The generalized opportunity process for utility
maximization and applications
Victor Nzengang
Humboldt-Universität zu Berlin
Joint work with Peter Imkeller (HU Berlin)
Berlin-Paris Young Researchers Workshop
Berlin, November 2016
The utility maximization problem
The opportunity process for power utility
The generalized opportunity process and main results
Stability of utility maximization
Financial Market
I
I
I
T
> 0 finite time horizon
Ω, G, F = (Ft )0≤t≤T , P filtered probability space.
n stocks with discounted semimartingale price process
S = (S 1 , · · · , S n ).
Me (S) := {Q ∼ P : S is a Q − local martingale} .
Assumption: Me (S) 6= ∅.
Utility maximization problem
I
x > 0: initial capital of an agent in the market.
I
U : (0, +∞) → R a strictly increasing, concave and
0
C 1 -function with AE [U] = lim supz→+∞
0
U (0) = +∞,
I
zU (z)
U(z)
< 1 and
0
U (+∞) = 0.
X (x) set of admissible wealth processes with initial capital x
Z ·
X (x) :=
X ≥0:X =x+
πdS .
0
b ∈ X (x) such that
Problem: Find X
h
i
bT )
u(x) = max E [U(XT )] = E U(X
X ∈X (x)
Duality approach
I
Convex conjugate to U: V (y ) = supz>0 (U(z) − zy ), y > 0.
I
Y: set of supermartingale deflators for S i.e.
Y := {Y : Y ≥ 0 and YX supermartingale ∀X ∈ X (1)} .
b ∈ Y such that
Dual problem: find Y
h
i
bT ) .
v (y ) := inf E [V (yYT )] = E V (y Y
Y ∈y Y
b and Y
b exist. Moreover,
Kramkov and Schachermayer’99: X
bT = U 0 (X
bT ) and X
bY
b a uniformly integrable (U.I.) martingale.
Y
Muckhenhoupt’s condition and reverse Hölder’s inequality
T (F): set of F-stopping times valued in [0, T ].
I Def: Let Y > 0 be a càdlàg semimartingale, r > 1 and
q ∈ (0, 1). Y satisfies:
I
I
I
Muckhenhoupt’s condition (Ar ) if there exists K1 > 0 s.t. for
every σ ∈ T (F) :
"
1 #
Yσ r −1 E
Fσ ≤ K1 .
YT
reverse Hölder’s inequality (bq− ) if there exists K2 > 0 s.t. for
every σ ∈ T (F):
q YT Fσ ≥ K2 .
E
Yσ Remark: (Ar ) is stronger than (bq− ). If Y satisfies (bq− ) and
E [YT ] < +∞, then Y is of class (D), i.e. {Yσ , σ ∈ T (F)} is
U.I.
Objective and motivation
b to
Objective: give necessary and/or sufficient conditions for Y
−
satisfy (Ar ) or (bq ) for some r > 1 or q ∈ (0, 1).
Motivation:
b is a local martingale satisfying (Ar ) for some
I Assume that Y
bT = Y
b0 d Qb for some Q
b ∈ Me (S).
r > 1. In this case, Y
dP
b is a pricing measure which reflects the agent risk preference
Q
and its initial position.
I
b provide a basis to investigate
The integrability properties of Y
its continuity properties w.r.t. risk preference and initial
capital in appropriate topologies:
bT in the topology of
1. continuity of the terminal values Y
L1 -convergence.
b in the topology of uniform
2. continuity of the process Y
convergence.
Objective and motivation
b to
Objective: give necessary and/or sufficient conditions for Y
−
satisfy (Ar ) or (bq ) for some r > 1 or q ∈ (0, 1).
Motivation:
b is a local martingale satisfying (Ar ) for some
I Assume that Y
bT = Y
b0 d Qb for some Q
b ∈ Me (S).
r > 1. In this case, Y
dP
b is a pricing measure which reflects the agent risk preference
Q
and its initial position.
I
b provide a basis to investigate
The integrability properties of Y
its continuity properties w.r.t. risk preference and initial
capital in appropriate topologies:
bT in the topology of
1. continuity of the terminal values Y
L1 -convergence.
b in the topology of uniform
2. continuity of the process Y
convergence.
Objective and motivation
b to
Objective: give necessary and/or sufficient conditions for Y
−
satisfy (Ar ) or (bq ) for some r > 1 or q ∈ (0, 1).
Motivation:
b is a local martingale satisfying (Ar ) for some
I Assume that Y
bT = Y
b0 d Qb for some Q
b ∈ Me (S).
r > 1. In this case, Y
dP
b is a pricing measure which reflects the agent risk preference
Q
and its initial position.
I
b provide a basis to investigate
The integrability properties of Y
its continuity properties w.r.t. risk preference and initial
capital in appropriate topologies:
bT in the topology of
1. continuity of the terminal values Y
L1 -convergence.
b in the topology of uniform
2. continuity of the process Y
convergence.
Opportunity process for power utility
Let p ∈ (−∞, 0) ∪ (0, 1) and U be given by
zp
, z > 0.
p
e ∈ X (x) : X
es = Xs , s ∈ [0, t] .
For X ∈ X (x), let Xt (X ) := X
U(z) =
Proposition (Nutz’10)
I
There exists a unique càdlàg semimartingale L(p) such that
1
eT )Ft = u(t, Xt ), ∀X ∈ X (x).
Lt (p) Xtp = ess sup E U(X
p
e ∈Xt (X )
X
I 1 Lt (p)x p = u(t, x) and LT (p) =
p
I L(p) is a supermartingale for p ∈
for p ∈ (−∞, 0).
1.
(0, 1) and a submartingale
Opportunity process for power utility (continued)
I
I
b =X
b p−1 L(p) = U 0 (X
b )L(p).
Y
For p ∈ (0, 1) and σ ∈ T (F),

(Lσ (p))
1
1−p
bσ
Y
= E
bT
Y
! 11 
"
11 #
−1
p
−1 Y
p
σ
Fσ  ≤ E
Fσ ∀Y ∈ Y,
YT
L(p) is bounded away from 0 and ∞ if and only if A 1
p
b or some Y ∈ Y.
holds for Y
I
For p ∈ (−∞, 0), q =
(Lσ (p))
1
1−p
"
=E
bT
Y
bσ
Y
p
p−1
∈ (0, 1) and σ ∈ T (F)
!q #
q Y
T
Fσ ≥ E
Fσ , ∀ Y ∈ Y.
Yσ b satisfies
L(p) is bounded away from 0 and ∞ if and only if Y
−
−
(bq ) or (bq ) holds for some Y ∈ Y.
Opportunity process for power utility (continued)
I
I
b =X
b p−1 L(p) = U 0 (X
b )L(p).
Y
For p ∈ (0, 1) and σ ∈ T (F),

(Lσ (p))
1
1−p
bσ
Y
= E
bT
Y
! 11 
"
11 #
−1
p
−1 Y
p
σ
Fσ  ≤ E
Fσ ∀Y ∈ Y,
YT
L(p) is bounded away from 0 and ∞ if and only if A 1
p
b or some Y ∈ Y.
holds for Y
I
For p ∈ (−∞, 0), q =
(Lσ (p))
1
1−p
"
=E
bT
Y
bσ
Y
p
p−1
∈ (0, 1) and σ ∈ T (F)
!q #
q Y
T
Fσ ≥ E
Fσ , ∀ Y ∈ Y.
Yσ b satisfies
L(p) is bounded away from 0 and ∞ if and only if Y
−
−
(bq ) or (bq ) holds for some Y ∈ Y.
Extension to general utility functions
Def: Let a, b, C > 0 with a ≤ b. U satisfies condition (Ga,b,C ) if
0
y b
1 y a
U (x)
≤ 0
≤C
, y ≥ x.
C x
x
U (y )
00
zU (z)
a≤− 0
≤ b for all z > 0 =⇒ (Ga,b,1 ) is satisfied.
U (z)
Theorem ( Kramkov and Weston’16)
Assume that (Ga,b,C ) holds for U with a ∈ (0, 1) and there exists
b satisfies (Ar ) with r = 1 + b .
Z ∈ Y satisfying (A 1 ). Then Y
1−a
1−a
Issues:
I
The necessity for the condition (Ar ) is not addressed.
I
The condition (Ar ) becomes too restrictive for a > 1 and if
b.
one is only interested in the class (D) property of Y
Extension to general utility functions
Def: Let a, b, C > 0 with a ≤ b. U satisfies condition (Ga,b,C ) if
0
y b
1 y a
U (x)
≤ 0
≤C
, y ≥ x.
C x
x
U (y )
00
zU (z)
a≤− 0
≤ b for all z > 0 =⇒ (Ga,b,1 ) is satisfied.
U (z)
Theorem ( Kramkov and Weston’16)
Assume that (Ga,b,C ) holds for U with a ∈ (0, 1) and there exists
b satisfies (Ar ) with r = 1 + b .
Z ∈ Y satisfying (A 1 ). Then Y
1−a
1−a
Issues:
I
The necessity for the condition (Ar ) is not addressed.
I
The condition (Ar ) becomes too restrictive for a > 1 and if
b.
one is only interested in the class (D) property of Y
Generalized opportunity process
b admits a multiplicative decomposition
Observation: Y
b
b
bT = U 0 (X
bT ) =⇒
X Y martingale and Y
h 0
i
bT )X
bT Ft
E
U
(
X
b
b
0
bt = Yt Xt = U 0 (X
bt )
bt )Lt .
Y
= U (X
0 b
b
b
Xt
Xt U (Xt )
I
L := 0Y b is refereed to as the generalized opportunity
U (X )
process.
I
L is càdlàg, strictly positive and LT = 1.
b U 0 (X
b ) = 1.
For U = log, L = 1 since X
I
b
b is based
−→ Our analysis on the conditions (Ar ) and (bq− ) for Y
on L.
Muckhenhoupt’s (Ar )-condition
Theorem
Suppose that there exist a, b, C > 0 with a ≤ b s.t. U satisfies
(Ga,b,C ).
1. If there exists Y ∈ Y satisfying (Ak ) for some k > 1 and L is
b satisfies (Ar ) with
bounded away from 0 and ∞. Then Y
r = 1 + bk.
b satisfies (Ar ) for some r > 1, then L is
2. Assume that Y
bounded away from 0.
3. Assume that a = b ∈ (0, 1). The following are equivalent:
3.1 L is bounded away from 0 and ∞.
b satisfies (A 1 ).
3.2 Y
1−a
1
Comments: If a ∈ (0, 1) and k = 1−a
, then r = 1 +
recover the result of Kramkov and Weston’16.
b
1−a
and we
Muckhenhoupt’s (Ar )-condition
Theorem
Suppose that there exist a, b, C > 0 with a ≤ b s.t. U satisfies
(Ga,b,C ).
1. If there exists Y ∈ Y satisfying (Ak ) for some k > 1 and L is
b satisfies (Ar ) with
bounded away from 0 and ∞. Then Y
r = 1 + bk.
b satisfies (Ar ) for some r > 1, then L is
2. Assume that Y
bounded away from 0.
3. Assume that a = b ∈ (0, 1). The following are equivalent:
3.1 L is bounded away from 0 and ∞.
b satisfies (A 1 ).
3.2 Y
1−a
1
Comments: If a ∈ (0, 1) and k = 1−a
, then r = 1 +
recover the result of Kramkov and Weston’16.
b
1−a
and we
Boundedness of L
Proposition
Assume that U satisfies (Ga,b,C ). Consider the following assertions:
B1. a ∈ (0, 1) and there exists Z ∈ Y satisfying (A
1
1−a
).
B2. a ≥ 1 and there exists Z ∈ Y satisfying (Ak ) for some k > 1.
If either B1. or B2. hold, then there exist K1 > 0, K2 > 0
depending only on a, b, C and k such that
K1 ≤ L ≤ K2 .
Idea of the proof:
1. For a ∈ (0, 1), the growth of L is controlled by L(1 − a).
2. For a ≥ 1, L is bounded from above.
b
Boundedness of L and reverse Hölder’s inequality for Y
Theorem
Suppose that U satisfies (Ga,b,C ) with a > 1. Let q ∈ (0, 1), the
following assertions are equivalent:
i) L is bounded away from 0 and ∞.
b.
ii) (b − ) holds for Y
iii)
q
(bq− )
holds for some Y ∈ Y.
iv) L(p) is bounded away from 0 and ∞ for all p < 0.
Idea of the proof:
I
i) implies ii)
relies on an estimate of the form
b q q
YT
Fσ ≥ KLσa−1 where K > 0.
E
b
Yσ
I
iv) implies i) is based on inequalities between L and L(p).
Continuous BMO-martingales
All local martingales are assumed to be continuous.
Definition: A local martingale M with M0 = 0, is a BMO
martingale if
||M||BMO := sup ||E hMiT − hMiσ Fσ ||∞ < +∞.
σ∈T (F)
Kazamaki’94: Let M be an F-local martingale with M0 = 0 and
1
E(M) := exp M − hMi .
2
The following assertions are equivalent:
i) M is a BMO martingale.
ii) Γ = E(M) satisfies (Ar ) for some r > 1.
iii) Γ = E(M) satisfies (bq− ) for all q ∈ (0, 1).
Example: continuous stock price process
I
Let W be a Rn -valued Brownian motion. We assume that
dSt = St (dWt + µt dt), where µ ∈ L2loc .
I
I
Z µ = E(−µ · W ) ∈ Y.
b is a local martingale and Y
b =E −µ·W +N
b with N
b a
Y
b = 0.
local martingale orthogonal to W , i.e. W , N
Proposition: Suppose that U satisfies (Ga,b,C ) with a > 1. The
following are equivalent:
1. µ · W ∈ BMO.
b satisfies (Ar ) for some r > 1.
2. Y
b satisfies (b − ) for all q ∈ (0, 1).
3. Y
q
4. L is bounded away from 0 and ∞.
Stability of utility maximization
Theorem
Let U and S be as in the example. Let (Um )m∈N and
(xm )m∈N ⊆ (0, +∞) be a sequence of utility functions and initial
b m and Lm be the dual minimizer
capitals. For each m ∈ N, let Y
and gen. opportunity process associated to Um and xm .
Assume that:
1. limm→+∞ Um = U pointwise and limm→+∞ xm = x.
2. for each m ∈ N, Um satisfies (Ga,b,C ).
3. a > 1 and µ · W ∈ BMO.
0
Then, there exists r > 1 such that
"
#
0
bm − Y
bt |r = 0.
lim E sup |Y
m→+∞
t
t∈[0,T ]
Sketch of the proof
bm → Y
bT in probability and Y
bm → Y
b0 by Kardaras and
1. Y
0
T
Žitković’11.
0
0
bm − Y
bT |r
is U.I.
2. Show that there exists r > 1 s.t. |Y
T
m∈N
I
As a > 1 and Z µ = E(µ · W ) satisfies (Ak ). There exists
K > 1 s.t. with r = 1 + bk



1
! r −1
! 1 
m
bσ r −1 Y
Y
σ
Fσ  + E 
F σ  ≤ K .
E
bT
bm
Y
Y
T
I
By Gehring’s lemma, there exist r > 1 and K2 , K3 such that
h
0i
0
bm − Y
bT |r ≤ K2 (|Y m + Y
b0 |r ) ≤ K3 .
E |Y
0
T
0
3. Apply Doob’s maximal inequality to the submartingale
0
bm − Y
b | and use L1 -convergence of |Y
bm − Y
bT |r .
|Y
T
Thank you for your attention!