Economics 101A
Section Notes
GSI: David Albouy
Consumer Theory and the Envelope Theorem
1
Utility Maximization Problem
The consumer problem looked at here involves
• Two goods: x and y with prices px and py .
• Conusumers facing a budget constraint px x + py y ≤ I, where I is income.Consumers maximize utility
U (x, y) which is increasing in both arguments and quasi-concave in (x, y). Also the non-negativity
constraints x ≥ 0 and y ≥ 0 must hold as consumption cannot be negative.
• Competition in this world is "perfect" or "pure" so consumers are price takers, i.e. px and py are fixed
for them. This may be justified by postulating a large number of consumers.
Since utility is increasing in both x and y we can safely assume that the budget constraint px x + py y ≤ I
is satisfied with equality. We will also assume that the non-negativity constraints are slack so that x > 0
and y > 0. Setting the inequalities to equality, combining the two constraints, and rearranging a bit we get
the standard utility maximization problem (UMP for short)
max U (x, y)
x,y
s.t. px x + py y = I
(UMP)
The solution to this problem is typically found by writing the Lagrangean
L (x, y, l) = U (x, y) + α (I − px x − py y)
and taking the first order conditions1
¢
¡
∂U xd , y d
∂L
=
− αd p x = 0
∂x
∂x
¢
¡
∂U xd , y d
∂L
=
− αd p y = 0
∂y
∂y
∂L
= I − px xd − py y d = 0
∂α
The superscript "d " is used to refer to the fact that these solutions are the consumer’s demands.
Solving yields the Lagrange multiplier αd = αd (px , py , I) and the demand functions
xd (px , py , I) y d (px , py , I)
To be more general we call these the uncompensated (or Marshallian or Walrasian) demand functions. These functions are "uncompensated" since price changes will cause utility changes: a situation that
does not occur with compensated demand curves. Substituting these solutions back into the utility function,
1 If
the we did not assume the non-negativity constraints held the first two first order conditions would be
¢
¡
∂U xD , yD
∂L
=
− α∗ px < 0, with " = " if xD > 0
∂x
∂x
¡ D D¢
∂U x , y
∂L
=
− α∗ py < 0, with " = " if y D > 0
∂y
∂y
Note that this is a simplification of the Kuhn-Tucker conditions described previously.
1
the maximand, we get the actual utility achieved as a function of prices and income. This function is known
as the indirect utility function
£
¤
V (px , py , I) ≡ U xd (px , py , I) , y d (px , py , I)
(Indirect Utility Function)
This function says how much utility consumers are getting when they face prices (px , py ) and have income
I.
An interesting question is how much utility changes when either prices or income change. As shown in
the "Calculus and Optimization" hand-out we should expect that giving the person one dollar in income
should increase her utility by the Lagrange multiplier αd
∂V
= αd
∂I
Taking the total derivative of Indirect Utility Function with respect to px yields
∂U ∂xd ∂U ∂y d
∂V
=
+
∂px
∂x ∂px
∂y ∂px
(1)
This expression does not tell us much, however it can be simplified with the two useful substitutions. First
the the first two first order conditions tell us that
∂U
∂U
= αd px and
= αd py
∂x
∂y
so substituting these in to 1 we have
µ
¶
∂xd
∂y d
∂xd
∂y d
∂V
d
d
d
= α px
+ α py
= α px
+ py
∂px
∂px
∂px
∂px
∂px
(2)
Second take the total derivative of the budget constraint px xd + py y d = I with respect to px to get
xd + px
∂xd
∂y d
∂xd
∂y d
+ py
= 0 ⇒ px
+ py
= −xd
∂px
∂px
∂px
∂px
Substituting this in to 2 we get a result known as Roy’s Identity
∂V
= −αd xd
∂px
(Roy’s Identity)
A basic (albeit somewhat flawed) intuition for this identity is straightforward: if px goes up by one dollar
then the consumer will lose xd number of dollars, which each have utility value αd , so that utility drops
by αd xd . The more correct intuition is that xd and y d are changing in response to the price increase, but
because the consumer is optimizing the entire time and facing a budget constraint, these indirect effects sum
up to zero. A similar result holds for y that
∂V
= −αd y d
∂py
which you can prove for yourself. Note that if you were just given the indirect utility function you could solve
for xd , y d , and αd , just be taking combining the derivatives of V appropriately, e.g. xd = − (∂V /∂p) /αd =
− (∂V /∂p) / (∂V /∂I).
Example 1 Quasilinear utility takes a form where one of the goods consumed enters linearly into the
utility function, which in this case we take to be√U (x, y) = x + h (y) where h0 (y) > 0 and h00 (y) < 0. (The
reader may like
¡ ¢to try the case where h (x) = x) The first order conditions for this problem simplify to
1 = αd px , h0 y d = αd py , and px xd + py y d = I . Solving yields
αd = 1/px
µ ¶
¡ d ¢
py
0 −1
y = (h )
α py = (h )
px
µ ¶
µ ¶
d
py
I
−
p
I
p
y
−1
y
y
· (h0 )
=
−
xd =
px
px
px
px
2
d
0 −1
Notice that y d does not depend on income I, only on relative prices py /px . Indirect utility is given by
¡ ¢
V (px , py , I) = xd + h y d
µ ¶
µ ¶
·
µ ¶¸
py
py
py
I
0 −1
0 −1
−
· (h )
+ h (h )
=
px
px
px
px
It is fairly easy to see that ∂V /∂I = 1/px = αd . Roy’s identity is more difficult to check
µ ¶ µ ¶
·
µ ¶¸
∂V
1
py
py
1
py
1
1
1
0 −1
0
0 −1
= − · (h )
+ h (h )
−
00
00
∂py
px
px
px h (py /px ) px
px
h (py /px ) px
µ ¶
µ ¶ µ ¶
1
1
1
p
p
p
1
1
−1
y
y
y
= − · (h0 )
+
−
00
00
px
px
px h (py /px ) px
px h (py /px ) px
µ ¶
1
p
−1
y
= − · (h0 )
px
px
= −αd y d
The reader is invited to check Roy’s Identity for themselves in the case of x.
2
Expenditure Minimization Problem
The consumer problem can be approached in a different way which produces some useful tools. Instead of
maximizing utility given a certain income, imagine how much income it would take to achieve a certain level
of utility. In other words consider the following expenditure minimization problem (EMP for short),
which as always take prices as given
min px x + py y
x,y
s.t. U (x, y) ≥ u
(EMP)
This problem looks very much like the UMP above except that the objective function and constraint have
been switched around. We wish to minimize the income I = px x+py y needed achieve a fixed level of income
u, for given prices (px , py ). Our third parameter in parameter in this problem (after px and py ) is no longer
I, but u. This problem can typically be solved by writing the Lagrangean
L (x, y, β) = px x + py y + β [u − U (x, y)]
Assuming the constraint holds with equality, the first order conditions are
∂U (xc , y c )
∂L
=0
= px − β c
∂x
∂xc
∂L
∂U (xc , y c )
=0
= py − β c
∂y
∂y c
∂L
= u − U (xc , y c ) = 0
∂β
The first two FOC are quite similar to above replacing β c with 1/αd , but the third constraint corresponding
to the constraint is very different. Solving these three equations in the three unknowns yields the Lagrange
multiplier β c = β c (px , py , u), the shadow price in dollars of having to provide an extra unit of utility to this
consumer, as well as the compensated demands
xc (px , py , u)
y c (px , py , u)
which are a function now of the required utility u, not income I. Note here that even though utility stays
the same, quantities demanded will change as px and py change since the individual is trying to minimize
her expenditures on consumption. Levels of required income I are assumed to automatically adjust to let
make sure that individual can still achieve utility u, although not necessarily the bundle of goods previously
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consumed. The individual is fully compensated for changes in price which could otherwise affect her utility
if I were held fixed. Substituting in the solutions back into the objective function, the minimand, we get
the expenditure function
E (px , py , u) ≡ px xc (px , py , u) + py y c (px , py , u)
(Expenditure Function)
which is precisely the amount I needed to maintain utility level u, for given prices px and py .
As before we can differentiate E with respect to u to get ∂E/∂u = γ̂. Taking the total derivative of E
with respect to px we get
∂E
∂xc
∂y c
= xc + px
+ py
(3)
∂px
∂px
∂px
We can simplify this expression with two substitutions. First we substitute in the first order conditions
which imply, px = β c ∂U/∂x and py = β c ∂U/∂y into 3 to get
·
¸
c
c
c
∂U ∂y c
∂E
c
c ∂U ∂x
c ∂U ∂y
c
c ∂U ∂x
=x +β
+β
=x +β
+
(4)
∂px
∂x ∂px
∂y ∂px
∂x ∂px
∂y ∂px
Second we differentiate the constraint U (xc , y c ) = u totally with respect to px to get
∂U ∂xc
∂U ∂y c
+
∂x ∂px
∂y ∂px
which implies that the second term in 4 is zero. This implies the result known as Shepard’s Lemma (the
analogue to Roy’s Identity) that
∂E
= xc
(Shepard’s Lemma)
∂px
Again the (somewhat misleading) intuition for this is clear. If px changes by a small amount then xc will
not change by very much and so the increased cost of consuming these units is precisely xc . The better
intuition is that there are changes in xc and y c , but because of optimizing behavior, the consumer avoids
spending any more than xc , although since she was optimizing before she cannot avoid spending any less.
The case for y is identical
∂E
= yc
∂py
Thus, given the expenditure function you can derive the compensated demands just be taking the partial
derivatives of E.
The UMP and EMP are mathematically known as "dual" problems. What is a constraint in one is
the objective in the other and vice-versa. Because of this a useful identities relating compensated and
uncompensated demands holds, namely
xd (px , py , I) = xc (px , py , V (px , py , I))
c
d
x (px , py , u) = x (px , py , E (px , py , u))
(ID1)
(ID2)
Similarly there are two identities relating the indirect utility function with the expenditure function
V (px , py , E (px , py , u)) = u
E (px , py , V (px , py , I)) = I
(ID3)
(ID4)
These identities can be proven formally (the proof is a little bit beyond the scope of this course), but a slow
reading of them should show them to be quite intuitive. These identities can be used to simplify problem
solving. For instance, solving the UMP one gets xd , y d , αd and V . Setting V (px , py , I) = u, and solving
for I yields the expenditure function E which can then be differentiated to get xc , y c and β c using Shepard’s
Lemma. Similarly one can proceed from the EMP, and solve the equation E (px , py , u) = I for u to get V ,
which can then be used to get xd , y d , and αd , using Roy’s Identity.
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Example 2 Continuing with quasilinear utility, the FOC imply that px = β c , py = β c h0 (y c ), and xc +
h (y c ) = u. Solving the system we get
β c = px
c
0 −1
y = (h )
µ
py
px
¶
·
µ ¶¸
py
0 −1
x = u − h (h )
px
c
Notice that y c = y d , as neither involves either u or I, so that the identity holds trivially. The expenditure
function is given by
E (px , py , u) = px xc + py y c
·
µ ¶¸
µ ¶
py
py
−1
−1
= px u − px h (h0 )
+ py (h0 )
px
px
Solving for u in this equation will yield the indirect utility function derived above.
also be verified rather similarly to how Roy’s Identity was verified.
3
Shepard’s Lemma can
The Envelope Theorem
The derivations of Roy’s Identity and Shepard’s Lemma, as well as the interpretation of the Lagrange
multipliers are all special cases of what is known as the envelope theorem. Stated generally, say we wish to
solve the maximization problem
max f (x, y, ξ)
s.t. g (x, y, ξ) ≤ c
x,y
where x and y are control variables and ξ is a given parameter which effects f and/or g, but over which we
do not maximize over, i.e. it is given exogenously. The Lagrangean is then written as
L (x, y, λ, ξ) = f (x, y, ξ) + λ [c − g (x, y, ξ)]
for which the first two first order conditions are given by2
∂f (x∗ , y ∗ , ξ)
∂g (x∗ , y ∗ , ξ)
∂L
=
−λ
=0
∂x
∂x
∂x
∂L
∂f (x∗ , y ∗ , ξ)
∂g (x∗ , y ∗ , ξ)
=
−λ
=0
∂y
∂y
∂y
Do not take the derivative with respect to ξ, as ξ is given exogenously. The solution to this problem, given
by x∗ (ξ) and y ∗ (ξ), depends on ξ as different values of ξ will imply different solutions.
Substituting the solutions into the function f gives the value function
F (ξ) = f [x∗ (ξ) , y ∗ (ξ) , ξ]
2 Equivalent
is the minimization problem
min f (x, y, α)
x,y
s.t. g (x, y, α) ≥ c
as this is equivalent to the maximization problem
max −f (x, y, α)
s.t.g (x, y, α) ≥ c
x,y
Writing out the Lagrangean one gets
L (x, y, λ) = −f (x, y, α) + λ [g (x, y, α) − c]
which gives equivalent first order conditions as the negative signs cancel.
5
which is the maximized value of f , which ultimately depends on ξ. Taking the total derivative of F (ξ) we
get
dF (ξ)
∂f dx∗
∂f dy ∗
∂f
=
+
+
(5)
dξ
∂x dξ
∂y dξ
∂ξ
The last term represents the direct effect of ξ on f , while the first two terms represent the indirect effect of
ξ on f by changing x∗ and y ∗ . This expression can be simplified in two steps. First substituting in the
first order conditions ∂f /∂x = λ∂g/∂x and ∂f /∂y = λ∂g/∂y into 5 yields
µ
¶
∗
∗
∂g dx∗
∂f
∂g dy ∗
∂f
dF (ξ)
∗ ∂g dx
∗ ∂g dy
∗
=λ
+λ
+
=λ
+
+
(6)
dξ
∂x dξ
∂y dξ
∂ξ
∂x dξ
∂y dξ
∂ξ
Second, differentiating the constraint g (x∗ , y ∗ , ξ) = c totally with respect to ξ yields
∂g dx∗
∂g dy ∗
∂g
∂g dx∗
∂g dy ∗
∂g
+
+
=0⇒
+
=−
∂x ∂ξ
∂y ∂ξ
∂ξ
∂x ∂ξ
∂y ∂ξ
∂ξ
which substituting into 6 and rearranging yields the envelope theorem
dF (ξ)
∂f
∂L
∂g
=
− λ∗
=
dξ
∂ξ
∂ξ
∂ξ
(Envelope Theorem)
Therefore the change in the value function is given by the partial derivative of the Lagrangean with respect
to ξ - a helpful simplification.3 This covers Roy’s Identity, Shepard’s Lemma, and the interpretation of the
Lagrange multiple (check!).
3 In
the unconstrained case or in the case where the constraint is slack (λ∗ = 0) is just a special case with
dF (ξ)
∂f
=
dξ
∂ξ
The total effect is given by just the partial, direct effect seen in f .
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