The Stable Limit Theorem
Thomas Roy∗
September 16, 2012
1
Introduction
Stable distributions are a class of probability distributions that have many useful
mathematical properties such as skewness and heavy tails. They have been used
as a model for different types of physical and economic systems. In practice it
has been difficult to use stable distributions due to the fact that the densities
and distribution functions have closed formulas only in specific cases (Gaussian,
Cauchy and Lévy). However, thanks to recent reliable computer programs, it is
possible to compute stable densities, distribution functions and quantiles. The
focus of this paper will be the Stable Limit Theorem. In the course of this
paper, we will try to rewrite the proof written by William Feller [1] and add
certain details that were omitted in his book. Afterwards, we will give some
examples where the Stable Limit Theorem can be applied.
2
Stable Distributions
Definition 2.1 (Definition VI.3.1 of [1]). A distribution F is infinitely divisible
if for each n it can be represented as the distribution of the sum Sn = X1,n +
... + Xn,n of n independent random variables with common distribution Fn .
We denote by ϕY the characteristic function of a random variable Y , i.e.
ϕY (u) = E(eiuY ), where u ∈ R. The next result gives the representation of
the characteristic function of an infinitely divisible random variable using an
integral with respect to a canonical measure. (See Definition A.2, Appendix A,
for the definition of a canonical measure.)
Theorem 2.2 (Theorem XVII.2.1 of [1]). The class of characteristic functions
of infinitely divisible distributions coincides with the class of functions of the
form:
Z ∞ iux
e − 1 − iu sin x
ϕY (u) = E(eiuY ) = exp iub +
M
(dx)
(1)
x2
−∞
∗ Project supported by an NSERC Undergraduate Research Award at University of Ottawa,
during the summer of 2012. (Supervisor: Raluca Balan)
1
where M is a canonical measure and b is a real number. In this case, we write
X ∼ IDF eller (b, M ).
−u2
Remark 2.3. In the right-hand side of (1), the integrand is defined to be
2
when x = 0. Hence, if M = Cδ0 , where δ0 denotes the Dirac measure at 0, then
Y has a normal N (b, C) distribution.
Remark 2.4. There is an alternative way of writing the characteristic function
of an infinitely divisible distribution as:
Z
σ 2 u2
ϕY (u) = exp iua −
+ (eiux − 1 − iux1{|x|≤1} )ν(dx)
2
R
where a ∈ R, σ 2 ≥ 0 and ν is a Lévy measure on R, i.e. ν satisfies:
Z
(x2 ∧ 1)ν(dx) < ∞ and ν({0}) = 0.
R
This is called the Lévy representation. Here a ∧ b denotes the minimum between
a and b. In this case, we write Y ∼ ID(a, σ 2 , ν). Note that if Y ∼ ID(a, σ 2 , ν)
then Y ∼ IDF eller (b, M ) where
Z
b = a + (sin x − x1{|x|≤1} )ν(dx) and M (dx) = x2 ν(dx) + σ 2 δ0 (dx)
R
Definition 2.5 (Definition VI.1.1 of [1]). P
Let F be the law of a random variable
∞
X, (Xi )i≥1 be i.i.d. copies of X and Sn = n=1 Xi . The distribution F is stable
if for each n there exist constants cn > 0, dn ∈ R such that
d
Sn = cn X + dn
(2)
and F is not concentrated at one point.
Theorem 2.6 (Theorem VI.1.1 of [1]). The norming constants in (2) are of
the form cn = n1/α with 0 < α ≤ 2.
Definition 2.7. A random variable Y with characteristic funtion ϕY (u) =
eψY (u) , u ∈ R, where ψY (u) is given by

πα iuδ − |u|α γ α 1 − isgn(u)β tan
, α 6= 1
2
(3)
ψY (u) =
2
iuδ − |u|γ(1 + isgn(u)β ln |u|),
α=1
π
for some α ∈ (0, 2], γ > 0, β ∈ [−2, 2] and δ ∈ R is said to have an α-stable
Sα (γ, β, δ) distribution. In this case, we write Y ∼ Sα (γ, β, δ).
Using Example XVII.3.(g)-(h) of [1], we obtain the following lemma.
2
Lemma 2.8. Assume that Y ∼ IDF eller (0, Mα ), i.e. its characteristic function
is given by (1) where 0 < α < 2 and
Mα (dx) = [c+ x−α+1 1(0,∞) (x) + c− (−x)−α+1 1(−∞,0) (x)]dx
for some c+ , c− ≥ 0. Then Y ∼ Sα (γ, β, δ) where the parameters (γ, β, δ)
given by:

πα
Γ(3 − α)

(c+ + c− )
cos
, α 6= 1
α
α(1 − α)(2 − α)
2
γ =
π

(c+ + c− ) ,
α=1
2
c+ − c−
β=
c+ + c−

Z ∞
sin x


dx,
0<α<1
−(c+ − c− )
α+1

x

0
Z ∞
x − sin x
δ = (c − c )
dx, 1 < α < 2
−
 +

xα+1

0


0,
α=1
(4)
are
(5)
(6)
(7)
If Y ∼ IDF eller (b, Mα ) where b ∈ R and Mα is given by (4), then Y ∼
Sα (γ, β, δ 0 ) where δ 0 = b + δ and (γ, β, δ) are the same as above.
Remark 2.9. The constant δ in (7) can also be expressed as:
 Z
sin x


−
Mα (dx),
0<α<1


Z R x 2
x − sin x
δ=
Mα (dx), 1 < α < 2


x2

R

 0,
α=1
(8)
To see this, note that if α ∈ (0, 1)
Z ∞
Z ∞
sin x −α+1
sin x −α+1
δ = − c+
x
dx
−
c
x
dx
−
x2
x2
0
0
Z ∞
Z 0
sin x −α+1
sin x −α+1
= − c+
x
dx + c−
x
dx
2
x2
0
−∞ x
Z
sin x
= −
Mα (dx)
2
R x
The cases α ∈ (1, 2) and α = 1 are similar.
Lemma 2.10. Assume that Y ∼ ID(0, 0, να ), where the measure να is given
by:
να (dx) = [c+ x−α−1 1(0,∞) (x) + c− (−x)−α−1 1(−∞,0) (x)]dx
(9)
3
for some c+ , c− ≥ 0 and α ∈ (0, 2). Then Y ∼ Sα (γ, β, δ 0 ) where (γ, β) are
given by (5) and (6), and
 Z


− x1{|x|≤1} να (dx)
0<α<1



Z R
x1{|x|≥1} να (dx),
1<α<2
δ0 =
(10)

ZR




 (sin x − x1{|x|≥1} )να (dx), α = 1
R
Proof. By Remark 2.4, Y ∼ IDF eller (b, Mα ), where M = Mα (Mα is given by
(4)) and
Z
b = (sin x − x1{|x|≤1} )να (dx)
R
By Lemma 2.8, Y ∼ Sα (γ, β, δ 0 ), where (γ, β) are given by (5) and (6), δ 0 = b+δ
and δ is given by (8) (see Remark 2.9). Hence, when α < 1
Z
Z
Z
δ 0 = (sin x − x1{|x|≥1} )να (dx) −
sin x να (dx) = − x1{|x|≤1} να (dx)
R
R
R
The desired expression for δ 0 in the case α ∈ (1, 2) or α = 1 follows similarly.
3
Domain of attraction
Definition 3.1. Let (Xi )i≥1 be independent
identically distributed (i.i.d.) ranPn
dom variables with law F and Sn = i=1 Xi . Let Y be a random variable with
law G. We say that X belongs to the domain of attraction of Y if there exist
an > 0 and bn ∈ R such that:
Sn
d
− nbn −
→Y
an
(11)
Remark 3.2. Note that (11) is equivalent to
n(ϕn (u) − 1) → ψY (u)
for any u ∈ R
(12)
where ϕn is the characteristic function of Fn , the probability distribution of
X
− bn , i.e.
an
u
ϕn (u) = ϕ X −bn (u) = ϕX
e−iubn
an
an
and ϕY (u) = eψY (u) . To see this, note that by the continuity of characteristic
functions, (11) is equivalent to:
ϕ Sn −nbn (u) → ϕY (u)
an
4
for any u ∈ R.
(13)
Sn
− nbn is given by:
an




n 
X
X
j
= E exp iu
− bn 


a
n
j=1
n
X
− bn
=
E exp iu
an
= [ϕn (u)]n .
Note that the characteristic function of
Sn
− nbn
E exp iu
an
Then (13) is equivalent to:
n
n(ϕn (u) − 1)
[ϕn (u)]n = 1 +
→ ϕY (u) = eψY (u)
n
for any u ∈ R
which is in turn equivalent to (12).
We consider the function:
U (x) = E X 2 1(|X|≤x) ,
x > 0.
(14)
Theorem 3.3. Let X,
Pn(Xi )i≥1 be i.i.d. random variables with common distribution F and let Sn = i=1 Xi . Suppose that there exist an > 0 and bn ∈ R such
that (11) holds. Then Y has a stable distribution with characteristic function
(1) where
X
b = lim nβn , βn = E sin
− bn
n→∞
an
M = Cδ0 or M = Mα and Mα is given by (4).
Proof. We denote
Mn (dx) = nx2 Fn (dx)
(15)
We consider separately two cases:
Case 1:
Assume that F is symmetric, (i.e. P (X < −x) = P (X > x) for any x > 0)
X
.
and (11) holds with bn = 0. In this case, Fn is the probability distribution of
an
Note that, if F is symmetric, we consider (Xi∗ )i≥1 , an independent copy of
(Xi )i≥1 and we let
X̃i = Xi − Xi∗ .
Then (X̃i )i is a sequence of i.i.d. random variables with a symmetric distribution
and
n
X
S̃n =
X̃i = Sn − Sn∗
i=1
5
where Sn∗ =
Pn
i=1
Xi∗ . By Theorem 2.8 of [2]
Sn
Sn∗
d
− nbn ,
− nbn −
→ (Y, Y ∗ )
an
an
where Y ∗ is an independent copy of Y . By applying the Continuous Mapping
Theorem of [2] to the map of h : R2 → R, h(x, y) = x − y, we get:
∗
Sn
Sn
S̃n
=
− nbn −
− nbn
an
an
an
d
−
→ Y − Y ∗ = Ỹ .
Hence, the symmetrized sequence (X̃i )i satisfies (11) with bn = 0 and the same
an as the original sequence (Xi )i≥1 .
Using Corollary A.4, with Cn = n and bn = 0, we get, by (12) that there
exist a canonical measure
M and b ∈ R such
Mn → M properly and
that
Z
X
. In this case
nβn → b, where βn =
sin xFn (dx) = E sin
an
R
Z +∞ iux
e − 1 − iu sin x
M (dx).
ψY (u) = iub +
x2
−∞
We now prove various properties of the sequence (an )n , which are consequences
Sn d
of the fact that
−
→Y.
an
Since Mn → M properly, we have (according to Definition A.2.(i))
Mn ([−x, x]) → M ([−x, x])
for any continuity point x of M.
Using the definition (15) of Mn and the fact that in this case Fn is the distriX
bution of
, we have:
an
Z
Mn ([−x, x]) =
1[−x,x] (y)ny 2 Fn (dy)
R
2
n
X
1
= 2 E X 2 1{|X|≤an x} .
= nE
|X|
≤x
2
{
}
an
an
an
We obtain that (an )n must satisfy:
n
U (an x) → M ([−x, x])
a2n
for any continuity point x of M.
From Definition A.2.(ii), we get:
Z ∞
Z ∞
1
1
+
+
Mn (dy) → M (x) =
(dy)
Mn (x) =
2
2
y
y
x
x
6
(16)
for any continuity point x of M. Note that:
Z ∞
Z ∞
Z
1
1 2
M
(dy)
=
ny
F
(dy)
=
1(x,∞) (y)nFn (dy)
n
n
y2
y2
x
x
R X
X
= nE 1(x,∞)
= nP
>x
an
an
= nP (X > an x) = n(1 − F (an x))
We obtain:
n[1 − F (an x)] → M + (x)
for any continuity point x of M.
(17)
Similarly,
nF (−an x) → M − (−x)
for any continuity point x of M.
Due to (12), we have ϕn (u)−1 = n(ϕn (u)−1) n1 → 0 i.e. ϕn (u) = ϕX ( aun ) →
u 1. This proves that ϕX ( aun ) = E ei an X → 1. Hence, Feller concludes in [1]
u
(he details of Feller’s reasoning are still not clear to us), that
→ 0 i.e.
an
an → ∞.
It remains to show the details of Feller’s reasoning.
We now prove that
an+1
→1
an
To see this, note that
Y
n
n
Xj
Sn
u
E exp iu
ϕ Sn (u) = E exp iu
=
= ϕX
= [ϕn (u)]n → ϕY (u).
an
an
a
a
n
n
j=1
Similarly,
n
u
1
= [ϕn+1 (u)]n = [ϕn+1 (u)]n+1
→ ϕY (u).
ϕ Sn (u) = ϕX
an+1
an+1
ϕn+1 (u)
Sn d
Sn d
an+1
−
→ Y and
−
→ Y . So, Feller concludes that
→ 1.
an
an+1
an
It looks intuitively right, but it remains to show the details of to prove this
statement.
Then we get
n
, χ(x) = M ([−x, x])
a2n
and D the set of continuity points of M, we obtain that U is regularly varying of
index ρ ∈ R and M ([−x, x]) = Cxρ , for some C > 0. Note that ρ ≥ 0. (If ρ < 0
Using (16) and Lemma A.7 (Appendix A) with λn =
7
then M (R) = limx→∞ M ([−x, x]) = limx→∞ Cxρ = 0, which is impossible). We
denote ρ = 2 − α for some α ≤ 2. Hence
U (x) = x2−α l(x)
(18)
for a slowly varying function l, and
M ([−x, x]) = Cx2−α
(19)
If α = 2, then M ([−x, x]) = C for any x > 0 i.e. M = Cδ0 .
If α < 2, then M is given by:
1 −α+1
1
M (dx) = C(2 − α)
x
1(0,∞) (x) + (−x)−α+1 1(−∞,0) (x) dx
2
2
because
Z x
1
1 −α+1
y
1(0,∞) (y) + (−y)−α+1 1(−∞,0) (y) dy
C(2 − α)
2
2
−x
(20)
=
Cx2−α
=
M ([−x, x]).
In this case, M has no atom at the origin, i.e. M ({0}) = 0.
Z ∞
1
+
Since M is a canonical measure, M (x) =
M (dy) < ∞, for any x > 0.
2
y
y
Using (20), we have
Z ∞
Z
1
1 −α+1
C(2 − α) ∞ −α−1
M + (x) =
C(2
−
α)
y
dy
=
y
dy
y2
2
2
x
x
C(2 − α) −α
=
x .
2α
We obtain that:
M + (x) =
C 2 − α −α
x ,
2 α
for any x > 0.
(21)
Case 2:
Assume now that F is not symmetric. Using (12) and Corollary A.4 with Cn = n
and b̃n = 0, we conclude that there exist a canonical measure M and b ∈ R such
that
Mn → M properly and nβn → b,
where
X
βn =
sin(x)Fn (dx) = E sin
− bn .
an
R
Z
In this case
Z
+∞
ψY (u) = iub +
−∞
eiux − 1 − iu sin x
M (dx).
x2
8
Since Mn → M properly, we have:
1. Mn ([−x, x])Z→ M ([−x, x])
Z ∞
∞
1
1
+
2. Mn (x) =
Mn (dy) →
M (dy) = M + (x) for any continuity
2
2
y
y
x
x
point
Z ∞ x of M . We note that, similarly to Case 1,
1
Mn (dy) = n[1 − F (an (x + bn ))].
y2
x
Z −x
Z −x
1
1
3. Mn− (−x) =
M
(dy)
→
M (dy) = M − (−x) for any continuity
n
2
2
y
y
−∞
−∞
Z −x
1
Mn (dy) = n[F (an (−x + bn ))].
point x of M . Similarly,
2
−∞ y
Using the definition (15) of Mn and the fact that, in this case, Fn is the distriX
bution of
− bn , we see that
an
#
"
2
Z
X
2
Mn ([−x, x]) =
1[−x,x] (y)ny Fn (dy) = nE
− bn 1{| X −bn |≤x}
an
an
R
Therefore from 1. we obtain that
#
"
2
X
nE
− bn 1{| X −bn |≤x} → M ([−x, x])
an
an
which is the analogue of (27) in the non-symmetric case. On the other hand:
Z ∞
Z
1 2
+
Mn (x) =
ny Fn (dy) =
1(x,∞) (y)nFn (dy)
y2
R
x
X
X
= nE 1(x,∞)
− bn
− bn > x
= nP
an
an
= nP (X > an (x + bn )) = n[1 − F (an (x + bn )].
Similarly, Mn− (−x) = nF [an (−x + bn )].
Using 2. and 3. we get:
n[1 − F (an (x + bn ))] → M + (x)
and
nF (an (−x + bn )) → M − (−x)
(22)
for any x > 0 such that −x and x are continuity points of M .
Note that
bn → 0
To see this, recall that ϕn (u) → 1 (due to (12)). But ϕn (u) = ϕX ( aun )e−iubn
and ϕX ( aun ) → 1 since an → ∞. (Note that we proved that an → ∞ in the
case when F is symmetric and bn = 0. In the general case, the symmetrized
sequence (X̃i )i≥1 satisfies (11) with the same an as (Xi )i . Hence by Case 1,
an → ∞. ) Hence e−iubn → 1, i.e. bn → 0.
9
We claim that, due to the fact that bn → 0, we have:
n[−F (an (x + bn )) + F (an x)] → 0 and n[F (an (−x + bn )) − F (−an x)] → 0.
It remains to show this claim is true. Hence,
n[1 − F (an x)] → M + (x) and n[F (−an x)] → M − (−x)
(23)
for any x > 0 such that −x and x are continuity points of M . This conclusion
also needs to be proven.
We now use Lemma A.7 three times:
1. Since n(1 − F )(an x) → M + (x) for any x > 0 continuity point of M ,
we have two situations: either M + (x) = 0 for any x or 1−F is regularly varying
of index −α+ and M + (x) = A+ x−α+ for some A+ > 0. Note that, because
0 = limx→∞ M + (x) = A+ limx→∞ x−α+ , it is necessary that α+ > 0.
2. Since n(−F )(an x) → M − (−x) for any x > 0 such that −x is a continuous point of M , we have two situations: either M − (−x) = 0 for any x or −F is
regularly varying of index −α− and M − (−x) = A− x−α− for some A− > 0.
Note that, similarly, α− > 0.
3.
nP (|X| > an x) → M + (x) + M − (−x)
(24)
for any x > 0 such that −x and x are continuity points of M . We denote
V (x) = P (|X| > x) (V is called the tail sum of X). We have two situations:
either M + (x) + M − (−x) = 0 for any x or V is regularly varying of index −α
and M + (x) + M − (−x) = Ax−α for some A > 0. Note that, similarly, α > 0.
If M + and M − do not vanish indentically (i.e. M + (x) = M − (−x) > 0 for
some x), we get:
A+ x−α+ + A− x−α− = Ax−α
for any x
Hence α+ = α− = α (i.e. we have the same exponent α for both tails) and
A+ + A− = A.
To summarize, we have two cases: either M + (x) = M − (−x) = 0 for any x
(i.e. M = Cδ0 ) or, M = Mα where Mα is given by (20) for some α > 0 and
c+ , c− > 0.
Note that if M
Z x= Mα , then α < 2. We get this by using the fact that
M ([0, x]) = c+
y −α+1 dy < ∞ for any x > 0.
0
10
c+ 2−α
Remark 3.4. When α ∈ (0, 2), Mα ([0, x]) =
x
and M ([−x, 0]) =
2−α
c−
. This gives us:
2−α
Mα ([−y, x]) = Cpx2−α + Cqy 2−α
where
C=
c+ + c−
,
2−α
p=
c+
c+ + c−
and q =
c−
.
c+ + c−
(25)
(26)
When α ∈ (0, 2), we also have:
1
2 − α −α
1
2 − α −α
Mα+ (x) = c+ x−α = Cp
x
x .
and Mα− (−x) = c− x−α = Cq
α
α
α
α
(27)
Relations (25) and (27) also hold when α = 2, replacing Mα by M = Cδ0 .
When M = Cδ0 , the characteristic function of Y is
E[eiuY ] = exp{iub −
u2
C}
2
i.e. Y has a normal distribution with mean b and variance C.
When M = Mα , the log-characteristic function of Y is given by:
Z +∞ iuα
e
− 1 − iu sin x
ψY (u) = iub +
Mα (dx), i.e. Y ∼ IDF eller (b, Mα )
x2
−∞
where b = limn→∞ nβn and βn = E sin( aXn − bn ) . In this case, according to
Lemma 2.8, Y has Sα (γ, β, δ 0 ) distribution, with δ 0 = δ + b.
Theorem 3.5 (Theorem XVII.5.1 of [1]). (0) A distribution G posesses a domain of attraction if and only if it is stable.
(i) The class of stable distributions coincides with the class of infinitely divisible
distributions with canonical measures given by M = M δ0 or M = Mα , where
Mα is given by (4) for some c+ , c− ≥ 0, 0 < α < 2.
(ii) The log-characteristic function of a stable distribution is given by (3) for
some (α, γ, β, δ) with α ∈ (0, 2], γ > 0, β ∈ [−1, 1], δ ∈ R.
(iii) If X has a stable distribution of index α, then
(
xα P (X > x) → Cp 2−α
as x → ∞
α
xα P (X < −x) → Cq 2−α
as x → ∞
α
where M ([−y, x]) = Cpx2−α + Cqy 2−α for some C > 0, p, q ≥ 0, p + q = 1.
Proof. (0) By Theorem 2.6, if a random variable Y possesses a domain of attraction, then Y has a stable distribution.
Conversely, suppose that Y has a stable distribution. We show that Y belongs to
its own domain of attraction. Let (Yi )i be i.i.d. (as Y ). Let Sn = Y1 + ... + Yn .
11
d
By the definition of the stable distribution, Sn = n1/α Y + dn for a constant
Sn − dn d
d
dn ∈ R. Hence
=Y −
→Y.
n1/α
(i) Suppose Y has a stable distribution. By (0), there exist i.i.d. random varid
→ Y . By Theorem 2.6 Y ∼ IDF eller (b, M ),
ables (Xi )i≥1 such that Sann − bn −
with M = Mα or M = Cδ0 .
Conversely, suppose that Y ∼ IDF eller (b, M ). By Lemma 2.8, Y has a logcharacteristic function (3).
(ii) We show that Y has a stable distribution if and only if Y has a logcharacteristic function (3).
Suppose Y has a stable distribution. Then by (i), Y ∼ IDF eller (b, M ) with
M = Mα or M = Cδ0 . By Lemma 2.8, Y has a log-characteristic function (3).
Suppose that Y has a log-characteristic function (3). Let Y1 , ...Yn be i.i.d. copies
of Y . We want to prove that there exist cn > 0 and dn ∈ R such that:
d
Y1 + ... + Yn = cn Y + dn
By Theorem 2.6, if such a cn exists, then cn = n1/α . We have:
h
i
ϕY1 +...+Yn (u) = E eiu(Y1 +...+Yn ) = [E(eiuY )]n
=
[ϕY (u)]n = enψY (u)
and
ϕcn X+dn (u)
h
i
= E eiu(cn Y +dn ) = eiudn E[eiucn Y ]
= eiudn ϕY (cn u) = eiudn +ψY (cn u)
It suffices to show that:
nψY (u) = iudn + ψY (n1/α u)
If α 6= 1, we have by (3) :
h
πα i
nψY (u) = niuδ − n|u|α γ α 1 − isgn(u)β tan
2
h
πα i
1/α α α
= iu(n − 1)δ + iuδ − |n u| γ 1 − isgn(n1/α u)β tan
2
= iudn + ψY (n1/α u)
with
dn := iu(n − 1)δ
The calculation for α = 1 is similar.
(iii) Suppose that Y has a stable distribution. By (0), Y belongs to its own
Sn
d
domain of attraction, i.e. 1/α − bn −
→ Y , where Sn = Y1 + ... + Yn and (Yi )i≥1
n
are i.i.d. copies of Y . By (23) and (27), it follows that, for any x > 0,
n[1 − F (n1/α x)] → M + (x) = Cp
12
2 − α −α
x
α
and
n[F (−n1/α x)] → M − (x) = Cq
2 − α −α
x .
α
Hence, for any x > 0,
(n1/α x)α P (Y > n1/α x) → Cp
2−α
α
and
(n1/α x)α P (Y < −n1/α x) → Cq
2−α
.
α
Denote n1/α x = xn . We get:
1/α
xα
x) → Cp
n P (Y > n
2−α
α
and
1/α
xα
x) → Cq
n P (Y < −n
2−α
α
If limx→∞ xα P (X > x) then limx→∞ xα P (X > x) = limn→∞ xα
n P (X > xn ) for
any sequence (xn )n , such that xn → ∞. Take xn = n1/α x.
We still need to see when the limit exists and why Feller makes these conclusions when it does.
4
Conditions for X belonging to the domain of
attraction of Stable(α)
Lemma 4.1. Let X be an arbitrary random variable. For any x > 0, define
U (x) = E X 2 1{|X|≤x} and V (x) = P (|X| > x). Suppose that E(X 2 ) = ∞,
i.e. limx→∞ U (x) = ∞.
(i) If U is regularly varying with index 2 − α for some 0 < α ≤ 2 or V is
regularly varying with index −α, for some 0 < α ≤ 2, then
x2 P (|X| > x)
2−α
=
.
x→∞
U (x)
α
lim
(28)
(ii) Conversely, if (28) holds for some 0 < α < 2, then there exists a slowly
varying function L0 such that:
U (x) ∼ αx2−α L0 (x) and P (|X| > x) ∼ (2 − α)x−α L0 (x).
If (28) holds for α = 2, then U is slowly varying.
Proof. Apply Theorem A.8 with a = 2 and b = 0.
Theorem 4.2 (Theorem XVII.5.2
of [1]). Let X be a random variable with law
F and U (x) = E X 2 1{|X|≤x} .
(i) If X belongs to some domain of attraction then:
U (x) = x2−α l(x)
13
(29)
for some 0 < α ≤ 2 and a slowly varying function l.
(ii) Assume that X is non-degenerate and U is slowly varying (i.e. (29) holds
for α = 2). Then X belongs to the domain of attraction of a normal distribution.
(iii) Assume that (29) holds for some 0 < α < 2. Then X belongs to the domain
of attraction of a stable law of index α if and only if the tails are balanced, i.e.
lim
x→∞
P (X > x)
=p
P (|X| > x)
and
lim
x→∞
P (X < −x)
=q
P (|X| > x)
(30)
for some p, q ≥ 0 with p + q = 1.
Sn
d
− nbn −
→Y,
Proof. (i) By Definition 3.1, there exist (an ) and (bn ) such that
a
n
Pn
where Sn =
i=1 Xi and (Xi )i≥1 are i.i.d. copies of X. By Theorem 3.3,
Y ∼ IDF eller (b, M ) where M = Mα or M = Cδ0 and Mα is given by (4).
Assume M = Mα with 0 < α < 2. By (23) and (27):
nP (|X| > an x) → M + (x) + M − (−x) = C
2 − α −α
x .
α
By Lemma A.7, it follows that V (x) = P (|X| > x) is regularly varying with
index −α. By Lemma 4.1.(i), (28) holds. By Lemma 4.1.(ii), U is regularly
varying with index 2 − α, i.e. (29) holds.
Assume that M = Cδ0 . In this case, M + (x) = M − (−x) = 0 for any x > 0 and
Y has a normal distribution. By (23),
nP (|X| > an ) → 0.
(31)
Sn
d
We now prove that U is slowly varying. Since
− nbn −
→ Y , it follows that
an
Sn
does not converge in probability to 0. Moreover,
an n≥1
|Xk |
> 1 for some k ≤ n → 0.
(32)
nP
an
This statement needs to be proven in details. We will prove that there exist
C1 > 0 and n0 ≥ 1 such that
2
n
Xk
nU (an ) X
=
E
1{|Xk |≤an } > C1 for any n ≥ n0 , n ∈ N
(33)
a2n
a2n
k=1
where N ⊂ N is a subsequence. Using (31) and (33), it follows that
U (an )
nU (an )
1
=
→ ∞.
a2n P (|X| > an )
a2n nP (|X| > an )
Hence
a2n P (|X| > an )
→0
U (an )
14
and relation (28) holds with α = 2. By Lemma 4.1.(ii),
U is slowly varying. It
Sn
does not converge
remains to prove (33). To see this, note that since
an n≥1
in probability to 0, there exists 0 > 0, C0 > 0 and a subsequence N ⊂ N such
that
P (|Sn | > 0 an ) > C0 for any n ∈ N.
(34)
Writing Xk = Xk 1{|Xk |≤an } + Xk 1{|Xk |>an } , we obtain that
n
n
X X
Sn X Xk
k
≤
+
1
1
{|X
|≤a
}
{|X
|>a
}
n
n k
k
an an
an
k=1
and hence
|Sn |
P
> 0 ≤ P
an
k=1
n
!
X X
k
0
.
1{|Xk |>an } >
an
2
k=1
k=1
(35)
By Chebyshev’s inequality (and assuming that Xk has a symmetric distribution)
n
!
X X
k
0
+P
1{|Xk |≤an } >
an
2
2
!
n
n
X
Xk
4 X Xk
4 nU (an )
0
P ≤ 2E .
1{|Xk |≤an } >
1{|Xk |≤an } = 2
an
2
0 an
0 a2n
k=1
k=1
(36)
On the other hand, for the second term on the right hand side of (35) we have
( n
)
X X
k
0
⊂ {|Xk | > an for some k ≤ n}
1{|Xk |>an } >
an
2
k=1
since if |Xk | < an for all k ≤ n then
n
X
Xk
k=1
an
1{|Xk |>an } = 0, which cannot be
0
. Hence, by (32)
2
n
!
X X
k
0
1{|Xk |>an } >
≤ lim P (|Xk | > an
n→∞
an
2
greater than
lim P
n→∞
for some k ≤ n) = 0.
k=1
(37)
Using (34), (35), (36) and (37) it follows that
|Sn |
4
nU (an )
C0 < limn→∞, n∈N P
> 0 ≤ 2 limn→∞, n∈N
an
0
a2n
nU (an )
2
≥ C0 0 =: C1 for any
2
an
4
n ≥ n0 , n ∈ N . This concludes the proof of (33).
Hence there exists an integer n0 ≥ 1 such that
(ii) This will follow from Theorem 4.12 (below).
15
(iii) Assume first that X belongs to the domain of attraction of a stable law of
index
α < 2. By (23), (24) and (27)
(
−α
nP (X > an x) → M + (x) = Cp 2−α
and
α x
+
−
−α
nP (|X| > an x) → M (x) + M (−x) = C 2−α
.
α x
We then get:
P (X > an x)
M + (x)
→ +
=: p
P (|X| > an x)
M (x) + M − (−x)
for any x > 0
P (X > x)
P (X > x)
P (X > xn )
exists, then limx→∞
= limn→∞
P (|X| > x)
P (|X| > x)
P (|X| > xn )
for any xn → ∞. We take xn = an .
If limx→∞
Assume next that (30) holds. The fact that X belongs to domain of attraction of a stable law will follow by Theorem 4.12 below.
We still need to know when the limit exists.
Theorem 4.2 has two important corollaries.
Corollary 4.3. A non-degenerate random variable X belongs to the domain of
attraction of the normal distribution (and we write X ∈ DAN ) if and only if U
is slowly varying, or equivalently:
x2 P (|X| > x)
=0
x→∞
U (x)
lim
(38)
Proof. We first show that X ∈ DAN if and only if U is slowly varying.
Assume X ∈ DAN . By Theorem 4.2.(i), U is slowly varying.
Assume that U is slowly varying. By Theorem 4.2.(ii), X ∈ DAN .
Note that, by Lemma 4.1, U is slowly varying if and only if (38) holds.
Corollary 4.4. A random variable X belongs to the domain of attraction of a
stable law of index α, with 0 < α < 2, if and only if the following conditions
hold:
(i) P (|X| > x) ∼ x−α L(x), for some slowly varying function L.
P (X > x)
P (X < −x)
(ii)
→ p and
→ q for some p, q ≥ 0, p + q = 1.
P (|X| > x)
P (|X| > x)
Condition (i) is equivalent to (28).
Proof. Assume first that X belongs to the domain of attraction of a stable
law of index α, with 0 < α < 2. Then U (x) ∼ x2−α l(x) (by Theorem
4.2.(i)). By Lemma 4.1.(i), (28) holds. By Lemma 4.1.(ii), with αL0 = l,
−α
l(x). Hence condition (i) holds with L = 2−α
P (|X| > x) ∼ 2−α
α x
α l. Condition (ii) holds by Theorem 4.2.(iii).
Assume (i) and (ii) hold. By (i), V (x) = P (|X| > x) is regularly varying
with index −α. By Lemma 4.1.(i), (28) holds. By Lemma 4.1.(ii), (29) holds.
16
By Theorem 4.2.(iii), (29) and (ii) imply that X belongs to the domain of attraction of a stable law of index α, with 0 < α < 2.
The fact that (i) is equivalent to (28) follows by Lemma 4.1.
The following result is known in the literature as Karamata Theorem.
Lemma 4.5. (i) Assume that U (x) = E X 2 1{|X|≤x} is regularly varying of
index 2 − α, where 0 < α ≤ 2. Then, for any β < α, E(|X|β ) < ∞ and
x2−β
2−α
E |X|β 1{|X|>x} =
.
x→∞ U (x)
α−β
lim
(39)
(ii) Assume that V (x) = P (|X| > x) is regularly varying of index −α, where
0 < α < 2. Then, for any β > α, E(|X|β ) = ∞ and
E |X|β 1{|X|≤x}
α
=
(40)
lim
x→∞
xβ P (|X| > x)
β−α
Proof. (i) Relation (39) follows using
b = β,
Theorem A.8 withβ a = 2 and
2
. Because
and
V
(x)
=
E
|X|
1
i.e. U (x) = U2 (x)
=
E
X
1
β
{|X|>x}
{|X|≤x}
<
∞,
we
get:
E |X|β 1{|X|≤x} ≤ xβ < ∞ and E |X|β 1{|X|>x}
E(|X|β ) = E |X|β 1{|X|≤x} + E |X|β 1{|X|>x} < ∞
(ii) Relation (40) follows using Theorem A.8 with a = β and b = 0, i.e.
Uβ (x) = E |X|β 1{|X|≤x} and V0 (x) = P (|X| > x). From (40), we get that:
α
E X β 1{|X|≤x} ∼
xβ P (|X| > x). But
β−α
α
xβ P (|X| > x) = ∞.
lim E X β 1{|X|≤x} = E(|X|β ) and
lim
x→∞
x→∞ β − α
It follows that E(|X|β ) = ∞.
The following result shows that if convergence (11) holds then the normalizing sequence (an )n must satisfy a very important condition.
Pn
Lemma 4.6. Suppose that (11) holds, where Sn = i=1 Xi and (Xi )i≥1 are
i.i.d. random variables, with common distribution F . Then the normalizing
sequence (an )n has to satisfy:
n
U (an x) → Cx2−α ,
a2n
(41)
n
U (an ) → C.
a2n
(42)
for any x > 0. In particular
Proof. In the proof of Theorem 3.3 the case when F is arbitrary, we have seen
that (an )n has to satisfy:
nP (|X| > an x) → M + (x) + M − (−x) = C
17
2 − α −α
x
α
for some 0 < α < 2,
(43)
(see relations (23) and (27)), or
nP (|X| > an x) → 0.
The measure M is given by M = Mα , respectively M = Cδ0 , where Mα is given
by (4).
To prove (41), in the case of α < 2, let V (x) = P (|X| > x). By Lemma A.7 and
(43), V is regularly varying of index −α. By Lemma 4.1.(i), (28) holds. This
2−α
a2 P (|X| > an x)
a2 x2 P (|X| > an x)
→
for any x > 0, i.e: n
→
gives us: n
U (an x)
α
U (an x)
2 − α −2
x . By multiplying the nominator and denominator by n, we get
α
nP (|X| > an x)
2 − α −2
x . Relation (42) follows by (43).
→
n
α
a2 U (an x)
n
In the case α = 2, relation (41) follows from (16) (assuming that F is symmetric) and the fact that M = Cδ0 , and hence M ([−x, x]) = C. (We still don’t
know why (41) holds in the non-symmetric case).
The following result gives an explicit recipe for constructing a sequence (an )n
satisfying (42).
Lemma 4.7. Assume that is U is regularly varying of index 2 − α, with 0 <
α ≤ 2. i.e. Let C > 0 arbitrary. Define
nU (x)
an = inf x > 0;
≤C
x2
where we use the convention inf ∅ = ∞.
Then: (i) an is finite, (ii) (an )n satisfies (42), (iii) limn→∞ an = ∞.
U (x)
C
Proof. (i) We have to prove that the set An = x > 0;
≤
is non
x2
n
empty. To see this, say U (x) = x2−α l(x) for a slowly varying function l. We
have to find some x > 0 such that
x−2 U (x) = x−2 x2−α l(x) = x−α l(x) ≤
C
.
n
l(x)
= 0.
xα
U (x)
U (x)
C
C
⊃
x;
.
(ii) Note that an ≤ an+1 since x;
≤
≤
x2
n
x2
n+1
U (x)
C
By the definition of an , we get:
>
for any x < an . Taking x = an−1 ,
x2
n
U (an−1 )
C
>
for any n. We apply this result with n + 1. We obtain:
we get
2
an−1
n
U (an )
C
>
.
(44)
a2n
n+1
This follows since limx→∞
18
On the other hand, by the definition of an :
C
U (an )
≤
2
an
n
(45)
By combining (44) and (45), we get:
C
C
U (an )
≤
<
2
n+1
an
n
which is equivalent to
Cn
nU (an )
≤ C.
<
n+1
a2n
n
Relation (42) follows from the fact that
→ 1.
n+1
(iii) Since (an )n is a non-decreasing sequence, lim an =: C0 ∈ [0, ∞] exists.
n→∞
Suppose C0 < ∞. By (42), we know that nU (an ) ∼ Ca2n . Note that U (an ) →
U (C0 ) as n → ∞. Hence nU (an ) → ∞, which is a contradiction.
Remark 4.8. Note that if U is regularly varying of index 2 − α, and (an )n is
chosen to satisfy (42), then it also satisfies (41).
To see this, note that
l(an x)
→ 1 (because l is slowly varying and an → ∞)
l(an )
and
na−α
n l(an ) =
l(an )
na2−α
nU (an )
n
=
→ C.
a2n
a2n
Hence
n
U (an x)
a2n
=
=
n 2−α 2−α
a
x
l(an x)
a2n n
l(an x) 2−α
(na−α
x
→ Cx2−α .
n l(an ))
l(an )
The following result specifies a more convenient way of choosing the sequence
(an )n , in the case α ∈ (0, 2).
Lemma 4.9. Suppose that α ∈ (0, 2) and (an )n satisfies:
nP (|X| > an ) → C1
(46)
for some C1 > 0. If V (x) = P (|X| > x) is regularly varying of index −α
(i.e. condition (i) of Corollary 4.4 holds), then (an )n satisfies (42) with C =
α
C1
.
2−α
19
Proof. By hypothesis, the function V (x) = P (|X| > x) is regularly varying of
index −α. By Lemma 4.1.(i) (28) holds. By Lemma 4.1.(ii) with L = (2 − α)L0
α
we obtain: U (x) ∼
x2−α L(x), i.e. U is regularly varying of index 2 − α.
2−α
a2 P (|X| > an )
nP (|X| > an )
2−α
By (28) n
=
→
. Using (46), we obtain:
nU (an )
U (an )
α
2
an
α
nU (an )
→ C1
= C, i.e. (42) holds.
2
an
2−α
Remark 4.10. Assume α ∈ (0, 2). An example of a sequence (an )n satisfying
(46) is:
C1
C1
an = inf x > 0; P (|X| > x) ≤
= inf x > 0; P (|X| ≤ x) ≥ 1 −
.
n
n
This can be proved similarly to Lemma 4.7. We denote by F0 the law of |X|.
Then
C1
C1
=: F0−1 1 −
,
an = inf x > 0; F0 (x) ≥ 1 −
n
n
where F0−1 denotes the generalized inverse of the non-decreasing function F0 . If
F0 is a continuous distribution, then this is equivalent to saying that:
F0 (an ) = 1 −
C1
,
n
C1
-quantile of F0 . (Recall that zα is called the α-quantile of a
n
random variable Z if P (Z > zα ) = α, i,e, P (Z ≤ zα ) = 1 − α.)
i.e. an is the
In the remaining part of this section, we concentrate on proving that relation
(11) holds, under suitable condition imposed on the underlying distribution of
the sequence (Xi )i≥1 of i.i.d. random variables. For this, we will assume that
(an )n is a non-decreasing sequence of constants with limn→∞ an = ∞, which
satisfies (42). We begin with a preleminary result.
X
and Mn (dx) = nx2 Fn (dx). Let U
an
be defined by (14). Assume that U is regularly varying of index 2−α, 0 < α ≤ 2.
Let (an )n be a sequence satisfying (42), and an → ∞. If α < 2, we assume in
addition that the tails of X are balanced, i.e. (30) holds. Then
Proposition 4.11. Let Fn be the law of
Mn → M properly
where M = Cδ0 if α = 2, and if α ∈ (0, 2), M = Mα , where Mα is given by (4)
with constants c+ , c− satisfying (26). Moreover,
Z iux
Z iux
e − 1 − iu sin x
e − 1 − iu sin x
M
(dx)
→
M (dx).
(47)
n
2
x
x2
R
R
20
Proof. Note that Mn is canonical measure on R because
Z
Z
Z
Z
1 2
1
2
Mn (dx) = n
Mn (dx) = n
x Fn (dx) < ∞.
x Fn (dx) < ∞ and
2
2
x
x
|x|≤1
|x|≤1
|x|≥1
|x|≥1
M is also a canonical measure.
By Remark 4.8, (41) holds. We claim that
Mn ([−x, x]) → M ([−x, x])
for any x > 0.
(48)
This follows from (41) and the definition of M since
2
nU (an x)
X
=
nE
1
{|X|≤an x}
a2n
a2n
Z
= n 1[−x,x] (y)y 2 Fn (dy) = Mn ([−x, x])
R
and M ([−x, x]) = Cx2−α .
From (41), we also obtain that, for any α ∈ (0, 2],
nP (|X| > an x) → M + (x) + M − (−x)
for any x > 0
(49)
where M + and M − are given by Definition A.1. To see this, note that since U
is regularly varying of index 2 − α,
x2 P (|X| > x)
2−α
→
,
U (x)
α
as x → ∞
by Lemma 4.1. Since an x → ∞, it follows that for any x > 0,
a2n x2 P (|X| > an x)
2−α
→
U (an x)
α
which is equivalent to saying that:
a2n
2 − α −2
nP (|X| > an x) →
x .
nU (an x)
α
By (41), we have:
a2n
1
→
. We then obtain:
nU (an x)
Cx2−α
nP (|X| > an x) → C
2 − α −2 2−α
x x
α
which is equivalent to (49), since due to the definiton of M ,
M + (x) + M − (−x) = C
2 − α −α
x
α
21
for any x > 0.
(50)
In particular, when α = 2, M + (x) = M − (−x) = 0 for all x > 0 (since M =
Cδ0 ), and relation (49) becomes:
nP (|X| > an x) → 0
for any x > 0.
When α < 2, the tails are balanced, (i.e. (30) holds) by hypothesis. Due to (49)
and (50), we obtain that
nP (X > an x) → Cp
2 − α −α
x
α
and nP (X < −an x) → Cq
2 − α −α
x
(51)
α
To see this, note that
nP (X > an x) =
2 − α −α
P (X > an x)
nP (|X| > an x) = Cp
x .
P (|X| > an x)
α
The second part of (51) is obtained similarly.
Due to the definition of Fn and (27), (51) can be written as:

Z ∞
Z ∞
1 2
1

+

M
(x)
=
ny
F
(dy)
→
M (dy) = M + (x)
 n
n
2
2
y
y
xZ
xZ
−x
−x
1 2
1

−

M
(−x)
=
ny
F
(dy)
→
M (dy) = M − (−x)
 n
n
2
2
y
y
−∞
−∞
(52)
Putting together (48) and (52) we get:
Mn+ (x) → M + (x),
Mn ([−x, x]) → M ([−x, x]),
Mn− (−x) → M − (−x)
i.e. Mn → M properly (according to Definition A.2)
Define
R
eiux − 1 − iu sin x
Mn (dx)
x2
R
eiux − 1 − iu sin x
M (dx).
x2
Z
ψn (u) =
Z
ψ(u) =
Since Mn → M properly (in the sense of definition A.2), by Theorem A.3, it
follows that ψn (u) → ψ(u). This proves (47).
The following result is called the Stable Limit Theorem, which is the focus
of the present project.
Theorem 4.12 (Theorem XVII.5.3 from [1]). Let Y be a random variable with
characteristic function ϕY (u) = eψY (u) , where


−|u|α C Γ(3 − α) cos πα 1 − isgn(u)(p − q) tan πα , α 6= 1

α(1
2
2
− α)
ψY (u) =
π
2


1 + isgn(u)(p − q) ln |u| ,
α=1
−|u|C
2
π
(53)
22
for some C > 0, p, q ≥ 0, p + q = 1, i.e. Y ∼ Sα (γ, β, 0), where


C Γ(3 − α) cos πα , α 6= 1
α
α(1 − α)
2
γ =
and β = p − q

C π ,
α=1
2
(54)
Let (Xi )i≥1 be i.i.d. random variables whose distribution F satisfies either one
of the following conditions:
(a) U (x) = E(X 2 1{|X|≤x} ) is slowly varying and X is non-degenerate
(b) U is regularly varying of index 2 − α for some 0 < α < 2, and the tails of
X are balanced, i.e. (30) holds.
Let (an )n be a sequence of constants satisfying (42).
(i) If 0 < α < 1, then
Sn d
−
→Y.
an
(ii) If 1 < α ≤ 2 and we assume that E(X 2 ) = ∞ when α = 2, then
nE(X) d
−
→Y
an
(iii) If α = 1, then
Sn
−
an
Sn
X
d
− nbn −
.
→ Y , where bn = E sin
an
an
Corollary 4.13. (0 < α < 2) Assume that P (|X| > x) = x−α L(x) where L is
slowly varying and the tails are balanced, i.e. (30) holds. Choose (an )n such that
α
L and the
(46) holds for some C1 > 0. Then U (x) = x2−α l(x), where l =
2−α
α
.
conclusion of Theorem 4.12 holds with ψY (u) given by (53) with C = C1
2−α
Proof. We apply Theorem 4.12. The fact that (an )n satisfies (42) follows by
Lemma 4.9.
Proof of Theorem 4.12. Define M such that M ([−y, x]) = Cpx2−α + Cqy 2−α ,
i.e.
M = Cδ0 if α = 2
or
M (dx) = (2 − α)Cpx−α+1 1(0,∞) (x)dx + (2 − α)Cq(−x)−α+1 1(−∞,0) (x)dx
if α < 2.
Assume first that α < 2. Then M = Mα where Mα is given by (4) with
c+ = (2 − α)Cp and c− = (2 − α)Cq. Note that
c+ + c−
=C
2−α
and
c+ − c−
=p−q
c+ + c−
(55)
Let W be an infinitely divisible random variable with ϕW (u) = eψW (u) where
Z iux
e − 1 − iu sin x
M (dx)
ψW (u) =
x2
R
23
According to Lemma 2.8, W ∼ Sα (γ, β, δ) where (γ, β, δ) are given by (5), (6),
(8). Due to (55), formulas (5) and (6) are exactly the same as (54). Hence,
d
d
W − δ ∼ Sα (γ, β, 0), i.e. Y = W − δ. Therefore W = Y + δ and
ψW (u) = iuδ + ψY (u)
Let Fn be the law of
X
an
and Mn = nx2 Fn (dx).
(i) Assume that α < 1. We claim that in this case,
Z iux
Z iux
e −1
e −1
M
(dx)
→
M (dx)
n
2
x
x2
R
R
(56)
To see this, recall that Mn → M properly (by Proposition 4.11). Since f (x) =
eiux − 1
is continuous on {x; |x| > δ},
x2
Z
Z
eiux − 1
eiux − 1
lim
M
(dx)
=
M (dx) for any δ > 0.
(57)
n
n→∞ |x|>δ
x2
x2
|x|>δ
This still needs to be shown in more details. We see that
Z
Z iux
eiux − 1
e −1
lim
M
(dx)
=
M (dx).
δ→0 |x|>δ
x2
x2
R
Hence, to prove (56), it suffices to show that
Z
Z
eiux − 1
eiux − 1
lim lim M
(dx)
−
M
(dx)
= 0.
n
n
2
δ→0 n→∞ R
x2
x
|x|>δ
Note that,
Z
eiux − 1
Mn (dx)
2
|x|≤δ
x
(58)
Z
Z
eiux − 1 2
iux
(e − 1)Fn (dx)
= nx Fn (dx) = n 2
|x|≤δ
|x|≤δ
x
Z
Z
√
≤ n
|eiux − 1|Fn (dx) ≤ 2n|u|
|x|Fn (dx)
|x|≤δ
=
√
|x|≤δ
|u| 2n E |X|1{|X|≤an δ} .
an
where we used the inequality
|eia − 1|2 = (cos a − 1)2 + sin2 a ≤ 2a2
for any a ∈ R.
Using Lemma 4.5.(ii) with β = 1 (since α < 1), we obtain that
E |X|1{|X|≤an δ} ∼
α
(an δ)P (|X| > an δ).
1−α
24
This gives us:
Z
√
eiux − 1
α
M
(dx)
δnP (|X| > an δ).
≤ 2|u|
n
|x|≤δ
x2
1−α
Taking the upper limit when n tends to infinity, we get:
Z
√
eiux − 1
α
lim M
(dx)
δn lim P (|X| > an δ)
≤ 2|u|
n
n→∞ |x|≤δ
x2
1 − α n→∞
2 − α −α
Using (49) and (50) we obtain limn→∞ P (|X| > an δ) = C
δ . This gives
α
us
Z
√
eiux − 1
2 − α 1−α
lim Mn (dx) ≤ 2|u|
Cδ
.
2
n→∞ |x|≤δ
x
1−α
Using the fact that 1 − α > 0, we get that when δ → 0, the term on the right
hand side of the previous inequality converges to zero. This concludes the proof
of (58) and the proof of (56).
Note that
Z iux
e −1
Mn (dx)
x2
R
and
Z
eiux − 1
M (dx)
x2
Z
eiux − 1 2
nx
F
(dx)
=
n
(eiux − 1)Fn (dx)
n
2
x
R
R X
u
= nE exp iu
− 1 = n ϕX
−1
an
an
Z
=
Z
eiux − 1 − iu sin x
sin x
M
(dx)
+
iu
M (dx)
2
2
x
R
R
R x
= ψW (u) − iuδ = ψY (u).
u
Using (56), we obtain: n ϕX
− 1 → ψY (u), which is equivalent to the
an
Sn d
−
→ Y (by Remark 3.2). This concludes the proof of the theorem
fact that
an
in the case α < 1.
Z
=
(ii) Assume that 1 < α ≤ 2. We claim that, in this case
Z iux
Z iux
e − 1 − iux
e − 1 − iux
Mn (dx) →
M (dx).
2
x
x2
R
R
To see this, we use again the fact that Mn → M properly, so
Z
Z
eiux − 1 − iux
eiux − 1 − iux
lim
M
(dx)
=
M (dx)
n
2
n→∞ |x|≤t
x
x2
|x|≤t
25
(59)
(60)
for any t > 0. We still need to show this in more details. We see that
Z
Z iux
eiux − 1 − iux
e − 1 − iux
lim
M (dx) =
M (dx).
t→∞ |x|≤t
x2
x2
R
Hence to show (59), it suffices to prove that:
Z
Z
eiux − 1 − iux
eiux − 1 − iux
M
(dx)
−
M
(dx)
lim lim = 0.
n
n
2
t→∞ n→∞ R
x2
x
|x|≤t
Note that,
Z
eiux − 1 − iux
M
(dx)
n
|x|>t
x2
=
=
≤
≤
Z
eiux − 1 − iux 2
nx
F
(dx)
n
|x|>t
x2
Z
n
(eiux − 1 − iux)Fn (dx)
|x|>t
Z
n
|eiux − 1 − iux|Fn (dx)
|x|>t
Z
√
( 2 + 1)n|u|
|x|Fn (dx)
|x|>t
=
√
( 2 + 1)n
|u|
E[|X|1{|X|>an t} ]
an
where we used the inequality
√
|eia − 1 − ia| ≤ |eia − 1| + |a| ≤ ( 2 + 1)|a|.
Using Lemma 4.5.(i) with β = 1 (since α > 1), we obtain that
E[|X|1{|X|>an t} ] ∼
2−α 1
U (an t).
α − 1 an t
This gives us:
Z
√
2 − α 1 nU (an t)
eiux − 1 − iux
M
(dx)
|u|
.
≤ ( 2 + 1)
n
|x|>t
x2
α−1 t
a2n
Taking the upper limit when n tends to infinity, we get:
Z
√
eiux − 1 − iux
2−α 1
nU (an t)
lim M
(dx)
|u| lim
.
≤ ( 2 + 1)
n
2
n→∞ |x|>t
n→∞
x
α−1 t
a2n
nU (an t)
= Ct2−α . This gives us
a2n
√
2−α
eiux − 1 − iux
M
(dx)
|u|Ct1−α .
≤ ( 2 + 1)
n
2
x
α−1
Using (41) we obtain lim
n→∞
Z
lim n→∞ |x|>t
26
Using the fact that 1 − α < 0, we get that when t → ∞, the right hand side of
the previous inequality converges to zero. This finishes the proof of (59).
Assume first that E(X) = 0. Then
Z iux
Z iux
e − 1 − iux
e − 1 − iux 2
Mn (dx) =
nx Fn (dx)
2
x
x2
R
R
X
X
− 1 − iu
= nE exp iu
an
an
u
u
u
= n ϕX
− 1 − i E(X) = n ϕX
−1
an
an
an
and
Z iux
e − 1 − iux
M (dx)
x2
R
Z
eiux − 1 − iu sin x
sin x − x
M
(dx)
+
iu
M (dx)
2
x
x2
R
R
= ψW (u) − iuδ = ψY (u).
Sn d
u
−
→ Y (by Remark
Using (59), we obtain n ϕX
− 1 → ψW (u), i.e.
an
an
3.2). This finishes the proof in the case α ∈ (1, 2] and E(X) = 0.
Z
=
Assume next that E(X) is arbitrary. We prove that
Sn − nE(X) d
−
→ Y.
an
(61)
Writing E(X) = E(X1{|X|>an } ) + E(X1{|X|≤an } ), and letting
1
bn =
E(X1{|X|≤an } ), we have
an
Sn − nE(X)
an
=
=
Sn − nE(X1{|X|≤an } )
n
−
E(X1{|X|>an } )
an
an
Sn
n
− nbn −
E(X1{|X|>an } ).
an
an
(62)
We claim that
n E X1{|X|>an } →
an
Z
|x|≥1
1
M (dx) := δ 0 .
x
To see this we use (57) with δ = 1. We get:
Z
Z
eiux − 1
eiux − 1
M
(dx)
→
M (dx).
n
2
x
x2
|x|≥1
|x|≥1
By taking the difference between (59) and (60) (with t = 1), we obtain:
Z
Z
eiux − 1 − iux
eiux − 1 − iux
M
(dx)
→
M (dx).
n
2
x
x2
|x|≥1
|x|≥1
27
(63)
Taking the difference between these two equations, we obtain:
Z
Z
Z
1
1
iu
Mn (dx) = iun
M (dx).
xFn (dx) → iu
x
x
|x|≥1
|x|≥1
|x|≥1
Due to (62) and (63), (61) is equivalent to
Sn
d
− nbn −
→ Y 0 := Y + δ 0 .
an
From (59), we know that
E(X)
u
− 1 − iu
n ϕX
an
an
eiux − 1 − iux
Mn (dx)
x2
R
Z iux
e − 1 − iux
→
M (dx)
x2
R
Z iux
e − 1 − iu sin x
=
M (dx)
x2
R
Z
sin x − x
+iu
M (dx)
x2
R
= ψW (u) − iuδ = ψY (u),
(64)
Z
=
(65)
d
using definition (8) of δ and the fact that Y = W −δ. Note that (64) is equivalent
to
n
u
−iubn
ϕ Sn −nbn (u) = ϕX
e
→ eψY 0 (u) = ϕY 0 (u)
an
an
To prove this, it suffices to show that
n
u
u
∼ exp n ϕX
−1
ϕX
an
an
(66)
since then
n
0
u
u
e−iubn
− 1 − iubn
→ eψY 0 (u) = eiuδ +ψY (u) .
ϕX
∼ exp n ϕX
an
an
For the last convergence, we used the fact that:
E(X1{|X|≤an } )
u
u
n ϕX
− 1 − iubn
= n ϕX
− 1 − iu
an
an
an
nE(X1{|X|>an } )
u
E(X)
= n ϕX
− 1 − iu
+ iu
an
an
an
0
= ψY (u) + iuδ ,
due to (65) and (63).
28
For the proof of (66), we refer to part (iii) below. This concludes the proof
in the case α ∈ (1, 2].
(iii) Assume that α = 1. Note that
Z iux
e − 1 − iu sin x
Mn (dx) =
x2
R
Z
=
n
eiux − 1 − iu sin x 2
nx Fn (dx)
x2
R
Z
(eiux − 1 − iu sin x)Fn (dx)
X
u
− 1 − iuE sin
n ϕX
an
an
u
n ϕX
− 1 − iubn
an
R
=
=
and, since in this case δ = 0, we have:
Z iux
e − 1 − iu sin x
M (dx) = ψY (u) = ψW (u).
x2
R
By Proposition 4.11, it follows that
u
n ϕX
− 1 − iubn → ψW (u).
an
Sn
d
− nbn −
→ W , we have to show that:
an
n
u
ϕ Sn −nbn = ϕX
e−iubn
→ eψW (u) = ϕW (u).
an
an
(67)
To show that
(68)
To show (68), it suffices to prove (66), since then
n
u
u
ϕX
e−iubn
∼ exp n ϕX
− 1 − iubn
an
an
and by (67), this last term converges to eψW (u) . The remaining part of the proof
is dedicated to (66).
n
nzn
Note
relation (66) can be written as (1 + zn ) ∼ e , where zn =
that
u
− 1 → 0, zn ∈ C, |zn | ≤ 2. We claim that n|zn |2 → 0 implies
ϕX
an
(1 + zn )n ∼ enzn . This implication still needs to be proven.
Hence it suffices to prove that
2
u
n ϕX
− 1 → 0
an
29
(69)
To prove (69), we use the fact that E|X|β < ∞ for β < 1 (by Lemma 4.5.(i),
since α ≥ 1). First note that
Z x
ϕX u − 1 = E exp iu u
exp iu
− 1 ≤
− 1 F (dx).
an
an
an
R
Now we use the fact that |eit − 1| ≤ 2|t|β , for any t ∈ R, β ∈ (0, 1]. We get:
Z β
Z
ux 2|u|β
2|u|β
ϕX u − 1 ≤
2 F (dx) ≤ β
|x|β F (dx) = β E|X|β .
an
an
an R
an
R
This gives us
2
u
n
2β
β 2 n
n ϕX
− 1 ≤ 4|u| (E|X| ) 2β = O
.
an
an
a2β
n
(70)
We claim that for any δ > 0, there exists Nδ such that
C1 aα−δ
≤ n ≤ C2 anα+δ
n
for any n ≥ Nδ
(71)
To see this, recall that U (x) = x2−α l(x), where l is slowly varying. By Remark
A.6, this implies that for any δ > 0, there exists xδ such that x−δ ≤ l(x) ≤ xδ
for any x ≥ xδ . The fact that an → ∞ implies that there exists Nδ such that
δ
an ≥ xδ for any n ≥ Nδ . We then get: a−δ
n ≤ l(an ) ≤ an for any n ≥ Nδ . Then
2−α+δ
2−α−δ
for any n ≥ Nδ .
≤ U (an ) ≤ an
an
nU (an )
→ C. This implies that there exist C1 , C2 > 0 such that
a2n
nU (an )
C1 ≤
≤ C2 , for any n > 0. Therefore, for any n ≥ Nδ
a2n
nU (an )
na2−α−δ
n
C2 ≥
≥
= nan−α−δ , which implies that C2 anα+δ ≥ n and
2
an
a2n
nU (an )
na2−α+δ
n
C1 ≤
≤
= nan−α+δ , which implies that C1 anα−δ ≤ n. This
a2n
a2n
concludes the proof of (71).
1/(α+δ)
1
1
From (71), we have that: anα+δ ≥
n, which gives us an ≥
n1/(α+δ) .
C2
C2
2β/(α+δ)
Hence, a2β
and
n ≥ C2 n
By (42),
n
a2β
n
≤
n
C2
n2β/(α+δ)
= C2−1 n1−2β/(α+δ) → 0
as n → ∞
(72)
α+δ
2β
i.e
< β. Since we also need β < 1, we must choose δ such
α+δ
2
α+δ
that
< 1, i.e. 0 < δ < 2 − α. Relation (69) follows from (70) and (72).
2
This concludes the proof in this case α = 1.
if 1 <
30
In the case α ∈ (0, 2), there is an alternative formulation of Theorem 4.12,
based on the representation of the characteristic function of a stable random
variable given by Lemma 2.10.
Theorem 4.14. Let (Xi )i≥1 be i.i.d. random variables such that P (|X| > x) =
x−α L(x) where L is slowly varying and 0 < α < 2, and the tails are balanced
n
X
i.e. (30) is satisfied for some p, q ≥ 0, p + q = 1. Let Sn =
Xi . Choose (an )n
i=1
such that (46) holds for some constant C1 > 0. Let Y 0 be an infinitely divisible
random variable with characteristic function
Z ∞
ϕY 0 (u) = exp
(eiux − 1 − iux1{|x|≤1} )να (dx)
−∞
where
να (dx) = C1 α[px−α−1 1(0,∞) (x) + q(−x)−α−1 1(−∞,0) (x)]dx.
Then
E(X1{|X|≤an } )
an
α
. Hence
Proof. By Lemma 4.9, (an )n satisfies (42) with C = C1
2−α
Sn
d
− nbn −
→ Y 0,
an
where
C1 =
bn =
2−α
C.
α
Note that the measure να has the form (9) with
c+ = C1 pα = (2 − α)Cp and c− = C1 qα = (2 − α)Cq.
By Lemma 2.10, Y 0 ∼ Sα (γ, β, δ 0 ) where (γ, β) are given by (5), (6) and δ 0 is
c+ − c−
c+ + c−
= C and
= p − q, formulas (5) and (6)
given by (10). Since
2−α
c+ + c−
are exactly the same as (54).
X
and Mn (dx) = nx2 Fn (dx). We will use some of
an
the results shown in the proof of Theorem 4.12 with Mα (dx) = x2 να (dx). We
consider three cases.
Sn d
−
→ Y ∼ Sα (γ, β, 0). We claim
1) Suppose that α < 1. By Theorem 4.12.(i),
an
that in this case,
Z
nE(X1{|X|≤an } )
→
x να (dx) = −δ 0 .
(73)
an
|x|≤1
Let Fn be the law of
To see this, by (60), with t = 1, we get
Z
Z
n
(eiux − 1 − iux)Fn (dx) →
|x|≤1
|x|≤1
31
(eiux − 1 − iux) να (dx).
(74)
Taking the difference between (56) and (57) with δ = 1, we obtain
Z
Z
n
(eiux − 1)Fn (dx) →
(eiux − 1) να (dx).
|x|≤1
(75)
|x|≤1
Taking the difference between (74) and (75), we get
Z
Z
X
nE
1{|X|≤an } = n
xFn (dx) →
x να (dx) = −δ 0 .
an
|x|≤1
|x|≤1
This proves (73). Hence
Sn − nE(X1{|X|≤an } )
→ δ0 + Y = Y 0 .
an
2) Suppose that α > 1. By relation (64) in the proof of Theorem 4.12
Sn − nE(X1{|X|≤an } )
→ δ0 + Y = Y 0 .
an
3) Suppose that α = 1. By Theorem 4.12.(iii),
Sn
X d
−
→ Y . We
− nE sin
an
an
claim that
Z
nE(X1{|X|≤an } )
X
0
nE sin
−
→ δ = (sin x − x1{|x|≤1} )να (dx). (76)
an
an
R
Z
To see this, note that by (57) with δ = 1, n
(eiux − 1)Fn (dx) →
|x|>1
Z
Z
(eiux − 1)να (dx). By (60) with t = 1, n
(eiux − 1 − iux)Fn (dx) →
R
|x|≤1
Z
(eiux − 1 − iux)να (dx). Taking the sum, we obtain:
R
Z
n
(eiux − 1 − iux1{|x|≤1} )Fn (dx) →
Z
R
(eiux − 1 − iux1{|x|≤1} )να (dx).
R
Z
By (47), n
(eiux − 1 − sin x)Fn (dx) →
R
Z
(eiux − 1 − sin x)να (dx).
R
Taking the difference between the two previous equations, we obtain:
Z
Z
n (sin x − x1{|x|≤1} )Fn (dx) → (sin x − x1{|x|≤1} )να (dx).
R
R
This proves (76). This gives us
Sn − nE(X1{|X|≤an } )
→ δ0 + Y = Y 0 .
an
32
Remark 4.15. Under the conditions of Theorem 4.14, we have:
Z
Z
X
sin x
nE sin
M
(dx)
=
sin x να (dx) if α ∈ (0, 1).
→
2
an
R\{0}
R x
(77)
To see this, note that
eiux − 1 − iu sin x
Mn (dx)
x2
R
Z iux
Z
e −1
sin x 2
M
(dx)
−
iu
nx Fn (dx)
n
2
2
x
R
R x
Z
ψn (u)
=
=
and
Z
ψY (u) =
R
eiux − 1 − iu sin x
M (dx) =
x2
Z
R
eiux − 1
M (dx) − iu
x2
Z
R
sin x
M (dx)
x2
Using (47) and (56), we obtain:
Z
Z
sin x
X
sin x 2
nE sin
nx
F
(dx)
→
M (dx).
=
n
2
2
an
x
R x
R
Remark 4.16. Under the conditions of Theorem 4.14
Z
Z
X
X
x − sin x
nE
− sin
M
(dx)
=
(x − sin x) να (dx)
→
an
an
x2
R\{0}
R
if α ∈ (1, 2]. To see this, note that
Z iux
e − 1 − iu sin x
Mn (dx)
ψn (u) =
x2
R
Z iux
Z
e − 1 − iux
x − sin x 2
=
Mn (dx) − iu
nx Fn (dx)
2
x
x2
R
R
and
eiux − 1 − iu sin x
M (dx)
x2
R
Z
Z iux
e − 1 − iux
x − sin x
M
(dx)
−
iu
M (dx)
2
x
x2
R
R
Z
ψY (u)
=
=
Using (47) and (59), we obtain:
Z
Z
X
X
x − sin x 2
x − sin x
nE
− sin
=
nx
F
(dx)
→
M (dx).
n
2
an
an
x
x2
R
R
33
5
Simulations on R
In this section we will use Theorem 4.12 to verify if certain distributions are
in the domain of attraction of a stable distribution. We consider a random
variable X with Pareto distribution of parameter α with distribution function
n
X
F . Let (Xi )i≥1 be i.i.d. copies of X. Let Sn =
Xi . By Corollary 4.13,
i=1
for 0 < α < 2, X is in the domain of attraction of a stable distribution, i.e.
Sn
d
− nbn −
→ Y ∼ Sα (γ, β, 0), (where the parameters are given in Theorem 4.12)
an
if
P (|X| > x) = x−α L(x) where L is a slowly varying function
(78)
and the tails are balanced, i.e. (30) holds. We choose an satisifying (46). By
Theorem 4.2, for α = 2, X is in the domain of attraction of a stable distribution
if U (x) = E(X 2 1{|X|≤x} ) is slowly varying and X is non-degenerate. We choose
an satisfying (42).
Because the Pareto distribution is always positive, we get that (78) is equivalent
to P (X > x) = x−α L(x), which is in turn equivalent to F (x) = 1 − P (X >
x) = 1 − x−α L(x). Another consequence of the positiveness of X is that
P (X < −x)
P (X > x)
= 1 and lim
= 0. So (30) holds with p = 1
lim
x→∞ P (|X| > x)
x→∞ P (|X| > x)
and q = 0. Therefore β = p + q = 1.
In this example, for 0 < α < 2 we will choose(L(x) = c, where c ∈ R. Then,
cαx−α−1 , x > c0
F (x) = 1 − cx−α . We then obtain: f (x) =
for some
0,
x ≤ c0
c0 ∈ R.
(
αx−α−1 , x > 1
Taking c0 = 1, we obtain X ∼ P areto(α), i.e. f (x) =
0,
x ≤ 1.
2
For α = 2, we have U (x) = E(X 1{|X|≤x} ) = l(x) for some slowly varying
function l. We have:
Z x
Z x
2
−α−1
U (x) =
y αy
dy = 2
y −1 dy = 2 ln x = l(x).
(79)
−x
1
We will now find the an for 0 < α < 2, i.e. satisfying (46). Taking C1 = 1,
and using the positiveness of X, we get: nP (X > an ) → 1. We will choose
1
an such that P (X > an ) = . Using the fact that, for X ∼ P areto(α),
n
1
. Hence, an = n1/α .
P (X > x) = x−α , we then obtain P (X > an ) = a−α
n =
n
α
α
By Corollary 4.13, C = C1
=
.
2−α
2−α
34
We will now find the an for α = 2. Using (79) and (42), we get:
a2n =
2n
ln(an ).
C
Taking C = 1 we obtain the following recurrence relation:
a2n = 2n ln(an )
(80)
1
Note that an = exp − W−1 (−1/n) where W−1 is the second branch of the
2
Lambert W function. To compute the value of W−1 we used the gsl package.
With n = 10000, we obtain: an ≈ 341.5903.
In the next examples we will use the parameter α with values 0.5, 1, 1.5 and 2.
In each case, using R, we will simulate k = 10000 Pareto distribution samples of
population n = 10000 using the VGAM package. We will then calcutate the sum
(i)
Sn
(i)
(i)
− nbn .
of values inside the sample to obtain Sn for 1 ≤ i ≤ k. Let Wn =
an
(i)
Let Wn be the vector with Wn as coordinates, then by Theorem 4.12, when
d
k is very large, Wn −
→ Sα (γ, β, 0). To illustrate this result, we compute the
density function of W by the histogram method on R, and the empirical distribution function of Wn . We then superpose the graphs obtained with the density
function and the distribution function of a stable distribution with parameters
(γ, β, 0). The graphs should converge.
Case 1: α = 0.5.
1/α
πα
Γ(3 − α)
cos
.
In this case, Theorem 4.12 gives us bn = 0 and γ = C
α(1 − α)
2
We know α = 0.5, then C = 1/3, hence γ ≈ 1.570796. We then compute
Wn and draw the two curves. To draw the stable curves, we use a Lévy distribution with parameters µ = 0 and c = 1.570796 (because as seen in [3],
Lévy(µ, c) ∼ Stable0.5 (1, c, µ)). Therefore, by Theorem 4.12, we obtain:
Wn =
1
d
Sn −
→ Y ∼ S0.5 (1.570796, 1, 0) ∼ Lévy(0, 1.570796).
2
n
We obtain the graphs (Y : red curve, Wn :black curve) in Figure 1
Case 2: α = 1.
π
X
and γ = C . We know
In this case, Theorem 4.12 gives us E sin
an
2
35
Figure 1: Illustration of Stable Limit Theorem for α = 0.5
π
α = 1, then C = 1, hence γ = . We now calculate nbn :
2
Z ∞
x X
nbn = nE sin
=
n
sin
αx−α−1 dx
n1/α
n1/α
1
Z ∞
Z ∞
x
x−2 dx = n
sin y(ny)−2 (ndy)
= n
sin
n
1/n
1
Z ∞
sin y
=
dy ≈ 9.63312
2
1/n y
x
using y = and computing the last integral with n = 10000 on www.wolframalpha.com.
n
Therefore, by Theorem 4.12, we obtain:
1
π
d
Sn − 9.63312 −
→ Y ∼ Stable1 ( , 1, 0)
n
2
To draw the stable curve, we use the package stabledist. The function uses
the approach of J.P. Nolan for general stable distributions. We obtain the graphs
(Y : red curve, Wn :black curve) in Figure 2
Wn =
Case 3: α = 1.5.
1/α
E(X)
Γ(3 − α)
πα
In this case, Theorem 4.12 gives us bn =
and γ = C
cos
.
an
α(1 − α)
2
We know α = 1.5, then C = 3, hence γ ≈ 1.84527. We now calculate nbn :
Z ∞
1
n
nbn = n E(X) = 1/α
xαx−α−1 dx
an
n
1
Z ∞
1/3
= 1.5n
x−1.5 dx = 3n1/3
1
36
Figure 2: Illustration of Stable Limit Theorem for α = 1
Therefore, by Theorem 4.12, we obtain:
Wn =
1
d
Sn − 3n1/3 −
→ Y ∼ Stable1.5 (1.84527, 1, 0)
n2/3
The stable curve is drawn using the package stabledist. We obtain the graphs
(Y : red curve, Wn :black curve) in Figure 3
Case 4: α = 2.
1/α
E(X)
Γ(3 − α)
πα
and γ = C
cos
.
an
α(1 − α)
2
√
2
. We now calculate nbn :
We know α = 2 and we took C = 1, hence γ =
2
Z ∞
1
n
nbn = n E(X) =
xαx−α−1 dx
an
341.5903 1
Z ∞
2n
n
=
2
x−2 dx =
.
341.5903 1
341.5903
In this case, Theorem 4.12 gives us bn =
by Theorem 4.12, we obtain:
√
1
2n
2
d
Wn =
Sn −
−
→ Y ∼ Stable2 (
, 1, 0)
341.5903
341.5903
2
The stable curve is drawn using the package stabledist. We obtain the graphs
(Y : red curve, Wn :black curve) in Figure 4
37
Figure 3: Illustration of Stable Limit Theorem for α = 1.5
Figure 4: Illustration of Stable Limit Theorem for α = 2
38
A
Auxiliary Results
Definition A.1. A measure M is called canonical if it attributes finite masses
to finite intervals and the integrals
Z ∞
Z −x
1
1
−
M + (x) =
M
(dy)
and
M
(−x)
=
M (dy)
(81)
2
2
y
x
−∞ y
are finite for some (and therefore all) x > 0.
Definition A.2. Let (Mn )n and M be canonical measures. We say that (Mn )n
converges properly to M (and we write Mn → M properly) if:
(i) Mn ([a, b]) → M ([a, b]) for any a, b ∈ R continuity points of M ,
(ii)Mn+ (x) → M + (x) and Mn− (−x) → M − (−x) for any x > 0 which is a
continuity point of M .
Theorem A.3 (Theorem XVII.2.2 of [1]). Let Mn be a canonical measure on
R, bn ∈ R and
Z +∞ iux
e − 1 − iu sin x
Mn (dx) ∀u ∈ R.
(82)
ψn (u) = iubn +
x2
−∞
There exists a continuous function ψ : R → C such that ψn (u) → ψ(u) ∀u ∈ R
if and only if there exist a canonical measure M on R and b ∈ R such that
Mn → M properly
and
bn → b.
In this case,
Z
+∞
ψ(u) = iub +
−∞
eiux − 1 − iu sin x
M (dx)
x2
∀u ∈ R.
(83)
Corollary A.4. Let
ψn (u) = Cn (ϕn (u) − 1 − iub̃n )
(84)
where ϕn is the characteristic function of a probability distribution Fn and
b̃n , Cn ∈ R. Then there exists a continuous function ψ : R → C such that
ψn (u) → ψ(u) for any u if and only if
Cn x2 Fn (dx) → M (dx) and Cn (βn − b̃n ) → b
Z
for some canonical measure M and b ∈ R where βn =
sin xFn (dx). In this
R
case, ψ(u) is given by (83).
Proof. Note that ψn (u) can be written in the form (82) with
Mn (dx) = Cn x2 Fn (dx)
and
bn = Cn (βn − b̃n )
. The result follows by Theorem A.3.
39
Definition A.5. (i) A function L is slowly varying (SV) if
L(λx)
=1
x→∞ L(x)
lim
for any λ > 0.
(ii) A function U is regularly varying with index ρ ∈ R (RVρ ) if
U (x) = xρ L(x)
for a slowly varying function L.
Remark A.6. Let L be a slowly varying function. Then for any δ > 0 there
exists xδ > 0 such that x−δ < L(x) < xδ for any x > xδ .
Lemma A.7 (Lemma VIII.8.3 of [1]). Suppose that
If U is a monotone function such that:
λn+1
→ 1 and an → ∞.
λn
lim λn U (an x) = χ(x) ≤ ∞ exists for any x ∈ D (D is a dense set)
n→∞
χ(x) is finite and positive for any x ∈ I (I is an interval),
then U is regularly varying of index ρ ∈ R and χ(x) = Cxρ ,
for some C > 0.
Theorem A.8 (Theorem VIII.9.2.(i) of [1]). Let a > 0 and −∞ < b < a.
Define the truncated moment functions Ua and Vb by
Ua (x) = E(|X|a 1{|X|≤x} )
and
Vb (x) = E(|X|b 1{|X|>x} )
Suppose that Ua (∞) = ∞.
(i) If either Ua or Vb varies regularly then
ua−b Vb (u)
=c
u→∞
Ua (u)
lim
(85)
for some 0 ≤ c ≤ ∞. We write this limit uniquely in the form
c=
a−α
α−b
(86)
for b ≤ α ≤ a with α = b if c = ∞.
(ii) If (85) holds for some c ∈ (0, ∞) then there exists a slowly varying function
L such that
Ua (x) ∼ (α − b)xa−α L(x)
and
Vb (x) ∼ (a − α)xb−α L(x)
where α ∈ [b, a] is chosen such that (86) holds.
40
References
[1] Feller, William. An Introduction to Probability Theory and Its Applications.
John Wiley, 1971.
[2] Billingsley, Patrick. Convergence of Probability Measures. John Wiley, 1999.
[3] P. Nolan, John. Stable Distributions: Models for Heavy Tailed Data. 2009.
[4] Gaudreau Lamarre, Pierre-Yves. Roots of x2 = 2n ln(x). 2012
41