Page 107 - 2022年第53卷第8期
P. 107

一步概率转移矩阵式(16)和(17)
                                         H(u,v) =P(X = k,A = vX= m,A= u)                               (15)
                                          t           t + 1   t + 1  t      t
                                           λ (1 - β ) + [1 - λ (1 - β )] π 0  (1 - β )(1 - λ ) π 1
                                      H =                                                              (16)
                                       0
                                                  β (1 - λ ) π 0         βλπ 0
                                                βλπ 1             β (1 - λ ) π 1
                                      H =                                                              (17)
                                        1
                                                             (
                                           (1 - β )(1 - λ ) π 0 λ 1 - β ) + [1 - λ (1 - β )] π 1
              3.3 DAR(1)和 DARMA(1,1)模型检验 DAR(1)和 DARMA(1,1)模型检验主要包括 ACF、干湿
              游程的理论值与样本值拟合效果评估。
                  (1)自相关系数 ACF检验。样本经验自相关函数是根据干日和湿日的序列计算的,即 0和 1序
              列  [26] ,计算公式为
                                               N- k                N
                                                                             - 1
                                                                           2
                                           k ∑
                                                      x)(x - 珋]∑
                                                                         x)
                                           r= [   (x - 珋  t + k  x)  [  (x - 珋 ]                       (18)
                                                                       t
                                                    t
                                               t =1               t =1
                                                   x为x序列的均值。
              式中:x为 0或 1序列;N为样本容量; 珋
                     t                                  t
                  将通过式(18)计算的样本序列 ACF值与式(5)和式(13)计算的理论 ACF值进行对比分析检验。
                  (2)干湿游程检验。游程定义为同一类型事件的持续,它在开始和结束时由其他类型的事件界定。
              在模拟日降水序列时,DAR(1)的理论干湿游程长度数学表示为
                                P(T = t) =P(X = 0 ,X = 1 ,…,X= 1 ,X = 0 X = 0 ,X = 1 )                 (19)
                                                                                     1
                                                                t
                                                    1
                                                                      t + 1
                                             0
                                                                              0
                                   1
                                P(T = t) =P(X = 1,X = 0,…,X= 0,X = 1 X = 0,X = 1)                      (20)
                                                    1
                                                                t
                                                                      t + 1
                                                                              0
                                   0
                                                                                     1
                                             0
                  采用 DAR(1)模型的一步转移概率矩阵简化游程计算,可得理论干湿游程长度概率分布为
                                                        t - 1
                                             P(T = t) =p (1,1)[1 - p(1,1)]                             (21)
                                                 1
                                                        t - 1
                                             P(T = t) =p (0,0)[1 - p(0,0)]                             (22)
                                                 0
                  DARMA(1,1)的理论湿游程长度数学表示为
                                P(T = t) =P(X = 0 ,X = 1 ,…,X= 1 ,X = 0X = 0 ,X = 1 )
                                    1        0      1           t     t + 1   0      1
                                          P(X = 0 ,X = 1 ,…,X= 1 ,X = 0 )                              (23)
                                             0
                                                                t
                                                                      t + 1
                                                    1
                                        =
                                                   P(X = 0,X = 1)
                                                              1
                                                       0
                  由X与A构成的一阶二元 Markov链可得
                          t
                      t
                            P(X = 0,X = 1) =P(X= 0,A= 0)P(X = 1,A = 0A= 0)
                                0      1         t      t        t + 1   t + 1  t
                                             + P(X= 0,A= 0)P(X = 1,A = 1A= 0)
                                                  t      t        t + 1   t + 1  t
                                             + P(X= 0 ,A= 1 )P(X = 1 ,A = 0A= 1 )
                                                  t      t        t + 1   t + 1  t
                                                                                                       (24)
                                             + P(X= 0 ,A= 1 )P(X = 1 ,A = 1A= 1 )
                                                  t      t        t + 1   t + 1  t
                                                                      ][H(0,0) + H(0,1)]
                                                0           0             1         1
                                            =[H(0,0) π 0  + H(1,0) π 1
                                                                       ][H(1,0) + H(1,1)]
                                             + [H(0,1) π 0 + H(1,1) π 1
                                                 0           0             1         1
                               P(X = 0 ,X = 1 ,…,X= 1 ,X = 0 )
                                                      t
                                          1
                                   0
                                                            t + 1
                                                                          n
                                                                                n + 1
                                                                      ][H(0) - H (0)]                  (25)
                                                0           0             1     1
                                            =[H(0,0) π 0  + H(1,0) π 1
                                                                                   n + 1
                                                                           n
                                                                       ][H (1) - H (1)]
                                            + [H(0,1) π 0 + H(1,1) π 1    1        1
                                                 0
                                                             0
                  化简式( 24)和(25)可得 DARMA(1,1)的理论湿游程公式为:
                                            1            1            1             1
                          P(X = 0 ,X = 1 ) = ∑  H(i,0) π i ∑  H(0,j) +  H(m,1) π m ∑   H(1,n)          (26)
                                                                     ∑
                              0      1          0            1           0              1
                                           i =0          j =0        m=0            n =0
                            P(X = 0 ,X = 1 ,…,X= 1 ,X = 0 )
                                                         t + 1
                                       1
                                                  t
                                0
                                                                                                       (27)
                                                                                n
                                                                                       n + 1
                                                     n
                                                            n + 1
                                        =P[W = 0 ][H(0) - H (0)] + P[W = 1 ][H(1) - H (1)]
                                              0      1      1           0       1      1
                                                                                                     n
                                                                                                 ];H(j) =
                         0        0           0                 0         0           0              1
              式中:P[W = 0] = [H(0,0) π 0     + H(1,0) π 1 ];P[W = 1] = [H(0,1) π 0   + H(1,1) π 1
                n
                         n
              H(j,0) + H(j,1),j = 0 ,1
                1        1
                     4
                —  9 9  —
   102   103   104   105   106   107   108   109   110   111   112