On the product of a singular Wishart matrix and a singular Gaussian - PowerPoint PPT Presentation

On the product of a singular Wishart matrix and a singular Gaussian vector in high dimension. Stanislas Muhinyuza Department of Mathematics, University of Rwanda Department of Mathematics, Stockholm University First Network Meeting for Sida- and ISP-funded PhD Students in Mathematics Stockholm 7–8 March 2017 1 / 13

My Advisors Taras Bodnar Marcel R.Ndengo Main advisor Assistant advisor Stockholm University University of Rwanda 2 / 13

Recall Asymptotics standard asymptotics 3 / 13

Recall Asymptotics standard asymptotics fixed dimension k and the sample size n goes to infinity; 3 / 13

Recall Asymptotics standard asymptotics fixed dimension k and the sample size n goes to infinity; classical limit theorems hold 3 / 13

Recall Asymptotics standard asymptotics fixed dimension k and the sample size n goes to infinity; classical limit theorems hold Large dimensional asymptotics (Asymptotic distribution under double asymptotic regime) 3 / 13

Recall Asymptotics standard asymptotics fixed dimension k and the sample size n goes to infinity; classical limit theorems hold Large dimensional asymptotics (Asymptotic distribution under double asymptotic regime) both the dimension k and the sample size n goes to infinity; 3 / 13

Recall Asymptotics standard asymptotics fixed dimension k and the sample size n goes to infinity; classical limit theorems hold Large dimensional asymptotics (Asymptotic distribution under double asymptotic regime) both the dimension k and the sample size n goes to infinity; the ratio k / n tends to a positive constant c > 0; 3 / 13

Recall Asymptotics standard asymptotics fixed dimension k and the sample size n goes to infinity; classical limit theorems hold Large dimensional asymptotics (Asymptotic distribution under double asymptotic regime) both the dimension k and the sample size n goes to infinity; the ratio k / n tends to a positive constant c > 0; classical limit theorems do not hold anymore. 3 / 13

Introduction Let X = ( X 1 , . . . , X n ) be a sample of size n from k -variate normal distribution, That is X i ∼ N k ( 0 , Σ ) for i = 1 , . . . , n . 4 / 13

Introduction Let X = ( X 1 , . . . , X n ) be a sample of size n from k -variate normal distribution, That is X i ∼ N k ( 0 , Σ ) for i = 1 , . . . , n . A = XX ′ ∼ W k ( n , Σ ); k ≤ n , Wishart distribution . The properties of A are detailed in [Muirhead, 1982]. 4 / 13

Introduction Let X = ( X 1 , . . . , X n ) be a sample of size n from k -variate normal distribution, That is X i ∼ N k ( 0 , Σ ) for i = 1 , . . . , n . A = XX ′ ∼ W k ( n , Σ ); k ≤ n , Wishart distribution . The properties of A are detailed in [Muirhead, 1982]. [Srivastava, 2003] defined A = XX ′ ∼ W k ( n , Σ ), k > n as a singular Wishart matrix. 4 / 13

Introduction Let X = ( X 1 , . . . , X n ) be a sample of size n from k -variate normal distribution, That is X i ∼ N k ( 0 , Σ ) for i = 1 , . . . , n . A = XX ′ ∼ W k ( n , Σ ); k ≤ n , Wishart distribution . The properties of A are detailed in [Muirhead, 1982]. [Srivastava, 2003] defined A = XX ′ ∼ W k ( n , Σ ), k > n as a singular Wishart matrix. Singular Gaussian vector is the normal distributed vector with a singular covariance matrix. 4 / 13

Introduction Let X = ( X 1 , . . . , X n ) be a sample of size n from k -variate normal distribution, That is X i ∼ N k ( 0 , Σ ) for i = 1 , . . . , n . A = XX ′ ∼ W k ( n , Σ ); k ≤ n , Wishart distribution . The properties of A are detailed in [Muirhead, 1982]. [Srivastava, 2003] defined A = XX ′ ∼ W k ( n , Σ ), k > n as a singular Wishart matrix. Singular Gaussian vector is the normal distributed vector with a singular covariance matrix. Wishart distributed matrix does not usually appear alone, but in combination with a normal random vector, its properties are detailed in [Bodnar et al., 2013]. 4 / 13

Introduction Example of application of the product The weights of the tangency portfolio are computed as products of the inverse sample covariance matrix multiplied by the sample mean vector using historical asset returns. 5 / 13

Introduction Example of application of the product The weights of the tangency portfolio are computed as products of the inverse sample covariance matrix multiplied by the sample mean vector using historical asset returns. Assumptions z ∼ N k ( µ , κ Σ ), κ > 0 with rank ( Σ ) = r < k ; 5 / 13

Introduction Example of application of the product The weights of the tangency portfolio are computed as products of the inverse sample covariance matrix multiplied by the sample mean vector using historical asset returns. Assumptions z ∼ N k ( µ , κ Σ ), κ > 0 with rank ( Σ ) = r < k ; A ∼ W k ( n , Σ ) with rank ( Σ ) = r < k ; 5 / 13

Introduction Example of application of the product The weights of the tangency portfolio are computed as products of the inverse sample covariance matrix multiplied by the sample mean vector using historical asset returns. Assumptions z ∼ N k ( µ , κ Σ ), κ > 0 with rank ( Σ ) = r < k ; A ∼ W k ( n , Σ ) with rank ( Σ ) = r < k ; M : p × k matrix of constants with rank ( M ) = p ≤ r ≤ min { n , k } such that MΣ � = 0 ; 5 / 13

Introduction Example of application of the product The weights of the tangency portfolio are computed as products of the inverse sample covariance matrix multiplied by the sample mean vector using historical asset returns. Assumptions z ∼ N k ( µ , κ Σ ), κ > 0 with rank ( Σ ) = r < k ; A ∼ W k ( n , Σ ) with rank ( Σ ) = r < k ; M : p × k matrix of constants with rank ( M ) = p ≤ r ≤ min { n , k } such that MΣ � = 0 ; A and z are independent. 5 / 13

Introduction Example of application of the product The weights of the tangency portfolio are computed as products of the inverse sample covariance matrix multiplied by the sample mean vector using historical asset returns. Assumptions z ∼ N k ( µ , κ Σ ), κ > 0 with rank ( Σ ) = r < k ; A ∼ W k ( n , Σ ) with rank ( Σ ) = r < k ; M : p × k matrix of constants with rank ( M ) = p ≤ r ≤ min { n , k } such that MΣ � = 0 ; A and z are independent. Aim: Distribution of MAz 5 / 13

Stochastic representation Theorem 1[Bodnar et al., 2016] 1 Let P ( rank (( M T , z ) T Σ ) = p + 1 ≤ r ) = 1 , and let Q = P T P with P = ( MΣM T ) − 1 / 2 MΣ 1 / 2 . 6 / 13

Stochastic representation Theorem 1[Bodnar et al., 2016] 1 Let P ( rank (( M T , z ) T Σ ) = p + 1 ≤ r ) = 1 , and let Q = P T P with P = ( MΣM T ) − 1 / 2 MΣ 1 / 2 .Then d ζ MΣ 1 / 2 t + � ζ ( MΣM T ) 1 / 2 = MAz √ � √ � � t T t − t T ( I k − Q ) t Ptt T P T × t T tI p − z 0 , t T Qt 6 / 13

Stochastic representation Theorem 1[Bodnar et al., 2016] 1 Let P ( rank (( M T , z ) T Σ ) = p + 1 ≤ r ) = 1 , and let Q = P T P with P = ( MΣM T ) − 1 / 2 MΣ 1 / 2 .Then d ζ MΣ 1 / 2 t + � ζ ( MΣM T ) 1 / 2 = MAz √ � √ � � t T t − t T ( I k − Q ) t Ptt T P T × t T tI p − z 0 , t T Qt where ζ ∼ χ 2 n , t ∼ N k ( Σ 1 / 2 µ , κ Σ 2 ), and z 0 ∼ N p ( 0 , I p ); ζ , t , and z 0 are mutually independent. 6 / 13

Stochastic representation Theorem 1[Bodnar et al., 2016] 1 Let P ( rank (( M T , z ) T Σ ) = p + 1 ≤ r ) = 1 , and let Q = P T P with P = ( MΣM T ) − 1 / 2 MΣ 1 / 2 .Then d ζ MΣ 1 / 2 t + � ζ ( MΣM T ) 1 / 2 = MAz √ � √ � � t T t − t T ( I k − Q ) t Ptt T P T × t T tI p − z 0 , t T Qt where ζ ∼ χ 2 n , t ∼ N k ( Σ 1 / 2 µ , κ Σ 2 ), and z 0 ∼ N p ( 0 , I p ); ζ , t , and z 0 are mutually independent. 2 Let m be a k -dimensional vector of constants such that m T Σm > 0 and P ( z T Σz = 0) = 0. 6 / 13

Stochastic representation Theorem 1[Bodnar et al., 2016] 1 Let P ( rank (( M T , z ) T Σ ) = p + 1 ≤ r ) = 1 , and let Q = P T P with P = ( MΣM T ) − 1 / 2 MΣ 1 / 2 .Then d ζ MΣ 1 / 2 t + � ζ ( MΣM T ) 1 / 2 = MAz √ � √ � � t T t − t T ( I k − Q ) t Ptt T P T × t T tI p − z 0 , t T Qt where ζ ∼ χ 2 n , t ∼ N k ( Σ 1 / 2 µ , κ Σ 2 ), and z 0 ∼ N p ( 0 , I p ); ζ , t , and z 0 are mutually independent. 2 Let m be a k -dimensional vector of constants such that m T Σm > 0 and P ( z T Σz = 0) = 0. Then z T Σz · m T Σm − ( m T Σz ) 2 � 1 / 2 z 0 , d m T Az = ζ m T Σz + � � ζ 6 / 13

Stochastic representation Theorem 1[Bodnar et al., 2016] 1 Let P ( rank (( M T , z ) T Σ ) = p + 1 ≤ r ) = 1 , and let Q = P T P with P = ( MΣM T ) − 1 / 2 MΣ 1 / 2 .Then d ζ MΣ 1 / 2 t + � ζ ( MΣM T ) 1 / 2 = MAz √ � √ � � t T t − t T ( I k − Q ) t Ptt T P T × t T tI p − z 0 , t T Qt where ζ ∼ χ 2 n , t ∼ N k ( Σ 1 / 2 µ , κ Σ 2 ), and z 0 ∼ N p ( 0 , I p ); ζ , t , and z 0 are mutually independent. 2 Let m be a k -dimensional vector of constants such that m T Σm > 0 and P ( z T Σz = 0) = 0. Then z T Σz · m T Σm − ( m T Σz ) 2 � 1 / 2 z 0 , d m T Az = ζ m T Σz + � � ζ where ζ ∼ χ 2 n and z 0 ∼ N (0 , 1); ζ , z 0 , and z are mutually indep. 6 / 13