6.2 A duality
6.2.1 The framework
Let Z be a real normed linear space with norm k·kZ and let Z∗ be its dual with norm k·kZ∗ . Denote the duality product of these spaces by <·,· >Z. For a subspace M ⊂ Z its orthogonal complement M⊥ ⊂ Z∗ is defined by M⊥ := {m∗ ∈ Z∗ :< m, m∗ >Z= 0 for all m∈ M}.
Theorem 6.7 Let Z and M be as above. Then for every z ∈ Z we have
m∈Minf kz−mkZ = max
m∗∈M⊥
km∗kZ∗ ≤1
< z, m∗ >Z . (6.22) The maximum on the right is achieved for some m∗0 ∈ M∗. If the infimum on the left is
achieved for some m0 ∈ M, then z −m0 and m∗0 are aligned , i.e. < z −m0, m∗0 >Z=
kz−m0kZ km∗0kZ∗.
We will address the maximization problem on the right hand side of (6.22) asdual problem.
It will turn out that, in the case of terminal cost, the dual problem essentially leads to the dual problem investigated in [GLP:MVH].
In the next subsection we will do some preparation so that we can apply this theorem to the problems (6.15) and (6.19). Let us sketch how the theorem fits in the situation of these problems. Choose some q > 1. Let M respectively Q and M be as in the Definitions 6.5 respectively 6.6. Define the normed linear spacesZT and ZRT by
ZT := (LqF
T(R),k·kZ
T) with kZkZ
T :=
1
qE[M|Z|q] 1q
, (6.23)
and
ZRT := (LqF(0, T;R)×LqF
T(R),k·kZ
RT) with k(z, Z)kZ
RT :=
1 qE[
Z T 0
Q(s)|z(s)|qds + M|Z|q] 1q
. (6.24)
(The use of the letters z and Z is not meant to allude to the solution of some FBSDE).
Further, let HT and HR be solution operators for the SDE (6.16) and (6.17) - actually integral operators - with initial value 0, i.e.
HT :V =Hq(0, T;R) −→ LqF
T(R) (6.25)
u 7→
Z T 0
B0(s)u(s)ds+ Z T
0
u0(s)σ(s)dw(s), (6.26) respectively
HR:V −→ LqF(0, T;R) (6.27)
u 7→
Z · 0
B0(s)u(s)ds+ Z ·
0
u0(s)σ(s)dw(s). (6.28) Of course we haveHTu= (HRu)(T). Note that the solution xof the state equation (6.16), (6.17) for the pair (x0, u)∈R× V can be written as
x=x0+HRu.
Finally, define the normed linear subspaces MT and MRT of ZT and ZRT by
MT := {HTu:u∈ V} ⊂ ZT, (6.29) MRT := {(HRu, HTu) :u∈ V} ⊂ ZRT. (6.30)
6.2. A DUALITY 91 With these notations, problem (6.15) can be equivalently formulated as
kx0+ZkZ
T =k−x0−ZkZ
T = min
Z∈MT!, (6.31)
and for (6.19) we get the equivalent problem k(x0+z, x0+Z)kZ
RT =k(−x0−z,−x0−Z)kZ
RT = min
(z,Z)∈MRT!. (6.32) We will call these problems primal problems. In a precise meaning, (6.31), (6.32) aren’t really identical to (6.15), (6.19), since the cost functional of the latter does not have an exponent 1q. But the optimal controls of course coincide, and the optimal values can be easily calculated form one another. It is essentially just a rephrasing if we shift from (6.15), (6.19) to (6.31), (6.32), in contrast to the shift from a dynamic to a static problem as de-scribed e.g. in [Ph:DLPH].
Looking at the duality statement in Theorem 6.7, this rephrasing indicates what we want to plug in the left hand side of (6.22). For evaluating the right hand side we must find out what M⊥T respectively M⊥RT is. This will be done in the next subsection. Note that the duality in Theorem 6.7 is as a special case of a more general theory of the minimization of convex functionals. See, for example, [ET:CA], Chapter 3, and [L:UDP],§§1.5.3 , 1.5.4.
6.2.2 The adjoint operators and orthogonal complements
As MT and MRT are images of the operatorsHT and HRT we may determine the orthog-onal complement of these spaces as the kernel of the adjoint operators. Thus we are lead to look for the adjoint operators. We start by noting what the dual spaces of ZT and ZRT are.
Remark 6.8 Consider the spaces ZT and ZRT. Set M˜ := (1qM)−q−11 andQ˜ := (1qQ)−q−11 . Consider the usual dual pairings < ·,· >ZT: LqF
T(R)×LqF0
T(R) −→R, (Z, Z∗) 7→ E[ZZ∗] and<·,·>ZRT: (LqF(0, T;R)×LqF
T(R))×(LqF0(0, T;R)×LqF0
T(R))−→R,((z, Z),(z∗, Z∗))7→
E[RT
0 z(s)z∗(s)ds+ZZ∗]. Based on these pairings we have ZT∗ = (LqF0
T(R),k·kZ∗
T) with kZ∗kZ∗
T =
E[ ˜M|Z∗|q0]q10
for Z∗ ∈LqF0
T(R), and ZRT∗ = (LqF0(0, T;R)×LqF0
T(R),k·kZ∗
RT) with k(z∗, Z∗)kZ∗
RT =
E[
Z T 0
Q(s)|z˜ ∗(s)|q0ds + ˜M|Z∗|q0] q10
for (z∗, Z∗)∈LqF0(0, T;R)×LqF0
T(R).
Proof: From the uniform boundedness and positivity of M and Qit is clear that ˜M and Q˜ are well defined and that the norms k·kZ
T respectively k·kZ pairings that we also use. Hence, every continuous linear functional on ZT respectively ZRT can be represented with some
Z∗ ∈LqF0
T(R) respectively (z∗, Z∗)∈LqF0(0, T;R)×LqF0
T(R) by
Z 7→< Z, Z∗ >ZT respectively (z, Z)7→<(z, Z),(z∗, Z∗)>ZRT . Let us calculate the dual norm of ZT. By definition, we have
kZ∗kZ∗
T(R). Using the criterion for equality in H¨olders’s inequality (see [DS:LO], Ex. III.9.42) in the second line, we get
E[Z0Z∗] = E[sign(M−
6.2. A DUALITY 93 Combined with the above estimate forkZ∗kZ
T, this shows thatkZ∗kZ∗
T =
E[ ˜M|Z∗|q0]q10
. Let us turn to the dual norm of k · kZRT. Using the same arguments, one can show that if we equip the linear space LqF(0, T;R) with the norm z 7→
1 qE[RT
0 Q|z|qds]1q
, the dual space isLqF0(0, T;R) with normz∗ 7→
E[RT
0 Q|z˜ ∗|q0ds]q10
(given the usual pairing). It then follows that the dual norm on the product space LqF0
T(R)×LqF0(0, T;R) is given by k·kZ∗
RT, see [A:LF], Chap. 4, Ex. 4.4 and 4.5. This finishes the proof of the remark.
Let us turn to the adjoint operators. Up to a technical detail, we draw the following result from [YZ:SC], Prop. 4.1 in § 6.4.
Lemma 6.9 Consider the operators HT and HR introduced in (6.26) and (6.28). Their adjoint operators HT∗ : LqF0
T(R) −→ Hq0(0, T;Rm) and HR : LqF0(0, T;R)−→ Hq0(0, T;Rm) are given as follows:
For Z∗ ∈LqF0
T(R) let (α, β) be the unique solution in LqF0(Ω, C([0, T];R))×Hq0(0, T;Rd) of the BSDE
dα=β0dw, α(T) =Z∗. (6.33) Then, HT∗Z∗ =B0α+σβ.
For z∗ ∈LqF0(0, T;R)let (α, β)∈LqF0(Ω, C([0, T];R))×Hq0(0, T;Rd) be the unique solution of the BSDE
dα =−z∗ds+β0dw, α(T) = 0. (6.34) Then, HRz∗ =B0α+σβ.
Proof: Let us first note that both BSDEs mentioned in the assertion are uniquely solvable. For the first one it’s just martingale representation, for the second one it follows from Theorem 5.1 in [EPQ:BSDE], though the present situation does not precisely match the hypothesis of this theorem, as it would require that z∗ ∈ Hq0(0, T;R). For q > 2 this does not hold true. The proof, though, shows that in our particular situation of a linear equation z∗ ∈LqF0(0, T;R) is sufficient.
The proof now consists in both cases of merely applying Itˆo’s formula to α·HRu and to evaluate E[RT
0 d(α·HRu)]. This leads to
E[(HTu)Z∗] = E[u(HT∗Z∗)] for all u∈ V, Z∗ ∈ ZT∗ E[
Z T 0
(HRu)z∗ds] = E[
Z T 0
u(HRz∗)ds] for all u∈ V, z∗ ∈LqF0(0, T;R).
We now can describe M⊥T and M⊥RT. The third part of the following lemma states a parametrization of the kernel kerσ. It will be helpful when stating dual problems as linear isoelastic stochastic control problems.
Lemma 6.10 Set θ :=σ0(σσ0)−1B0 . Then we have 1.
M⊥T = n
α1(T)∈LqF0
T(R) :α1 is the solution of the SDE
dα1 ={−θ0α1+k0}dw, α1(0) =α1,0, for some k∈Hq0(0, T;Rd) with σk= 0, Leb⊗P −a.s., and some α1,0 ∈R
o
, (6.35)
2.
M⊥RT = n
(z∗, α2(T))∈LqF0(0, T;R)×LqF0
T(R) :α2 is the solution of the SDE dα2 =−z∗ds+{−θ0α2+k0}dw, α2(0) =α2,0, for some
z∗ ∈LqF0(0, T;R), some k∈Hq0(0, T;Rd)
with σk= 0, Leb⊗P −a.s., and some α2,0 ∈R o
. (6.36)
3. If m < d choose a matrix-valued process eσ∈L∞F(0, T;R(d−m)×d) such that (i) eσeσ0 0, i.e eσσe0 is uniformly positive;
(ii) the matrix σ
eσ
is Leb⊗P −a.s. regular;
(iii) σσe0 = 0, Leb⊗P −a.s.;
Then,
k ∈Hq0(0, T;Rd) :σk= 0, Leb⊗P −a.s.
=
eσ0v ∈Hq0(0, T;Rd) :v ∈Hq0(0, T;Rd−m) . (6.37) Proof:
1. With the definition of M⊥T and MT we have M⊥T = n
Z∗ ∈LqF0
T(R) :< Z, Z∗ >ZT= 0 for all Z ∈ MTo
= n
Z∗ ∈LqF0
T(R) :< HTu, Z∗ >ZT= 0 for all u∈ Vo
=
Z∗ ∈LqF0
T(R) :E[
Z T 0
u0(HT∗Z∗)ds] = 0 for all u∈Hq(0, T;Rm)
= n
Z∗ ∈LqF0
T(R) :HT∗Z∗ = 0, Leb⊗P −a.s.o .
Hence,M∗T coincides with the kernel ofHT∗ and we have to show that the set described in the assertion is a parametrization of this kernel.
First assume that we are given an element α1(T) from (6.35). Then there is an α1,0 ∈ R and a k with σk = 0, such that dα1 ={−θ0α1+k0}dw, α1(0) = α1,0. We have to show thatα1(T)∈kerHT∗. Let (α, β) be the solution of equation (6.33) with
6.2. A DUALITY 95 Z∗ := α1(T). From unicity of solutions it is clear that α =α1, and β = −θα1 +k.
Hence,
HT∗ (α1(T)) = B0α+σβ =B0α1−σθα1+σk=σk= 0, Leb⊗P −a.s.
Now assume that some Z∗ belongs to kerHT∗. We have to find α1,0 and k such that for the resulting α1 we have α1(T) = Z∗. Let (α, β) be the solution of (6.33) for the Z∗ under consideration. Set k := β +θα ∈ Hq0(0, T;Rd) and α1,0 := α(0). As HT∗Z∗ = 0 we haveB0α+σβ= 0, hence
σk=σ(β+θα) =B0α+σβ= 0,
Leb⊗P −a.s.. By definition, α1 :=α satisfies dα1 ={−θα1+k}dw, α1(0) =α1,0. Hence, M∗T has the form as stated in the assertion.
2. As above we have M⊥RT =
n
(z∗, Z∗)∈LqF0(0, T;R)×LqF0
T(R) :
<(z, Z),(z∗, Z∗)>ZRT= 0 for all (z, Z)∈ MRTo
= {(z∗, Z∗) :<(HRu, HTu),(z∗, Z∗)>ZRT= 0 for allu∈ V}
=
(z∗, Z∗) :E[ Z T
0
u0(HR∗z∗+HT∗Z∗)ds] = 0 for all u∈Hq(0, T;Rm)
= {(z∗, Z∗) :HRz∗+HT∗Z∗ = 0, Leb⊗P −a.s.}. (6.38) We first want to characterize the set in (6.38) via a BSDE and claim that
{(z∗, Z∗) :HRz∗ +HT∗Z∗ = 0}
= {(z∗, Z∗) : the solution of dα=−z∗+β0dw, α(T) =Z∗,
satisfies B0α+σβ = 0}. (6.39)
To see this, assume HRz∗ + HTZ∗ = 0 for some (z∗, Z∗). Let (a1, b1) rectively (a2, b2) be the solution of (6.33) respectively (6.34) for this Z∗ respectively z∗ and set α :=a1 +a2 and β = b1 +b2. Then, (α, β) obviously satisfies the BSDE on the right hand side of (6.39) and
0 = HR∗z∗+HT∗Z∗ =B0a1+σb1 +Ba2+σb2 =B0α+σβ,
what shows the inclusion “⊂” in (6.39). Conversely, assume that we are given a pair (z∗, Z∗) from the set on the right hand side of (6.39). Define (a1, b1), (a2, b2) just as above. Again, set α =a1+a2, β = b1 +b2, and B0α+σβ = 0 now entails B0a1+σb1+B0a2+σb2 =HR∗z∗+HT∗Z∗ = 0. This proves (6.39).
Finally, it’s easy to see that the set on the right hand side of (6.39) coincides with the “parametrization” of M∗RT in (6.36): Given a triplet z∗, k and α0 from this
parametrization with solution (α2, β2) of the resulting SDE, set α := α2 and β :=
−θα2+k. (α, β) then satisfies the BSDE on the right hand side of (6.39), and due to the definition ofθ we have B0α+σβ= 0. Conversely, if (z∗, Z∗) belongs to the right hand side of (6.39) with corresponding (α, β), thenk :=β+θα andα2,0 :=α(0) will yield the desired representation of Z∗.
3. At first sight it may not be completely clear that a process eσ with the properties (i)−(iii) exists. A precise argument is given in Appendix A.
What follows is meant to holdLeb⊗P−a.s.. From (iii) we have that the image ofeσ0 is contained in kerσ, from (ii) we get that it actually coincides with kerσ. Hence, for every k ∈kerσ there is a v such that eσ0v =k. It is easily seen that v = (σeeσ0)−1eσk.
Passing to the consideration of processes (i) now shows that v ∈ Hq0(0, T;Rd−m) if k ∈Hq0(0, T;Rd).
Let α1(T) be as in the definition of M⊥T with α1(0) = 1. dR := α1(T)dP then defines a signed martingale measure on Ω for S(t) := Rt
0 B0ds +Rt
0 σdw, i.e. α1(t)S(t) is a P -martingale, where α1(t) = E[α1(T)|Ft]. If conversely α1(T) is a q0-integrable r.v. such thatα1(t)S(t) is a martingale (withα1(t) =E[α1(T)|Ft]), then arguments similar to those in the proof show that α1(T) belongs to M⊥T. Hence, the dual space of MT consists of signed martingale measures forS that are multiplied with some real. A characterization of equivalent martingale measures forSin terms of a SDE is given in [EQ:DPPC], Proposition 1.8. Martingale measures in different generality (equivalent, local, signed) play a central role in mathematical finance. See [L:UDP], Section 1.4 for a short overview.
The representation of M∗T and M∗RT in terms of SDEs will allow us to treat the dual problems as stochastic control problems that fit into the linear isoelastic framework.
So much for the objects involved in the dual problems for (6.15) and (6.19). The last lemma of this section indicates how to determine the solution of the primal problem, once the solution for dual problem is found. The result for the terminal cost problem are quite similar to those of [Ph:DLPH], Theorems 4.1 and 5.1, where Lq-hedging is considered in discrete time with cone constraints on the portfolio.
Lemma 6.11 Consider the recast forms of the hedging problems with terminal respectively running and terminal cost (6.31), (6.32), and their dual problems. Let M˜, Q˜ be as in Remark 6.8
1. Let u¯ be the solution for (6.15) and let Z¯∗ be the solution for the dual problem to (6.31). (the existence ofZ¯∗ follows from Theorem 6.7). Let α1,0 be the initial value in the representation ofZ¯∗ in (6.35). Then there is acT >0such that−x0−(HRu)(T¯ ) = cT M˜ sign( ¯Z∗)|Z¯∗|q0−1; more precisely,cT =|x0α1,0|(E[ ˜M|Z¯∗|q0])−1q. Z¯∗ P−a.snever has the sign of x0.
2. Letu¯ be the solution for (6.19) and let(¯z∗,Z¯∗)be the solution for the dual problem to (6.32) (see Theorem 6.7). Let α2,0 be the initial value in the representation of Z¯∗ in
6.2. A DUALITY 97 (6.36). Then, there constantscR, cT >0such that. −x0−HRu¯=cRQ˜sign(¯z∗)|¯z∗|q0−1 and −x0 − (HRu)(T¯ ) = cT M˜ sign( ¯Z∗)|Z¯∗|q0−1. z¯∗ and Z¯∗ Leb⊗P− respectively P −a.s. never have the sign of x0. Let Γ be the solution of dΓ =−θ0Γdw, Γ(0) = 1, θ :=σ0(σσ0)−1B0. Then cT can be calculated from x0+cTE[Γ(T)sign( ¯Z∗)|Z¯∗|q0−1] = 0. cR can be calculated from
−x0α2,0 =E[cqR Z T
0
Q|¯˜ z∗|q0ds+cqT M˜|Z¯∗|q0]. (6.40)
Proof: Let ¯x be the optimal state process for the problems (6.15) or (6.19). By Remark 4.4, ¯x(T) cannot be identical zero unless x0 = 0. In particular, the optimal cost for problems (6.15) and (6.19) is strictly positive. The representation of the optimal cost in Theorem 6.7 then shows that ¯Z∗ 6≡0 respectively (¯z∗,Z¯∗)6≡(0,0). Z¯∗ are identically zero, this expression is strictly greater than zero. Set M0 := 1qM. Since M
1 q
0M˜q10 = 1, the alignment (used in the second line of the following equation) means that H¨older’s inequality gives that the right hand side of the first line of this is less or equal the right hand side of the third line. The criterion for equality in H¨older’s inequality thus gives that there is a cT ∈Rsuch that
M
Noting that we have to modify the optimal values, we get from Theorem 6.7 that (−x0α1,0)q =J(T)(¯u). Now we can calculate cT from
(−x0α1,0)q = J(T)(¯u)
=
−x0−h(T)
q ZT
= |cT|qE[M0
M˜ |Z¯∗|q0−1q
]
= |cT|qE[ ˜M|Z¯∗|q0] =cqT Z¯∗
q0
ZT∗ , (6.44) sinceM0M˜q = ˜M. ¯x(T) =x0+h(T) is the optimal terminal state for problem (6.15), and Lemma 3.8 thus gives that this r.v. is only of one sign - that ofx0 (not counting 0 as an extra sign). So, from (6.42) it follows that ¯Z∗ never is of the sign of x0. 2. Again, set h = HRu. It is easily seen that the alignment of (−x¯ 0 −h,−x0 −h(T))
with (¯z∗,Z¯∗) guaranteed by Theorem 6.7 entails the alignment of −x0 −h with ¯z and of −x0−h(T) with ¯Z∗. With the same arguments as in part 1. of this proof it follows that there are constantscR, cT ∈Rsuch that
−x0−h=cRQ˜sign(¯z∗)|¯z∗|q0−1, −x0−h(T) = cT M˜ sign( ¯Z∗)|Z¯∗|q0−1. (6.45) By Remark 4.4, ¯x(T) =x0+h(T) is not identical zero, and by Lemma 3.8 it is only of one sign. Let us consider the SDE (6.16) for ¯x as an BSDE by setting S = ¯u0σ.
Then, ¯xsatisfies
dx¯=θ0S0ds+Sdw, x(T¯ ) = −cT M˜ sign( ¯Z∗)|Z¯∗|q0−1. The solution part ¯x of this BSDE is given by
¯
x(t) = Γ−1(t)E[Γ(T)¯x(T)|Ft],
(compare [EM:BSDE],§2.2, Prop. 2.4), hence x0 =−cT E[Γ(T) ˜Msign( ¯Z∗)|Z¯∗|q0−1], where the expectation does not vanish. This enables us to computecT. As ¯x(T)6≡0, the alignment shows that cT > 0. Further, we have ¯z∗ 6≡ 0, since otherwise from (6.45) it follows that x0+h= ¯x= 0, what would imply that ¯x(T)≡0. So again the alignment argument shows cR >0, and as ¯x never changes its sign (by Lemma 3.8), we see that ¯z∗ is Leb⊗P −a.s. of the opposite sign of x0.
Now let k∈kerσ be the process in the representation of ¯Z∗ in (6.36). Let α2 be the solution of the SDEdα2 =−¯z∗ds+{−θα2+k0}dw, α2(0) =α2,0, then ¯Z∗ =α2(T).
With this representation we get from Theorem 6.7 J(RT)(¯u) = <(−x0,−x0),(¯z∗,Z¯∗)>ZRT
= E[
Z T 0
−x0z¯∗ds−x0Z¯∗]
= −x0E[
Z T 0
¯
z∗ds+α2,0− Z T
0
¯ z∗ds+
Z T 0
−θα2+kdw]
= −x0α2,0. (6.46)
6.3. SOLUTIONS TO THE PROBLEMS 99