Asymptotic Methods in Statistical Inference - Statistics Centre

75
To see this take 0 = 0 for simplicity. Let be the
( 1 1) d.f. and the ( 2 1) d.f. Let =(1−
)+ for ∈ [0 1]; require 1 and 2 to satisfy
(1 − ) 1 + 2 =0 (8.1)
Then ∈ F. Represent the rejection region as ‘X ∈
’, then the level is
( ) = (X ∈ )
=
≥
Z
Y
=1
(1 − ) Z
{(1 − )( )+( )} 1 ···
Y
=1
( ) 1 ···
= (1− ) (X ∈ )
= (1− ) ³√
¯ ≥ ´
For any this may be made arbitrarily near 1 by choosing
sufficiently small, 1 sufficiently large (how?),
and 2 = −(1 − ) 1 to satisfy (8.1).
That inf ( ) = 0 may be shown similarly, by replacing
( )by1− ( )and by its complement,
and then proceeding as above to obtain
1 − ( ) ≥ (1 − ) ³ √
¯ ´
(Thanks to J. Sheahan for this second part.)