Log Ch.F. for Stable Distributions
General St(α, β, γ, δ):
(
iδω − γ|ω|α + iβγ tan πα
{|ω|α sgn ω − ω} α 6= 1
2
log φX (ω) =
iδω − γ|ω| − π2 iβγω log |ω|
α=1
(
πα
α
iω δ − βγ tan πα
−
γ|ω|
1
−
iβ
tan
sgn
ω
α 6= 1
2
2
=
iω (δ − 2βγ log |ω|/π) − γ|ω|
α=1
0 < α ≤ 2, −1 ≤ β ≤ 1, 0 < γ < ∞, −∞ < δ < ∞.
Standard St(α, β, 1, 0):
(
{|ω|α sgn ω − ω} α 6= 1
−|ω|α + iβ tan πα
2
log φX (ω) =
−|ω| − π2 iβω log |ω|
α=1
Symmetric St(α, 0, γ, 0):
log φX (ω) = −γ|ω|α
Skewed St(α, 1, γ, 0):
(
−γ|ω|α + iγ tan πα
{|ω|α sgn ω − ω} α 6= 1
2
log φX (ω) =
−γ|ω| − π2 iγω log |ω|
α=1
Standard Stable Lévy Measure
We now turn to computing the Lévy measure for the standard St(α, β, 1, 0)
stable distribution. For α < 0 and > 0 set ν+ (u) ≡ u−α−1 e−u on R+ and
1
compute
Z
∞
e
iωu
ν+ (du) =
0
=
∞
Z
eiωu u−α−1 e−u du
Z0 ∞
u−α−1 e−u(−iω) du
0
= Γ(−α)( − iω)α
ω
= Γ(−α)(2 + ω 2 )α/2 e−iα atan Z
∞
ν+ (du) = Γ(−α)α
0
Z ∞
Z ∞
−α−1 −u(−i)
u
e
du
sin u ν+ (du) = =
0
0
= Γ(−α)=( − i)α
= −Γ(−α)(2 + 1)α/2 sin(α atan 1 ).
Thus the ch.f. with Lévy measure given on R+ by ν+ (du) ≡ u−α−1 e−u du is
Z ∞
log φ+ (ω) =
eiωu − 1 − iω sin u ν+ (du)
0
n
ω ω
= Γ(−α) (2 + ω 2 )α/2 cos(α atan ) − i sin(α atan )
o
1
−α + iω(2 + 1)α/2 sin(α atan )
This is analytic in α (away from the nonpositive integers), so we first take
the analytic continuation to the segment 0 < α < 2, then the limit as → 0:
n
o
πα
πα
− i sin
|ω|α sgn ω − ω
(as → 0)
→ Γ(−α) |ω|α cos
2
2
n
o
−π
πα
α
α
=
|ω|
−
i
tan
|ω|
sgn
ω
−
ω
.
2αΓ(α) sin πα
2
2
Similarly the ch.f. with Lévy measure ν− (u) ≡ |u|−α−1 e−|u| du on R− is
Z
0
eiωu − 1 − iω sin u ν− (du)
−∞
n
o
−π
πα
α
α
→
|ω|
+
i
tan
|ω|
sgn
ω
−
ω
2αΓ(α) sin πα
2
2
log φ− (ω) =
2
so for −1 ≤ β ≤ 1 the ch.f. for Lévy measure
ν(du) =
πα
α
Γ(α) sin
(1 + β sgn u) |u|−α−1 du
π
2
(1)
is
log φ(ω) =
Z
∞
−∞
eiωu − 1 − iω sin u ν(du)
= −|ω|α + iβ tan
πα
|ω|α sgn ω − ω ,
2
exactly the Cheng and Liu parametrization of the St(α, β, 1, 0) distribution.
The Lévy measure and ch.f. for case α = 1 follow from a similar argument
or simply from continuity:
ν(du) =
1 + β sgn u
|u|−2 du
π
log φ(ω) = −|ω| −
2βi
ω log |ω|.
π
Four-Parameter Stable Distributions
Cheng and Liu (1997) take the remaining parameters γ and δ of the fourparameter family of stable distributions X ∼ St(α, β, γ, δ) to be scale and
location parameters, respectively, so X = γZ + δ for some Z ∼ St(α,P
β, 1, 0);
∗
∗
with this choice the parameters γ , δ governing the sum X+ ≡
Xi ∼
∗ ∗
St(α, β, γ , δ ) of independent stable random variables Xi ∼ St(α, β, γi , δi )
is exceedingly awkward, and it is difficult to describe the stable random
fields we will need below. We instead take γ to be a rate parameter (and δ
location), leading to the general form
πα
αγ
Γ(α) sin
(1 + β sgn u) |u|−α−1 du,
π
Z ∞ 2
eiωu − 1 − iω sin u ν(du)
log φ(ω) = iδω +
−∞
(
iδω − γ|ω|α + iβγ tan πα
|ω|α sgn ω − ω α 6= 1;
2
=
iδω − γ|ω| − iβγ π2 ω log |ω|
α = 1.
ν(du) =
(2)
(3)
(4)
P
With this choice the sum X+ ≡
Xi of independent stable random variables
Xi ∼ St(α, β, γi, δi ) has a stable distribution
P X+ ∼ St(α,
Pβ, γ+, δ+ ) with
parameters given simply by the sums γ+ = γi and δ+ = δi .
3
Location-Scale Family
If X is a stable random variable X ∼ St(α, β, γ, δ) and Y ≡ η + σX is in the
same location-scale family for some η, σ ∈ R (note σ need not be positive)
then Y ∼ St(α∗ , β ∗ , γ ∗ , δ ∗ ) with
α∗ = α
β ∗ = β sgn σ
γ ∗ = γ|σ|α
(
η + σδ + βγ tan πα
(|σ|α sgn σ − σ) α 6= 1
2
δ∗ =
η + σδ − βγ π2 σ log |σ|
α=1
It is straightforward to solve these for the parameters η and σ needed to
achieve standard form Y ≡ η + σX ∼ St(α, β, 1, 0) with γ ∗ = 1, δ ∗ = 0:
(
−σδ + β tan πα
(σ 1−α − 1) α 6= 1
2
σ = γ −1/α
η=
−σδ + β π2 log σ
α = 1.
Arbitrary Compensators
The Lévy-Khinchine representation
Z
iωu
∗
log φ(ω) = iωδ +
e − 1 − iωh(u) ν(du)
R
may be written with any bounded compensator h(u) = u + o(u2 ) near u ≈ 0.
The choice h(u) = sin u leads to δ ∗ = δ, as in Eqn. (3), while any other
bounded compensator h(u) = u + o(u2 ) may require a different shift,
Z
∗
δ = δ + [h(u) − sin u] ν(du).
For example, the function h(u) ≡ u 1{u2 <c2 } that fully compensates only
jumps of size |u| < c would lead to
Z
∗
δ = δ + [u 1{u2 <c2 } − sin u] ν(du)
1−α
αc
− Γ(2 − α) sin πα
2βγΓ(α) sin πα
2
2
,
α 6= 1
= δ+
π(1 − α)
[log c + γe − 1], for α = 1, where
and (by L’Hôpital’s rule) δ ∗ = δ + 2βγ
π
γe ≈ 0.577216 denotes the Euler-Mascheroni constant.
4
Explicit Examples with α = 3/2
Example St(3/2, 1, 1, 0):
For example, the fully-skewed Stable distribution with index α = 3/2 has
Lévy measure
3
ν(u) = √
u−5/2 ,
u > 0.
2 2π
With the standard compensator h(u) = sin u it has offset δ √
= 0, but with
∗
compensator h(u) = 1{u2 <1} it would have offset δ = 1 − 3/ 2π and, with
√
h(u) = 1{u2 <2 } , δ∗ = 1 − 3/ 2π.
ILM for St(3/2, 1, 1, 0):
A random variable X ∼ St(3/2, 1, 1, 0) may be generated by the ILM algorithm by fixing > 0, constructing a sequence τn of jumps of a standard
Poisson process, setting
ν + (u) = ν [u, ∞)
Z ∞
3
√ x−5/2 dx
=
2 2π
u
√
−3/2
= u
/ 2π,
ν − (t) = sup[u > 0 : ν + (u) > t]
√
= (2π)−1/3 t−2/3 = (t 2π)−2/3
√
υn = ν − (τn ) = (τn 2π)−2/3
Z ∞
X
X =
υn −
h(u) ν(du) + δ ∗
υn >
1
3
u √ u−5/2 du + δ ∗
2 2π
υn >
X
√
√
=
υn − 3(−1/2 − 1)/ 2π + [1 − 3/ 2π]
=
X
υn >
=
X
υn >
υn −
Z
√
υn + 1 − 3/ 2π.
5
√ iid
Equivalently, we may take J ∼ Po −3/2 / 2π and draw υj ∼ Pa(3/2, )
from the Pareto distribution with density (3/2)3/2 u−5/2 , u > , and set
X =
J
X
n=1
√
υn + 1 − 3/ 2π.
Note the series would not converge without compensation; but with compensation, X → X0 ∼ St(3/2, 1, 1, 0) as → 0.
A Stable process may be constructed similarly; if we draw σn ∼ Un(0, 1)
independently of τn , for example, we can set
X
3
−1
X (t) =
υn − t √
2π
υn >,σn ≤t
and again take → 0.
ILM for St(α, β, γ, δ):
A random variable X ∼ St(α, β, γ, δ) may be generated by the ILM algorithm
too, by fixing > 0, constructing a sequence τn of jumps of a standard Poisson
process and a sequence ρn of ±1-valued independent random variables equal
6
to +1 with probability (1 + β)/2, setting
ν + (u) = ν (−∞, −u] ∪ [u, ∞)
γ u−α
=
Γ(1 − α) cos πα
2
−
+
ν (t) = sup[u > 0 : ν (u) > t] =
υn
X
2Γ(α)γ sin πα
2
πt
1/α
2Γ(α)γ sin πα
2
= ν (τn ) =
πτ
Z ∞ n
X
=
υn ρn −
h(u) ν(du) + δ ∗
1/α
−
υn >
πα 1−α
2βγ
−1
Γ(1 + α) sin
π(α − 1)
2
υn >
πα Γ(1 + α)
πα 2
+ sin
+δ
−βγ tan
2
π
2 α−1
X
2αΓ(α) sin πα
πα
2 1−α
+ tan
=
υn ρn − βγ
+δ
π(α
−
1)
2
υ >
X
=
υn ρn −
n
A Stable process may be constructed similarly on [0, T ] by drawing σn ∼
Un(0, T ) independently of τn and setting
X
2αΓ(α) sin πα
πα
1/α
2 1−α
X (t) ≡ T
+ δt,
υn ρn − βγt
+ tan
π(α
−
1)
2
υ >,σ ≤t
n
n
0 ≤ t ≤ T . A d-dimensional Stable RF may be constructed on [0, T ]d by
X
X [φ] ≡ T d/α
υn ρn φ(σn )
υn >
Z
2αΓ(α) sin πα
πα
2 1−α
+ tan
− βγ
−δ
φ(s) dd s,
π(α − 1)
2
d
[0,T ]
for integrable φ ∈ L1 [0, T ]d . The distribution can be generalized somewhat
by replacing γ and δ by nonnegative measures (on an arbitrary measure
space), and α and β by measurable functions. The distribution really ought
to be parametrized by γ α , not γ, changing the standardization relation to
X = δ + γ 1/α Z,
Z ∼ St(α, β, 1, 0).
7
Truncation Error Bounds:
For η < < 1 the difference Xη − X is given by
πα X
α 2Γ(1 + α) sin 2
Xη − X =
υn ρn − βγ
η 1−α − 1−α ,
π(α − 1)
η<υ ≤
n
a mean-zero random variable with variance
Z 2
E(Xη − X ) =
u2 ν(du)
η
Z 2γ α Γ(1 + α) sin πα
2
u1−α
=
π
η
πα
α
2γ Γ(1 + α) sin 2 2−α
− η 2−α
=
π(2 − α)
α
2γ Γ(1 + α) sin πα
2
→
2−α
as η → 0,
π(2 − α)
so X converges in L2 as → 0 with the indicated L2 truncation bound.
SimilarRbounds apply to Stable processes and fields, with an additional factor
of t or φ2 respectively.
References
Cheng, R. C. H. and Liu, W. B. (1997) A continuous representation of the
family of stable law distributions. J. Roy. Stat. Soc. B, 59, 137–145.
8