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Appendix to the paper “A goodness-of-fit test for multinomial

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Appendix to the paper “A goodness-of-fit test for multinomial Powered By Docstoc
					  Appendix to the paper: “A goodness-of-fit
     test for multinomial logistic regression”
                        J. J. Goeman                      S. le Cessie

                                       February 8, 2006


    Numbers in brackets refer to equations in the paper.


A     Variance of the test statistic
                                  ˜                               ˜       g   g
We calculate the variance of Q as given in (5). Write Q = ∑s=1 ∑t=1 Qst , where
        n     n   ˜ st (yis − µis )(y jt − µ jt ) for s, t = 1, . . . g. We will calculate the
Qst = ∑i=1 ∑ j=1 Rij
g4 covariances of all Qst terms and sum them to find the variance of Q.               ˜
             st                                      st uv
   Define Sij = (yis − µis )(y jt − µ jt ). Then Cov(Sij , Skl ) = 0 unless i = k and
j = l or i = l and j = k, due to the independence of the samples under the null
hypothesis. Therefore

    Cov( Qst , Quv )   =      ∑ Rii Rii Cov(Sii , Sii ) + ∑ ∑ Rij Rij Cov(Sij , Sij )
                                ˜ st ˜ uv    st uv            ˜ st ˜ uv    st uv
                               i                                 i j =i

                              + ∑ ∑ Rij Ruv Cov(Sij , Suv ).
                                    ˜ st ˜
                                           ji
                                                 st
                                                       ji
                                   i j =i

             st
If i = j, E(Sij ) = 0, for all s and t, so that
      st uv                                                                         su tv
 Cov(Sij , Sij ) = E[(yis − µis )(yiu − µiu )] · E[(y jt − µ jt )(y jv − µ jv )] = Wii Wjj ,

while if i = j,
              st uv           st uv         st     uv        st uv       st uv
         Cov(Sii , Sii ) = E[Sii Sii ] − E[Sii ]E[Sii ] = E[Sii Sii ] − Wii Wii .

Using these expressions,
                                        n                    n    n
           Cov( Qst , Quv )        =   ∑ Rii Rii κistuv + ∑
                                         ˜ st ˜ uv               ∑ Rij Rvu Wii Wjj
                                                                   ˜ st ˜
                                                                          ji
                                                                             su tv
                                       i =1                 i =1 j =1
                                              n   n
                                        +∑        ∑ Rij Ruv Wii Wjj
                                                    ˜ st ˜
                                                           ji
                                                              sv tu
                                            i =1 j =1


                                                      1
                      st uv       st uv        su tv     sv tu
where κistuv = E(Sii Sii ) − Wii Wii − Wii Wii − Wii Wii . It is easy to check
that the value of κistuv does not depend on the order of s, t, u and v. Calculation
of the values of κistuv as given in (6) is straightforward but tedious.
    Taking all covariances of the Qst terms together, we have
                                                              g           g       g       g
                                    ˜
                               Var( Q) =                  ∑ ∑ ∑ ∑ Cov(Qst , Quv ).
                                                          s =1 t =1 u =1 v =1

The result (5) follows by rewriting
                       n         n
                      ∑ ∑ Rij Rvu Wii Wjj = trace( Rst W tv Rvu W su )
                          ˜ st ˜
                                 ji
                                    su tv          ˜        ˜
                     i =1 j =1

and
                 g    g        g          g
                ∑ ∑ ∑ ∑ trace( Rst W tu Ruv W sv ) = trace(RWRW).
                               ˜        ˜                  ˜ ˜
                s =1 t =1 u =1 v =1


B     Derivation of the test statistic
We derive the expression (9) for the score test statistic from the random ef-
fects model (7). The likelihood L(τ 2 ) = exp{ (τ 2 )} can be written L(τ 2 ) =
      n                                                             g
Ez ∏i=1 f i (z) , where f i (r) = exp{li (z)} and li (z) = ∑s=1 yis log{νis (z)};
compare (8). Note that f i (z) only depends on (zi1 , . . . , zig ). Therefore, Taylor
expanding L(τ 2 ) with respect to z at z = 0 gives
                                      n                               g       n
                                                                                          ∂ f i (0)
      L(τ 2 )    = Ez                ∏ fi (0) + ∑ ∑ zis                                    ∂zis       ∏ f j (0)
                                     i =1                         s =1 i =1                           j =i
                                       g          g       n
                                 1                    i (0)                   ∂2 f
                                 2 s∑ t∑ i∑ is it ∂zis ∂zit                                    ∏ f j (0)
                          +                  z z
                                    =1 =1 =1                                                   j =i
                                       g       n  g
                                 1                      ∂ f i (0) ∂ f j (0)
                          +        ∑ ∑ ∑ ∑ zis z jt ∂zis ∂z jt
                                 2 s =1 t =1 i =1 j = i                                                      ∏        f k (0) + o (zz )
                                                                                                             k =i,j
                           n                                          g       g       n
                                                      1                                         ∂2 f ( 0 )
                 =        ∏ fi (0) + 2 τ2 ∑ ∑ ∑ Rii ∂zisi∂zit ∏ f j (0)
                                                 st
                          i =1                                    s =1 t =1 i =1                             j =i
                                              g       g           n
                           1                      st ∂ f ( 0 )
                                                               ∂ f j (0)
                          + τ 2 ∑ ∑ ∑ ∑ Rij i
                                                      ∂zis ∂z jt k∏ k
                                                                              f ( 0 ) + o ( τ 2 ).
                           2 s =1 t =1 i =1 j = i                        =i,j




                                                                                  2
      ∂ f (z)
         i            i     ∂l (z)        ∂2 f i ( z )                     ∂2 li ( z )         ∂li (z) ∂li (z)
Using ∂z = f i (z) ∂z and                 ∂zis ∂zit      = f i (z)         ∂zis ∂zit     +      ∂zis ∂zit        , this expres-
           is           is
sion can be rewritten to
                                n                              g
                                                               n       g
                                                 1                st ∂l ( 0 )
            L(τ 2 )     =      ∏      f i (0) 1 + τ 2 ∑ ∑ ∑ Rii i
                                                 2 s =1 t =1 i =1   ∂zis ∂zit
                               i =1
                                     g g n n
                                1                                ∂l (0)
                                                      st ∂l ( 0 ) j
                               + τ 2 ∑ ∑ ∑ ∑ Rij i                      + o (τ 2 )
                                2 s =1 t =1 i =1 j =1     ∂zis ∂z jt

          ∂ (0)        1 ∂L(0)
Because    ∂τ 2
                  =   L(0) ∂τ 2
                                ,     the score function at τ 2 = 0 is

                         g     g      n                            g        g     n        n
      ∂ (0)   1                              ∂2 l ( 0 )                                                ∂li (0) ∂l j (0)
       ∂τ 2
            =
              2         ∑ ∑ ∑ Rii ∂zisi∂zit + ∑ ∑ ∑ ∑ Rij
                               st                      st
                                                                                                        ∂zis ∂z jt
                        s =1 t =1 i =1                         s =1 t =1 i =1 j =1

                                           ∂li (0)                                       ∂2 li ( 0 )       st
   The result (9) follows from              ∂zis         = yis − µis and                 ∂zis zit      = −Wii .




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