Page 450 - Probability and Statistical Inference
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8. Tests of Hypotheses 427
ψ* is given by
0
Of course, the test function ψ* corresponds to a level α test for H 0
versus H . Let us now compute the power function associated with ψ* and
1
show that its power coincides with the expression given in (8.5.5). Let us
write g(t) = E[ψ*(X) | T(X) = t] which can not depend upon θ because T is
sufficient for θ and thus we have
Now, ψ*(x) = 1 for any t > θ hence g(t) = 1 if t > θ . Thus, for any θ > θ ,
0 0 0
we can write
This last expression is the same as in (8.5.5). So, any test function ψ* defined
via (8.5.3) is UMP level a for testing H versus H . Now we are in position
0
0
to state and prove the following result.
Theorem 8.5.1 In the Uniform(0, θ) case, for testing the simple null hy-
pothesis H : θ = θ against the two-sided alternative H : θ ≠ θ where θ is
0
1
0
0
0
a fixed positive number, the test associated with
is UMP level α.
Proof First note that since P {T(X) ≥ θ } = 0, one obviously has
θ0 0
Along the lines of the Example 8.4.11, we can easily show that a UMP level a
test for deciding between H : θ = θ and would have its test
0 0
function as follows:
