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Flakqfqbq 2022-05-09 Answered
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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Answers (1)

Astok3mpd
Answered 2022-05-10 Author has 18 answers
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 "double counting" 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.
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