( ; 2
) -Se t i n t he Markowi tz Port fol i o Sel ect i on
Met hod J. K rien s 3 L .W .G. Str ij b osch 3 J. Voros 33 3
Tilburg Univers ity, P.O.Box 90 15 3, 500 0 LET ilburg, The Netherlands
DIFFERENT IABILITY PROPERTIES O F THE EFFICIENT (; 2
)
SET IN T HE MARKOW ITZ PORTFOLIO S EL EC TIO N METHO D
1 Introduct io n
The sta ndard portfolio selectio n problemwith linear cons traints may b e formula ted as
follow s. An inves torwants to investa n a mo unt o f o neun it in the securities 1 ;...;n. If
heinves ts an amount x
j
in s ecurity j(j =1;...;n ) the x
j
sho uld satisfy the conditio ns
AX =B; (1.1)
X O (1.2)
with A an (m2n)- ma trix w ith full rank, B an m- vecto ra nd X 0 = (x 1 ;...;x n ); (1.1)
includes the cond ition
n X j= 1 x j =1: (1.3)
The yea rly return on one do llar invested in security j equa ls r
j with j = Er j ; the
covariancematrix ofthe randomvariablesr
j
is C. Theyea rly return r (X)on ap ortfo lio
X equa ls r (X)= n X j= 1 x j r j ; (1.4) with M 0 = ( 1 ;...; n
), the exp ected yearly return Er(X) equals M 0 X and will b e denoted by (X),s o (X)=M 0 X; (1.5) the variance 2 (r(X))equals X 0
CX and willb e deno ted by 2
(X),so
For equivalent fo rmula tio nso f the conditio ns (1.1 ), (1.2) cf. H.M. Markowitz (198 7) p .
24 -27 , forno nlinear cons traints J. Kriens a nd J.Th. va nLieshou t (19 88).
A fea sible portfolio
X is ca lled ecientif it is aso lu tion of both
min X f 2 (X)j(X)( X)^A X =B ^ X O g (1.7) and max X f(X)j 2 (X) 2 ( X)^A X =B^ X O g: (1.8)
All ecient portfolio s can b e derivedby computing
min X fX 0 CX 0M 0 XjA X =B^ X O g (1.9)
for all 0; cf. H.M. Ma rkowitz (19 59) p. 31 5-3 16, or fo r a precise a nd more g en era l
sta tement of the theorem u nderlying the algorithm J. Kriens and J.Th. va n Lies hout
(19 88 ). WithU 0 =(u 1 ;...;u m )and V 0 =(v 1 ;...;v n
)as Lag rang emultipliers of(1.1 )a nd (1 .2 ),
resp ectively,the Kuhn -Tucker co nditions of (1 .9 ) run
02CX 0A 0 U+V =0M (1 .1 0) AX =B (1 .1 1) V 0 X =0;X O ;V O ;U free: (1 .1 2)
fol-andnextraisetog et(new)ecientp o rtfolio s. Forspeci cva lueso fthereisach ange
in th e ba sis . Let thes e va lues be 1 ;...; k
, the corresp o ndin g ecient solutio ns b e
X 1 ;...; X k
withmean-va ria nceco mbinations(( X 1 ); 2 ( X 1 ));...;(( X k ); 2 ( X k )). The sequen ce X 1 ;...; X k
is ca lledthe setof corner portfolios, the seto f all(( X); 2 ( X)) p o intsinthe (; 2
)-planecorresponding to ecientp ortfo lios
X isthe s et of ecient
(; 2
) combina tions of the pro blem,o r the ecien t frontier.
This las tset sa tis es the fo llowing prop erties:
a. b etween th e (; 2
) p oints o f two adjacentcorner portfolio s X i and X i+ 1 (6= X i ) it
is part of astrictly co nvex parabola;
b. on the interio r ofthe s eg mentsmentioned ina , the relatio n
d 2 d ! (; 2 ) = (1 .1 3)
holds; itis strictly increasing as afunction o f ;
c. in the (; 2
) points corresponding to corner p o rtfo lios, the left ha nd deriva tive
d 2 d L
and th eright ha nd derivative d 2 d R
exist and satisfy
d 2 d ! L d 2 d ! R : (1 .1 4)
Fromb it fo llows th at on tho se segments th ere is ao ne to one corresp o ndence between
the va lues of
a nd . In co rn er p o rtfo lios this is o nly true if d 2 d L = d 2 d R , which
implies di erentiability of the (; 2
) curve. For proofs cf. H.M. Ma rkowitz (1 987 ), p .
17 6 a nd J. Kriens and J.T h. van Lies hout (1 988 ).
Section 2 of this paper co nta ins a more precise discussion of the algorithm to so lve
(1.10 ),...,(1.12 ) for every 0 , s ectio n 4 neces sary and sucient conditio ns for the
equ alitysign in(1.14 ). In prepa rationfor thesecondtopic wepresentaslig htlya dapted
formo f the explicit formula e fo r X; ( X) a nd 2 ( X) a s derived by J.Kriens a nd J.TH.
van Lies hout (1 98 8) insectio n 3.
Section 5 compares with other literatureand section 6 considers the sta ndard p ortfo lio
2 The alg orithm
Ino rdertopres entamorep recis edis cus sio no fthea lg orithmwe rstprovethefo llowing
lemma.
Lemma 2.1
If in a portfolio selectio npro blem
1) 8 j 2 (r j )>0
2) therea re no linea r relations b etween the returns r
j , p o rtfolio s X 1 a nd X 2 (6= X 1 ) with ( X 1 )= ( X 2 ) and 2 ( X 1 ) = 2 ( X 2
) can not b e
e-cient. Proof Let X = X 1 +(10 ) X 2 (0 < <1); then r ( X)= r( X 1 )+(10 )r( X 2 ) ( X)=( X 1 )=( X 2 ) 2 (r ( X))= 2 2 (r( X 1 ))+2 (10 )(r ( X 1 ))(r ( X 2 ))+ (10 ) 2 2 (r ( X 2 ))= 2 ( X 1 )[ 2 +2 (10 )+(10 ) 2 ]= 2 ( X 1 )f( ): For 6=1 ;f( )<1fo r 0< <1,so 2 ( X)< 2 ( X 1 ) and X 1 and X 2 a ren otecient. (r ( X 1 );r( X 2 )) = 1 i all realizations of (r ( X 1 );r( X 2
)) are situated o n a s traig ht line,
so allp oints P n j=1 x j1 r j ; P n j= 1 x j2 r j
are on astra ig ht line. Thismea ns
9 a 9 d 8 R 0 @ n X j=1 x j2 r j 1 A =a+d 0 @ n X j=1 x j1 r j 1 A : Let 8 j a j =d x j1 0x j2 ;
8 R a+ n X j=1 a j r j =0: Wediscern fo urcases: a ) 8 j a j = 0 ) 8 j dx j1 = x j2 , leading with (1.3) to d = 1 an d X 1 = X 2 , which contradicts X 1 6= X 2 ; b ) a i 6= 0;8 j6= i a j = 0 ) 8 R a+a i r i = 0 a nd r i is xed, so 2 (r i ) = 0 , which contradicts conditio n1); c) a i 6=0;a k 6=0;8 j6= i;k a j =0)8 R a+a i r i +a k r k
=0 , whichcontradictscond ition
2);
d ) Morethan twoa
i
6=0 ; con clusion a sund er c).
So (r ( X 1 );r( X 2
))6=1 and the lemma is proved.
Remark 2.1. Fro m the proo f it fo llows that conditio ns 1) and 2) are also necess ary.
Moreoverthe conditio ns1 ) and 2) h oldi C is p os itivede nite.
Fromlemma2.1itisclea rth atfo rC positivede nitethecornerp o rtfo lios X 1 ;...; X k are
uniqu ely determin ed. However, there are not always as many di erent cornerportfolios
as there are di erentba ses du ring the computa tion s;di erent bases may yield the sa me
p o rtfolio a nd also di erent va lues
i
may yield th e sa me p o rtfolio. In this resp ect the
nota tio nin sectio n1 is mislea ding .
Starting the a lg orith m with = 0 and next raising , the a lg orith m produces a series
of ba ses. Ba ses which ho ld for just o ne va lue of a re dro pp ed so that only bases
corresp o nding to nond eg enerate -intervals a releft.
Denote fo ra g iven basis of th e system(1.10) ,..., (1.12), th e setof bas icx-va ria blesby
(X
b )
i
. In section 3we will s how tha t the values(
X
b )
i
of the basic x- variables sa tisfy
( X b ) i =A i +D i (2.1) fora ll
inthe corresp o ndin ginterval;thecons tantsA
i
a nd D
i
explic-Withco rn erp o rtfoliosthereco rresponda tlea sttwovectorsD
i
,thevectorcorresponding
to the "old" ba sis a nd the vector corresponding to the "new" ba sis . But there may b e
more a ssocia tedvectorsD
i
, either b eca use thereexis ts an equivalentbas isfor the " old"
or forthe "new"ba sispro d ucingthe sameco rnerportfolio(
X
b )
i
, orbecaus ethe seriesof
vectorsD
i
containso neo rmore vectors D =O. Inthe la tterca sethe s amevector(
X
b )
i
isproduced fordi erentvalues o f . If the" new "ba sis isuniquely determined, then for
ecient p o rtfo lios w hich are n otco rnerportfo liosthe vecto rD
i
3 Explicit expressions f or ecient po rtfolios
Starting fromthe Kuhn-Tucker co nditions fo r the s olutio no f (1 .9),J. Kriens and J.TH.
van Liesh out (19 88 ) derive an expression for the ba sic variables which, if C is p o sitive
de nite,ho lds foreveryecientp o rtfolio. Wepresenttheirresults inaslightlya dapted
form. For a xed value of (1.10 ) ,..., (1 .1 2) run 02CX 0A 0 U+V =0 M (3.1) AX =B (3.2) V 0 X =0;X O ;V O ;U free: (3.3)
The equations (3.1) a nd (3.2) ca n b e s ummarized as
X 0 U 0 V 0 02 C 0A 0 J 0 M A O O B (3.4) If Z 0 b =(X 0 b ;U 0 ;V 0 b ) (3.5)
denotesas eto fbasicva ria blesfo r agivenecientp o rtfolio(3 .4 )ca nb epartitio nedinto
X 0 b X 0 nb U 0 V 0 b V 0 nb 02C b 1 02 C nb 1 0A 0 b O J 0 M b 02C b2 02 C nb2 0A 0 nb J O 0 M n b A A O O O B (3.6)
Thematrix02C ispa rtitionedintothe s qua rematrices02 C
b1
a nd 02C
n b2
corresponding
to basic a nd no n-bas ic x-va ria bles and into 02 C
b 2 and 02C n b 1 with C b 2 = C 0 nb 1 :A b ;M b and A nb ;M nb
also co rres p o nd to ba sic and non- basic va ria bles res p ectively. T he matrix
of coecients o f bas ic va riab les is
B= 0 B B B @ 02C b1 0A 0 b O 02C b 2 0A 0 nb J A b O O 1 C C C A : (3.7)
To facilita te co mpu tatio ns Kriens a nd van Lies hout reshue (3.7) into
B v = 0 B B B @ 02C b 1 0A 0 b O A b O O 02C b2 0A 0 nb J 1 C C C A : (3.8)
The va lues ofthe ba sic va ria blesa re
Z b = 0 B B B @ X b U V b 1 C C C A =B 01 v 0 B B B @ O B O 1 C C C A 0 B 01 v 0 B B B @ M b O M nb 1 C C C A : (3.9)
We nd explicit expressions for these va lues by computingB 01 v : B 0 1 v = 0 B B B B B B B B B B B B B @ 0 @ 02C b 1 0A 0 b A b O 1 A 0 1 j O j 0000000000000000000000000 j (2C b2 A 0 nb ) 0 @ 02C b 1 0A 0 b A b O 1 A 01 j J 1 C C C C C C C C C C C C C A (3 .1 0) with 0 @ 02C b 1 0A 0 b A b O 1 A 0 1 = (3 .1 1) 0 @ 0 1 2 C 0 1 b 1 + 1 2 C 0 1 b 1 A 0 b (A b C 01 b 1 A 0 b ) 01 A b C 01 b 1 C 01 b 1 A 0 b (A b C 01 b 1 A 0 b ) 01 0(A b C 0 1 b A 0 b ) 0 1 A b C 01 b 02 (A b C 01 b A 0 b ) 01 1 A :
Subs tituting(3 .1 1) into (3 .10 ) an d the result into (3.9 ),we nd X b =A+D (3 .1 2) with A=C 0 1 b 1 A 0 b (A b C 0 1 b 1 A 0 b ) 0 1 B (3 .1 3) and D = 1 2 [C 0 1 b1 0C 01 b1 A 0 b (A b C 01 b1 A 0 b ) 01 A b C 0 1 b1 ]M b : (3 .1 4)
The corresponding valu es ( X b ) and 2 ( X b )are ( X b )=M 0 b A+M 0 b D (3 .1 5) 2 ( X b )=A 0 C b1 A+D 0 C b1 D 2 (3 .1 6)
(note tha t the co ecient of equa ls0). If the vecto r 0 @ M O 1 A
is linear indep endentof the ba sis (3.7),it can be shown that
M 0
b
:D 6=0 : (3 .1 7)
ToprovethisKriensandvanLieshoutstudy p roblem(1 .7 )withAX B. Withobvio us
ada ptationsin th en otation, the Kuhn- Tucker cond itions ofthis prob lemare in ourcase
02CX 0A 0
M 0 X 0y m+1 = (3 .2 0) X 0 V =y m+1 :=0;X O ;V O ;y m+ 1 0;0;U free: (3 .2 1) Becaus e 0 @ M O 1 A
isa ssumedtob elinearindependento fB
v
,thevecto rZ
b
(3.5)completed
with ,fo rmsaba sicsolutio no f(3 .18 ),...,(3 .21 ). Reorderinginthesameway asin(3.8)
the matrixof ba sicvectors cha nges into
B 3 v = 0 @ B v K L 0 O 1 A (3 .2 2) with L 0 =(M 0 b O 0 O 0 ) (3 .2 3) and K 0 =(M 0 b O 0 M 0 nb ): (3 .2 4)
Using the existenceof (B 3 v ) 0 1 , (3.17 ) can be proved. Rema rk 3.1 . T he co ndition 0 @ M O 1 A
linear indep endent of the ba sis (3.7) is
incor-rectly suppressedbyJ. Kriensa nd J.TH.va nLieshou t(198 8). J.Kriens(19 89 )provides
4 Necessary and sucient conditions fo r
di eren-tiability of the ecient f rontier
Becaus e o f p ro p erty b in section 1 we can restrict the dis cus sio n to the p o ints
(( X i ); 2 ( X i
)), in the sequel to be d eno ted by (
i ;
2
i
). Furthermore we o nly discuss
non deg en era temodels.
Con dition 1.
The ecientfrontier(e.f., forshort) is di erentia bleinth ep oint (
i ; 2 i )i on evalue corresp o nds to it. Proof.
Follow sdirectly fro m (1.13) and(1 .1 4).
Con dition 2.
The e.f. is di erentiable in the point (
i ; 2 i ) i no co rresponding X b - vecto r ca n b e re-presented by (2 .1) with D =O . Proof.
Necessary: D =O implies the same vecto r
X
b
and thus the samep o int
i ;
2
i
) for more
than one va lueo f . Sucient: D 6=O ; 1 6= 2 ) X( 1 )6= X( 2 ) a nd s o d i erentpoints (; 2 ), cf. lemma 2.1. Con dition 3.
The e.f. is di erentiable in the point (
i ; 2 i ) i no co rresponding X b - vecto r ca n b e re-presented by (2 .1) with M 0 b :D =0 . Proof. Follow sfrom D 6=O ! M 0 b :D 6=0. ! if chang es , X b chang es and ( X b
)mus t cha nge (lemma2 .1),so M 0
b
:D 6=0 (cf.
(3.15 )).
Con dition 4.
The e.f. is di erentia ble inthe point (
i ; 2 i ) i (B 3 v ) 0 1 exists. Proof. Follow sfrom (B 3 v ) 0 1 exis ts ! M 0 b :D6=0 .
! seeJ. Kriens a nd J.TH.van L iesho ut(19 88 ) p. 1 90- 191 .
if M 0 b :D 6=0 ,a llelementsof (B 3 v ) 01 exis t and B 3 v :(B 3 v ) 01 =J. Con dition 5.
The e.f. is di erentia ble in the p oint (
i ; 2 i ) i 0 @ M O 1 A
is linea r in depen dent of the
vectors of B v . Proof. 0 @ M O 1 A
linearindep endent of the vectors of B
v ! inverseo f B 3 v exists.
5 Relat io ns with statements on di erentiability in
the literature
The theorems tated by J. Voros (198 7) an d J.Krien s (19 89)a re eas ily checkedthro ugh
app lyin g the conditio ns o f section 4. We combine these theorems in one new theorem.
De ne min :=min i i ; max :=ma x i i , M=(m ij ):=C 01 b (5.1) f := k X i=1 k X j=1 m ij (5.2) d:= k X i=1 ( k X j=1 m ij j ): (5.3) Theorem 5.1
Ifintheinvestmentp roblems ubjectto (1.2)and(1.3),C pos itivede nite,acorner
port-folio with 2 (
min ;
max
) has k( 1)x-va ria bles in th e basis, then the set of ecient
(; 2
) points is no ndi erentiable in the co rresponding ( ; 2
) p o int if a nd onlyif there
exists arepresenta tio n of X : b =(x;...;x k ) with 8 1 i:jk i = j . Proof.
Wedisting uish between k =1 and k >1 .
Sucient. k = 1. Sup p os e x i > 0, then x i = 1;C b 1 = (c ii );A b = (1);M b = ( i ). Substitutio n of
these values into (3.14 ) lea ds to
D = 1 2 h C 0 1 b 1 0C 0 1 b 1 A 0 b (A b C 0 1 b 1 A 0 b ) 01 A b C 01 b 1 i M b = 1 2 c 0 1 ii h 10(c 0 1 ii ) 01 c 01 ii i i =0: (5.4)
X b = 0 B B B @ x 1 . . . x k 1 C C C A ;C b1 = 0 B B B @ c 11 ... c 1k . . . . . . c k 1 ... c k k 1 C C C A ;A 0 b = 0 B B B @ 1 . . . 1 1 C C C A ;M b = 0 B B B @ 1 . . . k 1 C C C A ; then (A b C 01 b 1 A 0 b ) 0 1 = 1 f (5.5)
and D can berew ritten a s
D = 1 2 M 2 6 6 6 4 J 0 1 f 0 B B B @ P i m i1 ... P i m ik . . . . . . P i m i1 ... P i m ik 1 C C C A 3 7 7 7 5 0 B B B @ 1 . . . k 1 C C C A : (5.6) If 1 =...= k
, then D =O and co ndition 2lea dsa ga in to no ndi erentia bility.
Necessary.
k =1. Trivia l.
k >1. Ifthereisno ndi erentia bility thenthereexistsarep resentationwithD =O . For
this vector (5 .6 ) is equ ivalent to
2 6 6 6 4 J 0 1 f 0 B B B @ P i m i1 ... P i m ik . . . . . . P i m i1 ... P i m ik 1 C C C A 3 7 7 7 5 0 B B B @ 1 . . . k 1 C C C A = 0 B B B @ 0 . . . 0 1 C C C A ; (5.7) or 0 B B B @ 1 . . . k 1 C C C A = 1 f 0 B B B @ P j ( P i m ij ) j . . . . . . P j ( P i m ij ) j 1 C C C A = 0 B B B @ d f . . . d f 1 C C C A ;
so nond i erentia bility implies
1
=...=
k .
Remark 5.1
(1.1) co nta ins two or mo re indep endent constra ints.
Remark 5.2
Theorem5.1 co mbines the theorems 5 .1 and 5.2 in J. Kriens (1 989 )a nd g eneralizesthe
case k >1 to situa tio ns in whichthe basis co nta ins x-va riab lesw ith value 0. T hetheo
-remals og en era lizesth eo rem2 by J.Voro s (1 987 ).
Remark 5.3
6 The standard portf olio select io n pro blem with
one riskless asset.
The stan dard p o rtfolio s election prob lem w ith conditio ns (1.2) an d (1 .3 ) ca n also b e
formula ted as min X 2 (X)=X 0 CX (6.1) subject to X O (6.2) n X j= 1 x j =1 (6.3) M 0 X =; (6.4)
using a s a para meter;the o ptimal so lution is denoted as
X( ).
Now,considerthesta ndard p ortfo lioca sewithone risklessass et: minimize(6 .1 )subject
to (6.2), n X j= 1 x j +y=1 (6.5) M 0 X +i y=; (6.6)
where y is the sha re of ca pita linvested in the riskless ass et and i is the rate o f interes t;
we allowy to be p os itive, 0o r negative.
We can eas ily sta te that for = i the o ptimal solutio n ru ns y = 1 ; X(i) = O w ith 2 (r(
X(i))) = 0. Thus we ca n res trict to the case > i; furthermo re we ass ume
i < ma x j f j g. L et a ga in X 0 = (x 1 ;...;x k
X 0 n b =(x k+1 ;...;x n
) the set of non -bas ic x- variables . Denote the La gran ge multipliers
of (6.5) and (6.6) by u
1
and respectively, and let I
n = 0 B B B @ 1 . . . 1 1 C C C A
with n elements. The
Kuhn-Tuckerequations for the pro blem(6.1), (6.2), (6.5), (6.6) are:
2C b 1 X b +I k :u 1 0M b =O (6.7) 2C b2 X b +I n 0k :u 1 0M nb O (6.8) 0u 1 +i =0 (6.9) X O (6.2) I 0 k X b +y=1 (6 .1 0) M 0 b X b +iy =: (6 .1 1) From(6.7) we have X b =0 1 2 u 1 C 01 b 1 I k + 1 2 C 01 b 1 M b : (6 .1 2) With(5.2), (5.3) and e:= k X i= 1 k X j= 1 m ij i j (6 .1 3)
we ca n derive from(6.10 ) and (6.11)
I 0 k X b =0 1 fu 1 + 1 d =10y (6 .1 4)
M 0 b X b =0 1 2 du 1 + 1 2 e=0iy: (6 .1 5) Lemma 6.1
The expres sio n fi 2
02d i+e isa lways positive,except in the cas e8
i2f1;...;k g i =i. Proof (M b 0iI k ) 0 C 01 b 1 (M b 0iI k )=fi 2 02d i+e=0i M b =iI k 0 ! 8 i2f1;...;kg i =i (cf. also J. Voros (19 87)). As8 i2f1;...;k g i
=iimplies =i,th ecase weexclu ded,fi 2
02 di+eisalways >0in our
mo del.
Lemma 6.2
For ag ivenseto fba sicx- variables X
b
the p roblem(6 .1 ),(6.2), (6.5 ),(6 .6) has aunique
so lution.
Proof
Elimin atin g y from(6 .1 4),(6.15) and u sing (6.9) we n d
= 2 (0i) fi 2 02di+e (6 .1 6) and u 1 = 2i(0i) fi 2 02di+e : (6 .1 7)
Fromthese equa tio ns a nd (6 .12 ) itfollow s that the solutio nis unique.
In the remainder of this s ectio n we exploit the well- known property that in the ( ;
)-plane the e.f. o f the model (6 .1 ), (6.2), (6.5), (6 .6 ) is a stra ight line thro ugh the p o int
=i; =0 which to uches the e.f. of the risky ass ets of the model (6.1),...,(6.4) if this
e.f. is di erentia ble (cf. e.g. Th.E . C opeland and J.F. Weston (198 8) p . 17 9-1 80 ). This
set of ba sic va riab les X
b
. Formu lae (6 .12 ) and (6 .8 ) provide a simple procedure for
deriving th e co rner p o rtfo lios of the risky a ssets. Therefo re we rewrite (6.12 ) and (6.8)
by substituting (6 .1 6) a nd (6.17 ) into C 0 1 b1 M b 0iC 0 1 b1 I k O (6 .1 8) C b 2 C 01 b 1 M b 0M n b +i(I n 0k 0C b 2 C 01 b 1 I k )O : (6 .1 9)
The a lg orith mru ns a s follow s.
Step 1: Determin e ma xf
j
g and ll up the sets X
b
and X
n b .
Step 2: Find the sma llest value of i fo r which (6.1 8) a nd (6 .1 9) ho ld.
Step3: Ifi=01then stop. Otherw iseremovethe variablefromX
b into X n b ifX b O
gives th es ma llest i,o r invers ely. Repea t step 2 .
If we apply this algo rithmto the well- know n Markow itz example
M = 0 B B B @ 1 3 5 1 C C C A C = 0 B B B @ 3 3 01 3 11 23 01 23 75 1 C C C A ; then 3 =ma xf j g;X b 1 =(x 3 ) and we nd su ccessively X b 2 = 0 @ x 2 x 3 1 A ;X b 3 =(x 2 );x b 4 = 0 @ x 1 x 2 1 A ;X b 5 = 0 B B B @ x 1 x 2 x 3 1 C C C A ;X b b = 0 @ x 1 x 3 1 A :
The expressions (6.7 ) (or (6.12)) a nd (6.8 ) are a sp ecia l form of the equations (3 .1 ).
Raising the value
of fro m 0 to the la rgest releva nt value in the s tanda rd a lg orithm
is the same as lowering i = u
1
fro m the largest releva nt value to 01 in the a lg orithm
justp resented. Bothalgo rithmsproduceexa ctlythesa mes teps,a lbeitinarevers eo rder.
Another pro of of theorem 5.1.
Ifthe e.f. of the riskyass ets isnon di erentia ble inapointP there ared i erent interest
ratesifro mw herewecandrawsubg radientstoP. Fo rthisintervalofva luesithereturn
on the p o rtfo lio of risky assets is the same i.e. in depen dento f i. Using (6.15)weg et
M 0 b X b =0 1 2 du 1 + 1 2 e= (6 .2 0)
with the exp ected return of the co rresponding p o rtfolio. We su bstitute (6 .1 6) and
(6.17 ) for a nd u
1
to n d
(d0f )i0(e0d)=0: (6 .2 1)
This can hold for the whole interval of i-valu es i
d0f=0a nd e0d =0 (6 .2 2)
fromwhich follows
f 2
02 d+e=0 (6 .2 3)
and with lemma 6.1:
i = j = fo r all x i ;x j 2X b .
Co nsid ering the ca se
i = j = fora ll x i ;x j 2X b we nd with (5.2), (5.3),(6.13 ) e= 2 f and d=f (6 .2 4) and with (6.12 ) X b =0 1 2 (u 1 0)C 01 b1 I k : (6 .2 5)
Subs titutiono f the equa tio ns(6 .1 6) a nd (6.17 ) lea dsto
X b = 1 f C 01 b 1 I k ; (6 .2 6)
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