TEST FOR INDEPENDENCE OF THE VARIABLES WITH MISSING ELEMENTS IN ONE AND THE SAME COLUMN OF THE EMPIRICAL CORRELATION MATRIX.

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2 Serdica Math J 34 (008, TEST FOR INDEPENDENCE OF THE VARIABLES WITH MISSING ELEMENTS IN ONE AND THE SAME COLUMN OF THE EMPIRICAL CORRELATION MATRIX Evelina Veleva Communicated by N Yanev Abstract We consider variables with joint multivariate normal distribution and suppose that the sample correlation matrix has missing elements, located in one and the same column Under these assumptions we derive the maximum likelihood ratio test for independence of the variables We obtain also the maximum likelihood estimations for the missing values 1 Introduction We consider variables with joint multivariate normal distribution and suppose that the sample correlation matrix has missing elements, located in one and the same column This might be due to a loss during the keeping or the transportation to the researcher Another reason is when a researcher A has the full matrix of the observations on n variables X 1,, X n, but he decides to include in considerations an additional variable X n+1 Assume that 000 Mathematics Subject Classification: 6H15, 6H1 Key words: Multivariate normal distribution, Wishart distribution, correlation matrix completion, maximum likelihood ratio test

3 510 Evelina Veleva his colleague B works in a competitive firm and has the data matrix for the variables X k+1,, X n, X n+1 Suppose the researcher A succeeds to get from B only the empirical correlation coefficients r i n+1, i = k + 1,, n, instead of the data vector of the observations on X n+1, which is m 1 and m is enough greater than n k (m n k In this situation A will have the complete empirical correlation matrix for the variables X 1,, X n, X n+1, with the exception of the coefficients r i n+1, i = 1,, k Recently, the question of evaluating the return to a given program or treatment has received considerable attention in the economics, statistics and medical literatures [, 5, 7, 8] The derived densities of the outcome gain distributions necessarily depend on the unidentified cross-regime correlation coefficient Roy s model of self-selection has been applied extensively in economics to explain individual choice between two alternatives or sectors Since an agent s earnings are observed only in the sector of choice under partial observability, the correlation coefficient between the disturbance terms in the earnings equations is not identified Similar investigations with one missing element in the correlation matrix have been made by Heckman and Honore [1], Vijverberg [11], Koop and Poirier [4], Sareen [6] In the present paper we assume that the researcher is interested in n + 1 random variables, which are multivariate normal distributed but he has only the empirical correlation matrix R, in which the elements r i n+1, i = 1,, k are missing In this case we obtain the maximum likelihood estimations for the correlation coefficients ρ i n+1, i = 1,, k, which are elements of the theoretical correlation matrix P We derive also the maximum likelihood ratio test for the hypothesis for the independence of the n + 1 random variables, ie where I is the identity matrix H 0 : P = I, The likelihood ratio test derivation We begin with one auxiliary Theorem Theorem 1 Let ω ij, 1 i j n be random variables with joint distribution - the Wishart distribution W n (m, Σ Consider the set of random variables V = {τ i, i = 1,, n, ν ij, 1 i < j n}, such that τ i = ω ii, i = 1,, n,

4 Test for independence of the variables 511 ν ij = ω ij ωii ω jj, 1 i < j n The joint density function of the random variables from V has the form f V (t 1,, t n, y ij, 1 i < j n (t 1 t n m = n nm π n(n 4 (det(σ m Γ ( (det(y n m n e 1 tr(tyntσ I E, m i+1 where Γ ( is the well known Gamma function; T and Y n are the matrices T = diag( t 1,, t n = t1 0 0 tn, Y n = 1 y 1 y 1n y 1 1 y n y 1n y n 1 ; E is the set of all points (t 1,, t n, y ij, 1 i < j n in the real space R n(n+1/, such that t i > 0, i = 1,, n and the matrix Y n is positive definite, and I E here and after denotes the indicator of a given set E P r o o f The inverse transformation formulas are ω ii = τ i, i = 1,, n, The Jacobian of the transformation is ω ij = ν ij τi τ j, 1 i < j n J = (ω 11,, ω nn, ω 1,, ω 1n, ω 3,, ω nn (τ 1,, τ n, ν 1,, ν 1n, ν 3,, ν nn

5 51 Evelina Veleva = τ 1 τ τn τ n Consequently det(j = (τ 1 τ n n The joint density function of ω ij, 1 i j n has the form f ωij,1 i j n(w ij, 1 i j n = (det(w m n e 1 tr(wσ n nm π n(n 4 (det(σ m Γ ( m i+1 I {W>0}, where W is the symmetric matrix with elements w ij, 1 i j n and {W > 0} here and after denotes the set of all values for the unknown elements of a given real matrix W, such that the matrix W is positive definite Hence, the joint density function of the random variables from the set V = {τ i, i = 1,, n, ν ij, 1 i < j n} will have the form f V (t 1,, t n, y ij, 1 i < j n = (det(ty nt m n e 1 tr(tyntσ n nm π n(n 4 (det(σ m (t 1 t n m = n nm π n(n 4 (det(σ m Γ ( m i+1 (t 1 t n n I E Γ ( (det(y n m n e 1 tr(tyntσ I E m i+1 Corollary 1 Under the conditions of Theorem 1, let Σ = D, where D is a diagonal matrix D = diag(σ1,, σ n Then the set of random variables {τ 1,, τ n } is independent of the set {ν ij, 1 i < j n} The variables τ 1,, τ n are mutually independent and τ i σi χ (m, i = 1,, n The random

6 Test for independence of the variables 513 variables ν ij, 1 i < j n has joint distribution Ψ(m, n with density function of the form f νij,1 i<j n(y ij, 1 i < j n = π n(n 4 [ Γ ( m n ] n Γ ( (det(y n m n I {Yn>0} m i Let x 1,, x m be a sample from N n+1 (µ, Σ, µ is unknown Assume that we wish to test the hypothesis H 0 : Σ = diag(σ 1,, σ n+1, σ i are unknown, i = 1,, n + 1 which is equivalent to H 0 : P = I against the alternative H 1 : no constraints on Σ Suppose at first, that we don t hold the observations themselves, but we have m and the empirical covariance matrix S = {s ij } n+1 i,j=1, in which the elements s in+1, i = 1,, k are unidentified It is well known, that the joint distribution of the elements of the matrix (m 1S is Wishart - W n+1 (m 1, Σ Consider the variables and ν ij = τ i = (m 1s ii, i = 1,, n + 1 (m 1s ij (m 1sii (m 1s jj = r ij, 1 i < j n + 1, where r ij is the corresponding element of the empirical correlation matrix R Let the hypothesis H 0 be true According to Corollary 1, the distribution of the elements of the empirical correlation matrix R is Ψ(m 1, n + 1 The random variables τ i = (m 1s ii, i = 1,, n + 1 are mutually independent, they are independent of the correlation coefficients r ij, 1 i < j n + 1 and τ i = (m 1s ii σ i χ (m 1, i = 1,, n + 1 The next proposition can be found in [9] Proposition 1 Let ν ij, 1 i < j n be random variables with distribution Ψ(m, n The joint density function of the random variables from the set V = {ν ij, 1 i < j n}\{ν 1n,, ν kn }, where k is an integer, 1 k n,

7 514 Evelina Veleva has the form f V (y ij, 1 i < j n, y k+1n,, y nn = π n(n k [ Γ ( m ] n 4 Γ ( m n+k+1 n Γ ( m i (det(y n({n} m n (det(y n ({1,, k} m n+k I (det(y n ({1,, k, n} m n+k {Yn({n}>0} I {Yn({1,,k}>0}, where A({i 1,, i l } here and after denotes the matrix, which is obtained from the matrix A, after deleting the rows and columns with numbers i 1,, i l From this Propositions it follows that under H 0, the joint density function of the empirical correlation coefficients from the set U = {r ij, 1 i < j n, r k+1n+1,, r nn+1 } will have the form f U (y ij, 1 i < j n, y k+1n+1,, y nn+1 [ ( Γ m ] n = π n(n+1 k 4 Γ ( m n+k n Γ ( I {Y n+1 ({n+1}>0} I {Yn+1 ({1,,k}>0} m i (det(y n+1({n + 1} m n (det(y n+1 ({1,, k} m n+k 3 (det(y n+1 ({1,, k, n + 1} m n+k Consequently, under H 0 the joint density function of our observations : (m 1s 11,, (m 1s n+1n+1, r 1,, r nn, r k+1n+1,, r nn+1, ie the likelihood function L 0, will have the form L 0 = (m 1 (n+1(m 3 (n+1(m π n(n+1 k 4 Γ ( m n+k n Γ ( m i (s 11 s n+1n+1 m 3 (σ 1 σ n+1 m e n+1 (m s ii σ i m n (det(r({n + 1} (det(r({1,, k} m n+k 3 (det(r({1,, k, n + 1} m n+k

8 Test for independence of the variables 515 The maximum likelihood estimations for the unknown parameters σ1,, σ n+1 are s 11,, s n+1n+1 respectively The maxima L 0 of the likelihood function is L 0 = (m 1 (n+1(m 3 (n+1(m π n(n+1 k 4 Γ ( m n+k n Γ ( m i (n+1(m e (det(r({n + 1} m n (det(r({1,, k} m n+k 3 (s 11 s n+1n+1 (det(r({1,, k, n + 1} m n+k Let the hypothesis H 1 be true Applying Theorem 1 for m = m 1 and n = n + 1, we get the joint density function f 1 of the observations : (m 1s 11,, (m 1s n+1n+1, r 1,, r nn, r k+1n+1,, r nn+1 and the unknown realizations r 1n+1,, r kn+1 of the correlation coefficients ρ 1n+1,, ρ kn+1 : f 1 = (n+1(m 3 (m 1 (s 11 s n+1n+1 (m 3 (n+1(m π n(n+1 4 (det(σ (m n+1 Γ ( m i (det(r m n 3 e (m tr(trtσ I {R>0}, where T is the diagonal matrix T = diag( s 11,, s n+1n+1 The likelihood function L 1 under H 1 is an integral of f 1 with respect to r 1n+1,, r kn+1 : L 1 = f 1 dr 1n+1 dr kn+1 Let us denote by σ ij, 1 i j n + 1 the elements of the matrix Σ It is easy to see that the trace where tr(trtσ = Ξ + n+1 n (1 Ξ = s ii σ ii + n j=i+1 k σ in+1 s ii s n+1n+1 r in+1, σ ij sii s jj r ij + n σ in+1 sii s n+1n+1 r in+1 i=k+1

9 516 Evelina Veleva Hence ( L 1 = K(det(Σ (m e (m Ξ J, where K = (n+1(m 3 (m 1 (s 11 s n+1n+1 (m 3 (n+1(m π n(n+1 4 n+1 Γ ( m i, J = (det(r m n 3 e (m k (σ jn+1 sjj s n+1n+1 r jn+1 j=1 I {R>0} dr 1n+1 dr kn+1 The maximum likelihood estimations ˆσ ij for σ ij, 1 i j n+1 satisfy the system of equations (3 L 1 = 0, i = 1,, n + 1 σii L 1 = 0, 1 i < j n + 1 σij It is easy to see that det(σ σ 11 = det(σ ({1} Hence the first equation of the system (3 is L 1 σ (m 1 = K J 11 = 0 Consequently, (det(σ (m 3 e (m Ξ [ det(σ ({1} det(σ s 11 ] det(σ ({1} det(σ = s 11

10 Test for independence of the variables 517 The left hand side of the above equation equals to the (1,1 element of the inverse matrix of the matrix Σ, ie the (1,1 element of the matrix Σ Consequently we have We conclude similarly that ˆσ 11 = s 11 ˆσ ii = s ii, i =,, n + 1 In the further considerations we need the following Theorem Theorem Let A = {a ij } n i,j=1, a ij = a ji be a n n symmetric matrix For each element a pq, p q of the matrix A, the determinant det(a can be written as (4 det(a = det(a({p, q}a pq + ( q p det(a({p}, {q} 0 a pq + det(a 0 p,q, where A({p}, {q} 0 is the matrix, which is obtained from the matrix A after deleting its p th row, q th column and replacing the element a qp with zero; A 0 p,q is the matrix, which is obtained from the matrix A, after replacing the elements a pq and a qp with zero P r o o f It is easy to see that (5 det(a = a pq ( q p det(a({p}, {q} + det(a 0, where A 0 is the matrix, which is obtained from the matrix A after replacing the element a pq with zero Analogically (6 det(a 0 = a qp ( p q det(a({q}, {p} 0 + det(a 0 p,q and (7 det(a({p}, {q} = a qp ( p q det(a({p, q} + det(a({p}, {q} 0 Since the matrix A is symmetric, from (5 (7 we obtain (4 Put in the above Theorem A = Σ, n = n + 1, p = 1, q = Then det(σ = det(σ ({1, }(σ 1 det(σ ({1}, {} 0 σ 1 +det((σ 0 1,

11 518 Evelina Veleva Consequently, det(σ σ 1 = det(σ ({1, }σ 1 det(σ ({1}, {} 0 Hence = det(σ ({1}, {} L 1 σ 1 = K J(m 1(det(Σ (m 3 e (m Ξ Therefore we get [ det(σ ({1}, {} det(σ s 11 s r 1 ] = 0 det(σ ({1}, {} det(σ = s 11 s r 1 = s 1 The left hand side of the above equation equals to the (1, element of the inverse matrix of the matrix Σ, ie the (1, element of the matrix Σ Consequently we get We conclude by analogy that ˆσ 1 = s 1 ˆσ ij = s ij, 1 i < j n, ˆσ in+1 = s in+1, i = k + 1,, n Let us apply Theorem for every one of the elements σ in+1, i = 1,, k of the matrix Σ We get det(σ = det(σ ({i, n + 1}(σ in+1 + ( n i+1 det(σ ({i}, {n + 1} 0 σ in+1 + det((σ 0 i,n+1, i = 1,, k Hence det(σ σ in+1 = det(σ ({i, n+1}σ in+1 +( n i+1 det(σ ({i}, {n+1} 0

12 Test for independence of the variables 519 = ( n i+1 [det(σ ({i}, {n + 1} 0 + ( n i det(σ ({i, n + 1}σ in+1 ] The integral J depends of σ in+1, i = 1,, k and where J i = s ii Therefore J σ in+1 = (m 1 s n+1n+1 J i, (det(r m n 3 r in+1 e (m = ( n i+1 det(σ ({i}, {n + 1} k (σ jn+1 sjj s n+1n+1 r jn+1 j=1 I {R>0} dr 1n+1 dr kn+1, i = 1,, k L 1 σ in+1 = K (m 1(det(Σ (m 3 e (m Ξ [( n i+1 det(σ ({i}, {n + 1} J det(σ s n+1n+1 J i ] = 0, i = 1,, k Consequently, for i = 1,, k we have ( n i+1 det(σ ({i}, {n + 1} det(σ = sn+1n+1 J i J The left hand side of the above equation equals to the (i, n + 1 element of the matrix Σ, ie the element σ in+1 Consequently for ˆσ in+1, i = 1,, k we get the equations (8 ˆσ in+1 = Ĵ i s n+1n+1, i = 1,, k, Ĵ where Ĵi and Ĵ are the integrals J i and J respectively, in which we are substituted the maximum likelihood estimations ˆσ in+1, i = 1,, k for the unknown parameters σ in+1, i = 1,, k

13 50 Evelina Veleva Let us denote by ˆΣ the maximum likelihood estimation of the unknown matrix Σ For the matrix ˆΣ we obtained that s 11 s 1k s 1n ˆσ 1n+1 ˆΣ = s 1k s kk s kn ˆσ kn+1 s 1n s kn s nn s nn+1 ˆσ 1n+1 ˆσ kn+1 s nn+1 s n+1n+1 Let us denote by A({i 1,, i l }, {j 1,, j l } the matrix, which is obtained from the matrix A after deleting the rows with numbers i 1,, i l and the columns with numbers j 1,, j l The matrix A({i 1,, i l }, {i 1,, i l, j} 0 is the matrix, which is obtained from the matrix A({i 1,, i l }, {i 1,, i l, j} after replacing the element a jil with zero We shall prove that for ˆσ in+1, defined by (9 ˆσ in+1 = (n i+1 det( ˆΣ({1,, i}, {1,, i 1, n + 1} 0, i = 1,, k, det( ˆΣ({1,, i, n + 1} the equations (8 holds The formulae (9 are recurrent They determine at first ˆσ kn+1, then ˆσ kn+1 and so on, and finally - ˆσ 1n+1 The next Theorem gives an equivalent presentation of ˆσ in+1, i = 1,, k Then Theorem 3 Let ˆσ in+1, i = 1,, k are defined by the equalities (9 (10 ˆσ in+1 = (n k+1 det( ˆΣ({1,, k}, {1,, i 1, i + 1,, k, n + 1} 0, det( ˆΣ({1,, k, n + 1} i = 1,, k P r o o f The proof is by induction The presentation (10 holds for i = k Suppose that it holds for i = k, k,, q+1, where q is an integer, 1 q k We shall prove that it holds for i = q too From (9 we have that (11 ˆσ qn+1 = (n q+1 det( ˆΣ({1,, q}, {1,, q 1, n + 1} 0 det( ˆΣ({1,, q, n + 1}

14 Test for independence of the variables 51 It can be easily seen that (1 det( ˆΣ({1,, q}, {1,, q 1, n + 1} 0 = det(a + ˆσ q+1n+1 ( n q+1 det( ˆΣ({1,, q, n + 1}, {1,, q 1, q + 1, n + 1}, where A is the matrix A = s qq+1 s q+1q+1 s q+1n s qn s q+1n s nn 0 0 s nn+1 If we move in the matrix A the last row after its first row, the determinant will eventually change its sign, more precisely (13 det(a = ( n q+1 det s qq+1 s q+1q+1 s q+1n 0 0 s nn+1 s qq+ s q+1q+ s q+n s qn s q+1n s nn Let us denote the last matrix by B Applying the Sylvester s determinant identity to the matrix B (Karlin, 1968 we have (14 det(b det( ˆΣ({1,, q + 1, n + 1} = ( n q+1 det( ˆΣ({1,, q + 1}, {1,, q, n + 1} 0 det( ˆΣ({1,, q, n + 1}, {1,, q 1, q + 1, n + 1} ( n q+1 det( ˆΣ({1,, q + 1}, {1,, q 1, q + 1, n + 1} 0 det( ˆΣ({1,, q, n + 1} Replacing (13 and (14 in (1 and using the representation (9 for ˆσ q+1n+1, we obtain that (15 det( ˆΣ({1,, q}, {1,, q 1, n + 1} 0 = det( ˆΣ({1,, q, n + 1} det( ˆΣ({1,, q + 1}, {1,, q 1, q + 1, n + 1} 0 det( ˆΣ({1,, q + 1, n + 1}

15 5 Evelina Veleva Consequently from the equality (11 it follows that (16 ˆσ qn+1 = (n q det( ˆΣ({1,, q + 1}, {1,, q 1, q + 1, n + 1} 0 det( ˆΣ({1,, q + 1, n + 1} According to the representation (9 for ˆσ q+1n+1, (17 ˆσ q+1n+1 = (n q det( ˆΣ({1,, q + 1}, {1,, q, n + 1} 0 det( ˆΣ({1,, q + 1, n + 1} The right hand sides of formulas (16 and (17 are almost the same The difference is only in the first column of the matrix in the numerator The other elements of the matrices are identical and do not depend of the elements in the first columns of the two matrices s qq+,, s qn and s q+1q+,, s q+1n respectively According to the induction assumption ˆσ q+1n+1 = (n k+1 det( ˆΣ({1,, k}, {1,, q, q +,, k, n + 1} 0 det( ˆΣ({1,, k, n + 1} Therefore we conclude that ˆσ qn+1 = (n k+1 det( ˆΣ({1,, k}, {1,, q 1, q + 1,, k, n + 1} 0 det( ˆΣ({1,, k, n + 1} Consequently the presentation (10 holds for i = q, hence it is true by induction Let us determine for ˆσ in+1, i = 1,, k, given by (9 the corresponding elements ˆσ in+1, i = 1,, k of the matrix ˆΣ By definition It is easy to see that ˆσ in+1 = (n i+1 det( ˆΣ({i}, {n + 1}, i = 1,, k det( ˆΣ ( n det( ˆΣ({1}, {n + 1} = ( n [ˆσ 1n+1 ( n det( ˆΣ({1, n + 1} + det( ˆΣ({1}, {n + 1} 0 ] [ ] = ( n ( det( ˆΣ({1}, {n + 1} 0 det( ˆΣ({1, n + 1} + det( ˆΣ({1}, {n + 1} det( ˆΣ({1, 0 n + 1} = 0

16 Test for independence of the variables 53 Consequently ˆσ 1n+1 = 0 We shall prove by induction that (18 ˆσ in+1 = 0, i = 1,, k Assume that ˆσ in+1 = 0 for i = 1,, q 1, where q is an integer, q k and consider the matrices A = ˆΣ({q,, n + 1}, D = ˆΣ({1,, q 1}, B = ˆΣ({q,, n + 1}, {1,, q 1}, C = ˆΣ({1,, q 1}, {q,, n + 1} The matrix ˆΣ can be written as a block matrix ( A B ˆΣ = C D It can be easily verified that ( ˆΣ (A BD C = (A BD C BD D C(A BD C D + D C(A BD C BD ( X Y = Z U The element in the lower left corner of the matrix U = D C(A BD C BD + D is exactly the element ˆσ qn+1 and we have to prove that it equals to zero At first we shall show that the element d 1n q in the lower left corner of the matrix D equals to zero Indeed, d 1n q = (n q det(d({1}, {n q} det(d

17 54 Evelina Veleva and it is easy to see that det(d({1}, {n q} = det( ˆΣ({1,, q}, {1,, q 1, n + 1} = ˆσ qn+1 ( n q det( ˆΣ({1,, q, n + 1} + det( ˆΣ({1,, q}, {1,, q 1, n + 1} 0 = det( ˆΣ({1,, q}, {1,, q 1, n + 1} 0 + det( ˆΣ({1,, q}, {1,, q 1, n + 1} 0 = 0 According to the induction assumption, all elements in the last row of the matrix Z = D C(A BD C are equal to zero Hence the elements in the last row of the matrix D C(A BD C BD = ZBD are equal to zero too Adding the matrix D to the last one we obtain the matrix U and conclude that the element ˆσ qn+1 in its lower left corner is equal to zero Consequently the equalities (18 are true by induction The integrals Ĵ and Ĵi then become and Ĵ = J i = s ii (det(r m n 3 I {R>0} dr 1n+1 dr kn+1 (det(r m n 3 r in+1 I {R>0} dr 1n+1 dr kn+1 Applying Proposition 1 for m = m 1, n = n + 1 we get that (19 [ ( Γ 1 ] k ( Γ m n Ĵ = Γ ( (det(r({n + 1} m n (det(r({1,, k} m n+k 3 m n+k (det(r({1,, k, n + 1} m n+k Let i is an integer, 1 i k If i, from Proposition 1 we obtain that (0 [ ( (det(r m n 3 Γ 1 ] i ( Γ m n I {R>0} dr 1n+1 dr in+1 = Γ ( m n+i m n (det(r({n + 1} (det(r({1,, i 1} m n+i 4 (det(r({1,, i 1, n + 1} m n+i 3 I {R({1,,i}>0}

18 Test for independence of the variables 55 For 1 i k 1 let us move in the matrix R the i th row after the k th row and do the same with the i th column Denote the new matrix by R It is easy to see that det(r ({1,, i 1} = det(r({1,, i 1} It can be easily shown that the matrix R ({1,, i 1} is positive definite iff the matrix R({1,, i 1} is so The proof uses the fact that the two matrices have one and the same eigenvalues From (0 it follows that (det(r ({1,, i 1} m n+i 4 I {R ({1,,i}>0} dr i+1n+1 dr kn+1 [ ( Γ 1 ] k i ( Γ m n+i = Γ ( I m n+k {R({1,,i,i+1,,k}>0} m n+i 3 (det(r({1,, i, n+1} (det(r({1,, i, i+1,, k} m n+k 4 (det(r({1,, i, i+1,, k, n+1} m n+k 3 Consequently for 1 i k, (det(r m n 3 I {R>0} dr 1n+1 dr in+1 dr i+11n+1 dr kn+1 [ ( Γ 1 ] k ( Γ m n = Γ ( I m n+k {R({1,,i,i+1,,k}>0} m n (det(r({n + 1} (det(r({1,, i 1, i + 1,, k} m n+k 4 (det(r({1,, i 1, i + 1,, k, n + 1} m n+k 3

19 56 Evelina Veleva Hence Ĵ i = [ ( Γ 1 ] k ( Γ m n s ii Γ ( m n+k (det(r({n + 1} m n (det(r({1,, i 1, i + 1,, k, n + 1} m n+k 3 r in+1 (det(r({1,, i, i+1,, k} m n+k 4 I {R({1,,i,i+1,,k}>0} dr in+1 Let us denote the above integral by H The next Proposition can be found in [10] Proposition Let A be a real symmetric matrix of size n, and let i, j be fixed integers such that 1 i < j n The matrix A is positive definite iff the matrices A({i} and A({j} are positive definite and the element a ij satisfies the inequalities ( j i det(a({i}, {j} 0 det(a({i} det(a({j} det(a({i, j} < a ij < (j i det(a({i}, {j} 0 + det(a({i} det(a({j} det(a({i, j} Let in the integral H we do the substitution r in+1 = (n k+1 det(r({1, i 1, i + 1,, k, n + 1}, {1,, k} 0 det(r({1,, k, n + 1} det(r({1, i 1, i + 1,, k, n + 1} det(r({1,, k} + t det(r({1,, k, n + 1} From Proposition it follows that the new variable t will runs from 1 to 1 Applying Theorem we obtain the representation (1 det(r({1, i 1, i + 1,, k} = det(r({1,, k, n + 1}r in+1 + ( n k+1 det(r({1,, i 1, i + 1,, k, n + 1}, {1,, k} 0 r in+1 + det(r({1,, i 1, i + 1,, k} 0 i,n+1

20 Test for independence of the variables 57 From the Sylvester s determinant identity we have ( det(r({1,, i 1, i + 1,, k} 0 i,n+1 det(r({1,, k, n + 1} = det(r({1,, i 1, i + 1,, k, n + 1} det(r({1,, k} (det(r({1,, i 1, i + 1,, k, n + 1}, {1,, k} 0 Using (1 and ( it can be shown that det(r({1, i 1, i + 1,, k} = det(r({1,, i 1, i + 1,, k, n + 1} det(r({1,, k} (1 t det(r({1,, k, n + 1} The integral H now can be found and we obtain that Ĵ i = [ ( Γ 1 ] k ( s ii ( n k+1 Γ m n Γ ( m n+k det(r({1,, i 1, i + 1,, k, n + 1}, {1,, k} 0 m n (det(r({n + 1} (det(r({1,, k} m n+k 3 (det(r({1,, k, n + 1} m n+k At last it can be easily checked that ˆσ in+1, i = 1,, k, defined by (9, or equivalently by (10, give a solution of the equations (8 To prove that the likelihood function L 1 reaches exactly maxima, one can calculate the Hess matrix of the second derivatives of L 1 with respect to σ in+1, i = 1,, k and ascertain that it is a negative definite matrix for ˆσ in+1, i = 1,, k, defined by (9 We now have to substitute in L 1 the obtained maximum likelihood estimations for the unknown parameters The determinant of the matrix ˆΣ can be found using the next statement Theorem 4 Let A be a symmetric matrix of size n Let the element a 1n satisfies the equality (3 a 1n = (n det(a({1}, {n} 0 det(a({1, n} Then (4 det(a = det(a({1} det(a({n} det(a({1, n}

21 58 Evelina Veleva P r o o f From Theorem it follows that (5 det(a = det(a({1, n}a 1n + ( n det(a({1}, {n} 0 a 1n + det(a 0 1,n From the Sylvester s determinant identity we have (6 det(a 0 1,n det(a({1, n} = det(a({1} det(a({n} (det(a({1}, {n} 0 Applying (6 and the representation (3 for a 1n in (5, we get (4 In the matrix ˆΣ the element ˆσ 1n+1, according to the presentation (9 satisfies the condition (3 Therefore from Theorem 4 it follows that det( ˆΣ = det( ˆΣ({1} det( ˆΣ({n + 1} det( ˆΣ({1, n + 1} Consider the matrix ˆΣ({1} The element ˆσ n+1, according to the presentation (9 satisfies the equality (3 Hence Consequently det( ˆΣ({1} = det( ˆΣ({1, } det( ˆΣ({1, n + 1} det( ˆΣ({1,, n + 1} Proceeding this way, we obtain finally det( ˆΣ = det( ˆΣ({1, } det( ˆΣ({n + 1} det( ˆΣ({1,, n + 1} det( ˆΣ = det( ˆΣ({1,, k} det( ˆΣ({n + 1} det( ˆΣ({1,, k, n + 1} det(r({1,, k} det(r({n + 1} = s 11 s n+1n+1 det(r({1,, k, n + 1} Using the equalities (18, the expression Ξ, defined by (1 can be written in the form n+1 n Ξ = s ii ˆσ ii + n j=i+1 = tr{ ˆΣ ˆΣ } = n + 1 s ij ˆσ ij + n i=k+1 s in+1ˆσ in+1 + k ˆσ in+1ˆσ in+1

22 Test for independence of the variables 59 So for the maxima L 1 of the likelihood function L 1 we obtain L 1 = (m 1 (n+1(m 3 (n+1(m π n(n+1 k 4 Γ ( m n+k (n+1(m e (s 11 s n+1n+1 n Γ ( m i (det(r({1,, k, n + 1} n k+1 (det(r({1,, k} n k+ (det(r({n + 1} n+1 Finally we calculate the likelihood ratio L 0 L 1 ( det(r({1,, k} det(r({n + 1} = det(r({1,, k, n + 1} m Obviously, for the testing of H 0 against H 1 the elements s 11,, s n+1n+1 of the covariance matrix are unnecessary So instead of the covariance matrix S with missing elements we need only the correlation matrix R with the same elements missing Instead of the likelihood ratio we can use the log-likelihood ( ( L log 0 = (m 1 log L 1 det(r({1,, k, n + 1} det(r({1,, k} det(r({n + 1} Then an asymptotic rejection ( region can be given by computing the 1 α quantile n(n + 1 χ of the χ distribution: 1 α; n(n+1 ( det(r({1,, k, n + 1} (m 1 log > χ det(r({1,, k} det(r({n + 1} 1 α; n(n+1 R E F E R E N C E S [1] J Heckman, B Honore The empirical content of the Roy Model Econometrica 58 (1990, [] J Heckman, J Tobias, E Vytlacil Simple Estimators for Treatment Parameters in a Latent Variable Framework Review of Economics and Statistics 85, 3 (003,

23 530 Evelina Veleva [3] S Karlin Total Positivity, Volume 1 Stanford University Press, Stanford, CA, 1968 [4] G Koop, D J Poirier Learning About the Across-Regime Correlation in Switching Regression Models J Econometrics 78 (1997, 17 7 [5] D J Poirier, J L Tobias On the Predictive Distribution of Outcome Gains in the Presence of an Unidentified Parameter J Bus Econom Statist 1 (003, [6] S Sareen Reference Bayesian Inference in nonregular models J Econometrics 113, (003, [7] J L Tobias Estimation, Learning and Parameters of Interest in a Multiple Outcome Selection Model Econometric Rev 5, 1 (006, 1 40 [8] J L Tobias, D J Poirier On the Predictive Distributions of Outcome Gains in the Presence of an Unidentified Parameter J Bus Econom Statist 1, (003, [9] E I Veleva Conditional independence of joint sample correlation coefficients of independent normally distributed random variables Math and Edication in Math 35 (006, [10] E I Veleva Positive definite random matrices Compt Rend Acad Bulg Sci 59, 4 (006, [11] W P M Vijverberg Measuring the Unidentied Parameter of the Extended Roy Model of Selectivity J Econometrics 57 (1993, Rousse University A Kanchev Department of Numerical Methods and Statistics 8 Studentska str, room Rouse, Bulgaria eveleva@abvbg Recieved July 7, 007 Revised December 13, 007