Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

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1 Eon : Fall 8 Suggested Solutions to Problem Set 8 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.

2 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

3 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,

4 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, ).

5 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