Let X be a sample space, T a test statistic and G be a finite group of transformations (with M elements) from X onto itself. Under the null-hypothesis the distribution of the random variable X is invariant under the transformations in G. Let p ^ = 1 M ∑ g ∈ G I { T ( g X ) ≥ T ( X ) } .Show that P ( p ^ ≤ u ) ≤ u for 0 ≤ u ≤ 1 under the null hypothesis

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Let   X be a sample space,   T a test statistic and   G be a finite group of transformations (with M elements) from   X onto itself. Under the null-hypothesis the distribution of the random variable   X is invariant under the transformations in   G. Let      p    ^    =      1    M        ∑          g      ∈      G            I          {      T      (      g      X      )      ≥      T      (      X      )      }        .Show that   P  (      p    ^    ≤  u  )  ≤  u for   0  ≤  u  ≤  1 under the null hypothesis

Astok3mpd · Accepted Answer

Clearly, the distribution of       p    ^   is supported by   S  :=  {  0  ,      1    M    ,      2    M    ,  …  ,  1  }. Since  P  (      p    ^    ≤  u  )  =  P        (        p    ^    ≤            ⌊      M      u      ⌋        M          )    ,it is enough to prove the statement for   u  ∈  S.To prove the claim, we will use kind of &quot;double counting&quot; argument.Let   u  =      k    M    ∈  S. What does       p    ^    (  x  )  ≤  u mean? It means that the value   T  (  x  ) is among the   k largest values in the orbit       O    x    :=  {  h  x  ,  h  ∈  G  } of   x. That said, for any   x  ∈  X,        |    {  y  ∈      O    x    :      p    ^    (  y  )  ≤  u  }        |    ≤  k  .(In the orbit, there are no more than   k values, which are among the   k largest. There can be less: e.g. in   {  1  ,  1  ,  1  ,  2  ,  2  ,  2  } there are only   3 values which are among the   4 largest.) In other words,      ∑          h      ∈      G              1                      p        ^            (      h      x      )      ≤      u        ≤  k  .Substitute   x  =  X and take the expectation:  k  ≥  E      [          ∑              h        ∈        G                        1                              p          ^                (        h        X        )        ≤        u              ]    =      ∑          h      ∈      G        E      [              1                              p          ^                (        h        X        )        ≤        u              ]      =      ∑          h      ∈      G        P        (        p    ^    (  h  X  )  ≤  u        )    =      ∑          h      ∈      G        P        (        p    ^    (  X  )  ≤  u        )    =  M    P        (        p    ^    (  X  )  ≤  u        )  (in the penultimate equality we have used the invariance of distribution under   G). Equivalently,  P  (      p    ^    (  X  )  ≤  u  )  ≤      k    M    =  u  ,as required.

Let <mi class="MJX-tex-caligraphic" mathvariant="script">X be a sample space, T a test s

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