The Envelope Theorem
3
604 CHAPTER A2 THEOREM A2.22 The Envelope Theorem C onsider the problem (A2.35) when there is just one constraint and suppose the objective function, f , and the constraint function, g, are continuously differentiable in (x, a) on an open subset W × U of R n × A. For each a ∈ U , suppose that x(a) ∈ W uniquely solves (A2.35), is continuously differentiable in a on U , and that the constraint g(x(a), a) ≤ 0 is binding for every a ∈ U . L et L(x, a, λ) be the problem’s associated Lagrangian function and let (x(a), λ(a)) solve the K uhn-Tucker conditions in Theorem A2.20. F inally, let V (a) be the problem’s associated value function. Then, the Envelope theorem states that for ever y a ∈ U, ∂ V (a) ∂ a j = ∂ L ∂ a j x(a),λ(a) j = 1,..., m. where the right-hand side denotes the partial derivative of the Lagrangian function with respect to the parameter a j evaluated at the point (x(a), λ(a)). T he theorem says that the total effect on the optimised value of the objective function when a parameter changes (and so, presumably, the whole problem must be reoptimised) can be deduced simply by taking the partial of the problem’s L agrangian with respect to the parameter and then evaluating that derivative at the solution to the original problem’s first-order K uhn-Tucker conditions. A lthough we have confined ourselves in the statement of the theorem to the case of a single constraint, the theorem applies regardless of the number of constraints, with the usual proviso that there be fewer constraints than choice variables. B ecause of the importance of this theorem, and because it is not so obviously true, we will work through a rather extended proof of the version given here. Proof: First, form the L agrangian for the maximisation problem: L ≡ f (x, a) − λ[g(x, a)]. B y hypothesis, x(a) and λ(a) satisfy the first-order K uhn-Tucker conditions given in T heorem A 2.20. T herefore for every a ∈ U , and because the constraint is binding, we have, ∂ f (x(a), a) ∂ x i − λ(a) ∂ g(x(a), a) ∂ x i = 0, i = 1,..., n g(x(a), a) = 0. (P.1) T he partial derivative of L with respect to the parameter a j would be ∂ L ∂ a j = ∂ f (x, a) a j − λ ∂ g(x, a) ∂ a j . CALCULUS AND OPTIMISATION 605 If we evaluate this derivative at the point (x(a), λ(a)), we obtain ∂ L ∂ a j x(a),λ(a) = ∂ f (x(a), a) ∂ a j − λ(a) ∂ g(x(a), a) ∂ a j . (P.2) If we can show that the partial derivative of the maximum-value function with respect to a j is equal to the right-hand side of (P.2), we will have proved the theorem. We begin by directly differentiating V (a) with respect to a j . B ecause a j affects f directly and indirectly through its influence on each variable x i (a), we will have to remember to use the chain rule. We get ∂ V (a) ∂ a j = n i =1 ∂ f (x(a), a) ∂ x i ∂ x i (a) ∂ a j chain rule + ∂ f (x(a), a) ∂ a j . Now, go back to the first-order conditions (P.1). R earranging the first one gives ∂ f (x(a), a) ∂ x i ≡ λ(a) ∂ g(x(a), a) ∂ x i , i = 1,..., n. Substituting into the bracketed term of the summation, we can rewrite the partial derivative of V (a) as ∂ V (a) ∂ a j = λ(a) n i =1 ∂ g(x(a), a) ∂ x i ∂ x i (a) ∂ a j + ∂ f (x(a), a) ∂ a j . (P.3) T he final ‘trick’ is to go back again to the first-order conditions (P.1) and look at the second identity in the system. B ecause g(x(a), a) ≡ 0, we can differentiate both sides of this identity with respect to a j and they must be equal. Because the derivative of the constant zero is zero, we obtain n i =1 ∂ g(x(a), a) ∂ x i ∂ x i (a) ∂ a j chain rule again + ∂ g(x(a), a) ∂ a j ≡ 0. R earranging yields ∂ g(x(a), a) ∂ a j ≡− n i =1 ∂ g(x(a), a) ∂ x i ∂ x i (a) ∂ a j .
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The envelope Theorem Microeconomic
Transcript of The Envelope Theorem
604
CH
APT
ERA
2
THEO
REM
A2.
22Th
eEn
velo
peTh
eore
m
Con
side
rtheprob
lem
(A2.35
)whe
nthereis
just
oneco
nstraint
andsupp
osetheob
jective
func
tion,
f,an
dtheco
nstraint
func
tion,
g,areco
ntinuo
usly
diffe
rentiablein
(x,a)
onan
open
subset
W×
Uof
Rn×
A.Fo
rea
cha
∈U,supp
osethat
x(a)
∈W
unique
lysolves
(A2.35
),is
continuo
usly
diffe
rentiablein
aon
U,a
ndthat
theco
nstraint
g(x(a)
,a)
≤0is
bind
ingforeverya
∈U
.Le
t L(x
,a,
λ)be
theprob
lem’s
associated
Lagran
gian
func
tion
andlet(x(a)
,λ(a
))solvetheKuh
n-Tu
cker
cond
ition
sin
Theo
rem
A2.20
.Finally,let
V(a
)
betheprob
lem’s
associated
valuefunc
tion.
Then
,theEnv
elop
etheo
rem
states
that
for
everya
∈U,
∂V
(a)
∂a j
=∂L
∂a j
� � � � x(a)
,λ(a
)
j=1,
...,
m.
whe
retherigh
t-ha
ndside
deno
testhepa
rtiald
erivativeof
theLa
gran
gian
func
tionwith
respecttothepa
rameter
a jevalua
tedat
thepo
int(x(a)
,λ(a
)).
The
theorem
says
that
thetotaleffect
ontheop
timised
valueof
theob
jectivefunctio
nwhenaparameter
changes(and
so,p
resumably,the
who
leprob
lem
mustb
ereop
timised)
canbe
dedu
cedsimplyby
taking
thepartialo
ftheprob
lem’s
Lagrang
ianwith
respectto
theparameter
andthen
evaluatin
gthat
derivativ
eat
thesolutio
nto
theoriginal
prob
lem’s
first-order
Kuh
n-Tu
cker
cond
ition
s.Alth
ough
wehave
confi
nedou
rselvesin
thestatem
ent
ofthetheorem
tothecase
ofasing
leconstraint,thetheorem
appliesregardless
ofthe
numberof
constraints,with
theusualp
roviso
that
therebe
fewer
constraintsthan
choice
variables.Because
oftheim
portance
ofthis
theorem,a
ndbecauseitis
notso
obviou
sly
true,w
ewill
workthroug
harather
extend
edproo
fof
theversiongivenhere.
Proo
f:First,form
theLagrang
ianforthe
maxim
isationprob
lem:
L≡
f(x,
a)−
λ[g(
x,a)
].
Byhy
pothesis,x(a)
and
λ(a
)satisfy
thefirst-order
Kuh
n-Tu
cker
cond
ition
sgivenin
Theorem
A2.20
.Therefore
foreverya
∈U,andbecausetheconstraint
isbind
ing,
we
have,
∂f(x(a)
,a)
∂x i
−λ(a
)∂g(x(a)
,a)
∂x i
=0,
i=1,
...,
n
g(x(a)
,a)
=0.
(P.1)
The
partiald
erivativeof
Lwith
respecttotheparameter
a jwou
ldbe
∂L
∂a j
=∂f(x,
a)a j
−λ
∂g(x,
a)∂a j
.
CA
LCU
LUS
AN
DO
PTIM
ISAT
ION
605
Ifweevaluate
thisderivativ
eat
thepo
int(x(a)
,λ(a
)),w
eob
tain
∂L
∂a j
� � � � x(a)
,λ(a
)
=∂f(x(a)
,a)
∂a j
−λ(a
)∂g(x(a)
,a)
∂a j
.(P.2)
Ifwecanshow
that
thepartiald
erivativeof
themaxim
um-value
functio
nwith
respectto
a jisequaltotherigh
t-hand
side
of(P.2),wewill
have
prov
edthetheorem.
Webeginby
directly
differentia
ting
V(a
)with
respectto
a j.Because
a jaffects
fdirectly
andindirectly
throug
hits
influ
ence
onea
chvariable
x i(a
),wewill
have
toremem
berto
usethechainrule.W
eget
∂V
(a)
∂a j
=n � i=1
� ∂f(x(a)
,a)
∂x i
� ∂x i
(a)
∂a j
���
�chainrule
+∂f(x(a)
,a)
∂a j
.
Now
,goback
tothefirst-order
cond
ition
s(P.1).Rearranging
thefirston
egives
∂f(x(a)
,a)
∂x i
≡λ(a
)∂g(x(a)
,a)
∂x i
,i=
1,..
.,n.
Substitutinginto
thebracketedterm
ofthesummation,
wecanrewritethepartiald
erivative
ofV
(a)as
∂V
(a)
∂a j
=λ(a
)
n � i=1
� ∂g(x(a)
,a)
∂x i
∂x i
(a)
∂a j
�+
∂f(x(a)
,a)
∂a j
.(P.3)
The
final
‘trick’is
togo
back
againto
thefirst-order
cond
ition
s(P.1)andlook
attheseco
ndidentityin
thesystem
.Because
g(x(a)
,a)
≡0,
wecandifferentia
tebo
thsides
ofthis
identitywith
respectto
a jandthey
mustbe
equal.Because
thederivativ
eof
the
constant
zero
iszero,w
eob
tain
n � i=1
� ∂g(x(a)
,a)
∂x i
∂x i
(a)
∂a j
�
���
�chainrule
again
+∂g(x(a)
,a)
∂a j
≡0.
Rearranging
yields
∂g(x(a)
,a)
∂a j
≡−
n � i=1
� ∂g(x(a)
,a)
∂x i
� ∂x i
(a)
∂a j
.
606
CH
APT
ERA
2
Mov
ingtheminus
sign
into
thebrackets,w
ecansubstitutetheleft-handside
ofthisidentity
fortheentiresummationterm
in(P.3)to
get
∂V
(a)
∂a j
=−λ
(a)∂g(x(a)
,a)
∂a j
+∂f(x(a)
,a)
∂a j
.(P.4)
The
righ
t-hand
side
of(P.4)isthesameas
therigh
t-hand
side
of(P.2).Thu
s,
∂V
(a)
∂a j
=∂L
∂a j
� � � � x(a)
,λ(a
)
aswewantedto
show
.
EXA
MPL
EA
2.11
Let
usseeif
wecanverify
theEnvelop
etheorem.Su
pposewehave
f(x 1
,x 2
)≡
x 1x 2
and
asimpleconstraint
g(x 1
,x 2
)≡
2x1+
4x2−
a.Wearegiven
the
prob
lem
max
x 1,x
2x 1x 2
s.t.
2x1+
4x2−
a=
0,
andwou
ldlik
eto
know
how
themaxim
umvalueof
theob
jectivefunctio
nvaries
with
the(single,
scalar)parameter
a.Wewill
dothis
twoways:
first,w
ewill
derive
thefunc-
tionV
(a)explicitlyanddifferentia
teitto
geto
uransw
er.T
henwewill
usetheEnvelop
etheorem
toseeifwegetthe
samething.
Toform
V(a
),wemustfi
rsts
olve
fortheop
timal
values
ofthechoice
variablesin
term
sof
theparameter.Wewou
ldthen
substitutetheseinto
theob
jectivefunctio
nas
in(A
2.36
)to
getan
expression
forV
(a).Noticethat
this
prob
lem
differsslightly
from
the
onein
(A2.35
)because
wedo
notrequire
non-negativ
ityon
thechoice
variables.Thu
s,we
candispense
with
theKuh
n-Tu
cker
cond
ition
sandjustusethesimpleLagrang
ianmetho
d.Fo
rmingtheLagrang
ian,
weget L=
x 1x 2
−λ[2x
1+
4x2−
a],
with
first-order
cond
ition
s:
L 1=
x 2−
2λ=
0
L 2=
x 1−
4λ=
0(E.1)
L λ=
a−
2x1−
4x2
=0.
These
canbe
solved
tofin
dx 1
(a)=
a/4,
x 2(a
)=
a/8,
and
λ(a
)=
a/16
.Weform
the
maxim
um-value
functio
nby
substitutingthesolutio
nsforx 1
and
x 2into
theob
jective
CA
LCU
LUS
AN
DO
PTIM
ISAT
ION
607
functio
n.Thu
s,
V(a
)=
x 1(a
)x2(a)
=�a 4
��a 8
�=
a2 32.
Differentiatin
gV
(a)with
respecttoawill
tellus
howthemaxim
ised
valueof
theob
jective
functio
nvaries
with
a.Doing
that
weget
dV(a
)
da=
a 16.
Now
letu
sverify
thisusingtheEnvelop
etheorem.T
hetheorem
tells
usthat
tosee
how
themaxim
ised
valueof
thefunctio
nvaries
with
aparameter,sim
plydifferentia
tethe
Lagrang
ianforthemaxim
isationprob
lem
with
respecttotheparameter
andevaluate
that
derivativ
eat
thesolutio
nto
thefirst-order
cond
ition
s(E.1).App
lyingthetheorem,w
efirst
obtain
dV(a
)
da=
∂L ∂a
=λ.
Wethen
evaluate
thisat
thesolutio
nto
(E.1),where
λ(a
)=
a/16
.Thisgivesus
dV(a
)
da=
λ(a
)=
a 16,
which
checks.
Besides
verifyingthat
theEnvelop
etheorem
‘works’,this
exam
plehasalso
given
ussomeinsigh
tinto
whatinterpretatio
nwecangive
tothose‘incidental’variables,
the
Lagrang
ianmultip
liers.T
hisispu
rsuedfurtherintheexercises.
Alth
ough
wehave
confi
nedattentionhere
tomaxim
isationprob
lemsandtheira
sso-
ciated
valuefunctio
ns,itsho
uldbe
clearthat
wecouldalso
constructv
alue
functio
nsfor
minim
isationprob
lemsanalog
ously,andthat
theEnvelop
etheorem
wou
ldapplyforthem
aswell.
A2.
5SE
PARA
TIO
NTH
EORE
MS
Weendthismathematical
append
ixwith
alook
atwhata
recalle
d‘separation’
theorems.
The
idea
isgeom
etrically
simpleenou
gh.Fig.
A2.14
show
stw
odisjoint
convex
sets,A
andB,in
R2 .
Itis
obviou
sthat
wecandraw
alin
ebetw
eenthem
.Suchalin
eis
said
to‘separate’
thetw
osets.Ifwethinkof
thelin
eas
beingdescribedby
theequatio
n,
p 1x 1
+p 2
x 2=
I,
608
CH
APT
ERA
2
x 1
x 2
A
B Figur
eA2.14
.Se
paratin
gconvex
sets.
where
p 1,p 2
,andIarepo
sitiv
econstants,then
everypo
int(
a 1,a 2
)∈
Aissuch
that
p 1a 1
+p 2
a 2>
I,
andeverypo
int(
b 1,b 2
)∈
Bissuch
that
p 1b 1
+p 2
b 2<
I.
Con
sequ
ently
,if
p=
(p1,
p 2),
then
weseethat
thegeom
etricno
tionof
separatio
nis
expressedanalytically
as,
p·a
>p
·b,foreverya
∈Aandeveryb
∈B.
Imagineno
wtw
odisjoint
convex
setsin
R3 ,
sayasphere
andabo
xwith
thesphere
entirelyou
tsidethebo
x.Again,itis
obviou
sthat
wecanseparate
thetw
osets,thistim
ewith
aplane,andan
identic
alanalyticexpression
,but
nowwith
allv
ectorsin
R3 ,
describes
thesituation.
The
separatio
ntheoremsbelowgeneralisethisto
anynu
mbero
fdim
ension
s.Wewill
prov
idetw
otheorems.The
second
theorem
strictly
generalises
thefirstandallowsthesets
AandBto
be,for
exam
ple,op
enand‘tangent’toon
eanother.
THEO
REM
A2.
23A
Firs
tSep
arat
ion
Theo
rem
Supp
osethat
Cis
aclosed
andco
nvex
subset
ofR
nthat
does
notco
ntaintheorigin,0.
Then
,the
reexists
avector
p∈
Rnof
leng
thon
ean
dα
>0such
that,
p·c
≥α,foreveryc
∈C
.
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