Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test of H : θ against H : θ. (Your test should be expressed in terms of α.) We apply the Neyman-Pearson Lemma. Within the support [, ] of f, the likelihood ratio is f(x, ) LR(x) f(x, ) x, whih inreases with x. By the Neyman-Pearson Lemma, the ritial region for the test (ϕ (x) ) must have the form x >. Using the level onstraint we get α. ϕ (x)dx dx α, Therefore, a most powerful level α test of H against H is ϕ (x) [x > α]. (It does not matter what value we hoose for ϕ (x) outside of the support [, ] of f.) (b) Is there a uniformly most powerful level α test of H : θ against H : θ >? If so, what is it? For any θ >, the likelihood ratio for H : θ against H : θ θ over the support [, ] is LR(x) f(x, θ ) f(x, ) θ x + θ. Sine LR(x) θ x >, LR(x) inreases with x. Therefore, ϕ (x) as found in part (a) is a most powerful level α test of H : θ against H : θ θ. Sine this holds for all θ >, we onlude that ϕ (x) as found in part (a) is a uniformly most powerful level α test of H : θ against H : θ >. () To test H : θ against H : θ > the test ϕ(x) [x > /] is used. Find the level and power funtion of the test. The level is ϕ(x)f(x, )dx ϕ(x)dx. Thanks to previous teahing assistant Rihard Chiburis for his solutions.
The power funtion is π(θ) ϕ(x)f(x, θ)dx ( (θx + θ) dx (θ + θ) θ + ) ( θ) +θ. Problem. Let X i, i, be iid with density f(x, θ) θx θ [ x ]. (a) To test H : θ against H : θ >, a ritial region W {(x, x ) : x x } was used. Find the size and power of this test. The size is f(x, )f(x, )dx dx W x dx dx ( ) x dx 5 8. The power funtion is π(θ) W f(x, θ)f(x, θ)dx dx x θx θ θx θ θx θ θx θ θx θ dx dx ( ) θx θ x dx dx ( ( ) ) θ dx ( ) θ. x dx ) θ ( θx θ dx (b) Find the most powerful size α ( ln ) test of H : θ against H : θ. The likelihood ratio over the support [, ] is LR(x) f(x, )f(x, ) f(x, )f(x, ) x x. The ritial region will then have the form x x >. To find, we use the level
onstraint: [x x > ] dx dx α dx dx ( ln ) /x ) ( x dx ( ln ) ( ln ) ( ln ) + ln + ln. (Sine + ln is stritly dereasing on (, ), this is the only solution.) Therefore, our test is ϕ (x, x ) [ x x > ]. () Are the tests that you obtained in parts (a) and (b) unbiased? The test is part (a) is unbiased beause π(θ) is inreasing, so for all θ >, π(θ) > π(). For part (b), the alternative hypothesis was H : θ, so we need only hek the rejetion probability for θ : π() Therefore, the test is unbiased. ( /x x x dx dx x x dx dx /x ( ) ) x ( dx x ( )) ln ln > α. Loally Best Tests. Let X be ontinuous with pdf f(x, θ). Consider the problem of testing the hypothesis H : θ θ against the alternative H : θ > θ. Let π ϕ (θ) E θ [ϕ(x)], θ > θ be the power funtion of the test ϕ. Assume that f(x, θ) is suh that π ϕ (θ) admits one ontinuous derivative,
whih an be passed inside the expetation, i.e. π ϕ(θ) dπ ϕ(θ) ϕ(x) dx dθ where f(x,θ) dx < for all θ. A test ϕ is alled loally best if for any other test ϕ of the same level, π ϕ (θ ) π ϕ(θ ). The loally best test is therefore one that maximizes the slope of the power funtion at θ over all level α tests. (a) Show that the general form of the loally best test ϕ under these assumptions is for f(x,θ) θθ > λf(x, θ ) ϕ (x) κ for f(x,θ) θθ λf(x, θ ) for f(x,θ) θθ < λf(x, θ ) for some λ and κ suh that E θ [ϕ (X)] α. We only need to slightly modify the proof of the Neyman-Pearson Lemma. do the proof for the ontinuous ase only. Let ϕ(x) be any other test of level α, i.e. ϕ(x)f(x, θ )dx α. We need to show that π ϕ (θ ) π ϕ(θ ). Note that ( ) (ϕ (x) ϕ(x)) θθ λf(x, θ ) dx, sine ϕ (x) ϕ(x) for all x suh that f(x,θ) θθ λf(x, θ ) > and ϕ (x) ϕ(x) for all x suh that f(x,θ) θθ λf(x, θ ) <. Therefore, using α ϕ(x)f(x, θ )dx ϕ (x)f(x, θ )dx, We ( ) (ϕ (x) ϕ(x)) θθ λf(x, θ ) dx (ϕ (x) ϕ(x)) π ϕ (θ ) π ϕ(θ ), θθ dx so π ϕ (θ ) π ϕ(θ ), as required. (b) Let X be distributed Cauhy with median θ, i.e. the pdf of X is f(x, θ). Show that no uniformly most powerful test of H π(+(x θ) ) : θ against H : θ > exists. Find the form of the loally best test. The most powerful level α test of H : θ against H : θ θ is given by ϕ (x) { for f(x, θ ) > λf(x, ) for f(x, θ ) λf(x, ).
Observe that f(x, θ ) f(x, ) + x + (x θ ) By taking the derivative( with respet to x and setting it equal to zero, we find that this funtion peaks at θ + ) θ +. Then it is lear that for different values of θ lose to zero and small α that the rejetion region varies with θ. Therefore, there is no UMP test. The loally best test has the form { for f(x,θ) ϕ (x) θθ > λf(x, ) for f(x,θ) θθ λf(x, ) with ϕ (x)f(x, )dx α. Observe that f(x,θ) θθ f(x, ) x π(+x ) π(+x ) x + x. Then, where λ satisfies { for x > λ ϕ (x) +x x for λ, +x [ ] x + x > λ dx α. π( + x ) 5