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8. Tests of Hypotheses  403

                           of ψ in (8.3.2). Also for such x one obviously has ψ(x) - ψ*(x) ≤ 0 since
                           ψ*(x) ∈ (0, 1). Again (8.3.4) is validated.. Now, if x ∈ χ  is such that 0 <
                                                                             n
                           ψ(x) < 1, then from (8.3.2) we must have L(x; θ ) - kL(x; θ ) = 0, and
                                                                       1
                                                                                  0
                           again (8.3.4) is validated. That is, (8.3.4) surely holds for all x ∈ θ . Hence
                                                                                    n
                           we have













                           Now recall that Q(θ ) is the Type I error probability associated with the test
                                            0
                           ψ defined in (8.3.2) and thus Q(θ ) = α from (8.3.3). Also, Q*(θ ) is the
                                                                                     0
                                                         0
                           similar entity associated with the test ψ* which is assumed to have the level
                           α, that is Q*(θ ) ≤ α. Thus, Q(θ ) – Q*(θ ) ≥ 0 and hence we can rewrite
                                                               0
                                                       0
                                        0
                           (8.3.5) as
                           which shows that Q(θ ) = Q*(θ ). Hence, the test associated with ψ is at least
                                             1
                                                     1
                           as powerful as the one associated with ψ*. But, ψ* is any arbitrary level a test
                           to begin with. The proof is now complete.¾
                              Remark 8.3.1 Observe that the Neyman-Pearson Lemma rejects H  in
                                                                                        0
                           favor of accepting H  provided that the ratio of two likelihoods under H  and
                                                                                       1
                                            1
                           H  is sufficiently large, that is if and only if L(X; θ )/L(X; θ ) > k for some
                                                                       1
                                                                              0
                            0
                           suitable k(≥ 0).
                                   Convention: The ratio c/0 is interpreted as infinity if c > 0
                                                    and one if c = 0.
                              Remark 8.3.2 In the statement of the Neyman-Pearson Lemma, note
                           that nothing has been said about the data points x which satisfy the equa-
                           tion L(x; θ ) = kL(x; θ ). First, if the X’s happen to be continuous random
                                    1
                                              0
                           variables, then the set of such points X would have the probability zero. In
                           other words, by not specifying exactly what to do when L(x; θ ) = kL(x;
                                                                                   1
                           θ ) in the continuous case amounts to nothing serious in practice. In a
                            0
                           discrete case, however, one needs to randomize on the set of X’s for
                           which L(x; θ ) = kL(x; θ ) holds so that the MP test has the size
                                        1          0
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