Derive the Cramer von Mises test statistic n C n = 1 12 n + ∑ i = 1 n ( U ( i ) − 2 i − 1 2 n ) 2 where U ( i ) = F 0 ( X ( i ) ) the order statistics

Question

Derive the Cramer von Mises test statistic  n      C          n        =      1          12      n        +      ∑          i      =      1              n                  (              U                  (          i          )                    −                        2          i          −          1                          2          n                    )        2  where       U          (      i      )        =      F          0        (      X          (      i      )        ) the order statistics

Patrick Arnold · Accepted Answer

Let       X    1    ,  …  ,      X    n   be a random sample of size   n from   f  (  x  ). The empirical CDF is                    F        ^              n    (  x  )  =      1    n        ∑          i      =      1        n    I  (      X    i    ≤  x  )  =      {                            0          ,                          x          &amp;lt;                      X                          1              :              n                                ,                                      i                      /                    n          ,                                      X                          i              :              n                                ≤          x          &amp;lt;                      X                          i              +              1              :              n                                ,                                      1          ,                                      X                          n              :              n                                ≤          x          .                        Let       U    i    =      F    0    (      X          i      :      n        ), then                              C          n                                    ≜        ∫                              [                                                            F                  ^                                            n                        (            x            )            −                          F              0                        (            x            )            ]                    2                                  d                                      F            0                    (          x          )                                                  =                  ∫                      −            ∞                                              X                              1                :                n                                                                          [            −                          F              0                        (            x            )            ]                    2                                  d                                      F            0                    (          x          )                +                  ∑                      i            =            1                                n            −            1                                    ∫                                    X                              i                :                n                                                                        X                              i                +                1                :                n                                                                          [                          i              n                        −                          F              0                        (            x            )            ]                    2                                  d                                      F            0                    (          x          )                +                  ∫                                    X                              n                :                n                                                          ∞                                                [            1            −                          F              0                        (            x            )            ]                    2                                  d                                      F            0                    (          x          )                                                  =                  1          3                          U          1          3                +                  1          3                          ∑                      i            =            1                                n            −            1                                    [                                    (                              U                                  i                  +                  1                                            −                              i                n                            )                        3                    −                                    (                              U                                  i                                            −                              i                n                            )                        3                    ]                −                  1          3                (                  U          n                −        1                  )          3                                                  =                  1          3                          U          1          3                +                  1          3                          ∑                      i            =            1                                n            −            1                                    [                      U                          i              +              1                        3                    −                      U                          i                        3                    +                                    3                              i                2                                                    n              2                                            (                          U                              i                +                1                                      −                          U              i                        )                    −                                    3              i                        n                                (                          U                              i                +                1                            2                        −                          U              i              2                        )                    ]                −                  1          3                (                  U          n                −        1                  )          3                                                  =                  1          3                          U          1          3                +                  1          3                          U          n          3                −                  1          3                          U          1          3                +                  (                      U            n                    −                      ∑                          i              =              1                        n                                              2              i              −              1                                      n              2                                            U            i                    )                −                  (                      U            n            2                    −                      ∑                          i              =              1                        n                                1            n                                U            i            2                    )                −                  1          3                (                  U          n                −        1                  )          3                                                  =                  1          3                +                  1          n                          ∑                      i            =            1                    n                          (                      U            i            2                    −                                    2              i              −              1                        n                                U            i                    )                                                  =                  1          3                +                  1          n                          ∑                      i            =            1                    n                                      (                          U              i                        −                                          2                i                −                1                                            2                n                                      )                    2                −                  1          n                          ∑                      i            =            1                    n                                      (                                          2                i                −                1                                            2                n                                      )                    2                                                  =                  1                      12                          n              2                                      +                  1          n                          ∑                      i            =            1                    n                                      (                          U              i                        −                                          2                i                −                1                                            2                n                                      )                    2                .                          [          by                                 ∑                          i              =              1                        n                    (          2          i          −          1                      )            2                    =                      1            3                    n          (          4                      n            2                    −          1          )          ]

Derive the Cramer von Mises test statistic nC_n=1/12n+sum_i=1^n(U_(i)−2i−1/2n)^2 where U_(i)=F_0(X_(i)) the order statistics

Answered question

Answer & Explanation

New Questions in College Statistics