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Appendix: Math Preparation
1. Lagrange Method for Constrained Optimization
The following classical theorem is from Takayama (1993, p.114).
Theorem A-1. (Lagrange). For
n
f : → and
nm
G : → , consider the following
problem
max ( )
st ( ) 0.
x
fx
Gx.. =
(Α.1)
Let ( ) () ()Lx f xGxλλ, ≡ + ? (Lagrange function).
? If x
?
solves (A.1) and if ()DGx
?
has full rank, then there exists
k
λ∈ (Lagrange
multiplier) such that
FOC ( ) 0
x
DL xλ
?
:,=,
and
2
SONC ( ) 0 for satisfying () 0hD f x h h DG x h
′? ?
: ≤ ,=.
? If the FOC is satisfied, () 0Gx
?
=,and
2
SOSC ( ) 0 for 0 satisfying ( ) 0hD f x h h DG x h
′? ?
:<,≠ =,
then
*
x is a unique local maximum.
What is the intuition for FOC and SOC?
How to verify the SOC?
Theorem A-2. Let
nn
A
×
∈ be symmetric,
mn
C
×
∈ has full rank, ,mn< and
1
,,
mn
bb
+
… be
the principal minors of
0
T
C
B
CA
??
??
?≡
?
?
? ?
??
. Then,
' 0 for 0 satisfying 0(1) 0,21
km
k
xAx x Cx b k m
?
< ≠ = ?? > ?≥ + .
Example A-1. For 0a > and 0b >,consider
12
22
12
12
()max
st 1.
xx
Fab ax bx
xx
,
, ≡??
.. + =
Is the solution indeed optimal?
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Theorem A-3 (Kuhn-Tucker). For differentiable
n
f : → and
nm
G : → , let
() () ()Lx f xGxλλ, ≡ + ? . If x
?
is a solution of
max ( )
st ( ) 0
x
fx
Gx.. ≥ ,
(Α.2)
then there exists
m
λ
+
∈ such that (Kuhn-Tucker condition) λ () 0Gx
?
? = and
FOC ( ) 0
x
DL xλ
?
:,=.
Theorem A-4 (Sufficiency). Let f and
i
g , 1…im=, , , be quasi-concave, where
1
(… )
T
m
Gg g=,, .Let x
?
satisfy the Kuhn-Tucker condition and the FOC for (A.2). Then, x
?
is a global maximum point if
(1) () 0Df x
?
≠ , and f is locally twice continuously differentiable, or
(2) f is concave.
Kuhn-Tucker Theorem is not useful. We usually use Lagrange Theorem only.
A mapping
nk
H : → is linear if it can be written as ( )Hx Ax B=+,where A is
matrix and B is a vector.
Theorem A-5. Let
n
f : → and
nm
G : → be concave, and
nk
H : → be linear.
Also,
0
x? s.t.
0
() 0Gx >.Then, x
?
is a solution of
max ( )
.. ( ) 0,
() 0.
x
fx
st G x
Hx
≥
=
if and only if ( ) 0 ( ) 0Gx hx
??
≥ ,=,and there exist
m
λ
+
∈ and
k
μ∈ such that
λ () 0Gx
?
? = and x
?
is a solution of
max ( ) ( ) ( ) ( )
x
Lxf xGxHxλ μ λ μ,, ≡ + ? + ? .
Example A-2. Given utility function
1
12 12
()
aa
ux x xx
?
,= ,consider
12
12 12
0
11 2 2
()max()
st
xx
vp p m ux x
p x p xm
, ≥
,, ≡ ,
.. + ≤ .
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2. Hamilton Method for Dynamic Optimization
A good reference for this section is Kamien–Schwartz (1991). See also Chiang (1992).
The following theorem is from Kamien-Schwartz (1981, p.16).
Theorem A-6. For :
kk
H ××→ , consider problem
0
00
max [ ( ) ( )]
st ( ) ( ) ,
T
uA t
T
Htut ut dt
ut u uT u
∈
,,
.. = , =
∫
null
(Α.3)
where the set of admissible controls is
{ }
0
continuously differentiable functions [ ]
k
Aut≡ :, → .
If H is continuous w.r.t. its first argument, continuously differentiable w.r.t. its second and
third arguments, then the solution u
?
must satisfy the Euler equation:
[ ( ) ( )] [ ( ) ( )]
uu
d
Htut t Htut t
uu
dt
??
,, =,, .
null
nullnull (Α.4)
If the terminal value ()uT is free, the transversality condition is
[()()]0
u
HTuT T
u
? ?
,, =.
null
null (Α.5)
If the initial value
0
()u t is free, the transversality condition is
00 0
[()()]0
u
Htut t
u
? ?
,, =.
null
null (Α.6)
If the terminal condition is () 0uT ≥ , the transversality conditions are
() [ () ()] 0 [ () ()] 0
uu
uTHTuT T HTuT T
uu
?? ???
,, =, ,, ≤ .
nullnull
nullnull (Α.7)
Conversely, if ()Htuu,,null is concave in ()uu,,null then any u A
?
∈ satisfying the Euler equation
(A.4) and the initial and terminal conditions is a solution of (A.3).
There is no SOSC; there is only a SONC:
**
Legendre Condition: [ , ( ), ( )] 0.
uu
H tutut≤
nullnull
null
Example A-3. The principal’s problem is
()
[ ]
[]
max ( ) ( )
st ( ) ( ) ( )
as
vy sy f yady
usyfyady ca u
, ?
? ,
.. , ≥ +.
∫
∫
(Α.8)
Example A-4. Consider
()
() (){ } ()
( )
max
st 0
x
ux vx dF
x
θ
θ
θθ θθ θ
θ
?
′
????
,+ ,
????
.. ≥ .
∫
(Α.9)
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Theorem A-7 (Special Model I). Let
nknk n
xug∈ , ∈ ,: × ×→
and
nk
f :××→ . For problem
0
00
00
( ) max [ () () ]
st ( ) [ ( ) ( ) ]
() () 0
T
T
u t
Jx x t f xt ut tdt
x t g xt ut t
xt x xT
,, ≡ ,,
.. = , ,
=, ≥
∫
null
define the Hamiltonian as
() ()Hfxut g xutλ=,,+? ,, .
Under certain differentiability conditions, if u
?
is a solution, then there exists a function
0
[]
n
tTλ :, → such that u
?
is a solution of
0
u
H =, (Α.10)
x
Hλ =? ,
null
Α.11)
with transversality conditions:
lim 0 ( ) 0
tT
xTλλ
→
=, ≥ . (Α.12)
Theorem A-8 (Special Model II). Let
nknk n
xug∈ , ∈ ,: × ×→ and
nk
f :××→ . For problem
0
0
()
00
00
()max[()()]
st ( ) [ ( ) ( ) ]
() () 0
T
tt
T
u t
J xxt f xt ut e dt
xt gxt ut t
xt x xT
θ??
,, ≡ ,
.. = , ,
=, ≥
∫
null
define the Hamiltonian as
() ( )H f xu g xutλ=,+? ,, .
Under certain differentiability conditions, if u
?
is a solution, then there exists a function
0
[]
n
tTλ :, → such that u
?
is a solution of
0
u
H =, (Α.13)
x
Hλθλ= ? ,
null
Α.14)
with transversality condition
lim 0 ( ) 0
t
tT
xe T
θ
λλ
?
→
=, ≥ . (Α.15)
Example A-5. Consider consumer’s problem:
0
0
0
()max ()
st
(0)
t
c
J Auced
ArAy c
AA
ρ?
≡
.. = + ?
=.
∫
null
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3. Envelope Theorem
Theorem A-9 (Envelope). Suppose fX A:×→ is differentiable,
n
X ? ,
k
A? , and
()x a
?
is an interior maximum point of
() max ( )
xX
Fa f xa
∈
≡ ,.
Then,
()
() ( )
.
xxa
dF a f x a
da a ?
=
? ,
=
?
Example A-6. For Example A-1, find
()Fab
a
? ,
.
?
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4. The Nash Bargaining Solution
There are two players 1i = and 2. Let X be a set of potential bargaining outcomes and
D be the outcome (the disagreement outcome) if the bargaining fails.
An agreement x X
?
∈ is the Nash solution if
( )
( )
*
*
If , ; ,1 for some , [0, 1] and ,
then , ; ,1 for .
i
j
x pD p x i N p x X
xpD p x j i
?∈∈∈
?≠
(Α.16)
Theorem A-10. There exists a unique Nash solution. And, an agreement x X
?
∈ is a Nash
solution iff it is the solution of the following problem:
[ ][ ]
112 2
max ( ) ( ) ( ) ( )
xX
ux uD ux uD
∈
??, (Α.17)
where
i
u is the expected utility index of
i
V for 12i =,.
Example A-7. Suppose the two players are risk neutral and ( )
ii
uD r=,where
i
r is the
reservation value of player i. Assume the size of the pie is R, called the revenue. Then, for
12
()x tt=, with
12
tt R+=,the utility values are
11
()ux t= and
22
()ux t=.The Nash
solution is
()
12
1
2
ii
tr Rrr
?
=+ ?? .
A generalized Nash solution is
( )
12iii
tr Rrrθ
?
=+ ?? ,
where 0
i
θ ≥ and
12
1θθ+=.
i
θ is the bargaining power of player i. Obviously, we can
extend this formula to a case with n individuals.
