)

c x U  U c yU
U
x
y


U
c*
U
c*
U
*




* c*
c*
* c*
*
*
*
c*
c*
 x c*
U  p x   U x c   y U p y   U yc     U U   U  U  x , y 

* c*
c*
* c*
*
c*
*
c*
c*
 x c*
U  p x   U x c   y U p y   U yc     U U  U  x , y 

L*
x c*  x c  p x , p y , U 
*
 x 0  y 0     0 ,

U
y c*  y c  p x , p y , U 
E*
*
c*
c*
Because U-U  x , y   0, L*=E * and
  p x , p y , U *    p x , p y , U 
U
In other words, if you want to increase utility by 1 util, you need to
c*
U
c*
U
*
c*
U

increase expenditure by *.


Finding
E*
p x
: Envelope Result


L*  p x x c (p x , p y , U)  p y y c (p x , p y , U)  (p x , p y , U) U  U x c (p x , p y , U), y c (p x , p y , U)
Differentiate with respect to p x
L*
c*
c*
*
*
c*
*
c* 
*
*
c*
c*


 p x x c*

x

p
y



U
x

U
y


U

U
x
,
y


c
c
px
y px
px
px 
px 
x
y


p x


L*
* *
c*
c*
c*
*
c*
c*


  p x  * U*xc  x c*

p


U
y

x


U

U
x
,
y


c
px
y
px
px
px 
y

p x
L*
c*
c*
*
  0  x c*
px   0  y px  x px   px  0  ,
p x
Because U - U  x , y
c*
c*

L*
 x c*
p x
E*
 0, L=E ,
 x c*
p x
x c*  x c  p x , p y , U 
y c*  y c  p x , p y , U 
*    p x , p y , U 
In other words, if px increases by \$1, you need to increase expenditure by x c* .

Expenditure Minimization
• Comparative Statics
Plug the optimization functions into the FOC
x c*  x c  p x , p y , U 
y c*  y c  p x , p y , U 
 c*   c  p x , p y , U 
FOC
L*x : p x   c  p x , p y , U   U*x  x c  p x , p y , U  , y c  p x , p y , U    0
L* y : p y   c  p x , p y , U   U* y  x c  p x , p y , U  , y c  p x , p y , U    0
L* : U  U*  x c  p x , p y , U  , y c  p x , p y , U    0
Comparative Statics
Differentiate with respect to p x
x c*
yc*
 *
*
1   U xx
  U xy
 Ux
0
p x
p x
p x
*
x c*
yc*
*c
c
0   U yx
  U yy
 Uy
0
p x
p x
p x
*
x c*
yc*
0  Ux
 Uy
0
p x
p x
Rearrange
x c*
yc*
 Ux
 Uy
0
p x
p x
*
x c*
y c*
*
*
U x
  U xx
  U xy
 1
p x
p x
p x
 c*
x c*
yc*
*
U y
  U yx
 U yy
0
p x
p x
p x
Comparative Statics: Effect of a change in px
Put in Matrix Notation
• Solve for x c*
p x
 0

U x
U
 y
  
 p 
U y   x   0 
  x c   
U xy   
   1

p
 x
U yy   c   0 
y


 p x 
U x
U xx
U yx
Assuming SOC are satisfied H  2U x U y U xy  U x 2 U yy  U y 2 U xx  0
0
U x
x c*  U y

p x
0
U y
1 U xy
0
H
U yy

U y2
( )
0
Compensated demand curves must be downward sloping.
Expenditure Minimization: Example
E  p x x  p y y, U  xy.5 , M  60, p x  1, p y  2
L  p x x  p y y    U  xy.5 
FOC
L x : p x  y.5
x
L y : p y  .5
2y
L  U  xy.5
With the result that
2p y y
p x 2y
px x

x
and y 
py x
px
2p y
Expenditure Minimization
• Combining with
L  U  xy.5
yields the Hicksian Demand Functions
1/3
 px U 
 2p y U 
*
x 

 and y  
 px 
 2p y 
2
*
2/3
Expenditure Minimization
• Expenditure Function
E*  p x x c*  p y yc*
1/3
 px U 
 2p y U 
E  px 

  py 
 px 
 2p y 
2
2/3
*
• And solving this for U would yield U* = V *(px,py,M)
Properties of Expenditure
Functions
• Homogeneity
– a doubling of all prices will precisely double the
value of required expenditures
• homogeneous of degree one
• Nondecreasing in prices
– E*/pi  0 for every good, i
• Concave in prices
– When the price of one good rises, consumers
respond by consuming less of that good and more
of other goods. Therefore, expenditure will not rise
proportionally with the price of one good.
Concavity of Expenditure Function
If the consumer continues to buy a fixed
bundle as p1’ changes (e.g. goods are
perfect compliments), the expenditure
function would be Ef
Ef
E(px,py,U*)
E(p1,…)
E(px’,py…U*)
px’
px
Since the consumption
pattern will likely change,
actual expenditures will be
less than portrayed Ef such as
E(px,py,U*). At the px where
the quantity demanded of a
good becomes 0, the
expenditure function will
flatten and have a slope of 0.
Max and Min Relationships
Utility Max
L = U(x) + λ(M-g(x))
x* = x(px, M)
Expenditure Min
L = g(x) + μ(U-U(x)))
xc* = xc (px, U)
Indirect Utility
U* = U *(x*)
V * = V *(px, M)
Expenditure Function
E* = E *(xc*)
E * = E *(px, U)
Expenditure Function
Solve V * for M (M=E)
E * = E *(px, U)
Indirect Utility
Solve E * for U (E=M)
U * = V *(px, M)
E  g(x)  px x
Shephards Lemma and Roy’s Identity
• Two envelope theorem results allow:
– Derivation of ordinary demand curves from the
expenditure function
– Derivation of compensated demand curves from
the indirect utility function
Envelope Theorem
• Say we know that y = f(x; ω)
– We find y is maximized at x* = x(ω)
• So we know that y* = y(x*=x(ω),ω)).
• Now say we want to find out
*
*
*
*
dy
dy dy dx

 *
d d dx d
• So when ω changes, the optimal x changes,
which changes the y* function.
• Two methods to solve this…
Envelope Theorem
• Start with: y = f(x; ω) and calculate x* = x(ω)
• First option:
• y = f(x; ω), substitute in x* = x(ω) to get y* = y(x(ω); ω):
*
dy
x(), 

dy

 y* ()
d
d
*
• Second option, turn it around:
then substitute x* = x(ω)
f  x, 

 y (x; )


• First, take y
*
dy
 x(),   y* ()
into yω(x ; ω) to get dy 

d
d
*
•And we get the identity
y* ()  y* ()
This is the basis for…
• Roy’s Identity
– Allows us to generate ordinary (Marshallian)
demand curves from the indirect utility function.
• Shephard’s Lemma
– Allows us to generate compensated (Hicksian)
demand curves from the expenditure function.
Roy’s Identity: Envelope Theorem 1
L  U  x, y     I  p x x  p y y 
Derive
x*  x  p x , p y , M  ; y*  y  p x , p y , M  ; *    p x , p y , M 
Substitute



L*  U* x* , y*  * M  p x x*  p y y*

Option 1: Plug the
optimal choice
variable equations
into the Lagrangian
and THEN
differentiate
L*  V* (x  p x , p y , M  , y  p x , p y , M )    p x , p y , M   I  p x x  p x , p y , M   p y y  p x , p y , M  
L*  V*  p x , p y , M     p x , p y , M  I  p x x  p x , p y , M   p y y  p x , p y , M  
Note that  I  p x x  p x , p y , M   p y y  p x , p y , M    0
So L*  V*  p x , p y , M 
L*  p x , p y , M 
p x

V*  p x , p y , M 
p x
Roy’s Identity: Envelope Theorem 2
x *  x  p x , p y , M  , y*  y  p x , p y , M  ,  *    p x , p y , M 
Option 2:
Differentiat
L(x, y, p x , p y , I, )
e the
   x
p x
Lagrangian
NOW plug in x*  x  p x , p y , M  , and *    p x , p y , M  to get and THEN
plug the
*
L  p x , p y , M 
optimal
   p x , p y , M   x  p x , p y , M 
p x
choice
* *
*
variable

U
x
,
y

V
p
,
p
,
M




x
y
Remember that * =
i.e. * =
into the
M
M
derivative
*
*
L  p x , p y , M 
V  p x , p y , M 
equation

 x p ,p ,M
Start with L  U(x, y)  (I  p x x  p y y)
p x
M
x
y
Envelope Theorem and Roy’s Identity
Option 1 yields:
Option 2 yields:

V* (p x , p y , I)
I
L* (p x , p y , I)
p x
L* (p x , p y , I)
p x
 x(p x , p y , I) 

V* (p x , p y , I)

p x
x  x(p x , p y , I)   *
V (p x , p y , I)
I
V* (p x , p y , I)
I
V* (p x , p y , I)
V* (p x , p y , I)
*
p x
p x
 x(p x , p y , I)
Shephard’s Lemma: Envelope
Theorem 1
Minimize
L  p x x  p y y    U  U  x, y  
Derive
x *  x c  p x , p y , U  , y *  y c  p x , p y , U  , *    p x , p y , U 
Plug them in to get
L*  p x x c  p x , p y , U   p y y c  p x , p y , U     p x , p y , U   U  U *  x c  p x , p y , U  , y c  p x , p y , U   
Since U  U  x c* , y c*   0 this reduces to
E*  E*  p x , p y , U 
L*  p x , p y , U 
p x

E*  p x , p y , U 
p x
Option 1: Plug the optimal
choice variable equations
into the Lagrangian and THEN
differentiate
Shephard’s Lemma: Envelope
Theorem 2
x*  x c  p x , p y , U  , y*  y c  p x , p y , U  , *    p x , p y , U 
L  p x x  p y y    U  U(x, y) 
Take the derivative:
L  x, y, p x , p y , M,  
p x
NOW plug in x*  x c  p x , p y , U  to get:
L*  p x , p y , U 
p x
 xc  px , py , U 
x
Differentiate the
Lagrangian and THEN
plug the optimal choice
variable into the
derivative equation
Shephard’s Lemma
• Bring results of Option 1 and Option 2
together:
L*  p x , p y , U 
p x
L*  p x , p y , U 
p x

E*  p x , p y , U 
p x
 xc  px , py , U 
xc  px , py , U  
E*  p x , p y , U 
p x
The Relationships
Primal
Max U(x), s.t. M = px
L=U(x)-λ(p•x-M)
Marshallian Demand
x* = x(p,M’)
λ=UM
When E* = M’
And U* = Ū
xc (p,Ū)
x(p,M’) = x* =
when
E* = M’ and U* = Ū
Dual
Min E=p•x, s.t. Ū=U(x)
L=px-μ(Ū=U(x))
Hicksian Demand
x*=xc(p, Ū)
μ=EU
x* = x(p,M)
x*=xc(p,E(P,U))
Indirect Utility Function
U* = U(x*)
U* = U(x*=x(p,M’))
U* = V(p, M’)
xc (p,U)
x* =
x*=x(p,V(p,M))
Expenditure Function
E* = p•x*
where x*=xc(p, Ū)
M’=E* = E(p, Ū)
U* =V(p,M’) when solved for M’ is E*= E(p, Ū)
The Relationships
Primal
Indirect Utility Function
U* = V*(p, M)
Roy’s
Identity
∂V*(p,M)
∂pi
xi* = xi(p,M)= ∂V*(p,M)
∂M
Dual
Expenditure Function
E* = E*(p, U)
Shephard’s
Lemma
xi* = xci (p,U) =
∂E*(p,U)
∂pi
Ordinary (Marshallian) Demand
Slope of budget
line from px/py to
steeper px’/py
y
Income is fixed at M’, but
utility falls
px/py
Ū
U2
px’/py
x*=x(px,py,M’)
px/py
xb
xa
Qd falls from xa to xb
x
xb
xa
Qd falls from xa to xb
x
Compensated (Hicksian) Demand
y
Slope of budget
line from px/py to
steeper px’/py
px/py
Utility is fixed at Ū, but
expenditure rises
x*=xc(px,py, Ū)
px’/py
U1
xc xa
Qd falls from xa to xc
x(px,py,M’)=xc(px,py,Ū)
px/py
x
xc xa
Qd falls from xa to xc
x
Ordinary (Marshallian) Demand
Slope of budget
line from px/py to
flatter px’’/py
y
Income is fixed at M’, but
utility rises
px/py
U0
Ū
px/py
x*=x(px,py,M’)
px’’/py
xa
xb
Qd falls from xa to xb
x
xa
xb
Qd falls from xa to xb
x
Compensated (Hicksian) Demand
Slope of budget
line from px/py to
flatter px’’/py
y
Utility is fixed at Ū, but
expenditure falls
px/py
Ū
x*=xc(px,py,Ū)
px/py
x(px,py,Ī)=xc(px,py,Ū)
px’’/py
xa
xc
Qd rises from xa to xc
x
xa
xc
Qd rises from xa to xc
x
Ordinary and Compensated
• If price changes and Qd changes along the
ordinary demand curve, then utility changes