CAPÍTULO 3. RESULTADOS Y DISCUSIÓN
3.9 Comportamiento del riego durante el ciclo
The sampling approach described in Section 6.1 does not extend to the algebraic constraints as the set of draws from the posterior satisfying an equality constraint will have zero probabil- ity. Thus an alternative approach is taken using sample distributions of minors of a covariance matrix.
6.2.1 Means and variances
Considering Theorem5.3.4, it is clear that information on whether a Gaussian distribution lies in someM(T)for these particular constraints is recorded in the sign of tetrad constraintsσikσjl−
σilσjk and other quadratic binomials of the formσiiσjk −σijσik. Here we use the covariance
matrix instead of correlation matrix as the distributional results associated with correlations are notably more difficult to deal with. Both type of constraints can be realised as minors of the
covariance matrixΣ
det(Σij,kl) and det(Σij,ik). (6.2.1)
However, we are particularly interested in the former as the algebraic constraints cannot be treated in the same way as the semi-algebraic constraints in Section6.1.
Definition 6.2.1. A matrixAvanishes if thedet(A) = 0.
LetA|Bbe a split of the set of leaves that is induced byT by removing an edge and considering two resulting components. Consider the following block matrix
ΣA,A ΣA,B ΣB,A ΣB,B .
Proposition 6.2.2. The equality constraints in (5.3.3) imply that all2×2minors ofΣA,Bmust vanish ifA∪B =V forT = (V, E)⊂M(T).
Proof. Without loss of generality, consider i, j ∈ A and k, l ∈ B, i, j, k, l distinct. Then
{i, j}|{k, l}and soσikσjl =σilσjk as a consequence of Theorem (5.3.4). Thus,det(Σij,kl) =
σikσjl−σilσjk = 0, i.e.Σij,klvanishes.
Corollary 6.2.3. The rank of all2×2 minors ofΣA,B must equal 1 ifA∪B = V forT = (V, E)⊂M(T).
Proof. Without loss of generality, consideri, j∈Aandk, l∈B,i, j, k, ldistinct. Then
Σij,kl= σik σil σjk σjl .
Multiplying the second row ofΣij,klby σik
σjk produces: σik σil σik σikσjkσjl . Sinceσikσjl−σilσjk = 0 =⇒ σikσjl
σjk = σil, rows 1 and 2 ofΣij,kl are linearly dependent.
We wish to understand the distributional properties of the algebraic tetrad constraints for the Gaussian setting. Suppose that a sample X ∈ Rn×m was observed and let S = XTX so thatS/nis the sample covariance matrix. IfCis the covariance matrix of the data generating distribution thenS has Wishart distributionWm(n, C). Denote by
m
2 the set of all subsets
of{1, . . . , m}of cardinality two. We are going to use the following estimator of the value of
det(CI,J)forI, J ∈ m
2
QI,J :=
1
n(n−1)det(SI,J). (6.2.2)
In this section we provide the first two moments for the random m2 × m2
-matrix with entries given byQI,J. The following result, that shows thatQI,Jis an unbiased estimator, follows from
Drton et al.[2008, Corollary 4.2].
Proposition 6.2.4. IfI, J ∈m
2 thenE[QI,J] = det(CI,J).
It is convenient to introduce the following notation. For anm×mmatrixA let A(2) denote the matrix with rows and columns indexed by elements m2 whose (I, J)-th element is the corresponding minordet(AI,J). The rows and columns are ordered in the natural order of
m 2
given by
{1,2} ≺ {1,3} ≺ · · · ≺ {1, m} ≺ {2,3} ≺ · · · ≺ {m−1, m}.
With this notation, the matrix, whose elements are estimatorsQI,J is given byS(2)/(n(n−1)).
Proposition6.2.4now reads:E[S(2)] =n(n−1)C(2).
Remark6.2.5. This given ordering ofm2 is induced from the natural ordering on{1, . . . , m}
given by1<2<· · ·< m. A different ordering of{1, . . . , m}will lead to a different ordering ofm2 .
We next provide the variance forQI,J. The formulae depend on the cardinality ofI∩J. Proposition 6.2.6. IfI, J ∈m
2 are disjoint, then
var(QI,J) = 1
n(n−1)
(n+ 2) det(CI,I) det(CJ,J)−ndet(CI∪J,I∪J) + 3ndet(CI,J)2 IfI =J ∈m 2 then var(QI,I) = 4n+ 2 n(n−1)det(CI,I) 2.
Moreover, ifI, J ∈m
2 such thatI ={i, j},J ={i, k}andj 6=k, then
var(QI,J) = 4n+ 2 n(n−1)det(CI,J) 2+ n+ 2 n(n−1)C 2
i,idet(Cjk,jk−Cjk,iCi,i−1Ci,jk).
Remark6.2.7. The result forI =J ∈m
2 cannot (and is not intended to) be derived directly
from result for theI, J ∈m
2 disjoint.
Proof. IfI, Jare disjoint, this is a classical result ofWishart[1928b], see alsoDrton et al.[2008, Corollary 5.6]. The second part of the result, covering the case|I∩J| ≥ 1, follows from the main theorem ofDrton et al.[2008] (seeDrton and Goia[2012] for the corrected version).
6.2.2 Covariances between minors
There is no simple explicit formula for covariances of various2-minors but they can be computed if the true distribution ofCis known.
ByDrton et al.[2008, Proposition 3.3]
cov[S(2)] = [(C1/2)(2)⊗(C1/2)(2)]·(cov(W(2)))·[(C1/2)(2)⊗(C1/2)(2)], (6.2.3)
where W has standard Wishart distribution Wm(n, I) and⊗is the Kronecker product. This follows from
S=C1/2W C1/2 ∼ Wm(n, C1/2ImC1/2) =Wm(n, C)
and the Cauchy–Binet formula [Olkin and Marshall,2014]
(AB)(k)= (A)(k)(B)(k).
In the rest of this section we provide a complete description of the covariance matrixcov(W(2)). This matrix has many symmetries that we want to exploit both in the exposition below and in the computations. First note thatdetWI,J = detWJ,Ifor allI, J ∈
m
2 and hence
This enables us to take the following convention. Using the natural order of m2 we always assume that
I J, KL and I K. (6.2.4)
If we can findcov(detWI,J,detWK,L)for allI, J, K, Lsatisfying (6.2.4) we can then fill out
the rest of the matrixcov(W(2))using basic symmetries.
LetA∆B := (A\B)∪(B\A)be the symmetric difference of two sets.
Proposition 6.2.8. Suppose that I, J, K, L ∈ m
2 satisfy (6.2.4) and (I, J) 6= (K, L). If
I∆J 6=K∆L, thencov(detWI,J,detWK,L) = 0.
Proof. Note that necessarily eitherI 6=J orK 6=L(otherwiseI∆J =∅=K∆L) and hence eitherdetCI,J = 0ordetCK,L = 0, whereCis the identity matrix. By Proposition6.2.4
E(detWI,J)E(detWK,L) =n2(n−1)2detCI,JdetCK,L= 0,
ByDrton et al.[2008, Proposition 3.4] alsoE(detWI,JdetWK,L) = 0.
In the rest of this section we assume that I∆J = K∆L. If(I, J) = (K, L) then trivially
cov(detWI,J,detWK,L) = var(detWI,J)and by Proposition6.2.6
1 n2(n−1)2var(detWI,J) = 4n+2 n(n−1) ifI =J 2 n(n−1) ifI∩J =∅ n+2 n(n−1) otherwise. (6.2.5)
For the remaining cases we are going to use the following result, which deals with general minors of sizer.
Theorem 6.2.9(Drton et al.[2008], Theorem 4.5). SupposeI, J, K, L∈m
r such thatI∆J =
satisfies
(I∩J)\(K∩L) I\J J\I (K∩L)\(I∩J)
(I\J)∩(K\L) (I\J)∩(L\K), (6.2.6)
(J\I)∩(K\L) (J\I)∩(L\K).
If any terms result in an order comparison with the empty set then by convention this is said to hold. If the ordering here holds and ifW ∼ Wm(n, I), then
E[det(WI,J) det(WK,L)] =
n! (n−r)!· (n+ 2)! (n+ 2− |I∩J∩K∩L|)!· ·(n−r+|(I∩J)\(K∩L)|)! (n−r)! · |(I\J)∩(K\L)|!· |(I\J)∩(L\K)|!
In the rest of this section we analyse the special case of Theorem 6.2.9, when r = 2. Our motivation is to obtain a more concrete version, with a more algorithmic approach to the ordering constraint (6.2.6). If(I, J)6= (K, L)andI∆J =K∆Lthen we have three cases
(i) I =J andK=Land either|I ∩K|= 0or|I∩K|= 1
(ii) I ={i, j},J ={i, k},K ={j, l}L={k, l}for some distincti, j, k, l
(iii) I ={i, j},J ={k, l},K={i, k},L={j, l}for some distincti, j, k, l
Proposition 6.2.10. IfI, J, K, L ∈ m
2 satisfy (6.2.4) andI∆J = K∆Lthen givenI = J,
K =LandI 6=K, we have
1
n2(n−1)2cov(detWI,J,detWK,L) = 0 if|I∩K|= 0 2 n if|I∩K|= 1.
Proof. IfI =J andK =Lthen, by Proposition6.2.4,E(detWI,J) =E(detWK,L) = 1. By Theorem6.2.9E(detWI,JdetWK,L)is equal, up to sign, to 1if|I ∩K| = 0and to n+2n if |I∩K|= 1. To show that the sign is always positive we use the fact thatI ≺K by (6.2.4). If
|I∩K|= 0, the ordering constraint (6.2.6) is trivially satisfied. If|I∩K|= 1then it is satisfied becauseI\K≺K\I.
Obtaining remaining covariances is more subtle because they can take different signs depending on whether (6.2.6) holds or not. In our special case these technical conditions can be translated into checking parity of certain permutations.
Definition 6.2.11. We say that a sequencei1, . . . , ikdefinesa partitionσof the set{i1, . . . , ik} ifσ(oj) =ij forj = 1, . . . , k, whereo1, . . . , okis the sequence ofij’s given in an increasing order. This is also sometimes called the one-line notation for permutations.
For example ifi1 = 2,i2 = 1,i3 = 4,i4 = 3then{i1, i2, i3, i4}={1,2,3,4}and the sequence 2,1,4,3defines permutation such thatσ(1) = 2,σ(2) = 1,σ(3) = 4andσ(4) = 3.
Proposition 6.2.12. IfI, J, K, L∈m
2 satisfy (6.2.4) andI∆J =K∆Lthen for some distinct
i, j, k, l, whereI ={i, j}, J ={i, k}, K ={j, l}, L={k, l}then
1
n2(n−1)2cov(detWI,J,detWK,L) = (−1) sgn(σ)1
n
whereσ is a permutation of the set{i, j, k, l}defined by the sequencei, j, k, l.
Proof. SinceI 6=J thenE(detWI,J) = 0and hence to compute this covariance it suffices to computeE(detWI,JdetWK,L). By Theorem 6.2.9this second order moment is equal (up to
sign) to1/nand the sign is positive ifi < j < k < l. Because I, J, K, Lsatisfy (6.2.4), we necessarily havej < kandi < lso there are six possible orderings
i < j < k < l, j < k < i < l, j < i < l < k i < l < j < k, j < i < k < l, i < j < l < k.
Becausei < j < k < l gives a positive sign and the sign ofdetWij,klchanges if you swap
rows or columns, we check directly that only the last two situations lead to negative values of E(detWI,JdetWK,L). Again, a direct check shows that only the last two sequences define
permutations with negative parity.
Proposition 6.2.13. IfI, J, K, L∈m
2 satisfy (6.2.4) andI∆J =K∆Lthen for some distinct
i, j, k, l, whereI ={i, j}, J ={k, l}, K ={i, k}, L={j, l}then
1
n2(n−1)2cov(detWI,J,detWK,L) = (−1)
sgn(σ) 1
whereσ is a permutation of the set{i, j, k, l}defined by the sequencei, j, k, l.
Proof. We proceed in a similar way as in the proof of Proposition6.2.12. To determine the sign note that, because I, J, K, Lsatisfy (6.2.4), necessarilymin{i, j} < min{k, l}, min{i, k} <
min{j, l}andj < k. There are three possible orderings that satisfy these constraints
i < j < k < l i < l < j < k i < j < l < k.
To see this note that j < k and that there are three possible ways to positionl: j < k < l,
l < j < k andj < l < k. But now for each of these orderings the position ofiis already determined. Since the sign ofE(detWij,kldetWik,jl)is positive ifi < j < k < l, then it is
also positive if i < l < j < kand it is negative fori < j < l < k. Again, a direct check confirms that the sign coincides with the parity of the corresponding permutation.