Portfolio Optimization for Exponential L´ evy Processes
4.2 The q -Optimal Signed Martingale Measure
4.2 The q -Optimal Signed Martingale Measure
In this section, we derive an explicit form of theq-optimal signed martingale measure. We relax assumptionCq and only require assumptionCq−.
4.2.1 Relaxing Assumption Cq
This section is devoted to a thorough analysis of conditionCq−. In particu-lar, we derive an explicit form of theq-optimal signed martingale measure, qSMM. Moreover, we will see that Cq− is a very weak condition. However, the qSMM will be an equivalent measure if and only ifCq+ holds. Moreover, ifCq+ does not hold, the qEMM will not exist. This last case is discussed in Corollary 4.3.4 and Section 4.6. Finally, the qSMM will coincide with the q-optimal absolutely continuous martingale measure, qAMM, ifCq0 holds.
Before we start dealing with the qSMM, we like to stress that ZTq is no longer equal to|ZT|q forZ ∈ Dsq:
Lemma 4.2.1. Let E¯T(f, h −1) be a stochastic exponential with time-independent and deterministic Girsanov parameters satisfying
Z
RN\{0}
(|h(x)|q−1−q(h(x)−1))ν(dx)<∞ (4.8) for some q >1. Then
|E¯t(f, h−1)|q = E¯t(qf,|h|q−1)et
q(q−1) 2 f′f+R
RN\{0}(|h(x)|q−1−q(h(x)−1))ν(dx)
and so
For the last assertion it remains to prove that Z
RN\{0}
(|egq(x)|κ−1−κ(egq(x)−1))ν(dx)<∞
for 1≤κ≤q. For small jumps, we choose a constantK >0 such that egq is non-negative for kxk ≤K. Integrability follows thanks to Taylor’s formula.
For large jumps, integrability follows as for an arbitrary constant K > 0 R
kxk>K|egq(x)|qν(dx)<∞.
It is now time to present one of the main theorems of this chapter:
Theorem 4.2.2. Suppose that Cq− holds. Then Z(q) = ¯E(θq′σ,egq−1)
is the density process of q-optimal signed martingale measure, qSMM. If in addition Cq0 holds, then the qSMM is equal to the qAMM and equal to the qEMM underCq+.
As already mentioned in Section 3.2.2, we cannot extend the proof in Jeanblanc et al. [69] as it uses that all equivalent martingale measures can be represented by stochastic exponentials. Instead we will prove The-orem 4.2.2 via a verification procedure by a hedging argument and duality.
The proof will be given in Section 4.2.2.
It remains to analyze how restrictive Cq− actually is. In the one-dimensional case, Proposition 4.2.3 gives a necessary and sufficient char-acterization of condition Cq−. Roughly speaking, Cq− holds, if and only if the tails of upward jumps are moderately sized, which is much weaker than Cq:
Proposition 4.2.3. Suppose N = 1, P is not a martingale measure, and the set of equivalent martingale measures is nonempty, i.e. P /∈ Me and Me6=∅. Then,Cq− holds for q >1, the conjugate of p, if and only if
Z
x≥1
epxν(dx)<∞. (4.11) Before we start with the proof, we present an example.
Example 4.2.4. SupposeN = 1.
(i) Ifν(dx) behaves (up to a slowly varying function) ase−γ+xdxforx→ ∞, thenCq− holds for p= q−1q < γ+ and fails forp > γ+. However, Cq fails for allq, if
Z
R\{0}
((ex−1)−x1|x|≤1)ν(dx) + (b+1
2σ2)>0. (4.12) As explained in Section 1.1 this tail behavior is inherent in the Kou and the variance gamma model. But also most of generalized hyperbolic models satisfy this assumption see [32, Table 4.7].
(ii) If there are constantsη0, η1 >0 such that Z
x≥1
eη0x1+η1ν(dx)<∞, (4.13) thenCq−holds for all q >1. However Cq fails for allq, if the upward jumps are not bounded and
Z
R\{0}
((ex−1)−x1|x|≤1)ν(dx) + (b+1
2σ2)>0.
Another popular model introduced in Section 1.1 is the Merton model. It satisfies (4.13) while possessing unbounded upward jumps.
We need some preparation to prove Proposition 4.2.3. Recall, forN = 1 which are, for fixedq >1, now defined on the domain
Dom(s)q :=
With these renewed definitions at hand, we are able to extend Lemma 3.2.6.1:
Lemma 4.2.5. (i) Let N = 1. Suppose σ 6= 0 or ν(R\ {0}) 6= 0. Then the mappings Φe andΦ(q,·) are continuous and strictly increasing onDome resp. Dom(s)q . Proof of Lemma 4.2.5. (i) Φ(q,·) is strictly monotonic forq >1. The asser-tions again follow directly from the monotone convergence theorem.
(ii) The assertion is trivial as the parts dependent onq orevanish.
(iii) Since q−11 =p−1, there is a constant Kp >0 such that for x >1 and θ∈R
(ex−1)|(q−1)θ(ex−1) + 1|q−11 ≤ Kp
p−1|θ|(ex−1)p+Kp(ex−1),
which isν-integrable over (1,∞) by assumption.
(iv) By a direct calculation, as in the proof of Lemma 3.2.5.1, θ′Φ(q, θ)
= Z
RN\{0}
θ′(ex−1)|(q−1)θ′(ex−1) + 1|q−11 sgn((q−1)θ′(ex−1) + 1)
−θx1kxk≤1
ν(dx)−θ′β+θ′σσ′θ
≥ θ′( Z
RN\{0}
((ex−1)−x1kxk≤1)ν(dx)−β) +θ′σσ′θ
≥ −kθkk Z
RN\{0}
((ex−1)−x1kxk≤1)ν(dx)−βk+kσθk2, (4.14) as (q−1)θ′(ex−1) + 1≥1 for θ′(ex−1)>0,
0≤sgn((q−1)θ′(ex−1) + 1)|(q−1)θ′(ex−1) + 1|q−11 ≤1 forθ′(ex−1)<0 and sgn((q−1)θ′(ex−1) + 1) = 1,and
θ′(ex−1)≤0≤θ′(ex−1)sgn((q−1)θ′(ex−1) + 1)|(q−1)θ′(ex−1) + 1|q−11 forθ′(ex−1)<0 and sgn((q−1)θ′(ex−1) + 1) =−1.
Proof of Proposition 4.2.3. Recall, the set of equivalent martingale mea-sures is non-empty and P is not a martingale measure. As discussed in Section 1.2 this immediately implies thatσ 6= 0 or ν(R\ {0})6= 0.
⇒: SupposeCq−holds for some fixedq. Then, by definition withsg(x) = sgn((q−1)θq(ex−1) + 1)
R
x>1
|(q−1)θq(ex−1) + 1|q−1q −1
−q sg(x)|((q−1)θq(ex−1) + 1|q−11 −1)
ν(dx)<∞. Note that p=q/(q−1). Hence, for large x the term epx dominates in the above integrand and we conclude that (4.11) holds.
⇐: Suppose that condition (4.11) holds for some fixedq = p−1p >1. Recall that Dom(s)q = R thanks to Lemma 4.2.5, (iii). Now, in view of Lemma 4.2.5, (i), it suffices to show that
θ↓−∞lim Φ(q, θ) =−∞ and lim
θ↑∞Φ(q, θ) =∞. (4.15) The existence of an equivalent martingale measure implies that σ 6= 0 or ν(R\ {0}) 6= 0. We first treat the caseσ 6= 0. Dividing (4.14) by θ implies (4.15) immediately. Ifσ = 0, thenν(R\{0})6= 0. Note that, for every given x, (ex−1)|(q−1)θ′(ex−1) + 1|q−11 sgn((q−1)θ′(ex−1) + 1) is monotone
inθ and converges to ±∞ forθ→ ±∞. Hence, the monotone convergence theorem yields (4.15) in the latter case. Obviously, condition (4.11) ensures that (4.3) holds.
We close the section by presenting a very simple example when Cq− is satisfied, butCq+ fails:
Example 4.2.6. Suppose an exponential of a Poisson process with jump size a6= 0 and a Brownian component is considered:
St=S0eXˇt, Xˇt=bt+σWt+Ut,
where b∈R, U is a Poisson process with intensity γ > 0 and jump size a, without loss of generality assume |a|> 1 (otherwise set ˜b = b−γa1|a|<1), andW is a standard Brownian motion, independent ofU. Moreover, set the riskless interest rate r= 0. To establish Cq− we have to find a solution of
σ2θq+γ(ea−1)|(q−1)θq(ea−1)+1|q−11 sgn((q−1)θq(ea−1)+1) =β, (4.16) whereβ =−(b+12σ2).
Case 1: Setσ = 0 andq = 32,then:
γ(ea−1) 1
2θq(ea−1) + 1 2
sgn 1
2θq(ea−1) + 1
=−b We start with two candidates:
θ±q =
±2 r
−bsgn(12θq(ea−1)+1) γ(ea−1) −2
ea−1 ,
but we need to guarantee that sgn(12θq(ea−1) + 1) = sgn(a)sgn(−b):
1
2θq(ea−1) + 1 =± s
−bsgn(12θq(ea−1) + 1) γ(ea−1) ,
hence Cq+ can be satisfied by θq+ only, but the square root has to be real-valued:
• If sgn(a) = −sgn(b), we have sgn(12θq(ea−1) + 1) = 1 and thus, Cq+ andCq− hold. θq+ is the right solution.
• If sgn(a) =sgn(b), we have sgn(12θq(ea−1) + 1) =−1 and thus, Cq− holds, butCq+ fails and we have arbitrage. θq− is the right solution.
Case 2: Set σ = 0 and q = 2, i.e. the variance case, then we have one unique solution of (4.16):
θq=
−b γ(ea−1) −1
ea−1 and eg2(a) =θq(ea−1) + 1 = −b γ(ea−1). θq(ea−1) + 1 = γ(e−ba−1) >0 if and only if sgn(a) =−sgn(b).Hence:
• Cq− always holds and a unique solutionθ2 exists.
• Cq+ holds if and only if sgn(a) =−sgn(b).
• If sgn(a) = sgn(b), we have arbitrage although the qSMM still exists as Cq− is valid.
Case 3: Setσ6= 0, sg:= sgn(12θq(ea−1) + 1), and b=−12σ2+µ, then σ2θq+γ(ea−1)|(q−1)θq(ea−1) + 1|q−11 sg=β=−(b+ 1
2σ2) =−µ, by setting again q= 1.5 :
σ2θq+γ(ea−1)sg 1
2θq(ea−1) + 1 2
=−µ
⇔ σ2θq+sg
4 θq2γ(ea−1)3+sgθqγ(ea−1)2+sgγ(ea−1) +µ= 0
⇔ γ
4sg(ea−1)3θ2q+ (σ2+sgγ(ea−1)2)θq+sgγ(ea−1) +µ= 0 We have
θq± = −(σ2+γ(ea−1)2)
γ
2(ea−1)3
±
p(σ2+sgγ(ea−1)2)2−γ2(ea−1)4−sgµγ(ea−1)3
γ
2(ea−1)3
= −(σ2+γ(ea−1)2)±p
σ4+ 2sgσ2γ(ea−1)2−sgµγ(ea−1)3
γ
2(ea−1)3 .
Ifσ4+sgσ2γ(ea−1)2−sgµγ(ea−1)3 ≥0, we have again two candidates, but which one is right?
1
2θq±(ea−1) + 1
= −(σ2+γ(ea−1)2)±p
σ4+ 2sgγσ2(ea−1)2−µsgγ(ea−1)3
γ(ea−1)2 + 1
= −σ2±p
σ4+sg(2σ2γ(ea−1)2−µγ(ea−1)3) γ(ea−1)2
Ifσ2 ≤µ(ea−1),thensg=−1 and soσ4+sgγ(ea−1)2(σ2−µ(ea−1))≥0, henceCq− is satisfied, butCq+ fails.
• If sgn(a) = sgn(µ), thenCq− holds. Cq+ is only satisfied ifσ2 > µ(ea− 1), in this case θq+ is the unique solution of (4.16). If σ2≤µ(ea−1), then sg=−1 and θq− is the right solution.
• If sgn(a) =−sgn(µ), then sg= 1, as σ2 ≥0≥µ(ea−1), hence Cq+ andCq− hold. θq+ is the right solution.
Case 4: Setσ 6= 0 andb=−12σ2+µ, andq = 2. We need to solve:
σ2θq+γ(ea−1)((ea−1)θq+ 1) =−µ Hence,
θq = −(µ+γ(ea−1)) σ2+γ(ea−1)2 and
eg2(a) = −µ(ea−1) +σ2
σ2+γ(ea−1)2 >0⇔σ2 > µ(ea−1).
• Cq− always holds and a unique solution θ2 exists. We do not have arbitrage in all cases as σ2 >0.
• Cq+ holds if and only if σ2 > µ(ea−1),similar to the case q= 32.
4.2.2 Verification of the q-Optimal Signed Solution
This section is devoted to the proof of Theorem 4.2.2. As a first step we derive a verification theorem for the q-optimal signed martingale measure, i.e. we suggest a density with certain properties and prove by a duality relation that it is actually the density of the qSMM. This theorem does not rely on the L´evy setting, but holds for general semimartingale models (i.e. for stock processesSwith unique Doob-M´eyer-decomposition S =S0+M+A).
Theorem 4.2.7. Suppose Z˘ ∈ Dsq, q = p−p1 and, for some x < p, the˜ contingent claim
X(p,˜x)( ˘Z) :=p− sgn( ˘ZT)|Z˘T|p−11
p−x˜ E(|Z˘|
p p−1
T )
(4.17)
is replicable with initial wealth x˜ and a predictable strategy ϑ, such that ϑ∈ Ap,i.e. ϑsatisfies
kϑkLp(M) := k( Z T
0
ϑ′td[M]tϑt)12kLp(Ω,P)<∞, (4.18) kϑkLp(A) := k
Z T
0 |ϑ′tdAt|kLp(Ω,P)<∞. (4.19) Then Z˘ is the density process of theq-optimal signed martingale measure.
Before, we proceed with a proof we like to mention a property similarly applied in the proof of Lemma 3.3.2:
Remark 4.2.8. Suppose q = p−p1, ϑ ∈ Ap, x ∈ R, and consider the wealth
Then, by the inequalities of H¨older and Burkholder-Davis-Gundy as de-scribed in Section 1.2.2 (Lemma 1.2.8), we have sufficient integrability such thatZtYt(ϑ, X) is a martingale for every Z∈ Dqs. In particular,
E[ZTYT(ϑ, x)] =E(Z0x) =x.
Therefore, a direct calculation shows that the initial capital required to repli-cateX(p,˜x)( ˘Z) in (4.17) with a strategy of classApis ˜x(provided replication, in the sense of Definition 1.2.7, is possible).
Proof of Theorem 4.2.7. Letq=p/(p−1). We consider the following max-imization problems with utility functionup(x) =−|1−xp|p: From Remark 4.2.8 we derive that
E(up(X(1)))≥E(up(X(2)))≥E(up(X(3)))≥E(up(X(p,˜x)( ˘Z))). (4.20) Due to Lemma 1.2.8, Max2 dominates Max3,i.e.V(˜x)0,dyn ≤ V(˜x)s,stat. From Section 1.2.5, we know that the convex dual of up is given by
ˇ
up(y) = (p−1)|y|p−1p −py. (4.21) The dual problem reduces to an optimization over the non-negative real axis. Standard duality theory can be applied to verify thatX(p,˜x)( ˘Z) is the maximizer of problem Max1, see Section 1.2 for details. In particular, we observe that all inequalities turn into identities in (4.20). In a next step we observe that the problem Max2 is dominated by its dual minimization problem [see e.g. 88, p. 225], and due to (4.21), we have
From (1.33), we know that
X(p,˜x)( ˘Z) =p−psgn( ˘ZT)(˘yp|Z˘T|)p−11 = (u′p)−1(˘ypZ˘T), where
˘ yp=
(p−x)˜ pE
|Z˘T|p−1p −1p−1 , is the optimal claim of Max1,ifp >x. Moreover, as˜
ˇ
up(y) =up((u′p)−1(y))−(u′p)−1(y)y, and E( ˘Z(u′p)−1(˘ypZ˘)) = ˜x ifp >x, we have˜
E(up(X( ˘Z))) =E(ˇup(˘yp·Z˘T)) + ˜xy˘p.
Consequently, the pair ( ˘Z,y˘p) is the minimizer of the problem on the right hand side of (4.22). This immediately implies that ˘Z is the density process of the p/(p−1)-optimal signed martingale measure.
We next apply the above verification theorem to the L´evy case and sup-pose thatCq− holds for someq >1. We can therefore define
Z˘= ¯E(θ′qσ,egq−1).
Z˘ is q-integrable due to (4.3), i.e. E|Z˘|q < ∞. Moreover, applying the Dol´eans-Dade formula and integration by parts, one easily verifies that ˘ZS is a local martingale thanks to (4.2). Hence, ˘Z ∈ Dqs. For this reason, (4.2) is referred to as the martingale condition, compareKunita[84].
In order to prove Theorem 4.2.2, it remains to show that X(p,˜x)( ˘Z) is replicable within the class Ap for q = p/(p−1) and some ˜x < p. We will again derive a replicating strategy constructively. This closely follows the approach in Chapter 3, but for the reader’s convenience we present a shortened version stemming the additional difficulties coming from the missing positivity condition Cq+. We fix some ˜x < p.
As the Girsanov parameters are deterministic, Lemma 4.2.1 yields
|E¯T(θq′σ,egq−1)|q= ¯ET(qθq′σ,|eg|qq−1)E[|E¯T(θ′qσ,egq−1)|q].
Moreover, thanks to Lemma 4.2.9 below we obtain X(p,˜x)( ˘Z) = ˘YT, where Y˘ is the following wealth process with a multiplicative decomposition into a process of bounded variation and a stochastic exponential (see F¨ollmer [47];M´eyer[98] for multiplicative decompositions in general):
Y˘t := p−(p−x) ¯˜ Et((q−1)θq′σ,|eg|q−1q sgn(egq)−1) (4.23)
×et
R
RN\{0}(egq(x)−|egq|q(x)+|egq|q−1(x)sgn(egq(x))−1)ν(dx)−(q−1)kθ′qσk2
=: p−(p−x) ˘˜ MtA˘t.
Recall, ifX(p,˜x)( ˘Z) is replicable, then its conditional expectation under every martingale measure is equal to the price process of X(p,˜x)( ˘Z). This fact leads to the above representation of the price process, see (3.32) and the following discussion:
Our aim is now to translate the above multiplicative decomposition of ˘Y into an additive semimartingale decomposition of Doob-M´eyer-type. Taking the definition of egqin conditionCq−and the martingale condition (3.11) into account, we obtain as in Chapter 3
A˘t = et
Hence, applying the Dol´eans-Dade formula to ˘M and integration by parts, Thus, we have proved the following lemma.
Lemma 4.2.10. Suppose Cq− holds. Define ϑ(p)t = −p−x˜ is replicable with initial wealth x˜ and the predictable strategy ϑ(p).
It remains to show that ϑ(p) ∈ Ap. The respective proof in Chapter 3 cannot be generalized as it needs boundedness of upward jumps or the assumption that θe>0.
Lemma 4.2.11. Suppose Cq− holds. Then the hedge ϑ(p) constructed in Lemma 4.2.10 belongs to the class Ap.
Proof. Recall that by the definition of egq E|E¯t((q−1)θ′qσ,(q−1)θq′(e·−1))|p
thanks to Taylor’s formula for small jumps and due to (4.3) for large jumps.
Hence, ¯Et((q−1)θq′σ,(q−1)θ′q(e·−1)) is ap-integrable martingale underCq− thanks to (4.3) (also see Lemma 4.2.1). By Doob’s inequality, we immedi-ately observe thatϑ(p) satisfies (4.19). To prove (4.18), we first notice that, for some constant Kq>0,
applying the Dol´eans-Dade formula for the last identity. By the Burkholder-Davis-Gundy inequality and since ¯E((q −1)θ′qσ,(q − 1)θ′q(e· − 1)) is p-integrable, the term on the right hand side is finite.
In analogy to Chapter 3, one can prove thatE|E¯t((q−1)θ′qσ,(q−1)θ′q(e·− 1))|p < ∞ slightly different: by the definition of egq and the fact that sgn( ¯Et(θ′qσ,egq−1)) =Q is ap-integrable martingale under Cq−.