Page 41 - 2021年第52卷第7期
P. 41

把 x ˉ = u 、 η ˉ = 0 、 x = η + u 代入上式,可得:
                                            t
                                        t
                                     t
                                                      n      n
                                                                    )
                                                     å  η + (u - u η  t
                                                            å
                                                         2
                                                                t
                                                         t
                                                 r =  t = 1  t = 1                                     (4)
                                                      n       2  n     2
                                                             )
                                                     å( x - x ˉ å(η - η ˉ  )
                                                         t
                                                                   t
                                                     t = 1    t = 1
               式中:
                                    n
                                           )
                                                            )
                                                                  u
                                                                                η
                                                                       η
                                   å(u - u η = nE (u η - uη = nE ( )E ( ) - nuE ( ) = 0                (5)
                                                                        t
                                                                                 t
                                                      t
                                                           t
                                                                   t
                                        t
                                                     t
                                             t
                                    t = 1
                   因此,由式(4)可得:
                                                             σ  2
                                                         r =   η                                       (6)
                                                          2
                                                             σ x 2
                     2
               式中 σ 与 σ 分别为 η 和 x 的方差,它们之间具有以下关系:
                           2
                     η
                          x
                                    t
                                        t
                                                      σ  2  = σ  2  + σ  2                             (7)
                                                        x   η    u
               式中 σ  2  为纯随机序列的方差,根据式(6)和式(7)可得:
                    u
                                                              σ  2
                                                        r = 1 -  u                                     (8)
                                                         2
                                                              σ  2
                                                                x
                   将式(1)进行移项并整理得:
                        x - u = φ ( x  t - 1 - u ) + φ ( x  t - 2 - u ) + ⋯ + φ ( x t - p  - u ) + ε - θ ε  t - 1 - θ ε  t - 2 - ⋯ - θ ε t - q  (9)
                                                                       t
                         t
                                                                                             q
                                                           p
                                                                           1
                                1
                                           2
                                                                                  2
                   将式(9)两边同乘以 x - u 并取数学期望得:
                                      t
                                                                                    é
                              2
                      E ( x - u ) = φ E [( x  - u )( x - u  ] ) + φ E [( x  - u )( x - u  ] ) + ⋯ + φ E ( x  - u )( x - u ) ù
                          t       1    t - 1   t       2    t - 2   t            p  ë  t - p  t   û   (10)
                                             ]
                                                                         )
                                          )
                                                                                              )
                                                         )
                                 +E [( x - u ε - θ E [( x - u ε  ] - θ E [( x - u ε  ] - ⋯ - θ E [( x - u ε  ]
                                       t    t   1    t     t - 1  2   t    t - 2     q    t     t - q
                                  2
                   上式两端除以 σ 后考虑到自相关系数的定义:
                                 x
                                               E [( x t - i  - u )( x - u  ] )
                                                           t
                                                                 = ρ ,ρ = 1                           (11)
                                                                   i
                                                      σ x 2            0
                   可得:
                                                           ]
                                                                                            )
                                                                       )
                                                         )
                                                E [( x - u ε - θ E [( x - u ε  ] - ⋯ - θ E [( x - u ε  ]
                        1 - ρ φ - ρ φ - ⋯ - ρ φ =    t    t   1    t     t - 1     q    t     t - q   (12)
                                                                       σ x
                            1  1  2  2     p  p                          2
                   式(12)两端再同乘以 σ 可得:
                                        2
                                        x
                                                             ]
                                             )
                                                                                              )
                                                                         )
                                                          )
                       ( 1 - ρ φ - ρ φ - ⋯ - ρ φ σ = E [( x - u ε - θ E [( x - u ε t - 1 ] - ⋯ - θ E [( x - u ε t - q ]  (13)
                                                2
                                                                                          t
                                                                     t
                                                            t
                                                                                     q
                                            p
                                          p
                                               x
                                                       t
                                                                1
                             1
                           1
                                   2
                                2
                   将式(9)两边同乘以 ε 并取数学期望得:
                                      t
                                    ]
                                 )
                                                                                é
                                                                                          ù
                         E [( x - u ε = φ E [( x  t - 1 - u )ε t ] + φ E [( x  t - 2 - u )ε t ] + ⋯ + φ E ( x t - p  - u ) ε +
                                                                                ë
                                                                                          û
                                                                             p
                                   t
                             t
                                                                                         t
                                                        2
                                       1
                                        ε
                                      E ( ) - θ E (ε ε  t - 1 ) - θ E (ε ε t - 2 ) - ⋯ - θ E ( ε ε t - q ) = E ( ) = σ ε 2  (14)
                                                                                         ε
                                          2
                                                                                          2
                                                                           q
                                                               t
                                                  t
                                         t
                                                                               t
                                                                                          t
                                                          2
                                              1
                   将式(9)两边同乘以 ε          并取数学期望得:
                                      t - 1
                            )
                                                                               é
                    E [( x - u ε t - 1 ] = φ E [( x  t - 1 - u )ε t - 1 ] + φ E [( x t - 2 - u )ε  t - 1 ] + ⋯ + φ E ( x t - p  - u ) ε  t - 1 ù û +
                                                                               ë
                        t
                                                                            p
                                                      2
                                    1
                                E (ε ε t - 1 ) - θ E ( ε  t - 1 ) - θ E (ε  t - 1 ε  t - 2 ) - ⋯ - θ E ( ε  t - 1 ε t - q ) = -θ E ( ε  t - 1 ) = -θ σ ε 2  (15)
                                                2
                                                                                           2
                                                                       q
                                   t
                                           1
                                                                                       1
                                                     2
                                                                                                  1
                   同理,依此类推,直至将式(9)两边同乘以 ε                  t - q  再取数学期望得:
                                )
                                                                                   é
                        E [( x - u ε t - q ] = φ E [( x t - 1 - u )ε t - q ] + φ E [( x t - 2  - u )ε t - q ] + ⋯ + φ E ( x t - p  - u ) ε t - q ù û +
                                                                                   ë
                                                                                p
                            t
                                       1
                                                          2
                           E ( ε ε t - q ) - θ E ( ε  t - 1 ε t - q ) - θ E ( ε  t - 2 ε t - q ) - ⋯ - θ E ( ε t - q ) = -θ E ( ε t - q ) = -θ σ ε 2  (16)
                                                                           2
                                                                                       2
                                                                      q
                                                                                              q
                                                                                  q
                              t
                                                    2
                                      1
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