Existence of a critical point for the infinite divisibility of squares of Gaussian vectors in $R^{2}$ with non--zero mean

Let $G=(G_{1},G_{2})$ be a Gaussian vector in $R^{2}$ with $EG_{1}G_{2}\neq 0$. Let $c_{1},c_{2}\in R^{1}$. A necessary and sufficient condition for $G=((G_{1}+c_{1}\alpha)^{2},(G_{2}+c_{2}\alpha)^{2})$ to be infinitely divisible for all $\alpha\in R^{1}$ is that \[ \Ga_{i,i}\geq \frac{c_{i}}{c_{j}}\Ga_{i,j}>0\qquad\forall 1\le i\ne j\le 2.\] In this paper we show that when this does not hold there exists an $0<\alpha_{0}<\ff $ such that $G=((G_{1}+c_{1}\alpha)^{2},(G_{2}+c_{2}\alpha)^{2})$ is infinitely divisible for all $|\alpha|\leq \alpha_{0}$ but not for any $|\al|>\al_{0}$.


Introduction
Let η = (η 1 , . . ., η n ) be an R n valued Gaussian random variable.η is said to have infinitely divisible squares if η 2 := (η 2  1 , . . ., η 2 n ) is infinitely divisible, i.e. for any r we can find an R n valued random vector Z r such that where {Z r,j }, j = 1, . . ., r are independent identically distributed copies of Z r .We express this by saying that η 2 is infinitely divisible.
Paul Lévy proposed the problem of characterizing which Gaussian vectors have infinitely divisible squares.It is easy to see that a single Gaussian random variable has infinitely divisible squares.However, even for vectors in R 2 this is a difficult problem.It seems that Lévy incorrectly conjectured that not all Gaussian vectors in R 2 have infinitely divisible squares.If he had said R 3 his conjecture would have been correct.
Theorem 1.1 Let G = (G 1 , . . ., G n ) be a mean zero Gaussian random variable with strictly positive definite covariance matrix Γ = {Γ i,j } = {E(G i G j )}.Then G 2 is infinitely divisible if and only if there exists a signature matrix N such that N Γ −1 N is an M matrix. (1.2) We need to define the different types of matrices that appear in this theorem.Let A = {a i,j } 1≤i,j≤n be an n × n matrix.We call A a positive matrix and write A ≥ 0 if a i,j ≥ 0 for all i, j.We say that A has positive row sums if n j=1 a i,j ≥ 0 for all 1 ≤ i ≤ n.The matrix A is said to be an M matrix if (1) a i,j ≤ 0 for all i = j.
A matrix is called a signature matrix if its off-diagonal entries are all zero and its diagonal entries are either one or minus one.The role of the signature matrix is easy to understand.It simply accounts for the fact that if G has an infinitely divisible square, then so does (ǫ 1 G 1 , . . ., ǫ n G n ) for any choice of ǫ i = ±1, i = 1, . . ., n.Therefore, if (1.2) holds for N with diagonal elements n 1 , . . ., since the inverse of an M matrix is positive.Thus (n 1 G 1 , . . ., n n G n ) has a positive covariance matrix and its inverse is an M matrix.(For this reason, in studying mean zero Gaussian vectors with infinitely divisible squares one can restrict ones attention to vectors with positive covariance.) The natural next step was to characterize Gaussian processes with infinitely divisible squares which do not have mean zero.We set η i = G i + c i , EG i = 0, i = 1, , . . ., n.Let Γ be the covariance matrix of (G 1 , . . ., G n ) and set c := (c 1 , . . ., c n ).Results about the infinite divisibility of (G + c) 2 when c 1 = • • • = c n , are given in the work of N. Eisenbaum [2,3] and then in joint work by Eisenbaum and H. Kaspi [4], as a by product of their characterization of Gaussian processes with a covariance that is the 0-potential density of a symmetric Markov process.We point out later in this Introduction how Gaussian vectors with infinitely divisible squares are related to the local times of the Markov chain that is determined by the covariance of the Gaussian vector.It is this connection between Gaussian vectors with infinitely divisible squares and the local times of Markov chains, and more generally, between Gaussian processes with infinitely divisible squares and the local times of Markov processes, that enhances our interest in the question of characterizing Gaussian vectors with infinitely divisible squares.Some of the results in [2,3,4] are presented and expanded in [9,Chapter 13].The following theorem is taken from [9, Theorem 13.3.1 and Lemma 13.3.2].
In [8], Theorem 1.2 is generalized so that the mean of the components of G + c in (1.5) need not be the same.In this generalization certain trivial cases spoil the simplicity of the final result.We avoid them by requiring that the covariance matrix of the Gaussian process is irreducible.
Then the following are equivalent: (1) G + cα has infinitely divisible squares for all α ∈ R 1 ; (2 infinitely divisible squares for some b = 0. Furthermore, if this holds for some b = 0, it holds for all b ∈ R 1 ; (3) C Γ −1 C is an M matrix with positive row sums.
By definition, when (G + c) 2 is infinitely divisible, it can be written as in (1.1) as a sum of r independent identically distributed random variables, for all r ≥ 1.Based on the work of Eisenbaum and Kaspi mentioned above and the joint paper [5] we can actually describe the decomposition.We give a rough description here.For details see [2,3,4] and [9,Chapter 13].
Assume that (1), ( 2) and ( 3) of Theorem 1.3 hold.Let Let Γ c denote the covariance matrix of G/c.Theorem 1.2 holds for G/c and Γ c , so Γ −1 c is an M matrix with positive row sums.To be specific let G/c ∈ R n .Set S = {1, . . ., n}.By [9, Theorem 13.1.2],Γ c is the 0-potential density of a strongly symmetric transient Borel right process, say X, on S. We show in the proof of [9,Theorem 13.3.1]that we can find a strongly symmetric recurrent Borel right process Y on S ∪ {0} with P x (T 0 < ∞) > 0 for all x ∈ S such that X is the process obtained by killing Y the first time it hits 0. Let L x t = {L x t ; t ∈ R + , x ∈ S ∪ {0}} denote the local time of Y .It follows from the generalized second Ray-Knight Theorem in [5], see also [9,Theorem 8.2.2] for all t ∈ R + , where τ (t) = inf{s > 0|L 0 s > t}, the inverse local time at zero, and Y and G are independent.Consequently x ∈ S} are independent.G 2 is infinitely divisible and for all integers r ≥ 1 where {L • τ (α 2 /(2r)),j }, j = 1, . . ., r are independent.Note that in (1.9) we identify the components of the decomposition of {(G x + c x α) 2 ; x ∈ S} that mark it as infinitely divisible.
In Theorem 1.3 we have necessary and sufficient conditions for ((G 1 + c 1 α) 2 , (G 2 + c 2 α) 2 ) to be infinitely divisible for all α ∈ R 1 .There remains the question, can ((G 1 + c 1 α) 2 , (G 2 + c 2 α) 2 ) have infinitely divisible squares for some α > 0 but not for all α ∈ R 1 ?When we began to investigate this question we hoped that such points α do not exist.This would have finished off the problem of characterizing Gaussian random variables with infinitely divisible squares and, more significantly, by (1.9), would show that when a Gaussian random variable with non-zero mean has infinitely divisible squares, it decomposes into the sum of two independent random variables.The Gaussian random variable itself minus its mean, and the local time of a related Markov process.This would be a very neat result indeed, but it is not true.
To conclude this Introduction we explain how we approach the problem of showing that (G + cα) 2 is infinitely divisible for all α ∈ R 1 or only for some α ∈ R 1 .Since we can only prove Theorem 1.4 for Gaussian random variables in R 2 we stick to this case, although a similar analysis applies to R n valued Gaussian random variables.
Let Γ be the covariance matrix of G and where Consider the Laplace transform of (( where 2 ), with the change of variables λ 1 = t(1 −s 1 ) and λ 2 = t(1 −s 2 ).It is easy to see from Theorem 1.1 that all two dimensional Gaussian random variables are infinitely divisible.Therefore, for all t sufficiently large, all the coefficients of the power series expansion of U in s 1 and s 2 are positive, except for the constant term.This is a necessary and sufficient condition for a function to be the Laplace transform of an infinitely divisible random variable.See e.g.[9,Lemma 13.2.2].Now, suppose that for all t sufficiently large, V s 1 , s 2 , t, c 1 , c 2 , Γ has all the coefficients of its power series expansion in s 1 and s 2 positive, except for the constant term.Then the right-hand side of (1.15) is the Laplace transform of two independent infinitely divisible random variables.It is completely obvious that this holds for all α ∈ R 1 .
On the other hand suppose that for all t sufficiently large, the power series expansion of V s 1 , s 2 , t, c 1 , c 2 , Γ in s 1 and s 2 has even one negative coefficient, besides the coefficient of the constant term.Then for all α sufficiently large, (1.16) is not the Laplace transform of an infinitely divisible random variable.In other words ((G 1 +c 1 α) 2 , (G 2 +c 2 α) 2 ) is not infinitely divisible for all α ∈ R 1 .But it may be infinitely divisible if α is small, since the positive coefficients of U may be greater than or equal to α 2 times the corresponding negative coefficients of V .Clearly, if this is true for some |α| = α 0 > 0, then it is true for all |α| ≤ α 0 .
The preceding paragraph explains how we prove Theorem 1.4.We consider vectors ((G 1 + c 1 α) 2 , (G 2 + c 2 α) 2 ) that are not infinitely divisible for all α ∈ R 1 , (this is easy to do using (1.12)), and show that for |α| sufficiently small the coefficients in the power series expansion of in s 1 and s 2 are positive, except for the constant term.Our proof only uses elementary mathematics, although it is quite long and complicated.In the course of the proof we show that the coefficients of the power series expansion of U in s 1 and s 2 are positive, except for the constant term.This provides a direct elementary proof of the fact that the Gaussian random variable (G 1 , G 2 ) always has infinitely divisible squares.
As we have just stated, and as the reader will see, the proof of Theorem 1.4 is long and complicated.So far we have not been able to extend it to apply to Gaussian random variables in R 3 .One hopes for a more sophisticated and much shorter proof of Theorem 1.4 that doesn't depend on the dimension of the Gaussian random variable.
2 Gaussian squares in R 2 and their Laplace transforms ) be a mean zero Gaussian process with covariance matrix where ab = d + 1 > 1, and let and and where (We use repeatedly the fact that ab = d + 1.) Note that by (2.2) we have that .
A simple computation shows that 2 )/2 and by [8, Corollary 1.1, 2.] it is the Laplace transform of an infinitely divisible random variable.The exponential term may or may not be a Laplace transform.In fact, by (1.12), we know it is the Laplace transform of an infinitely divisible random variable, for all {c 1 α, c 2 α}, for all α ∈ R 1 , if and (2.12) To prove Theorem 1.4 we must show that when (2.12) does not hold, there exists an 0 < α 0 < ∞ such that (2.9) is the Laplace transform of an infinitely divisible random variable when c 1 and c 2 are replaced by c 1 α and c 2 α for any |α| ≤ α 0 .Actually, as we see in Section 8, the general result follows from the consideration of three cases, In these three cases the numerator of the fraction in the exponential term on the right-hand side of (2.9) is Note unless det Γ = 0, ab > 1.Since Theorem 1.4 obviously holds when det Γ = 0, we can exclude this case from further consideration.Thus we always have γ > 0.
3 Power series expansion of the logarithm of the Laplace transform of (( Bapat's proof of Theorem 1.1 involves the analysis of a certain power series expansion of the logarithm of the Laplace transform.We need a similar, but more delicate, analysis.(See Lemma 4.1 below).Using (2.9) and the remarks following Lemma 2.1 we can write ) is infinitely divisible, for some, but not for all, c > 0, we must consider a, b > 0 such that In the rest of this paper we assume that (3.3) holds. Let We consider P and Q as functions of and s 1 , s 2 , t, and write We expand these in a power series in s 1 , s 2 .
and set Consequently In this section we obtain explicit expressions for P j,k (t), Q j,k (t).We write where (3.10) and (3.12) Using these definitions we have We make some preliminary observations that enable us to compute the coefficients of the power series expansions of P (s 1 , s 2 , t) and Q(s 1 , s 2 , t).Let (u 1 , u 2 ) ∈ [0, 1) 2 , and θ ∈ [0, 1), and assume that (3.15) (It is clearly greater than zero.)Let (3.17) We give explicit expressions for C j,k and D j,k .To begin we give several equalities that are easy to verify.
We list the following equalities without proof.
Also, C 0,0 = 0, C j,0 = 1/j, C 0,k = 1/k, j, k = 0, and for Using this series we can determine D j,k .Since j ≤ k it is clear that each of the first j + 1 terms in the series immediately above can contribute a term in u j 1 u k 2 .For a given 0 ≤ p ≤ j we get (1 − θ) p times the coefficient of u j−p 1 in the power series expansion of 1/(1 − u 1 ) p+1 and the coefficient of u k−p 2 in the power series expansion of 1/(1 − u 2 ) p+1 .We see from Lemma 3.1 that they are j j−p = j p and k k−p = k p respectively.Thus we get (3.20).To obtain (3.21) we write This gives us C j,0 and C 0,k and, similar to the computation of D j,k , we can use the last series above to see that This gives us (3.21). Set Proof Note that since 0 < α, β, θ < 1, Consequently, for all t sufficiently large 0 ≤ αs 1 + βs 2 − θαβs 1 s 2 < 1.
Lemma 3.5 For all t sufficiently large, and j, k ≥ 1 Furthermore, for all t sufficiently large and for all 1 Proof It follows from (3.31) that for 0 ≤ s 1 , s 2 ≤ 1 Using this along with (3.14) we see that from which we get (3.35), and from which we get (3.36).
To obtain (3.37) we use (3.39), and the terms of D j,k defined in (3.16), to see that Using (3.42), (3.20) and Lemma 3.2 we see that Consequently, for all t sufficiently large, for all 1 Consider (3.44).We write to obtain Therefore, for each 1 ≤ p ≤ j we incorporate the term in p 2 /jk in (3.47) into the preceding term in the series in (3.47) to get (3.37).(Note that we can not add anything to the p = j term.The expression in (3.37) reflects this fact since j−p j k−p k = 0 when p = j.) 4 A sufficient condition for a vector in R 2 to be infinitely divisible.
We present a sufficient condition for a random vector in R 2 to be to be infinitely divisible, and show how it simplifies the task of showing that (( 2 and suppose that for all t > 0 sufficiently large, log ψ(t(1 − s 1 ), t(1 − s 2 )) has a power series expansion at s = 0 given by Suppose also that there exist an increasing sequence of finite subsets Then ψ(λ 1 , λ 2 ) is the Laplace transform of an infinitely divisible random variable on (R + ) 2 .
Proof It is clear from (4.2) that the power series in (4.1) converges absolutely for all s ∈ [0, 1] 2 .Let We show that for each (λ As we point out in [9, page 565], Ψ i t i ; e −λ 1 /t i , e −λ 2 /t i is the Laplace transform of a discrete measure.It then follows from the continuity theorem and the fact that ψ(0, 0) = 1 that ψ(λ 1 , λ 2 ) is the Laplace transform of a random variable.Furthermore repeating this argument with φ i (t i ; s 1 , s 2 ) replaced by φ i (t i ; s 1 , s 2 )/n shows that ψ 1/n (λ 1 , λ 2 ) is the Laplace transform of a random variable.This shows that ψ(λ 1 , λ 2 ) is the Laplace transform of an infinitely divisible random variable on (R + ) 2 . Let Clearly lim i→∞ δ i = 0.By (4.2) where Therefore, by the triangle inequality, (4.6) and (4.9) Using (4.6) we see that this is Thus we justify (4.5) and the paragraph following it.
Remark 4.1 In [9, Lemma 13.2.2]we present the well known result that the conclusion of Lemma 4.1 holds when log ψ(t(1 − s 1 ), t(1 − s 2 )) has a power series expansion at s = 0 with all its coefficients, except for the coefficient of the constant term, are positive.Lemma 4.1 is useful because it allows us to only verify this condition for a subset of these coefficients, (depending on t).
The following lemma enables us to apply Lemma 4.1.
Lemma 4.2 For any c 3 > 0 there exists a constant B = B(γ, d, c 3 ) for which In [6, page 42 ], Feller shows that p! (e/p) p is increasing in p.Since it is equal to e when p = 1, (and 1 when p = 0), we get (4.17).The first inequality in (4.15) follows from (4.14) the next one is obtained by maximizing the middle term with respect to p.
Proof of Lemma 4.2 By (3.8) and Lemmas 3.4 and 3.5 we see that for all t sufficiently large, for all 1 ≤ j ≤ k, where C depends on c, γ and d but not on j, k, t or t.Furthermore, C is bounded for all |c| ≤ T , for any finite number T .(We also use the fact that lim t→∞ αβ = 1.) For any δ > 0, for t sufficiently large, (4.20) Using these estimates along with (3.28) we see that for all t sufficiently large, for all 1 ≤ j ≤ k, where, for the inequality we take the minimum of aθ+b/θ and for the equality we use the fact that ab = d + 1.Combined with (4.15) this shows that when We note that the cardinality of A n is less than 3nd Clearly, there exists a constant B such that when M = B log t, this last term is o(1) as t → ∞.

5
Proof of Theorem 1.4 when (c 1 , c 2 ) = (c, c) and EG 1 G 2 > 0 In this section we prove Theorem 1.4 in case 1. and for EG 1 G 2 > 0, by establishing the positivity conditions on the coefficients R j,k (t, c), (when EG 1 G 2 = 1), as discussed in Remark 4.2.We pass to the case EG 1 G 2 > 0 on page 41.
To proceed we need several estimates of parameters we are dealing with as t → ∞.They follow from the definitions in (3.10)-(3.12).
The rest of the lemma follows similarly.
Proof of Theorem 1.4 when c 1 = c 2 = c.To begin let note that it is easy to see from Lemma 3.4, that P j,k (t) ≥ 0 for all 0 ≤ j, k < ∞, with the exception of P 0,0 (t).This must be the case because exp(P (a, b, λ 1 , λ 2 )) is the Laplace transform of an infinitely divisible random variable, as we remark following the proof of Lemma 2.1.By (3.35) (5.5) Thus we see that there exists a t 1 sufficiently large such that for all t ≥ t 1 , R j,0 (t, c) and R 0,k (t, c) are both positive for all j, k ≥ 1.
We now examine R j,k (t, c) for j ∧ k ≥ 1, j ≤ k.We write R j,k,p (t, c).
(5.6) Using (3.37) and (3.29) we see that When p = 0 we get which is independent of j, k.Using Lemma 5.1 we see that Using this and Lemma 5.1 again we get where the O (1/t) term is independent of j and k.
We now simplify the expression of the other coefficients R j,k,p (t, c), where (5.13) Using Lemma 5.1 we have In (5.14) and (5.15) the expressions O (1/t) are not necessarily the same from line to line.Nevertheless, it is important to note that they are independent of p, j and k.That is there exists an M > 0 such that all terms given as O (1/t) in (5.14) and (5.15) satisfy This is easy to see since the O (1/t) terms, in addition to depending on t, depend on a, b, p/j and p/j ≤ p/k ≤ 1.Using (5.14), (5.15) we have where, for the final equality we use (5.2).Note that (For the inequality use α 2 +β 2 ≥ 2αβ.)Therefore since ζ = a+b−(d+2) > 0, Thus we see that there exists a function ǫ t , depending only on a and b such that where lim and, as we point out above the O (1/t) is independent of p, j and k.
Remark 5.1 We interrupt this proof to make some comments which may be helpful in understanding what is going on.Note that if for all δ > 0. This follows from (5.10) and (5.20) since when (5.22) holds for all p ≥ 1, for all t is sufficiently large.Consequently when (5.22) holds R j,k (t, c) > 0 for all t is sufficiently large when (Here we also use (5.21).)(When ζ ≤ 0, (5.23) shows that R j,k (t, c) > 0 for all c ∈ R 1 .This is what we expect.(See the paragraph containing (2.12).) We use the next two lemmas to complete the proof of Theorem 1.4, in case 1.
Lemma 5.2 For any N 0 ∈ R + , we can find c 0 > 0 and t c 0 < ∞ such that for all t ≥ t c 0 R j,k,p (t, c) > 0 (5.26) for all |c| ≤ c 0 and all p, j and k for which jk/t ≤ N 0 .
Proof This follows from (5.10) when p = 0. Therefore, we can take p ≥ 1.
We first show that for any N ∈ R + , R j,k,p (t) > 0 when √ jk = Ndt, for all t sufficiently large.By Remark 5.1 we can assume that N ≥ 1 − ǫ.It follows from (5.20) that where ǫ t satisfies (5.21).Therefore when p ≥ ΛN for any Λ > 1, A j,k,p (t) > 0, and hence R j,k,p (t, c) > 0, for all t sufficiently large.Now suppose that p < ΛN. ( Since √ jk = Ndt we see that where the O(1/t) term is independent of p, j and k.Note that by (5.1) and (5.12) Therefore, if in addition to (5.28) we also have ΛN ≤ j/2, so that p < j/2, we see by (5.11), (5.17) and (5.29) that Therefore we can obtain R j,k,p (t) ≥ 0 by taking for some Λ ′ > Λ.Now suppose that ΛN > j/2.In this case we use (5.17) to see that It is easy to see that the right-hand side of (5.33) is greater than zero for all t sufficiently large since Thus we see that for any fixed N, R j,k,p (t) > 0 for all t sufficiently large.Since the O(1/t) terms are independent of p, j and k this analysis works for all j and k satisfying (5.26), and all 1 ≤ p ≤ j as long as (5.32) holds with N replaced by N 0 .Lemma 5.3 For all N 0 and B ∈ R + we can find a c ′ 0 > 0 and for all |c| ≤ c ′ 0 and all 0 ≤ p ≤ j ≤ k for which (The value of N 0 in Lemmas 5.2 and 5.3 can be taken as we wish.It will be assigned in the proof of this lemma.) Proof By adjusting N 0 and B we can replace (5.36) by the condition (5.37) Using (5.17), we see that if j ≤ ρ t Clearly, there exists a ρ > 0, independent of j and k such that this term is positive.Thus we can assume that j ≥ ρ t.
(5.39) Furthermore, when √ jk/ t = N, it follows from (3.28) that we can write (5.19) as Let δ N = (10 log N/N)  (5.45) Proof By Stirlings's formula for integers q, q! = √ 2πq q+1/2 e −q 1 + O(q −1 ) . (5.46) Therefore, since j is large and p/j is small, terms of the form (5.47) Using this we see that Since this also holds with j replaced by k we get (5.44).
To get (5.45)we multiply each side of (5.44) by t −2p and substitute for √ jk/ t = dN and use the fact that under the assumptions (5.39) and (5.41), (5.62) We also note that since −m/y − 2 log y is increasing for y > m/2, f ′ m (y) is positive for m/2 < y < y m and negative for y > y m .Consequently, f m (y) is unimodal.Now consider the last line of (5.54).The function being summed is f 2 (p).The above discussion shows that this function is unimodal, with a maximum at, at most, two points at which it is less than 2e 2N /N 2 .Consequently, to obtain (5.51) we can replace the sum in the last line of (5.54) by an integral and show that (5.63) Making the change of variables r = xN we have Recall that N 0 ≤ N ≤ 2 log t, and that we can take N 0 as large as we want, (but fixed and independent of t), and that δ N = (10 log N/N) 1/2 .Therefore (5.66) When |y| ≤ (10 log N/N) 1/2 , so that |y| It follows from this that when we make the change of variables x = 1 + y in (5.64) we get we see that (5.63) follows.Thus we have established (5.51).
Before proceeding to the proof of (5.52) we note that To prove this we use (5.45) and the same argument that enables us to move from a sum to an integral that is given in (5.54)-(5.63),except that we use (5.62) with m = 1.We continue and then use (5.67) to get We now obtain (5.52).When √ jk = tN, by (5.11) and (5.40), t2 By (5.71) we see that Therefore, to obtain (5.52), it suffices to show that for some D < ∞ (5.74) Here we use the fact that N ≤ B log t for some 0 < B < ∞.
Remark 5.2 Since the proof of (5.74) is rather delicate we make some heuristic comments to explain how we proceed.When √ jk = tN the term t −2p j p k p , as a function of p, is exp(2N) times values that are sort of normally distributed with mean p = N, and, roughly speaking, for all t sufFiciently large.(In fact the upper bound is given in (5.71).)This is too large to enable us to get (5.74)so we must make use of the factors p N − 1 , which is an odd function with respect to p = N, to get the cancellations that allow us to obtain (5.74).However, because we are canceling terms, we must take account of the error in Stirling's approximation; (see (5.46)).To do this we need to show that the estimate in (5.74) remains the same even when we eliminate the terms in the summand that are not close to N.
Proof of Lemma 5.3 continued Note that by (5.45)

.76)
The fact that f m (y) is unimodal on y > m/2 implies that f 1 (p) is increasing on the interval [p 0 , N(1 we see that a j,k + c 2 m b j,k ≥ 0 for each (j, k) = (0, 0).Letting c m ↑ c we therefore have a j,k + c 2 b j,k ≥ 0 for each (j, k) = (0, 0).This shows that exp ((P + c 2 Q)) is the Laplace transform of an infinitely divisible random variable.
Remark 5.3 In the remainder of this paper we continue to prove Theorem 1.4 for all c 1 , c 2 and arbitrary covariance EG 1 G 2 .In each case, as immediately above, because (1.12) does not hold, there exists a c ′ < ∞ such that (G 1 + cc 1 , G 2 + cc 2 ) does not have infinitely divisible squares for all c such that |c| > c ′ .Therefore, if we can show that there exists some c = 0 for which both We first assume that EG 1 G 2 > 0 and that (c 1 , c 2 ) = (c, −c).In this case we have where ρ = a + b + 2. This is exactly the same as (3.1) except that γ is replaced by ρ.We now trace the proof in Sections 3-5 and see what changes.Obviously much remains the same.In particular the power series P is unchanged.The basic expression for Q in (3.14) is essentially the same except that γ is replaced by ρ.Thus Lemma 3.5 is also essentially the same except that γ is replaced by ρ.
The analysis in Section 4 only uses the fact that γ < ∞, and since ρ < ∞, Lemma 4.1 also holds in this case.
In going through Section 5 we see the coefficients of Q change, but they still lead to essentially the same inequalities that allow us to complete the proof.In place of (5.We also see that we get (5.7) with γ replaced by ρ and consequently, in place of (5.10), we get Of course the key term in the proof is the analogue of A j,k,p (t).We get the third line of (5.17 Now EG 1 (−G 2 ) > 0 and we are in the case proved in the beginning of this section.
The term in the numerator of the exponential lacks the λ 2 that is present in (3.1) and (6.1).Therefore, the formulas for the coefficients of the power series for the analogue of Q, which we denote by Q, are different.It is easy to see that in place of (3.39) we get Furthermore, for all t sufficiently large and for all 1 ≤ j ≤ k The analysis in Section 4 only uses the fact that γ < ∞.Since b < ∞, Lemma 4.1 also holds in this case.
In going through Section 5 we see the coefficients of Q change, but they still lead to similar inequalities that allow us to complete the proof.Using (7.3), (7.4)We next consider he analogue of (5.6) which we denote by R j,k (t).We see that in computing this the first two lines of the analogue of (5.7) remain unchanged, except for replacing c by 1.The last two lines of (5. c := (G 1 + c 1 , . . ., G n + c n ) (1.5) and (G + c) 2 := ((G 1 + c 1 ) 2 , . . ., (G n + c n ) 2 ).(1.6)