2.5 Absolute Continuity and Singularity for Generalized Hyperbolic Lévy Processes
2.5.2 Allowed and Disallowed Changes of Parameters
Proposition (2.19) connects the change of the Lévy measure with the change of the underlying probability measure. In particular, for generalized hyperbolic Lévy processes it allows us to analyze whether a certain change of parameters can be achieved by a change of the probability measure.
Proposition 2.20. Let X be a generalized hyperbolic Lévy process under the measure P. Let (λ, α, β, δ, à) denote the parameters of the generalized hyperbolic distribution of X1. Then there is another measureP0 loc∼ P under whichXis again a generalized hyperbolic Lévy process, with parame- ters(λ0, α0, β0, δ0, à0), if and only ifδ0 =δandà0 =à.
Proof. Since generalized hyperbolic Lévy processes are purely discontinuous, the change of measure is determined by the densityk(x) := dK0/dK, where K0(dx) andK(dx) are the Lévy measures of the generalized hyperbolic distributions underP0 andP, respectively. By Proposition (2.19), there exists a measureP0 under whichXis a generalized hyperbolic Lộvy process with parameters (λ0, α0, β0, δ0, à0) if and only if this density satisfies the conditions
Z 1−p
k(x) 2
K(dx)<∞ (2.39)
and b+
Z
h(x)(k(x)−1)K(dx) =b0. (2.40)
Since K(dx) and K0(dx) are both continuous with respect to the Lebesgue measure on IR\{0}, the densityk(x)is equal to the quotient of the respective (modified) Lévy densities.
k(x) = 1
x2ρ(λ0,α0,β0,δ0,à0)(x) 1
x2ρ(λ,α,β,δ,à)(x)
= ρ(λ0,α0,β0,δ0,à0)(x) ρ(λ,α,β,δ,à)(x) .
For the case of generalized hyperbolic Lévy processes, the integrability condition (2.39) is therefore Z
1−p k(x)2
K(dx) = Z
1− q
ρ(λ0,α0,β0,δ0,à0)(x) q
ρ(λ,α,β,δ,à)(x) 2
1
x2ρ(λ,α,β,δ,à)(x)dx
= Z q
ρ(λ0,α0,β0,δ0,à0)(x)−q
ρ(λ,α,β,δ,à)(x) 2dx
x2 <∞.
Since the Lévy densities—modified or not—of generalized hyperbolic distributions satisfying|β| < α always decay exponentially as|x| → ∞, the only critical region for the boundedness of the integral is the neighborhood ofx= 0. Here we have, by Proposition 2.18,
ρ(x) = δ
π +λ+ 1/2
2 |x|+δβ
π x+o(|x|) = δ
π +O(|x|) (|x| →0).
Therefore q
ρ(λ,α,β,δ,à)(x) = δ π
p1 +O(|x|) = δ
π 1 +O(|x|)
(|x| →0).
since√
1 +y= 1 +y/2 +o(|y|)for|y| →0. Using this, the integrability condition (2.39) becomes Z
1−p k(x)
2
K(dx) = Z δ0
π 1 +O(|x|)
− δ
π 1 +O(|x|)2dx x2
=
Z δ0−δ
π +O(|x|) 2
dx x2
=
Z (δ0−δ)2
π2x2 +2(δ0−δ)
π O(1/|x|) +O(1)
dx <∞. Clearly, this is satisfied if and only ifδ0 =δ.
Now we consider the second condition (2.40). Under the assumption δ0 = δ, this will turn out to be equivalent toà0 =à, thus providing the second constraint stated above.
By Jacod and Shiryaev (1987), Theorem II.4.15, the constantbfrom the Lévy-Khintchine representation is indeed the drift coefficient from the characteristic triplet. We have
b=E[L1] + Z
(h(x)−x)K(dx) (2.41)
and b0=E0[L1] + Z
(h(x)−x)K0(dx), (2.42)
whereE0[ã]denotes the expectation under the measureP0. On the other hand, condition (2.40) requires b0 =b+
Z
h(x)(k(x)−1)K(dx), which, in view of (2.41), is equivalent to
b0 =E[L1] + Z
(h(x)−x)K(dx) + Z
h(x)(k(x)−1)K(dx)
=E[L1] + Z
(h(x)k(x)−x)K(dx).
Comparison with equation (2.42) yields that condition (2.40) is satisfied if and only if E0[L1]−E[L1] =
Z
x(1−k(x))K(dx).
(2.43)
For the generalized hyperbolic distribution, the expectation is known explicitly. So E[L1] =à+ βδãKλ+1(δp
α2−β2) pα2−β2Kλ(δp
α2−β2).
Hence ifL1has a generalized hyperbolic distribution with equal parameterδ under both the measureP andP0, we have
E0[L1]−E[L1] =à0−à+ β0δãKλ0+1(δp
α02−β02) pα02−β02Kλ0(δp
α02−β02) − βδãKλ+1(δp
α2−β2) pα2−β2Kλ(δp
α2−β2).
Hence we can rewrite condition (2.43) (and hence condition (2.40)) as a condition on the change of the parameterà:
à0−à= Z
x(1−k(x))K(dx) + βδãKλ+1(δp
α2−β2) pα2−β2Kλ(δp
α2−β2) − β0δãKλ0+1(δp
α02−β02) pα02−β02Kλ0(δp
α02−β02). (2.44)
It will turn out that the right-hand side vanishes, so that, under the assumption δ0 = δ, condition (2.40) is equivalent toà0 =à.
We consider two cases: Case 1:β0 =β = 0. Then(1−k(x))K(dx)is symmetric with respect tox= 0, and the integral in (2.44) vanishes. On the other hand, the remaining expressions on the right-hand side of (2.44) also vanish. Hence (2.44) is equivalent toà0−à= 0.
Case 2: βarbitrary andβ0 = 0, with the remaining parametersλ0 =λandα0 =α. Then alsoà0 =à.
Before we prove this, we will make clear why these two cases indeed suffice to complete the proof. We can decompose a change of parameters(λ, α, β, δ, à);(λ0, α0, β0, δ, à0)into three steps:
(λ, α, β, δ, à);(λ, α,0, δ, à2), (λ, α,0, δ, à2);(λ0, α0,0, δ, à3), and (λ0, α0,0, δ, à3);(λ0, α0, β0, δ, à0).
In the second step the parameteràdoes not change by what we have proved for case 1 above, soà2 =à3. If we can, in addition, show that setting the parameterβto zero does not changeà—this is the statement in case 2—, then we also haveà=à2and, by symmetry,à3=à0. So indeed the situation considered in case 2 above is sufficiently general.
Now we prove the statement for case 2. We have Z
x(1−k(x))K(dx) = Z
x(1−k(x))K(dx)
= Z
x 1
x2ρβ(x)− 1
x2ρβ=0(x)
dx
= Z 1
x
ρβ(x)−ρβ=0(x)
dx
=− Z 1
2π Z 1
i ã i
x ãexiuã d du
χβ0(u)
χβ(u) −χβ=00(u) χβ=0(u)
du dx
= i 2π
Z Z i
x ãexiuã d du
χβ0(u)
χβ(u) −ià− χβ=00(u) χβ=0(u) +ià0
du dx
= i 2π
Z hi
x ãexiuG(u) iu=∞
u=−∞− Z
exiuG(u)du (2.45) dx,
with
G(u) := χβ0(u)
χβ(u) −ià−χβ=00(u) χβ=0(u) +ià0 (2.46)
By (2.14) and (2.18),
χβ=00(u)
χβ=0(u) =ià− ∂
∂u δp
α2+u2
ãKλ+1(δ√
α2+u2) Kλ(δ√
α2+u2)
=ià− δu
√α2+u2 ãKλ+1(δ√
α2+u2) Kλ(δ√
α2+u2) (2.47)
Analogously one can derive the relation χβ0(u)
χβ(u) =ià0+ iδ(β+iu) pα2−(β+iu)2
Kλ+1(δp
α2−(β+iu)2) Kλ(δp
α2−(β+iu)2) . (2.48)
Using relations (2.47) and (2.48), we can rewrite the expression forG(u)given in (2.46).
G(u) =
χβ0(u) χβ(u) −ià
−χβ=00(u) χβ=0(u) −ià0
= δi(β+iu)
pα2−(β+iu)2 ãKλ+1(δp
α2−(β+iu)2) Kλ(δp
α2−(β+iu)2) − δu
√α2+u2 ãKλ+1(δ√
α2+u2) Kλ(δ√
α2+u2)
= iβ
2 ã(λ+ 1/2)(λ+ 3/2)ã 1
u2 +O 1
|u|3
(|u| → ∞).
The last equality follows from the asymptotic expansion of the Bessel functionsKλ+1andKλ. Around u = 0, G(u) is bounded because it is a differentiable (and hence continuous) function. This, together with the limit behavior as |u| → ∞, shows that in (2.45) the term in square brackets vanishes and the integral converges absolutely. Consequently we can continue the chain of equalities in (2.45).
Z
x(1−k(x))K(dx) =− i 2π
Z Z
exiuG(u)du dx
=− i 2π
Z Z
exiuG(u)du eiã0ãxdx
=− δβ
pα2−β2 ãKλ+1(δp
α2+β2) Kλ(δp
α2+β2) − δ
αãKλ+1(δα) Kλ(δα)
Here we have used the fact that the integration with respect toxcan be interpreted as a Fourier integral taken at the point u= 0. So since the functionG(u)is continuous, the Fourier transform of the inverse Fourier transform ofG(u)coincides withG(u). Substituting this into (2.44) completes the proof.
The main difficulty in the proof of Proposition 2.20 was to derive the local behavior of the generalized hyperbolic Lévy measure nearx= 0. For a distribution with a Lévy measure of a simpler structure, it is much easier to derive an analogous result. As an example, consider the class of CGMY Lévy processes.
(See Section A.3.) For this class the Lévy measure is known explicitly and has a simple form.9
Proposition 2.21. LetLbe a CGMY(C, G, M, Y)Lévy process under the measureP. Then the follow- ing statements are equivalent.
(i) There is a measureQloc∼ P under whichLis a CGMY(C0, G0, M0, Y0)Lévy process.
9In fact, the CGMY distributions are defined by giving their Lévy measure.
(ii) Either the CGMYparameters satisfy the relationsC0 =CandY0 =Y, orY, Y0<0.
Proof. The CGMY(C, G, M, Y)Lévy measure has Lebesgue density kCGMY(x) = C
|x|1+Y exp
G−M
2 x−G+M 2 |x|
. (2.49)
(Cf. eq. (A.9).) Conditions (i), (ii), and (iii) of Proposition 2.19 are obviously satisfied if we set k(x) := kCGMY(C0,G0,M0,Y0)(x)
kCGMY(C,G,M,Y)(x) .
Hence a measureQloc∼P with the desired property exists if and only if condition (iv) of Proposition 2.19
holds, that is, iff Z
1−p k(x)2
K(dx)<∞. Using the explicit form (2.49) of the Lévy density, this condition reads Z s C0
|x|1+Y0 exp
G0−M0
2 x−G0+M0 2 |x|
− s
C
|x|1+Y exp
G−M
2 x−G+M 2 |x|2
dx <∞.
It is easy to see that this condition holds ifC =C0 andY =Y0. On the other hand, it does not hold if any of these equalities is violated: First, it is clear that the finiteness of the integral depends only on the local behavior of the integrand aroundx= 0. Since the exponential factor tends to1forx →0, we can skip it for this discussion. So all amounts to the question whether
Z 1
−1
√ C0
|x|(1+Y0)/2 −
√C
|x|(1+Y)/2 2
dx <∞. (2.50)
ForY, Y0 < 0, this integral is obviously finite. If any of the relations Y ≥ 0,Y0 ≥ 0holds, then the integral can only be finite ifC =C0andY =Y0, as was stated above:
• If0 ≤ Y0,Y < Y0, then for|x| → 0the first summand in (2.50) grows faster than the second.
Therefore we have
1 2 ã
√C0
|x|(1+Y0)/2 ≥
√C
|x|(1+Y)/2 for|x| ≤,∈(0,1)small enough. Hence we can estimate
Z 1
−1
√ C0
|x|(1+Y0)/2 −
√C
|x|(1+Y)/2 2
dx≥ Z
−
1 2 ã
√C0
|x|(1+Y0)/2 2
dx= C0 4
Z
−
1
|x|1+Y0dx=∞. The case0≤Y,Y0 < Y may be treated analogously.
• If0≤Y0 =Y, butC0 6=C, then Z 1
−1
√ C0
|x|(1+Y0)/2 −
√C
|x|(1+Y)/2 2
dx= √
C0−√
C2Z 1
−1
1
|x|1+Y dx=∞.