INDIAN INSTITUTE OF SCIENCE STOCHASTIC HYDROLOGY Lecture -35 Course Instructor : Prof. P. P. MUJUMDAR Department of Civil Engg., IISc.
Summary of the previous lecture Multivariate stochastic models Multisite Markov model X EX Gε = + t+ t where X t is a p x vector of standardized values of the variable generated at time t, E is a p x p diagonal matrix whose j th diagonal element is ρ j (), G is a p x p diagonal matrix whose j th diagonal element is 2 ρ j ε is a p x vector of random variates Matalas model X = AX + Bε t+ t t+ 2
Multivariate Stochastic models Matalas (967) has given a multisite normal generation model that preserves the mean, variance, lag one serial correlation, lag one crosscorrelation and lag zero cross-correlation. X = AX + Bε where X t and X t+ are p x vectors representing standardized data corresponding to p sites at time steps t and t+ resp. t+ t t+ Assumption is that the model is multivariate normal. Ref.: Matalas, N.C. (967) Mathematical assessment of synthetic hydrology, Water Resources Research 3(4):937-945 3
Multivariate Stochastic models ε t+ is N(0,); p x vector with ε t+ independent of X t. A and B are coefficient matrices of size p x p. B is assumed to be lower triangular matrix M = E X X 0 t t M 0 is the cross-correlation matrix (size pxp) of lag zero m 0 ( i, j) Q Q Q Q = n s s n it, i jt, j t= i j 4
Multivariate Stochastic models The expectation of X t X t- is denoted by M M = E X X t t If m (i, j) is a element of M matrix (size p x p) in the i th row and j th column, m i, j = E X i, t X j, t n (, ) = x( i, t) x( j, t ) m i j n t= 2 Expected value of a matrix is matrix of expected values of individual elements n is no. of time periods 5
Multivariate Stochastic models (, ) m i j Q Q Q Q = n s s n it, i jt, j t= 2 i j Q is the original random variable before standardization e.g., stream flow i.e., m (i, j) represents lag one cross correlation between the data at sites i and j. Therefore M is the cross-correlation matrix of lag one. 6
Multivariate Stochastic models Considering the model, X = AX + Bε t+ t t+ Post multiplying with X t on both sides and taking the expectation,. E Xt X t AE XtXt BE ε + = + t+ X t M A 0 = = AM + M M 0 0 0 ε t+ and X t are independent 7
Multivariate Stochastic models Post multiplying with X t+ on both sides and taking the expectation,. X = AX + Bε t+ t t+ X X = AX X + Bε X t+ t+ t t+ t+ t+ E Xt X t AE XtXt BE ε + + = + + t+ X t+ M 0 8
Multivariate Stochastic models M M = E X X t t { } E XtX t = { } = E XtX t = E Xt Xt or M = E XtX t + 9
Multivariate Stochastic models ε X = ε { AX + Bε } t+ t+ t+ t t+ ε = XA+ ε ε t+ t t+ t+ Taking expectation on both sides, E εt X t E εt XtA ε ε + + = + + t+ t+ B B = E εt XtA E ε ε + + t+ t+ B = 0 + IB = B Since ε t+ has unit variance 0
Multivariate Stochastic models Substituting in the equation, E Xt X t AE XtXt BE ε + + = + + t+ X t+ M = AM + BB 0 M = M M M + BB 0 0 A = M M 0 BB = M M M M 0 0 If C = BB C = M M M M 0 0
Multivariate Stochastic models B does not have a unique solution. One method is to assume B to be a lower triangular matrix. BB b, 0 0.. 0 b, b 2,... b p, b 2, b 2, 2 0.. 0 0 b 2, 2... b p, 2 =................ b( p, ) b( p,2 )... b( p, p) 0 0... b( p, p) (,) (, 2) (,3 ).. (, ) ( 2,) ( 2, 2) ( 2,3 ).. ( 2, ) c c c c p c c c c p C =............ c( p, ) c( p,2 )... c( p, p) 2
Multivariate Stochastic models The diagonal elements of the B matrix are obtained as, b, = c, 2 2 2 { } b 2, 2 = c 2, 2 b 2,... These elements are obtained one by one, using also the expressions for the k th row elements given in the next slide (, ) = (, ) (, ) (, 2 )... (,) { } bkk ckk b kk b kk b k 2 2 2 2 3
Multivariate Stochastic models The elements in the k th row are obtained as, (, ) b k j b k, b k,2... = = c k, b, c k,2 b 2, b k, b 2, 2 (, ) (,) (,) (,2) (,2)... (, ) (, ) b( j, j) c k j b j b k b j b k b j j b k j = 4
Example The annual flow in MCM at two sites P and Q is given below. Generate the first two values of data from these two sites. Year 2 3 4 5 6 7 8 9 0 Annual flow at site P (MCM) Annual flow at site Q (MCM) 4946 707 6653 6355 5908 5327 4548 3556 3852 539 542 6240 5648 5977 6008 5045 4630 4604 4250 682 Year 2 3 4 5 6 7 8 9 Annual flow at site P (MCM) Annual flow at site Q (MCM) 463 5746 5 549 6060 7336 3736 3780 6034 4703 6582 546 5288 5440 7546 4634 4823 5577 5
Example (Contd.) Site P Q Mean 5333 5462 Std.dev. 25. 823.5 M 0 matrix is cross correlation matrix of lag zero P Q M 0 P r PP, 0 rpq, 0 = Q r QP, 0 rqq, 0 6
Example (Contd.) r PQ, 0 = n i= ( x ) Pi, xp xqi, xq n s s P Q M 0 0.796 = 0.796 M matrix is cross correlation matrix of lag one P Q M P r PP, rpq, = Q r QP, rqq, 7
Example (Contd.) r PQ, = n i= ( x ) Pi, xp xqi, + xq ( n ) s s P Q M 0.302 0.64 = 0.02 0.8 A = M M 0 M 0 2.73 2.7 = 2.7 2.73 8
Example (Contd.) A = M M 0 0.302 0.64 2.73 2.7 = 0.02 0.8 2.7 2.73 A 0.47 0.2 = 0.3 0.37 9
Example (Contd.) C = M M M M 0 0 M 0 M 0 0.796 = 0.796 2.73 2.7 = 2.7 2.73 M M 0.302 0.64 = 0.02 0.8 0.302 0.02 = 0.64 0.8 C 0.89 0.76 = 0.76 0.95 20
Example (Contd.) c, = 0.89, c, 2 = c 2, = 0.76, c b b 2, 2 = 0.95 2 2, = c, = 0.89 = 0.94 c( 2,) 0.76 2, = = = 0.8 b, 0.94 { } { } 2 2 b 2, 2 = c 2, 2 b 2, 2 2 = 0.95 0.8 = 0.54 (,) b k = (,) (,) c k b 2
Example (Contd.) b, = 0.94, b 2, = 0.8, b 2, 2 = 0.54 B 0.94 0 = 0.8 0.54 X = AX + Bε t+ t t+ xpt, + 0.47 0.2 xpt, 0.94 0 ε Pt, + x = 0.3 0.37 + x 0.8 0.54 ε Qt, + Qt, Qt, + 22
Example (Contd.) xpt, + 0.47 0.2 xpt, 0.94 0 ε Pt, + x = 0.3 0.37 + x 0.8 0.54 ε Qt, + Qt, Qt, + The initial value of x P,0 and x Q,0 are considered as zero ε P, = -0.34 and ε Q, = -0.268 xp, 0.94 0 0.34 x = Q, 0.8 0.54 0.268 0.26 = 0.254 23
Example (Contd.) xp,2 0.47 0.2 xp, 0.94 0 ε P,2 x = 0.3 0.37 + x 0.8 0.54 ε Q, Q, Q,2 ε P,2 =.639 and ε Q,2 = 0.34 xp,2 0.47 0.2 0.26 0.94 0.639 x = Q, 0.3 0.37 0.254 + 0.8 0.54 0.34.543 =.449 24
Example (Contd.) Generated annual flow at Site X : Q x x X s X, = X +, X Q X, = 5333 0.26*25. = 59.2 MCM Q x x X s X,2 = X +,2 X Q X,2 = 5333 +.543*25. = 7069 MCM Similarly at Site Y : Q Y, = 5462 0.254*823.5 = 5252.8 MCM Q Y,2 = 5462 +.449*823.5 = 6655.3 MCM Site X Y Mean 5333 5462 Std.dev. 25. 823.5 25
CASE STUDIES 26