Here, we prove the uniform positivity ofEµ,Σ, in the sense that infu∈(0,∞)nEµ,Σ(x,
u) > 0 for x in some open set containing µ 6= 0. This result, stated in Proposi- tion 5.4.11, is important to proving Theorem 5.3.3 (ii).
We can see that this result straightforwardly holds in some limited cases. Recall that Σ∈Rn×n is an invertible covariance matrix. For n = 1,E
µ,Σ(µ, u) =µ2/Σ >0, u > 0. For V GGn,1 processes, Eµ,Σ(µ, ue) = kµk2Σ−1 > 0, u > 0. For V GGn,n
processes, Eµ,Σ(µ,u) = kµk2Σ−1 > 0, u ∈ (0,∞)n. In each of these cases uniform
positivity holds. However, forW V AGn processes,Eµ,Σ(µ, gα) =hµ,αµi(αΣ)−1,
g >0, and determining whether this is positive is non-trivial.
We begin by proving some lemmas. Ifn = 1, set Ξn(x)≡2, x∈R. Otherwise, if
n≥2, let Ξn(x) = (Ξn,kl)∈Rn×n, x= (x1, . . . , xn−1)∈Rn−1, be defined by Ξn,kl(x) := 2 if k=l, xk if 1≤k < n and l 6=k, 1 if k=n and l≤l < n. (5.4.29)
Recall that ∂A denotes the boundary of A⊆Rn relative to
Rn.
Lemma 5.4.7. For n ≥ 1, infx∈[0,1]n|Ξn+1(x)| = 2 +n and supx∈[0,1]n|Ξn+1(x)|=
2n+1.
Proof. Ifn = 1, then the result is trivial. Otherwise, assumen ≥2.
Let hn(x) :=|Ξn+1(x)|, x= (x1, . . . , xn)∈Rn, which is a polynomial of degree
n in the variables x1, . . . , xn. Expanding the determinant along its first row yields
hn(x) = 2hn−1(xe) +x1rn−1(ex), x= (x1,ex), where x1 ∈R, ex= (x2, . . . , xn)∈R
n−1, and ex7→rn−1(ex) is a remainder polynomial .
Note that x1 7→ hn(x1,ex) is an affine function in its first variable, so that
∂2
x1hn(x1,xe) ≡ 0. Also, hn is invariant under coordinate permutations so that
hn(xP) = hn(x) for any permutation matrix P ∈ Rn×n. Thus, hn is a harmonic
5.4 Technical Results for Proving the Necessary Conditions 107 The maximum principle for harmonic functions (see Section 1.I.4 in [Doo01]) states that inf x∈[0,1]nhn(x) = x∈min∂[0,1]nhn(x), sup x∈[0,1]n hn(x) = max x∈∂[0,1]nhn(x).
Due to the permutation invariance, to determine the infimum and supremum ofhn,
we only need to check its value on the boundary points
Fn:={0,e} ∪
[
1≤k<n
{x= (x1, . . . , xn) :x1 =· · ·=xk = 0, xk+1 =· · ·=xn = 1}.
Obviously,hn(0) = 2n+1, while hn(e) = 2 +n (see Theorem 8.4.4 in [Gra83]). For
1≤k < n, we have hn(x) = 2khn−k(1, . . . ,1) = 2k(2 +n−k) when x= (x1, . . . , xn), x1 =· · ·=xk= 0, xk+1 =· · ·=xn= 1. Thus, inf x∈[0,1]nhn(x) = minx∈Fnhn(x) = 2 +n, sup x∈[0,1]n hn(x) = max x∈Fn hn(x) = 2n+1,
which completes the proof.
Remark 5.4.8. The matrix Ξn+1(e) is the covariance matrix of some n+ 1 equicor- related random variables.
Forn ≥2 andw= (w1, . . . wn−1)∈Rn−1, define the symmetric matrix Υn(w) :=
(Υn,kl(w))∈Rn×n by setting Υn,kl(w) := 2 if k =l, 1 +Q k≤m<lwm if 1≤k < l≤n, Υn,lk(w) := Υn,kl(w), 1≤k < l≤n. (5.4.30)
Lemma 5.4.9. If n≥2 and w∈[0,1]n−1, then Υ
n(w) is nonnegative definite.
Proof. We perform the following three operations on Υn(w) fork = 1 tok =n−1.
All three operations are completed before moving to the next iteration ofk. Firstly, multiply its (k+ 1)th column by wk and subtract this from itskth column. Secondly,
multiply its (k+ 1)th row by wk and subtract this from its kth row. Thirdly, factor
out xk:= 1−wk from thekth column if xk ∈(0,1].
If xk ∈ (0,1] for all 1≤k < n, this yields |Υn(w)| =|Ξn(x)|Q1≤k<nxk, where
Ξn(x)∈Rn×n is defined in (5.4.29). Thus, by Lemma 5.4.7, we have |Υn(w)|>0,
w ∈[0,1]n−1. Otherwise, if there exists some x
k = 0 for 1≤ k < n, then the kth
5.4 Technical Results for Proving the Necessary Conditions 108 Every other principal submatrix of Υn(w), formed by keeping the rows and
columns in the index set {j1, . . . jm}, 1≤ j1 <· · · < jm ≤ n, 1≤ m ≤ n−1 and
deleting the rest, is 2 if m= 1, otherwise it is given by Υm(w), where
w:= j2−1 Y k=j1 wk, . . . , jm−1 Y k=jm−1 wk ∈[0,1] m−1 .
Repeating the above argument on Υm(w) shows that |Υm(w)| ≥0. Hence, Υn(w) is
nonnegative definite for all w∈[0,1]n−1 by Sylvester’s criterion.
For n ≥ 2 and w = (w1, . . . , wn−1) ∈ Rn−1, define the matrix ∆n(w) :=
(∆n,kl(w))∈Rn×n by setting ∆n,kl(w) := Q k≤m<lwm if 1≤k < l≤n, 1 if 1≤l ≤k≤n. (5.4.31)
Recall that Σ = (Σkl)∈Rn×n an invertible covariance matrix and ∗ denotes the
Hadamard product.
Lemma 5.4.10. If n ≥2 and w∈[0,1]n−1, then ∆
n(w)∗Σ is invertible.
Proof. Recall that (0,∞)n
≤ := {u = (u1, . . . , un) ∈ (0,∞)n : u1 ≤ · · · ≤ un}. The
mapping
u= (u1, . . . , un)7→w= (u1/u2, . . . , un−1/un) (5.4.32)
defines a bijection from S∩(0,∞)n
≤ to (0,1]n−1 with inverse w−1 : (0,1]n−1 →
S ∩ (0,∞)n≤. Note that ∆n(w) ∗ Σ = (u Σ) diag(1/u), where diag(1/u) :=
diag(1/u1, . . . ,1/un). Thus, ∆n(w)∗Σ is invertible for all w ∈ (0,1]n−1 because
(uΣ) diag(1/u) is invertible due to (5.4.1).
It remains to show invertibility forw∈En, where
En:= n−1
[
k=1
{w= (w1, . . . , wn−1)∈∂[0,1]n−1 :wk = 0}. (5.4.33)
We use mathematical induction. If n= 2, note that En={0}, so that
∆2(w)∗Σ = ∆2(0)∗Σ =
Σ11 0 Σ12, Σ22
!
5.4 Technical Results for Proving the Necessary Conditions 109 which is invertible as the product of the diagonal is strictly positive. Next, assume n≥3 and w∈En. If w1 = 0, then, ∆n(w)∗Σ = Σ11 0 e σ0 ∆n−1(w2, . . . , wn−1)∗Σe ! , Σ = Σ11 σe e σ0 Σe ! . Ifwn−1 = 0, then ∆n(w)∗Σ = ∆n−1(w1, . . . , wn−2)∗Σe 00 e σ Σnn ! , Σ = Σe σe 0 e σ Σnn ! .
Otherwise, there exists a 1< k < n−1 such that wk = 0, and we have
∆n(w)∗Σ = ∆k(w1, . . . , wk−1)∗Σe11 0 e Σ21 ∆n−k(wk+1, . . . , wn−1)∗Σe22 ! , Σ = Σe11 Σe12 e Σ21 Σe22 ! ,
where Σe11∈Rk×k,Σe22 ∈R(n−k)×(n−k). In all these cases, ∆n(w)∗Σ is invertible as
the product of the determinants of the block diagonal is strictly positive due to the inductive hypothesis and Σ being positive definite. This completes the proof.
Now we introduce the set of points V+µ,Σ where uniform positivity of Eµ,Σ holds. Let V+µ,Σ := x∈Rn : inf u∈(0,∞)nEµ,Σ(x,u)>0 . (5.4.34)
We show that this set is nonempty and open.
For a matrix A∈ Rn×n, let sym(A) := (A+A0)/2 ∈
Rn×n denote the symmet-
risation ofA. In particular, sym(A) is always a symmetric matrix.
Proposition 5.4.11. Let µ∈ Rn. The set
V+µ,Σ is an open convex cone of Rn. If
µ6=0, then µ∈V+
µ,Σ 6=∅.
Proof. Open convex cone. If a, b >0 and x,y ∈V+
µ,Σ, then it immediately follows that ax+by∈Vµ+,Σ, so V+µ,Σ is a convex cone.
For all x,y∈Rn and u∈(0,∞)n, (5.4.23) gives
|Eµ,Σ(x,u)−Eµ,Σ(y,u)| ≤Ckx−yk, C := 1 + n X k=1 sup u∈(0,∞)n E2µ,Σ(ek,u) !1/2 ,
5.4 Technical Results for Proving the Necessary Conditions 110 where C is a finite constant due to (5.4.14), and it is positive due to the addition of 1. Now choosex∈V+µ,Σ so that E:= infu∈(0,∞)nEµ,Σ(x,u)>0, and choose y∈Rn
satisfying kx−yk ≤E/(2C). With these choices, we have
Eµ,Σ(y,u)≥Eµ,Σ(x,u)−Ckx−yk ≥
E
2.
Thus, infu∈(0,∞)nEµ,Σ(y,u)≥E/2, so y∈V+µ,Σ. This shows that V+µ,Σ is open.
Positivity. For the remainder of the proof, assume that µ 6= 0. We show that Eµ,Σ(µ,u) > 0, u ∈ (0,∞)n. Recall that S++ := S∩ (0,∞)n. By (5.3.8),
Eµ,Σ(x,u) =Eµ,Σ(x,u0), whereu0 :=u/kukandu∈(0,∞)n, so we assume without loss of generality thatu∈S++.
Let Σs(u,Σ) := sym((uΣ) diag(1/u)), where diag(1/u) := diag(1/u
1, . . . ,1/un). We have Eµ,Σ(µ,u) =µ(uΣ)−1diag(u)µ0 (5.4.35) =kµ((uΣ) diag(1/u))−1k2 (uΣ) diag(1/u) =kµ((uΣ) diag(1/u))−1k2 Σs(u,Σ), (5.4.36)
where the last line follows from Lemma A.2.3. Thus,Eµ,Σ(µ,u)>0 provided that Σs(u,Σ) is positive definite.
Introduce g(u) := (1∧u) + (1∧(1/u))∈(1,2], u >0, and the symmetric matrix Θn(u) := (Θn,kl(u)) defined by Θn,kl(u) :=g(uk/ul), 1≤k, l ≤n, u∈S++. We now prove that Θn(u) is nonnegative definite. For n= 1, this is clear as |Θ1(u)| ≡2.
Forn ≥2, note that Θn(uP) = P0Θn(u)P,u∈S++, for any permutation matrix P ∈Rn×n as the (k, l) elements of both sides are g(u
(k)/u(l)), whereh(1), . . . ,(n)i is the permutation associated withP. Thus, we can assumeu∈S∩(0,∞)n
≤ without loss of generality as proving that Θn(u) is nonnegative definite foru ∈S∩(0,∞)n≤ implies it for allu∈S++. This follows from the definition of a nonnegative definite matrix. Now with w ∈ (0,1]n−1 determined by the bijection in (5.4.32), we have Θn(u) = Υn(w), where Υn(w) is defined in (5.4.30) and nonnegative definite by
Lemma5.4.9.
Finally, note that 2Σs(u,Σ) = Θ
n(u)∗Σ,u ∈S++. Since Σ is positive definite by assumption and Θn(u) is nonnegative definite with positive diagonal elements
Θn,kk(u) ≡2, 1 ≤k ≤ n, Oppenheim’s inequality (see the LHS of (A.2.1)) imme-
diately implies that every leading principal minor of Θn(u)∗Σ is positive. Thus,
Σs(u,Σ) is positive definite, proving that Eµ,Σ(µ,u)>0,u ∈(0,∞)n.
Uniform positivity. Now we show that µ∈V+
5.5 Applications of Self-Decomposability Conditions 111