On Kesten’s counterexample to the Cramér-Wold
device for regular variation
Henrik Hult
School of ORIE
Cornell University
Ithaca NY 14853
USA
[email protected]
Filip Lindskog∗
Department of Mathematics
Royal Institute of Technology
10044 Stockholm
Sweden
[email protected]
Abstract
In 2002 Basrak, Davis and Mikosch showed that an analogue of the
Cramér-Wold device holds for regular variation of random vectors if the
index of regular variation is not an integer. This characterization is of
importance when studying stationary solutions to stochastic recurrence
equations. In this paper we construct counterexamples showing that for
integer-valued indices, regular variation of all linear combinations does
not imply that the vector is regularly varying. The construction is based
on unpublished notes by Harry Kesten.
Key words: heavy-tailed distributions; linear combinations; multivariate
regular variation
1
1
Introduction
For Rd -valued random vectors Xn and X, the well-known Cramér-Wold Theod
d
rem says that a necessary and sufficient condition for Xn →
X is that xT Xn →
T
d
d
x X for every x ∈ R . We use the convention that x ∈ R is a column vector
and xT its transpose. In Basrak et al. (2002a) it was shown that, for nonintegervalued indices of regular variation, there is a similar characterization of regular
variation for a random vector in terms of regular variation of its linear combinations; meaning that for some α > 0 and some function L which is slowly
varying at infinity,

 for every x 6= 0, limt→∞ tα L(t) P(xT X > t) = w(x) exists,
(1)

w(x) > 0 for some x 6= 0.
If (1) holds, then necessarily w(ux) = uα w(x) for all x 6= 0 and u > 0. The
interest in this condition originates from a classical result by Kesten (1973b)
which (briefly) says that, under mild conditions, the stationary solution X of a
multivariate stochastic recurrence equation Xn = An Xn−1 + Bn satsifies (1),
where L(t) = 1 and α is the unique solution to
lim
n→∞
1
E log kAn · · · A1 kα = 0.
n
Here k · k denotes the operator norm, kAk = sup|x|=1 |Ax|, and | · | is the
Euclidean norm on Rd . A popular example is the stationary GARCH model
which can be embedded in a stochastic recurrence equation (see Basrak et al.,
2002b). Other examples where the condition (1) appears are the stochastic recurrence equations with heavy-tailed innovations studied in Konstantinides and
Mikosch (2005) and the random coefficient AR(q) models of Klüppelberg and
Pergamenchtchikov (2004). Important and accessible papers on the extremal
behavior of solutions to univariate stochastic recurrence equations are de Haan
et al. (1989), Goldie (1991) and Borkovec and Klüppelberg (2001).
A random vector X is said to be regularly varying if there exist an α > 0
and a probability measure σ on B(Sd−1 ), the Borel σ-field of Sd−1 = {x ∈ Rd :
|x| = 1}, such that, for every x > 0, as t → ∞,
P(|X| > tx, X/|X| ∈ · ) v −α
→ x σ(·) on B(Sd−1 ).
P(|X| > t)
(2)
v
Here →
denotes vague convergence, and α and σ are respectively called the
index of regular variation and spectral measure of X. For α ∈ (0, 2) this formulation of multivariate regular variation is a necessary and sufficient condition for
the convergence in distribution of normalized partial sums of independent and
identically distributed random vectors to a stable random vector; see Rvačeva
(1962). It is also used for the characterization of the maximum domain of attraction of extreme value distributions and for weak convergence of point processes,
see for example Resnick (1987).
In Basrak et al. (2002a, Theorem 1.1) it was proved that (2) implies (1) and
the following statements hold:
(A) If X satisfies (1), where α is positive and noninteger, then (2) holds and
the spectral measure σ is uniquely determined.
2
(B) If X assumes values in [0, ∞)d and satisfies (1) for x ∈ [0, ∞)d \{0}, where
α is positive and noninteger, then (2) holds and the spectral measure σ is
uniquely determined.
(C) If X assumes values in [0, ∞)d and satisfies (1), where α is an odd integer,
then (2) holds and the spectral measure σ is uniquely determined.
In Section 2 we construct a counterexample which shows that (A) cannot be
extended to integer-valued indices of regular variation without additional assumptions on the distribution of X. In Section 3 we construct a counterexample
which shows that (B) cannot be extended to integer-valued indices of regular
variation without additional assumptions on the distribution of X. Whether
(C) is true or not in the case of α belonging to the set of even integers is, to the
best of the knowledge of the authors, still an open problem.
Let us point out that there are several equivalent formulations of (2); many
of them are documented in Basrak (2000) and Resnick (2004). See also Basrak et al. (2002a), Hult (2003), Lindskog (2004) and Resnick (1987) for more
on multivariate regular variation. For a detailed treatment of the concept of
regularly varying functions, see Bingham et al. (1987).
2
Construction of the counterexamples
The constructions of the counterexamples corresponding to (A) and (B) for
integer-valued indices of regular variation are rather similar and consist of two
steps. First note that due to the scaling property the limit function w in (1)
is determined by its values on Sd−1 . Therefore it is sufficient to consider linear
combinations xT X with x ∈ Sd−1 .
The first step consists of finding two bivariate regularly varying random
vectors X0 and X1 with index of regular variation α > 0 and spectral measures
σ0 and σ1 , with σ0 6= σ1 , such that for every x ∈ S and t > 1,
P(xT X0 > t) = P(xT X1 > t).
(3)
For the example corresponding to (B) we restrict x to S ∩ [0, ∞)2 .
In the second step we will construct the counterexamples by finding a random
vector X such that, for every x ∈ S and x ∈ S ∩ [0, ∞)2 , respectively,
lim tα P(xT X > t) = lim tα P(xT X0 > t) = lim tα P(xT X1 > t) =: w(x)
t→∞
t→∞
t→∞
(4)
and such that there are subsequences (un ), (vn ), un ↑ ∞, vn ↑ ∞, with the
property that for every S ∈ B(S) with σ0 (∂S) = σ1 (∂S) = 0,
P(|X| > un , X/|X| ∈ S)
= σ0 (S),
P(|X| > un )
P(|X| > vn , X/|X| ∈ S)
lim
= σ1 (S).
n→∞
P(|X| > vn )
lim
n→∞
(5)
(6)
The counterexamples are easily extended to Rd -valued random vectors. Take
X = (X (1) , X (2) )T as above and Y = (Y (1) , . . . , Y (d−2) )T independent of X with
3
Y satisfying (1) with the same α as X, L(t) = 1, and limit function wY . Put
Z = (X (1) , X (2) , Y (1) , . . . , Y (d−2) )T . Then, by independence (c.f. Davis and
Resnick, 1996, Lemma 2.1),
lim tα P(zT Z > t)
t→∞
= lim tα P((z (1) , z (2) )X > t) + lim tα P((z (3) , . . . , z (d) )Y > t)
t→∞
t→∞
= w(z (1) , z (2) ) + wY (z (3) , . . . , z (d) ).
Hence, Z satisfies (1). However, Z does not satisfy (2). Indeed, assume on the
contrary that Z satisfies (2) with spectral measure σZ . Then since X = T (Z)
with T : Rd → R2 is given by T (z) = (z (1) , z (2) )T it follows that (e.g. Basrak
et al., 2002b, Proposition A.1) X satisfies (2) for some spectral measure σ. This
is a contradiction.
2.1
Construction of X0 and X1
We will now focus on the construction of X0 and X1 in the counterexample
corresponding to (A) when α is a positive integer.
We will construct two regularly varying random vectors X0 and X1 with
index of regular variation α and different spectral measures such that (3) is
satisfied. A different construction in the case α = 1 can be found in Meerschaert
and Scheffler (2001, Example 6.1.35).
Let Θ0 be a [0, 2π)-valued random variable with density f0 satisfying, for
some ε > 0, f0 (θ) > ε for all θ ∈ [0, 2π). Take v ∈ (0, ε) and let Θ1 have density
f1 given by
f1 (θ) = f0 (θ) + v sin((α + 2)θ),
θ ∈ [0, 2π).
Let R ∼ Pareto(α), that is P(R > x) = x−α for x ≥ 1, be independent of Θi ,
i = 0, 1, and put
d
Xi =
(R cos Θi , R sin Θi )T .
Obviously Xi is regularly varying with σi (·) = P((cos Θi , sin Θi ) ∈ · ). Take
x ∈ S and let β ∈ [0, 2π) be given by x = (cos β, sin β)T . Then, for t > 1,
P(xT X1 > t) − P(xT X0 > t)
Z ∞ Z β+arccos(t/r)
αr−α−1 sin((α + 2)θ)dθdr
=v
t
β−arccos(t/r)
Z ∞
³
−α−1
vα
=−
α+2
r
cos{(α + 2)(β + arccos(t/r)}
t
´
− cos{(α + 2)(β − arccos(t/r))} dr
Z ∞
2vα
=
sin((α + 2)β)
r−α−1 sin{(α + 2) arccos(t/r)}dr.
α+2
t
Using standard variable substitutions and trigonometric formulae the integral
4
can be rewritten as follows:
Z ∞
r−α−1 sin{(α + 2) arccos(t/r)}dr
t
Z
= t−α
1
rα−1 sin{(α + 2) arccos(r)}dr
0
Z
= t−α
π/2
cosα−1 (r) sin((α + 2)r) sin(r)dr
0
Z
=t
−α
Z
π/2
cos
α−1
(r) cos((α + 1)r)dr − t
−α
0
π/2
cosα (r) cos((α + 2)r)dr.
0
The two last integrals equal zero for every α ∈ {1, 2, . . . }, see Gradshteyn and
Ryzhik (2000) p. 392. Hence, for t > 1, P(xT X1 > t) = P(xT X0 > t), which
proves (3).
The following construction of X satisfying (4)-(6) is based on unpublished
notes by Harry Kesten (1973a) relating to Remark 4, p. 245, in Kesten (1973b).
2.2
The counterexample
Consider the random vectors X0 and X1 above. Let g0 and g1 denote their
densities. We construct a random vector X satisfying (1) which is not regularly
varying; we will show that it satisfies (4) - (6).
Take y ∈ R2 with |y| > 1. There exist unique integers j, n ≥ 1 such that
|y| ∈ (j!, (j + 1)!]
and
j∈
n n−1
X
2k + 1, . . . ,
k=1
n
X
o
2k .
k=1
2
Let X be an R -valued random vector with density g given by
³
n−1
n−1
³
´
³
X ´
X ´
1− j−
2k 2−n gb(n) (y) + j −
2k 2−n gb(n+1) (y),
k=1
k=1
for |y| ∈ (j!, (j + 1)!], where
b(n) =
½
0,
1,
if n is odd,
if n is even.
That is, the density g is given by

g0 (y) = 0,







1
1


2 g0 (y) + 2 g1 (y),







g1 (y),





3
1
g(y) =
4 g0 (y) + 4 g1 (y),





1
1


2 g0 (y) + 2 g1 (y),






3
1


4 g0 (y) + 4 g1 (y),






etc.
5
|y| ∈ (0, 1],
|y| ∈ (1, 2],
|y| ∈ (2, 3!],
|y| ∈ (3!, 4!],
|y| ∈ (4!, 5!],
|y| ∈ (5!, 6!],
Note that in each disc |y| ∈ (j!, (j + 1)!] the density g is a convex combination
of the densities g0 and g1 . Therefore,
Z
∞ Z
X
g(y)dy =
g(y)dy
R2
=
j=1 |y|∈(j!,(j+1)!]
∞ Z (j+1)!
X
−α−1
αr
j=1
Z
∞
dr =
j!
αr−α−1 dr = 1,
1
so g is indeed a probability density. Take x ∈ S and t ∈ ((j − 1)!, j!]. Then
there are two possibilities:
(i) j − 1 ∈
n n−1
X
2k + 1, . . . ,
k=1
(ii) j − 1 =
n−1
X
n
X
o
2k − 1
or
k=1
2k .
k=1
³
Pn−1 ´
Suppose (i) holds. Then, with γ = j − k=1 2k 2−n ∈ [0, 1], we have
¡
¢
P(xT X > t) = 1 − γ + 2−n P(xT Xb(n) > t, |Xb(n) | ∈ (t, j!])
¡
¢
+ γ − 2−n P(xT Xb(n+1) > t, |Xb(n+1) | ∈ (t, j!])
+ (1 − γ) P(xT Xb(n) > t, |Xb(n) | ∈ (j!, (j + 1)!])
+ γ P(xT Xb(n+1) > t, |Xb(n+1) | ∈ (j!, (j + 1)!])
+ P(xT X > t, |X| > (j + 1)!).
Hence,
n
P(xT X > t) = 2−n P(xT Xb(n) > t, |Xb(n) | ∈ (t, j!])
− P(xT Xb(n+1) > t, |Xb(n+1) | ∈ (t, j!])
o
+ (1 − γ) P(xT Xb(n) > t) + γ P(xT Xb(n+1) > t)
{z
}
|
Bn
− (1 − γ) P(xT Xb(n) > t, |Xb(n) | > (j + 1)!)
− γ P(xT Xb(n+1) > t, |Xb(n+1) | > (j + 1)!)
+ P(xT X > t, |X| > (j + 1)!).
We have
n
o
2−n P(xT Xb(n) > t, |Xb(n) | ∈ (t, j!]) − P(xT Xb(n+1) > t, |Xb(n+1) | ∈ (t, j!])
n
o
≤ 2−n P(|Xb(n) | > t) + P(|Xb(n+1) | > t) = 2−n+1 t−α .
Moreover, since P(xT X1 > t) = P(xT X0 > t) we have Bn = P(xT X0 > t).
The absolute value of each of the remaining terms is less than or equal to
((j + 1)!)−α ≤ (j j!)−α ≤ (j t)−α , so we conclude that
tα | P(xT X > t) − P(xT X0 > t)| ≤ 2−n+1 + 3j −α .
6
Since j = j(t) → ∞ and n = n(t) → ∞ as t → ∞, we have
¯
¯
lim tα ¯ P(xT X > t) − P(xT X0 > t)¯ = 0.
t→∞
That is,
lim tα P(xT X > t) = lim tα P(xT X0 > t) = lim tα P(xT X1 > t).
t→∞
t→∞
t→∞
Suppose now that (ii) holds. Then
P(xT X > t) = P(xT Xb(n) > t, |Xb(n) | ∈ (t, j!])
+ (1 − 2−n ) P(xT Xb(n) > t, |Xb(n) | ∈ (j!, (j + 1)!])
+ 2−n P(xT Xb(n+1) > t, |Xb(n+1) | ∈ (j!, (j + 1)!])
+ P(xT X > t, |X| > (j + 1)!)
= P(xT Xb(n) > t) − P(xT Xb(n) > t, |Xb(n) | > (j + 1)!)
n
+ 2−n P(xT Xb(n+1) > t, |Xb(n+1) | ∈ (j!, (j + 1)!])
o
− P(xT Xb(n) > t, |Xb(n) | ∈ (j!, (j + 1)!])
+ P(xT X > t, |X| > (j + 1)!).
By arguments similar to those for case (i) we obtain
tα | P(xT X > t) − P(xT X0 > t)| ≤ 2(2−n + j −α ).
It follows that
lim tα P(xT X > t) = lim tα P(xT X0 > t) = lim tα P(xT X1 > t),
t→∞
t→∞
t→∞
which proves (4).
Finally, we find subsequences (un ) and (vn ) satisfying (5) and (6). Take
S ∈ B(S) with σ0 (∂S) = σ1 (∂S) = 0. Put
cn =
2n
X
2k
and
dn =
k=1
2n+1
X
2k .
k=1
Note that for cn ! < |y| ≤ (cn + 1)! we have g(y) = gb(2n+1) (y) = g0 (y), whereas
for dn ! < |y| ≤ (dn + 1)! we have g(y) = gb(2n+2) (y) = g1 (y). It follows that,
with un = cn !,
α
uα
n P(|X| > un , X/|X| ∈ S) = un P (cn ! < |X| ≤ (cn + 1)!, X/|X| ∈ S)
+ uα
n P (|X| > (cn + 1)!, X/|X| ∈ S) .
Since the second term is less than or equal to
α
uα
n P (|X| > (cn + 1)!) = (cn !) [(cn + 1)!]
−α
→ 0,
as n → ∞, it follows that
³
´
−α
α
−α
lim uα
P(|X|
>
u
,
X/|X|
∈
S)
=
lim
u
u
−
[(c
+
1)!]
σ0 (S)
n
n
n
n
n
n→∞
n→∞
= σ0 (S).
7
By a similar argument, with vn = dn !,
lim vnα P(|X| > vn , X/|X| ∈ S) = σ1 (S).
n→∞
Thus, we have found sequences (un ) and (vn ) satisfying (5) and (6) and the
counterexample is complete.
3
Vectors with nonnegative components
In this section we construct a counterexample corresponding to (B) in the case
of integer-valued indices of regular variation.
The following construction of X0 and X1 was given in Basrak et al. (2002a)
for α = 2 but can, as we will see, be extended to any positive integer α. Take
α ∈ {1, 2, . . . } and let Θ0 , Θ1 be two [0, π/2]-valued random variables with
unequal distributions satisfying
E(cosk Θ0 sinα−k Θ0 ) = E(cosk Θ1 sinα−k Θ1 ),
k = 0, 1, . . . , α.
(7)
−α
Let R ∼ Pareto(α), that is P(R > x) = x
for x ≥ 1. Suppose R is independ
dent of Θi , i = 0, 1 and put Xi =
(R cos Θi , R sin Θi )T . For x ∈ [0, ∞)2 \{0},
we have
tα P(xT Xi > t) = tα P(x1 R cos Θi + x2 R sin Θi > t)
Z ∞
= tα
P(x1 cos Θi + x2 sin Θi > t/r)αr−α−1 dr
1
tα
Z
P((x1 cos Θi + x2 sin Θi )α > v)dv
=
0
α µ ¶
X
α k α−k
=
E(cosk Θi sinα−k Θi )
x x
k 1 2
k=1
for t sufficiently large. We can now construct the vector X as in Section 2.2,
with x ∈ S+ = S ∩ [0, ∞)2 and the new densities g0 and g1 of X0 and X1 . It
remains to show that we can find unequal distributions of the [0, π/2]-valued
random variables Θ0 and Θ1 satisfying (7). Let Θ0 have density f0 satisfying,
for some ε > 0, f0 (θ) > ε for all θ ∈ [0, π/2]. We will show that the density f1
of Θ1 can be chosen as f1 (θ) = f0 (θ) + vf (θ), where v ∈ (0, ε) and f is chosen
R π/2
such that supθ∈[0,π/2] |f (θ)| = 1, 0 f (θ)dθ = 0, and (7) holds. Let
A := span{1, sinα (θ), cos(θ) sinα−1 (θ), . . . , cosα (θ)} ⊂ C2 ([0, π/2]),
where C2 ([0, π/2]) is the space of real-valued continuous functions on [0, π/2]
R π/2
equipped with the inner product (h1 , h2 ) = 0 h1 (s)h2 (s)ds, which makes it
an inner product space. For any nonzero fe ∈
/ A with fe ∈ C2 ([0, π/2]) we can
choose
fe − ProjA (fe)
.
f :=
supθ∈[0,π/2] |{fe − ProjA (fe)}(θ)|
Then f ⊥ A, f1 is a density function and (7) holds. Since A is a finitedimensional subspace of the infinite-dimensional space C2 ([0, π/2]) it is clear
that its orthogonal complement is nonempty. Hence, the density f1 above exists.
8
4
Acknowledgements
An essential part of the construction of the counterexamples in this paper is
based on unpublished notes by Harry Kesten. The authors wish to thank Laurens de Haan for providing them with these notes and Harry Kesten for supporting this publication. The authors thank the anonymous referee for comments
that led to an improvement of the presentation of the material. The research of
the first author was supported by the Swedish Research Council.
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