Infinite divisibility of a class of two-dimensional vectors with components in the second Wiener chaos is studied. Necessary and sufficient conditions for infinite divisibility are presented as well as more easily verifiable sufficient conditions. The case where both components consist of a sum of two Gaussian squares is treated in more depth, and it is conjectured that such vectors are infinitely divisible.
Sums of Gaussian squaresinfinite divisibilitysecond Wiener chaos60E0760G1562H0562H10This research was supported by the Danish Council for Independent Research (Grant DFF-4002-00003).Introduction
Paul Lévy raised the question of infinite divisibility of Gaussian squares, that is, for a centered Gaussian vector (X1,…,Xn) when can (X12,…,Xn2) be written as a sum of m independent identically distributed random vectors for any m∈N? (See [11].) Several authors have studied this problem. We refer to [4–8, 13, 14] and references therein. These works include several novel approaches and give a great understanding of when Gaussian squares are infinitely divisible. In this paper we will provide a characterization of infinite divisibility of sums of Gaussian squares which to the best of our knowledge has not been studied in the literature except in special cases. This problem is highly motivated by the fact that sums of Gaussian squares are usual limits in many limit theorems in the presence of either long range dependence, see [2] or [17], or degenerate U-statistics, see [9]. In the following we will go in more details.
Let Y be a random variable in the second (Gaussian) Wiener chaos, that is, the closed linear span in L2 of {W(h)2−1:h∈H,‖h‖=1} for a real separable Hilbert space H and an isonormal Gaussian process W. For convenience, we assume H is infinite-dimensional. Then there exists a sequence of independent standard Gaussian variables (ξi) and a sequence of real numbers (αi) such that
Y=d∑i=1∞αi(ξi2−1),
where the sum converges in L2 (see for example [9, Theorem 6.1]). Since the ξi’s are independent, (ξ12,…,ξd2) is infinitely divisible for any d≥1 and therefore, Y is infinitely divisible. Such a sum of Gaussian squares appears as the limit of U-statistics in the degenerate case (see [9, Corollary 11.5]). In this case the αi are certain binomial coefficients times the eigenvalues of operators associated to the U-statistics. But even though any random variable in the second Wiener chaos is infinitely divisible, it is well known (cf. Theorem 1 below) that a vector with dimension greater than two and components in the second Wiener chaos needs not be infinite divisibility. In between the case of a random variable in the second Wiener chaos and the vector case with dimension greater than two, there is the open question of infinite divisibility of a two-dimensional vector with components in the second Wiener chaos. Let (X1,…,Xn1+n2) be a zero mean Gaussian vector for n1,n2∈N. The fact that any two-dimensional vector in the second Wiener chaos is infinitely divisible is equivalent to
(d1X12+⋯+dn1Xn12,dn1Xn1+12+⋯+dn1+n2Xn1+n22)
being infinitely divisible for any d1,…,dn1+n2=±1, any covariance structure of (X1,…,Xn1+n2), and any n1,n2∈N (something what follows by the definition of the second Wiener chaos).
The following theorem, which is due to Griffiths [8] and Bapat [1], is an important first result related to infinite divisibility in the second Wiener chaos. We refer to [12, Theorem 13.2.1 and Lemma 14.9.4] for a proof.
(Griffiths and Bapat).
Let(X1,…,Xn)be a zero mean Gaussian vector with a positive definite covariance matrix Σ. Then(X12,…,Xn2)is infinitely divisible if and only if there exists ann×nmatrix U of the formdiag(±1,…,±1)such thatUtΣ−1Uhas nonpositive off-diagonal elements.
This theorem resolved the question of infinite divisibility of Gaussian squares. For n≥3 there is an n×n positive definite matrix Σ for which there does not exist an n×n matrix U of the form diag(±1,…,±1) such that UtΣ−1U has nonpositive off-diagonal elements. Consequently, there are zero mean Gaussian vectors (X1,…,Xn) such that (X12,…,Xn2) is not infinite divisible whenever n≥3.
Eisenbaum [3] and Eisenbaum and Kaspi [5] found a connection between the condition of Griffiths and Bapat and the Green function of a Markov process. In particular, a Gaussian process has infinite divisible squares if and only if its covariance function (up to a constant function) can be associated with the Green function of a strongly symmetric transient Borel right Markov process.
When discussing the infinite divisibility of the Wishart distribution, [16] showed the following result.
(Shanbhag).
For anyn∈Nand any zero mean Gaussian vector(X1,…,Xn), the two-dimensional random vector(X12,X22+⋯+Xn2)is infinitely divisible.
Furthermore, it was found that infinite divisibility of any bivariate marginal of a centered Wishart distribution can be reduced to infinite divisibility of (X1X2,X3X4). By the polarization identity,
(X1X2,X3X4)=14((X1+X2)2−(X1−X2)2,(X3+X4)2−(X3−X4)2).
Consequently, infinite divisibility of any bivariate marginals of a centered Wishart distribution is again related to the question of infinite divisibility of a two-dimensional vector from the second Wiener chaos.
Key question and contributions. The main question of the paper is: Given a zero mean Gaussian vector (X1,…,Xn1+n2)when is(X12+⋯+Xn12,Xn1+12+⋯+Xn1+n22)infinitely divisible?
The vector in (2) is (1) in the case d1=⋯=dn1+n2=1. In Theorem 3 we give a necessary and sufficient condition for infinite divisibility of the random vector in (2); namely that (7) is satisfied for all k,l∈N0. Furthermore, in Theorem 4 we give sufficient conditions for infinite divisibility of (2) in the case n1=n2=2, We conjecture that (2) is always infinitely divisible in the case n1=n2=2, and in fact in Theorem 5 we show that our necessary and sufficient condition (7) is always satisfied for all k,l∈N0 with k+l≤7. The general case of infinite divisibility of (1), where di=−1 for at least one i, seems to require new ideas going beyond this paper.
Note that if (X12,…,Xn1+n22) is infinitely divisible then so is the vector in (2). Thus, under the conditions in Theorem 1, (2) is answered in the affirmative. However, we shall see that there are cases where the vector in (2) is infinitely divisible even though (X12,…,Xn1+n22) is not, see the comment after Theorem 4. Besides, based on our results we can give a short new proof of Shanbhag’s Theorem 2, see just below Theorem 3.
The paper is structured as follows. In Section 2 we introduce the required notation to state our results. The main results without proofs are presented in Section 3. Section 4 contains two examples and a small numerical discussion. We end with Section 5 where the proofs are given.
Notation
In the following two subsections we will introduce the notation used for our main results on infinite divisibility of two-dimensional vectors in the second Wiener chaos (2).
The general case <inline-formula id="j_vmsta160_ineq_045"><alternatives>
<mml:math><mml:msub><mml:mrow><mml:mi mathvariant="italic">n</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo mathvariant="normal">,</mml:mo><mml:msub><mml:mrow><mml:mi mathvariant="italic">n</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub><mml:mo stretchy="false">∈</mml:mo><mml:mi mathvariant="double-struck">N</mml:mi></mml:math>
<tex-math><![CDATA[${n_{1}},{n_{2}}\in \mathbb{N}$]]></tex-math></alternatives></inline-formula>
Let n1,n2∈N and consider a zero mean Gaussian vector (X1,…,Xn1+n2) with a positive definite covariance matrix Σ. For a>00$]]>, let Q=I−(I+aΣ)−1 and write
Q=Q11Q12Q21Q22
where Q11 is an n1×n1 matrix, Q22 is an n2×n2 matrix, and Q12=Q21t (where Q21t is the transpose of Q21) is an n1×n2 matrix. Note that if λ is an eigenvalue of Σ, aλ1+aλ is an eigenvalue of Q. Since Q is symmetric and has positive eigenvalues, it is positive definite.
The following definition is a natural extension to the present setup of the terminology used by [1].
Let n1,n2∈N. An (n1+n2)×(n1+n2) orthogonal matrix U is said to be an (n1,n2)-signature matrix if
U=U100U2
where U1 is an n1×n1 matrix and U2 is an n2×n2 matrix, both orthogonal, and for 0’s of suitable dimensions.
The special case <inline-formula id="j_vmsta160_ineq_066"><alternatives>
<mml:math><mml:msub><mml:mrow><mml:mi mathvariant="italic">n</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo>=</mml:mo><mml:msub><mml:mrow><mml:mi mathvariant="italic">n</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub><mml:mo>=</mml:mo><mml:mn>2</mml:mn></mml:math>
<tex-math><![CDATA[${n_{1}}={n_{2}}=2$]]></tex-math></alternatives></inline-formula>
The following notation is only used in the special case where n1=n2=2 in (2). Consider a 2×2 symmetric matrix A. Let v1 and v2 be the eigenvectors of A, and λ1 and λ2 be the corresponding eigenvalues. We say that vi is associated with the largest eigenvalue if λi≥λj for j=1,2. Furthermore, whenever A is a multiple of the identity matrix, we fix (1,0) to be the eigenvector associated with the largest eigenvalue.
Let W be a (2,2)-signature matrix such that
WtQW=W1tQ11W1W1tQ12W2W2tQ21W1W2tQ22W2=q110q13q140q22q23q24q13q23q330q14q240q44,
where q11≥q22>00$]]> and q33≥q44>00$]]> which exist by Lemma 2. Note that qij is the (i,j)-th entry not of Q but of WtQW. Let (γ1,γ2) be the eigenvector of W1tQ12Q21W1 associated with the largest eigenvalue. If q11=q22 or q33=q44, any orthogonal W1 or W2 gives the desired form. In this case, we may always choose W1 or W2 such that γ1q13(γ1q13+γ2q23)≥0 (see the proof of Lemma 3, (ii) ⇒ (iii)), and we fix this choice.
Write
Σ−1=Σ11Σ12Σ21Σ22
where Σij is a 2×2 matrix for i,j=1,2. Let W be a (2,2)-signature matrix such that
WtΣ−1W=W1tΣ11W1W1tΣ12W2W2tΣ21W1W2tΣ22W2=σ110σ13σ140σ22σ23σ24σ13σ23σ330σ14σ240σ44
where σ11≥σ22>00$]]> and σ33≥σ44>00$]]> which exist by Lemma 2. Note that σij is the (i,j)-th entry not of Σ−1 but of WtΣ−1W. Let (ν1,ν2) be the eigenvector of W1tΣ12Σ21W1 associated with the largest eigenvalue. If σ11=σ22 or σ33=σ44, any orthogonal W1 or W2 gives the desired form. In this case, we may choose W1 or W2 such that ν2σ24(ν2σ24+ν1σ14)≥0, and we fix this choice.
Main results
In this section we will first consider the general case n1,n2∈N in Subsection 3.1, and thereafter, in Subsection 3.2, provide more specialized conditions for the case n1=n2=2.
The general case
The following result gives a necessary and sufficient condition for infinite divisibility as stated in the Key question (2).
Letn1,n2∈Nand(X1,…,Xn1+n2)denote a zero mean Gaussian vector with a positive definite covariance matrix Σ, and let Q be defined in (3). Then(X12+⋯+Xn12,Xn1+12+⋯+Xn1+n22)is infinitely divisible if and only if for allk,m∈N0and for alla>00$]]>sufficiently large,∑traceQ11k1Q12Q22m1Q21Q11k2⋯Q11kdQ12Q22mdQ21Q11kd+1+∑traceQ22m1Q21Q11k1Q12Q22m2⋯Q22md−1Q21Q11kdQ12Q22md+1≥0,where the first sum is over allk1,…,kd+1andm1,…,mdsuch thatk1+⋯+kd+1+d=kandm1+⋯+md+d=m,and the second sum is over allm1,…,md+1andk1,…,kdsuch thatm1+⋯+md+1+d=mandk1+⋯+kd+d=k.
Our proof of Theorem 3 relies on similar techniques used in the proof of Theorem 1, namely the series expansion of the Laplace transform together with an application of a result by Feller. By applying Theorem 3 we can give a new and simple proof of Shanbhag’s result, Theorem 2, which states that (X12,X22+⋯+X1+n22) is infinitely divisible.
Consider Theorem 3 in the case n1=1 and n2∈N. Then Q11 is a positive number and Q12Q22mQ21 is a nonnegative number for any m∈N. In particular, we have
traceQ11k1Q12Q22m1Q21⋯Q12Q22mdQ21Q11kd+1=Q11k1⋯Q11kd+1Q12Q22m1Q21⋯Q12Q22mdQ21≥0
for any k1,…,kd+1,m1,…,md∈N0. Consequently, the first sum in (7) is a sum of nonnegative numbers. A similar argument gives that the other sum is nonnegative, too. We thus conclude that (X12,X22+⋯+X1+n22) is infinite divisible by Theorem 3. □
Despite the fact that Theorem 3 implies Shanbhag’s result, in general it can be difficult to check condition (7). The following proposition yields a sufficient condition for infinite divisibility which might be more easy to check in some cases.
Let(X1,…,Xn1+n2)be a zero mean Gaussian vector with a positive definite covariance matrix Σ. Then(X12+⋯+Xn12,Xn1+12+⋯+Xn1+n22)is infinitely divisible if there exists an(n1,n2)-signature matrix U (cf. Definition1) such thatUtΣ−1Uhas nonpositive off-diagonal elements.
Write X=(X1,…,Xn1) and Y=(Xn1+1,…,Xn1+n2), and note that
(X12+⋯+Xn12,Xn1+12+⋯+Xn1+n22)=(‖X‖2,‖Y‖2)=(‖U1X‖2,‖U2Y‖2)
for any n1×n1 orthogonal matrix U1 and n2×n2 orthogonal matrix U2. Consequently, any property of the distribution of (8) is invariant under transformations of the form
U1t00U2tΣU100U2
of the covariance matrix Σ. Therefore, when there exists an (n1,n2)-signature matrix U such that UtΣ−1U has nonpositive off-diagonal elements, Theorem 1 ensures infinite divisibility of (9). □
In the following subsection we will specialize our setting to obtain conditions which are easier to apply.
The <inline-formula id="j_vmsta160_ineq_140"><alternatives>
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<tex-math><![CDATA[${n_{1}}={n_{2}}=2$]]></tex-math></alternatives></inline-formula> case
The following theorem provides easy to check conditions for infinite divisibility of vectors of the form (X12+X22,X32+X42). By “easy to check” we mean that the conditions (i) and (ii) of Theorem 4 can be explicitly calculated through a finite number of standard matrix operations, what is opposite to the general condition (7).
Let(X1,X2,X3,X4)denote a zero mean Gaussian vector with a positive definite covariance matrix Σ, and letqij,γi,σijandνibe defined as in Subsection2.2. Then the vector(X12+X22,X32+X42)is infinitely divisible if at least one of the following two conditions is satisfied:
The inequalityγ1q13(γ1q13+γ2q23)≥0holds.
The inequalityν2σ24(ν2σ24+ν1σ14)≥0holds.
Example 2, from the next section, shows that (4) of Theorem 4 holds in some cases where (X12,X22,X32,X42) is not infinitely divisible, and hence it cannot be deduced from a direct application of Griffiths and Bapat’s result (Theorem 1). The following result gives some insight to the question whether infinite divisibility of the vector (X12+X22,X32+X42) holds in general, which still remains an open problem. The result, in particular, says that the necessary and sufficient condition (7) for infinitely divisibility is always satisfied whenever k and m are not too large.
Consider a zero mean Gaussian vector(X1,X2,X3,X4)with a positive definite covariance matrix Σ. Recall that(X12+X22,X32+X42)is infinitely divisible if and only if condition (7) is satisfied for allk,m∈N0. We have that (7) is satisfied for anyk,m∈N0for which at least one of the following inequalities holds (i):k≤2, (ii):m≤2, or (iii):k+m≤7.
In the proof of Theorem 4 we show that the inequality γ1q13(γ1q13+γ2q23)≥0 holds if and only if for all integers d∈N0, k1,…,kd+1∈N0 and m1,…,md∈N0 we have
traceQ11k1Q12Q22m1Q21Q11k1⋯Q11kdQ12Q22mdQ21Q11kd+1≥0.
Hence, when γ1q13(γ1q13+γ2q23)<0, we know that there are k,m∈N0 such that (7) with n1=n2=2 contains negative terms. It is still an open problem to decide if the positive terms will always compensate for the negative terms such that (X12+X22,X32+X42) is always infinitely divisible, or this is only the case when k and m is small, e.g. k+m≤7, cf. Theorem 5.
Examples and numerics
We begin this section by presenting two examples treating the inequalities in Theorem 3.2 in special cases. Then we calculate the sums in (7) numerically with n1=n2=2 for a specific value of Q for k and m less than 60.
Fix a>00$]]> and assume that the matrix Q=I−(I+aΣ)−1 is of the form
Q=Q11Q12Q21Q22=q10εε0q2ε−δεεq30ε−δ0q4
where δ,ε>00$]]>, q1>q2>0{q_{2}}>0$]]>, and q3>q4>0{q_{4}}>0$]]>. Let γ=(γ1,γ2) be the eigenvector of
Q12Q21=2ε2ε(ε−δ)ε(ε−δ)ε2+δ2
associated with the largest eigenvalue λ1. We will argue that the inequality in Theorem 3.2(i), which reads
γ1(γ1+γ2)≥0
in this case, holds if and only if δ≤ε, and hence (X12+X22,X32+X42) is infinitely divisible whenever δ≤ε, cf. Theorem 3.2(i).
Since −γ also is an eigenvector of Q12Q21 associated with the largest eigenvalue, we assume γ1≥0 without loss of generality. Assume δ≤ε. If δ=ε, v=(1,0) and the inequality in (10) holds. Assume δ<ε. Since λ1 is the largest eigenvalue,
λ1=sup|γ|=1γtQ12Q21γ≥2ε2
which implies that
2ε2−λ1≤0≤ε(ε−δ).
Since γ is an eigenvector, (Q−λ1)γ=0 and we therefore have that
0=(2ε2−λ1)γ1+ε(ε−δ)γ2≤ε(ε−δ)(γ1+γ2).
We conclude that (10) holds.
On the other hand, assume δ>ε\varepsilon $]]> and γ1≥0. Since λ1 is the largest eigenvalue, λ1≥δ2+ε2>δε+ε2\delta \varepsilon +{\varepsilon ^{2}}$]]> and therefore
(λ1−2ε2)>ε(δ−ε).\varepsilon (\delta -\varepsilon ).\]]]>
Note that γ1 cannot be zero since the off-diagonal element in Q12Q21 is nonzero. We conclude that
0=(λ1−2ε2)γ1+ε(δ−ε)γ2>ε(δ−ε)(γ1+γ2).\varepsilon (\delta -\varepsilon )({\gamma _{1}}+{\gamma _{2}}).\]]]>
This implies that (10) does not hold.
Assume Σ−1 is of the form
Σ−1=Σ11Σ12Σ21Σ22=σ10−δε0σ2εε−δεσ30εε0σ4
where σ1>σ2>0{\sigma _{2}}>0$]]>, σ3>σ4>0{\sigma _{4}}>0$]]>, and δ,ε>00$]]>. Let ν=(ν1,ν2) be the eigenvector of Σ12Σ21 associated with the largest eigenvalue. We will argue that the inequality in Theorem 4(ii) holds if and only if δ≤ε. Then the same theorem implies that (X12+X22,X32+X42) is infinitely divisible whenever δ≤ε. On the other hand, Theorem 1 implies that (X12,X22,X32,X42) is never infinite divisible under (11) since there does not exist a matrix D of the form diag(±1,±1,±1,±1) such that DΣ−1D has nonpositive off-diagonal elements. Indeed, for any two matrices D1 and D2 of the form diag(±1,±1), D1Σ12D2 has either three negative and one positive or one negative and three positive entries.
In the following we will see that the inequality in Theorem 4(ii), which reads ν2(ν1+ν2)≥0 in this setting, holds if and only if δ≤ε. Let
P=0110
and Q12 be given as in Example 1. Then PΣ12P=Q12, implying that (ν2,ν1) is the eigenvector associated with the largest eigenvalue of Q12Q21. We have argued in Example 1 that ν2(ν1+ν2)≥0 holds if and only if δ≤ε, what is the desired conclusion.
Now we investigate infinite divisibility of (X12+X22,X32+X42) numerically. More specifically, we consider the sums in (7) with n1=n2=2 for a specific choice of a positive definite matrix and different values of k and m. We will scale Q to have its largest eigenvalue equal to one to avoid getting too close to zero. Due to Theorem 4(i) the case where γ1q13(γ1q13+γ2q23)<0 (in the notation of Theorem 4) is the only case where the question on the infinite divisibility of (X12+X22,X32+X42) is open.
Let
Q=1λ0.800.010.0100.30.01−0.20.010.010.800.01−0.200.3
where λ>00$]]> is chosen such that Q has its largest eigenvalue equal to 1. Note that by Example 1, γ1q13(γ1q13+γ2q23)<0. In the below Figure 1 the logarithm of the sums in (7) for k and m between 0 and 60 is plotted. It is seen that the logarithm seems stable and hence the sums in (7) seem to remain positive in this case. A similar analysis have been done for other positive definite matrices, and we have not encountered any k,m∈N0 such that (7) is negative. This, together with Theorem 4(i), leads us to the conjecture that (X12+X22,X32+X42) is infinitely divisible for any zero mean Gaussian vector (X1,X2,X3,X4).
The logarithm of the sums in (7) for k and m between 0 and 60
Proofs
In this section we will prove Theorems 3, 4, and 5.
Proof of Theorem <xref rid="j_vmsta160_stat_004">3</xref>
The following lemma will be useful in the proof of Theorem 3. A proof can be found in [12, Lemma 13.2.2].
Letψ:R+n→(0,∞)be a continuous function. Suppose that, for alla>00$]]>sufficiently large,logψ(a(1−s1),…,a(1−sn))has a power series expansion fors=(s1,…,sn)∈[0,1]narounds=0with all its coefficients nonnegative, except for the constant term. Then ψ is the Laplace transform of an infinitely divisible random variable inR+n.
We now give the proof of Theorem 3.
By [12, Lemma 5.2.1],
P(s1,s2)=Eexp{−12a((1−s1)(X12+⋯+Xn12)+(1−s2)(Xn1+12+⋯+Xn22))}=1|I+Σa(I−S)|1/2,
where S is the (n1+n2)×(n1+n2) diagonal matrix with s1 on the first n1 diagonal entries and s2 on the remaining n2 diagonal entries. Recall that Q=I−(I+aΣ)−1. Then
P(s1,s2)2=|I+aΣ−aΣS|−1=|(I−Q)−1−((I−Q)−1−I)S|−1=|I−Q||I−QS|−1,
from which it follows that
2logP(s1,s2)=log|I−Q|−log|I−QS|=log|I−Q|+∑n=1∞trace{(QS)n}n,
where the last equality follows from [12, p. 562]. Now assume that the vector (X12+⋯+Xn12,Xn1+12+⋯+Xn1+n22) is infinitely divisible, and write
(X12+⋯+Xn12,Xn1+12+⋯+Xn1+n22)=dY1n+⋯+Ynn
where Y1n,…Ynn are 2-dimensional independent identically distributed stochastic vectors. Let Yijn be the j-th component of Yin and note that Yijn≥0 a.s. for all i, j, n. Then
P(s1,s2)1/n=Eexp{−12a((1−s1)Y11n+(1−s2)Y12n)}.
From the expression
exp{−12a((1−sj)Y1jn)}=exp{−12aY1jn}∑k=0∞(sjaY1jn)k2kk!,
it follows that P1/n(s1,s2) has a power series expansion with all coefficient nonnegative. We have that
logP(s1,s2)=limn→∞(n(P1/n(s1,s2)−1)).
Note that (s1,s2)↦n(P1/n(s1,s2)−1) and all its derivatives converge uniformly on [0,1)×[0,1) by a Weierstrass M-test (see for example [15, Theorem 7.10]). Consequently, we may use [15, Theorem 7.17] to conclude that
∂α+β∂s1α∂s2βlimn→∞(n(P1/n(s1,s2)−1))=limn→∞∂α+β∂s1α∂s2β(n(P1/n(s1,s2)−1))
for any α,β∈N0. Thus, the fact that all the terms in the power series expansion of P1/n(s1,s2) are nonnegative implies that all the terms in the power series representation of logP(s1,s2) except the constant term are nonnegative by (13). By (12) we conclude that any coefficient in front of s1ks2m in trace{(QS)k+m} has to be nonnegative for all k,m∈N and a>00$]]>. Expanding the trace then gives that this is equivalent to nonnegativity of the sum in (7) for all k,m∈N0.
On the other hand, if the sum in (7) is nonnegative for all k,m∈N0 and a>00$]]> sufficiently large, (12) and Lemma 1 imply that
(X12+⋯+Xn12,Xn1+12+⋯+Xn1+n22)
is infinitely divisible. □
Proof of Theorem <xref rid="j_vmsta160_stat_008">4</xref>
We start this section with two lemmas on linear algebra. Lemma 3 will be very useful in the proofs that make up the rest of this section.
Let A be ann×npositive definite matrix. Letn1,n2∈Nbe such thatn1+n2=nand writeA=A11A12A21A22whereA11is ann1×n1matrix,A22is ann2×n2matrix, andA12=A21tis ann1×n2matrix. Then there exists an(n1,n2)-signature matrix W such thatWtAWhas the formA˜11A˜12A˜21A˜22whereA˜11=diag(a1,…,an1)andA˜22=diag(an1+1,…,an1+n2)withai>00$]]>fori=1,…,n1+n2, and whereA˜12=A˜21t. Furthermore, we may choose W such thata1≥a2≥⋯≥an1andan1+1≥an1+2≥⋯≥an1+n2.
Since A is positive definite, A11 and A22 are positive definite. Consequently, by the spectral theorem (see for example [10, Corollary 6.4.7]), there exist an n1×n1 matrix W1 and an n2×n2 matrix W2, both orthogonal, such that W1tA11W1 and W2tA22W2 are diagonal with positive diagonal entries. Since permutation matrices are orthogonal matrices, we may assume the diagonal is ordered by size in both W1tA11W1 and W2tA22W2. Consequently, letting
W=W100W2,
implies that WtAW has the right form. □
For a fixed eigenvector vi we will say that the system Avi=λivi is the system of eigenequations. The k-th equation in this system will be called the k-th eigenequation associated with vi. Let A be a 4×4 positive definite matrix, and let W be a (2,2)-signature such that
WtAW=W1tA11W1W1tA12W2W2tA21W1W2tA22W2=a110a13a140a22a23a24a13a23a330a14a240a44,
where a11≥a22>00$]]> and a33≥a44>00$]]> which exist by Lemma 2. Note that aij is the (i,j)-th entry not of A but of WtAW. Let v1=(v11,v21) be the eigenvector associated with the largest eigenvalue of W1tA12A21W1. If a11=a22 or a33=a44, any orthogonal W1 or W2 give the desired form. In this case, we may choose W1 or W2 such that v11a13(v11a13+v21a23)≥0, and we fix this choice. Then the lemma below will play a central role in the proofs of the previously stated results.
In the notation above, the following are equivalent.
There exists a(2,2)-signature matrix U such thatUtAUhas all entries nonnegative.
For anyd∈Nandk1,…,kd+1,m1,…md∈N0,traceA11k1A12A22m1A21A11k2⋯A11kdA12A22mdA21A11kd+1≥0.
The inequalityv11a13(v11a13+v21a23)≥0holds.
(i) ⇒ (ii). Let
U=U100U2
be such that Bij=UitAijUj has nonnegative entries for i,j=1,2. Then
traceA11k0A12A22m1A21A11k1⋯A11kd−1A12A22mdA21A11kd=traceB11k0B12B22m1B21B11k1⋯B11kd−1B12B22mdB21B11kd.
This trace is nonnegative since all matrices in the product only contain nonnegative entries.
(ii) ⇒ (iii). By the spectral theorem, we may write W1tA12A21W1=VΛVt where V is a 2×2 orthogonal matrix and Λ=diag(λ1,λ2) with λ1≥λ2≥0. Note that v1, the eigenvector associated with the largest eigenvalue of W1tA12A21W1, is the first column of V. If λ1=λ2, v1=(1,0) and the inequality holds. If a11=a22 or a33=a44, W1tA11W1=A11 or W2tA22W2=A22, and choosing W1 or W2 such that a23=0 then ensures that the inequality in (iii) holds.
Assume now that λ1>λ2{\lambda _{2}}$]]>, a11>a22{a_{22}}$]]>, and a33>a44{a_{44}}$]]>. It follows by assumption that
0≤1a11k1a33k1λ1ktraceA11kA12A22kA21(A12A21)k=trace100(a22a11)kW1tA12W2100(a44a33)kW2tA21W1V100(λ1λ2)kVt→trace1000W1tA12W21000W2tA21W1V1000Vt
as k→∞. This gives the inequality in (iii) since
trace1000W1tA12W21000W2tA21W1V1000Vt=v11a13(v11a13+v21a23).
(iii) ⇒ (i). To ease the notation and without loss of generality assume that W=I. We are then pursuing two 2×2 orthogonal matrices U1 and U2 such that U1tA11U1, U1tA12U2, and U2tA22U2 all have nonnegative entries. Initially consider D1 and D2 of the form diag(±1,±1). Then clearly, D1A11D1=A11 and D2A22D2=A22 since A11 and A22 are diagonal matrices. Next, note that it is possible to find D1 and D2 such that either D1A12D2 has all entries nonnegative or
D1A12D2=a13a14a23−a24
where a13,a23,a14,a24>00$]]>. Consequently, we will assume A12 is of the form in (14) since otherwise choosing U1=D1 and U2=D2 would be sufficient.
As one of two cases, assume a13a23−a14a24≥0, and define
U2=αa14a24a23βa23αa14−βa24
where α,β>00$]]> are chosen such that each column in U2 has norm one. Then U2 is orthogonal,
A12U2=α(a142+a13a14a24a23)β(a13a23−a14a24)0β(a232+a242),
and
U2tA22U2=α2a33a14a24a232+a44a142αβa14a24(a33−a44)αβa14a24(a33−a44)β2a232+β2a242.
Since a33≥a44, all entries in A12U2 and U2tA22U2 are nonnegative. Choosing U1=I then gives a pair of orthogonal matrices with the desired property.
Now assume a13a23−a14a24<0. Note that A12 of the form (14) cannot be singular and consequently, there exist λ1≥λ2>00$]]> and an orthogonal matrix V such that A12A21=VΛVt, where Λ=diag(λ1,λ2). Furthermore, since V contains the eigenvectors of A12A21 we may assume v11 and v12 have the same sign where vij is the (i,j)-th component of V. Define
W=A21V(Λ1/2)−1,
and note that this is an orthogonal matrix which, together with V, decomposes A12 into its singular value decomposition, that is, VtA12W=Λ1/2. Then
VtA11V=a11v112+a22v212v11v12(a11−a22)v11v12(a11−a22)a11v122+a22v222.
All entries in VtA11V are nonnegative since we chose v11 and v12 to have the same sign, and since a11≥a22>00$]]>.
To see that WtA22W also have all entries nonnegative, consider the first line in the eigenequations for A12A21 associated with the eigenvector (v12,v22), the eigenvector associated with the smallest eigenvalue λ2,
(a132+a142−λ2)v12+(a13a23−a14a24)v22=0.
Since λ2 is the smallest eigenvalue of A12A21,
λ2=inf|v|=1vtA12A21v,
and since the off-diagonal elements in A12A21 are nonzero, (1,0) and (0,1) cannot be eigenvectors. Consequently, λ2 is strictly smaller than any diagonal element of A12A21, and in particular a132+a142−λ2>00$]]>. Since we also have a13a23−a14a24<0, (16) gives that v12 and v22 need to have the same sign for the sum to equal zero. Let wij be the (i,j)-th component of W and note that by (15),
w11w12=v11a13+v21a23λ11/2v12a13+v22a23λ21/2.
The assumption v11a13(v11a13+v21a23)≥0 implies that v11a13+v21a23 and v11 have the same sign. Since v11 and v12 were chosen to have the same sign, and v12 and v22 have the same sign, we conclude that (v11a13+v21a23)(v12a13+v22a23) is nonnegative and, therefore, w11w12 is nonnegative, too. Then writing
WtA22W=a33w112+a44w212w11w12(a33−a44)w11w12(a33−a44)a33w122+a44w222
makes it clear that WtA22W has nonnegative elements. Thus, letting U1=V and U2=W completes the proof. □
Let A andv1be given as in Lemma3. Then there exists a(2,2)-signature matrix U such thatUtAUhas nonpositive off-diagonal elements if and only ifv21a24(v21a24+v11a14)≥0.
Let W be defined as in Lemma 3. Define
P1=0110andP=P100P1.
Then P1v1=(v21,v11) is the eigenvector of P1W1tA12A21W1P1 associated with the largest eigenvalue. Let
A˜=W1tA11W1P1W1tA12W2P1P1W2tA21W1P1W2tA22W2=a110a24a230a22a14a13a24a14a330a23a130a44.
By Lemma 3, there exists a (2,2)-signature matrix
U˜=U˜100U˜2
such that U˜tA˜U˜ has nonnegative entries if and only if v21a24(v21a24+v11a14)≥0. Define now the (2,2)-signature matrix U as
U=U100U2=−W1P1U˜100W2P1U˜2.
Let u˜ij be the (i,j)-th component of U˜1. Since U˜1 is orthogonal, u˜12u˜22=−u˜11u˜21 implying that
U1tA11U1=u˜112a22+u˜212a11u˜11u˜12(a22−a11)u˜11u˜12(a22−a11)u˜122a22+u˜222a11
and
U˜1tW1tA11W1U˜1=u˜112a11+u˜212a22u˜11u˜12(a11−a22)u˜11u˜12(a11−a22)u˜122a11+u˜222a22.
Consequently U˜1tW1tA11W1U˜1 has nonnegative elements if and only if U1tA11U1 has nonpositive off-diagonal elements. Similarly, U˜2tW2tA22W2U˜2 has nonnegative elements if and only if U2tA22U2 has nonpositive off-diagonal elements by a similar argument. Finally we note that
U1tA12U2=−U˜1tP1W1tA12W2P1U˜2,
and it follows that UtAU has nonpositive off-diagonal elements if and only if
U˜1tP1W1tA12W2P1U˜2,U˜1tW1tA11W1U˜1and,U˜2tW2A22W2U˜2
have all entries nonnegative. We conclude that we can find a (2,2)-signature matrix U such that UtAU has nonpositive off-diagonal element if and only if (17) holds. □
To prove (i) set Q=I−(I−aΣ)−1 for sufficiently large a>00$]]>. The implication (iii)⇒ (ii) of Lemma 3 used on A=Q, together with Theorem 3, show that (i) implies infinite divisibility of (X12+X22,X32+X42). To prove (ii) we use Corollary 1 on A=Σ−1, which together with Proposition 1, show that (ii) implies infinite divisibility of (X12+X22,X32+X42). □
Proof of Theorem <xref rid="j_vmsta160_stat_009">5</xref>
Now we set out to show that the sum in Theorem 3 is nonnegative for k,m∈N0 such that k≤2, m≤2, or k+m≤7 in the case n1=n2=2. To this end, consider a 4×4 positive definite matrix Q and write
Q=Q11Q12Q21Q22
where Qij is a 2×2 matrix for i,j=1,2. Let W1 and W2 be two 2×2 orthogonal matrices and define Pij=WiQijWj. Then
traceQ11k1Q12Q22m1Q21⋯Q12Q22mdQ21Q11kd+1=traceP11k1P12P22m1P21⋯P12P22mdP21P11kd+1.
Consequently (see Lemma 2), we may assume, without loss of generality, that Q11 and Q22 are diagonal with the first diagonal element greater than or equal to the other and all entries nonnegative.
There exist D1 and D2 of the form diag(±1,±1) such that either D1Q12D2 has all entries nonnegative or such that
D1Q12D2=q13q23q14−q24
where q13,q23,q14,q24>00$]]>. If D1Q12D2 has all entries nonnegative, writing as in (18) with Wi replaced by Di implies nonnegativity of each individual trace. We conclude that we may assume
Q=λ10q13q140λ2q23−q24q13q23λ30q14−q240λ4,
where λ1≥λ2≥0 and λ3≥λ4≥0 and q13,q23,q14,q24>00$]]>, without loss of generality.
We now write out the traces in (7) for specific values of k and m and show nonnegativity in each case.
Assume k=0 and fix some m∈N. Then the terms in the sum in Theorem 3 reduce to traceQ22m. Since Q22 is positive definite, Q22m is positive definite. Consequently, traceQ22m>00$]]>. Similarly, when m=0 and k∈N, the terms in the sum in Theorem 3 reduce to traceQ11k, which again is positive since Q11 is positive definite.
Assume k=1 and fix some m∈N. Then (7) reduces to
traceQ12Q22mQ21+∑m1=0m−1traceQ22m1Q21Q12Q22m−1−m1,
which equals
(m+1)traceQ12Q22mQ21.
Since Q12=Q21t and Q22 is positive definite, Q12Q22mQ21 is positive semidefinite. We conclude that traceQ12Q22mQ21≥0.
Assume m=1 and fix some k∈N. Similar to above, (7) reduces to
traceQ21Q11kQ12+∑k1=0k−1traceQ11k1Q12Q21Q11k−1−k1.
Nonnegativity of this trace follows by arguments similar to those above.
Assume that k=2 and let m∈N. The case m=1 is discussed above. Assume m≥2. Then (7) reduces to
traceQ11Q12Q22m−1Q21+∑m1+m2+1=mtraceQ22m1Q21Q11Q12Q22m2+∑m1+m2+2=mtraceQ12Q22m1Q21Q12Q22m2Q21+∑m1+m2+m3+2=mtraceQ22m1Q21Q12Q22m2Q21Q12Q22m3.
All the traces above are nonnegative. To see this, consider for example
traceQ22m1Q21Q12Q22m2Q21Q12Q22m3
for some m1,m2,m3∈N0. Since Q22 is positive definite it has a unique positive definite square root Q221/2. We conclude that
traceQ22m1Q21Q12Q22m2Q21Q12Q22m3=traceQ22(m1+m3)/2Q21Q12Q22m2Q21Q12Q22(m1+m3)/2.
Note that
Q22(m1+m3)/2Q21Q12=(Q21Q12Q22(m1+m3)/2)t,
which implies that (19) is the trace of a positive semidefinite matrix and therefore nonnegative.
Nonnegativity of the traces when m=2 and k∈N follows by symmetry.
In the following we will need to expand traces, and we therefore note that
traceQ11kQ12Q22mQ21=λ1kλ3mq132+λ1kλ4mq142+λ2kλ3mq232+λ2kλ4mq242
for any k,m∈N, and
traceQ11k1Q12Q22m1Q21Q11k2Q12Q22m2Q21=λ1k1+k2λ3m1+m2q134+λ1k1+k2λ4m1+m2q144+λ2k1+k2λ3m1+m2q234+λ2k1+k2λ4m1+m2q244+λ1k1+k2(λ3m1λ4m2+λ4m1λ3m2)q132q142+λ2k1+k2(λ3m1λ4m2+λ3m2λ4m1)q232q242+λ3m1+m2(λ1k1λ2k2+λ1k2λ2k1)q132q232+λ4m1+m2(λ1k1λ2k2+λ1k2λ2k1)q142q242−(λ1k1λ2k2+λ1k2λ2k1)(λ3m1λ4m2+λ3m2λ4m1)q13q23q14q24
for any k1,k2,m1,m2∈N.
Assume now k=3 and m=3 and consider the sum in Theorem 3. The sum contains all terms of the form
traceQ11k1Q12Q222Q21Q11k2
where k1+k2=2 and
traceQ22m1Q21Q112Q12Q22m2
where m1+m2=2. All these traces equal
traceQ112Q12Q222Q21,
and there are alltogether 6 of these terms. Next, the sum in Theorem 3 also contains all terms of the form
traceQ11k1Q12Q22m1Q21Q11k2Q12Q22m2Q21Q11k3
where k1+k2+k3=1 and m1+m2=1, and
traceQ22m1Q21Q11k1Q12Q22m2Q21Q11k2Q12Q22m3
where m1+m2+m3=1 and k1+k2=1. Using both that traceAB=traceBA and traceAt=traceA for any two square matrices A and B of the same dimensions we get that all these traces share the common trace
traceQ11Q12Q21Q12Q22Q21.
Alltogether there are 12 of these terms. Finally, the sum in Theorem 3 contains the two terms
trace(Q12Q21)3andtrace(Q21Q12)3,
which share a common trace. We conclude that the sum in Theorem 3 reads
trace{6Q112Q12Q222Q21+12Q11Q12Q21Q12Q22Q21+2(Q12Q21)3}.
Since Q12=Q21t, the matrix Q12Q21 is positive semidefinite and consequently, trace(Q12Q21)3≥0. Furthermore, we have
traceQ112Q12Q222Q21=traceQ11Q12Q222Q21Q11≥0.
Contrarily, there exists a positive definite matrix Q such that
traceQ11Q12Q21Q12Q22Q21<0.
(To see this, consider Q of the form in Example 1 with ε small and δ large relative to ε.) We will now argue that despite this, (22) remains nonnegative. Initially we note that
Q11kiQ12Q22miQ21=λ1ki(λ3miq132+λ4miq142)λ1ki(λ3miq13q23−λ4miq14q24)λ2ki(λ3miq13q23−λ4miq14q24)λ2ki(λ3miq232+λ4miq242)
and
Q22miQ21Q11kiQ12=λ3mi(λ1kiq132+λ2kiq232)λ3mi(λ1kiq13q14−λ2kiq23q24)λ4mi(λ1kiq13q14−λ2kiq23q24)λ4mi(λ1kiq142+λ2kiq242).
Since λ1≥λ2 and λ3≥λ4, we see that if q13q14≥q23q24 or q13q23≥q14q24, then one of two matrices above have only nonnegative entrances for any ki,mi∈N0. Consequently,
traceQ11k1Q12Q22m1Q21Q11k2Q12Q22m2Q21=traceQ22m1Q21Q11k1Q12Q22m2Q21Q11k2Q12
would be nonnegative if this was the case. Especially, we would have
traceQ11Q12Q21Q12Q22Q21≥0.
Assume now that q13q14≤q23q24 and q13q23≤q14q24. By (20) and (21),
trace{12Q112Q12Q222Q21+Q11Q12Q22Q21Q12Q21}=12λ12λ32q132+12λ12λ42q142+12λ22λ32q232+12λ22λ42q242+λ1λ3q134+λ1λ4q144+λ2λ3q234+λ2λ4q244+λ1(λ3+λ4)q132q142+λ2(λ3+λ4)q232q242+λ3(λ1+λ2)q132q232+λ4(λ1+λ2)q142q242−(λ1+λ2)(λ3+λ4)q13q23q14q24.
We are going to bound the term (λ1+λ2)(λ3+λ4)q13q23q14q24 by the positive terms to show nonnegativity of this trace. We recall that λ1≥λ2>00$]]> and λ3≥λ4>00$]]>. Initially, note that
λ2λ3q13q23q14q24≤λ2λ3q132q232λ2λ4q13q23q14q24≤λ1λ4q132q142λ1λ4q13q23q14q24≤λ1λ3q132q142.
This leaves only λ1λ3q13q23q14q24 to be bounded. If λ1λ3q13q23q14q24≤12λ12λ32q132, we have a bounding term in (23). Therefore, assume 2q23q14q24≥λ1λ3q13. Since Q was assumed positive definite, λ2λ4≥q242. Consequently,
λ1λ3q13q23q14q24≤2q232q142q242≤2λ2λ4q232q132≤λ2λ4(q234+q134)≤λ2λ3q234+λ1λ3q134.
We conclude that (23) and hence (22) is nonnegative.
Now consider k,m∈N such that k+m=7. Whenever k,m=1,2, we already know that the sum in Theorem 3 is nonnegative. Let k=3 and m=4. Then the sum in Theorem 3 reads
trace{14Q11Q12Q21Q12Q222Q21+7Q112Q12Q223Q217Q11(Q12Q22Q21)2+7Q12Q22Q21(Q12Q21)2}.
Initially we note that
traceQ11(Q12Q22Q21)2≥0andtraceQ12Q22Q21(Q12Q21)2≥0
since they both can be written as the trace of positive semidefinite matrices (see above for more details). Next, by (20) and (21),
trace{12Q112Q12Q223Q21+Q11Q12Q222Q21Q12Q21}=12λ12λ33q132+12λ12λ43q142+12λ22λ33q232+12λ22λ43q242+λ1(λ32+λ42)q132q142+λ2(λ32+λ42)q232q242+λ32(λ1+λ2)q132q232+λ42(λ1+λ2)q142q242+λ1λ32q134+λ1λ42q144+λ2λ32q234+λ2λ42q244−(λ1+λ2)(λ32+λ42)q13q23q14q24.
Again we bound the negative term by positive terms. Recall that λ1≥λ2 and λ3≥λ4, and that we may assume q23q24≥q13q14 and q14q24≥q13q23 without loss of generality. Consequently,
λ1λ42q13q23q14q24≤λ1λ42q142q242λ2λ32q13q23q14q24≤λ2λ32q232q242λ2λ42q13q23q14q24≤λ2λ42q142q242,
leaving λ1λ32q13q23q14q24 to be bounded. First note that
12λ12λ33q132−λ1λ32q13q23q14q24=λ1λ32q13(12λ1λ3q13−q23q14q24),
so that nonnegativity holds if 12λ1λ3q13≥q23q14q24. Assume λ1λ3q13≤2q23q14q24 and recall that λ2λ4≥q242 since Q is positive definite. Then
λ1λ32q13q23q14q24≤2λ3q232q142q242≤2λ2λ3λ4q232q142≤λ2λ32q234+λ2λ42q144≤λ2λ32q234+λ1λ42q144
so we have found bounding terms for the last expression. We conclude that (25) is nonnegative and therefore, (24) is nonnegative, too. The case k=4 and m=3 follows by symmetry. It follows that the sum in Theorem 3 is nonnegative for k+m=7. □
Acknowledgement
We thank the referee for an insightful and constructive report.
ReferencesBapat, R.B.: Infinite divisibility of multivariate gamma distributions and M-matrices. Sankhyā, Ser. A51(1), 73–78 (1989). MR1065560Dobrushin, R.L., Major, P.: Non-central limit theorems for nonlinear functionals of Gaussian fields. Z. Wahrscheinlichkeitstheor. Verw. Geb.50(1), 27–52 (1979). MR0550122. https://doi.org/10.1007/BF00535673Eisenbaum, N.: On the infinite divisibility of squared Gaussian processes. Probab. Theory Relat. Fields125(3), 381–392 (2003). MR1964459. https://doi.org/10.1007/s00440-002-0245-zEisenbaum, N.: Characterization of positively correlated squared Gaussian processes. Ann. Probab.42(2), 559–575 (2014). MR3178467. https://doi.org/10.1214/12-AOP807Eisenbaum, N., Kaspi, H.: A characterization of the infinitely divisible squared Gaussian processes. Ann. Probab.34(2), 728–742 (2006). MR2223956. https://doi.org/10.1214/009117905000000684Evans, S.N.: Association and infinite divisibility for the Wishart distribution and its diagonal marginals. J. Multivar. Anal.36(2), 199–203 (1991). MR1096666. https://doi.org/10.1016/0047-259X(91)90057-9Griffiths, R.C.: Infinitely divisible multivariate gamma distributions. Sankhyā, Ser. A32, 393–404 (1970). MR0295406Griffiths, R.C.: Characterization of infinitely divisible multivariate gamma distributions. J. Multivar. Anal.15(1), 13–20 (1984). MR0755813. https://doi.org/10.1016/0047-259X(84)90064-2Janson, S.: Gaussian Hilbert spaces. Cambridge Tracts in Mathematics, vol. 129. Cambridge University Press, Cambridge (1997). MR1474726. https://doi.org/10.1017/CBO9780511526169Leon, S.J.: Linear algebra with applications. Macmillan, Inc., New York, Collier-Macmillan Publishers, London (1980)Lévy, P.: The arithmetic character of the Wishart distribution. Proc. Camb. Philos. Soc.44, 295–297 (1948). MR0026261. https://doi.org/10.1017/S0305004100024270Marcus, M.B., Rosen, J.: Markov processes, Gaussian processes, and local times. Cambridge Studies in Advanced Mathematics, vol. 100. Cambridge University Press, Cambridge (2006). MR2250510. https://doi.org/10.1017/CBO9780511617997Moran, P.A., Vere-Jones, D.: The infinite divisibility of of multi-gamma distributions. Sankhyā, Ser. A31, 191–194 (1969). MR0226704Paranjape, S.R.: Simpler proofs for the infinite divisibility of multivariate gamma distributions. Sankhyā, Ser. A40(4), 393–398 (1978). MR0589292Rudin, W.: Principles of mathematical analysis, 3rd edn. International Series in Pure and Applied Mathematics. McGraw-Hill Book Co., New York, Auckland, Düsseldorf (1976). MR0385023Shanbhag, D.N.: On the structure of the Wishart distribution. J. Multivar. Anal.6(3), 347–355 (1976). MR0420785. https://doi.org/10.1016/0047-259X(76)90044-0Taqqu, M.S.: Convergence of integrated processes of arbitrary Hermite rank. Z. Wahrscheinlichkeitstheor. Verw. Geb.50(1), 53–83 (1979). MR0550123. https://doi.org/10.1007/BF00535674