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STAT 611 Solutions to Selected Problems from Assignment #9 8.5 c. Under H0 , T has the same distribution as " log n Y # Yi / min Yi , i i=1 where Y1 , . . . , Yn are iid from density f (y) = y −2 I[1,∞) (y). Letting Y(1) < Y(2) < · · · < Y(n) £Qn ¤ denote the ordered Yi ’s, T can be expressed as log i=2 (Y(i) /Y(1) ) . The probability distribution of T can thus be found from the joint distribution of the ordered Yi ’s. The Qn joint density of Y(1) , . . . , Y(n) is n! i=1 f (ui ), where 1 < u1 < · · · < un . Define the random variables Z1 , . . . , Zn by Z1 = Y(1) and Zi = Y(i) /Y(1) , i = 2, . . . , n. Using the standard change of variables technique, the joint density of Z1 , . . . , Zn is fZ1 ,...,Zn (z1 , . . . , zn ) = −(n+1) n!z1 n Y zi−2 , for z1 > 1, 1 < z2 < · · · < zn , i=2 and 0 otherwise. Integrate out z1 to get the joint density of Z2 , . . . , Zn . Upon doing so, we see that Z2 , . . . , Zn have the same density as the order statistics in a random sample of size n − 1 from the density f defined above. Together with previous conclusions, the last statement implies that T has the same probability distribution as S= n−1 X log W(i) = i=1 n−1 X log Wi , i=1 where W(1) , . . . , W(n−1) are the order statistics of a random sample W1 , . . . , Wn−1 from density f . Finally, it is easy to verify (using independence of the Wi ’s) that S has a χ2 distribution with 2(n − 1) degrees of freedom. 8.33 a. It is easy to show that k = 1 − α1/n . b. Clearly, if θ ≥ 1 − α1/n , then the power is equal to 1. So, let θ < 1 − α1/n , and consider Power = Pθ (Y1 ≥ k ∪ Yn ≥ 1) = Pθ (Y1 ≥ k) + Pθ (Yn ≥ 1) − Pθ (Y1 ≥ k ∩ Yn ≥ 1). Now, Pθ (Y1 ≥ k) = Pθ (all Xi ’s ≥ k) = (θ + 1 − 1 + α1/n )n = (θ + α1/n )n , 1 and Pθ (Yn ≥ 1) = Pθ (at least one Xi ≥ 1) = 1 − Pθ (all Xi ’s < 1) = 1 − (1 − θ)n . Using the expression for the joint density of Y1 and Yn on p. 230 of Casella and Berger, Z θ+1 Z θ+1 Pθ (Y1 ≥ k ∩ Yn ≥ 1) = Z k 1 1 θ+1 Z = k n(n − 1)(yn − y1 )n−2 dyn dy1 1 Z n(n − 1)(yn − y1 )n−2 I(y1 ,θ+1) (yn ) dyn dy1 θ+1 Z θ+1 + 1 n(n − 1)(yn − y1 )n−2 dyn dy1 y1 = (θ + α1/n )n − α. It follows that ½ Power(θ) = 1 + α − (1 − θ)n , θ < 1 − α1/n , 1, θ ≥ 1 − α1/n . c. When θ ≥ 1, the power of the test in question is 1, and hence must be UMP for θ ≥ 1. Now let 0 < θ1 < 1 and consider testing the hypotheses H0 : θ = 0 vs. H1 : θ = θ1 . Show that the rejection region of the Neyman-Pearson test of these hypotheses has the form Y1 ≥ k or Yn ≥ 1. Since k is chosen to make the test have size α, k has nothing to do with θ1 . It follows from Theorem 8 of the class notes that the test is UMP for all θ. d. The power of the test is 1 for every θ ≥ 1 − α1/n . Since 1 − α1/n ≤ 1, it is true by default that the power is always greater than 0.8 when θ > 1. 2