STAT 514 - HW 11 Solution Key - m'ieme'mmfimj sup WU my )...

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Unformatted text preview: m'ieme'mmfimj sup WU my ) 6500. 11 a m sgp L(9|y1.m.ym) ' 8.3 The LRT statistic is Air) == Let yI= lyi and note that the MLE in the numerator is the minimum of (firm, 30) 1:: _ (see Exercise 7.12) while the denominator hes y/m an the MLE (see Example 7.2.5). Thus 19 if 34m 5 901 (99)’(1 " 5’0)”,y (i’lmiS (1 — VIMEH' . 1' .. m'y . . (90) (1 90) < c. To show that this is e"-l‘li‘mlfim' to Iquims d we re'ect B if .. a" J 0 (was) Sky/m) _ . if y > b, we could show My 15 decreasmg m is easier to work with 103 My) (which is equivalent) and we have 103%!) = W139“ + (111 ~ y} 103(1 ' 99) - ¥1°g(%)‘- (:11 *- 3r)1°s( Ay : () y so that My) < (2 occurs for y > b > 12199. It m—Y T and F 3 5%?) + (m - 105:? 1 d log My) 2:: log 60 —— 10g (1 - 60) - iogG—é) —- yy + log( "<5 my 59 ("W ) (In—yum < 1 - 99, so eadi fraction above is less than i, and the i For y/m > 90. i —y/m = 103 A < 0, which shows that A is decreasing in y and My) <1: if Elogiaieestimnfl. Thnsac-‘f andonlyify>b. 8.5 a) The log-likelihood is 7 log L(9'-p|x) = n 1030 + n6 logy -— (0+1)log( Hxi), v 5 1(1), where 3(1) = For any vaiue of 6, is an increasing function of v for v 5 1(1). Se both the restricted and unrestricted MLE of v is 5' = :0). To find the MLE of 0, set 3 ' _., 9. _, ', _ wiog L(9,x(1)|x) —- a + niogxu) iog( H35) — O and solving for 9 yields - Ii - _ n L " “’i'o'stnxi/Egy T 7 (62/892) iog L(6,x(1)]x) .1: "11/92 < o, for oil 9. So a is a maximum. b) Under Ho, the MLE of 9 is £20 = 1 and theMLE of v is still 9 23:”). So the likelihood ratio statistic is - ~ aria,“ {1:92 A E (n/T)“x?1)/T/ ( [hon/T *1 (6/8T)log M1) = (ll/T) - 1. So M1) is increasing if T s n and decreasing if '1'; n. Thus, T s c is equivalent to T _<_ :1 or T 3 c2, for appropriateiy chosen constants c1 and c2. 8.6 a) 11) “sup L(9li) su‘p fiie'xi/a filc'Vi/e 9 i=1‘9 i=1” a o Kuhn L9 = n “a ,_ 9p (lwi flup.111 :57 111317;: _._‘ -e 0,}: 1:19 i=1 n 'm .. 9 sup 1 8 (1:1 121 1/ 9 9111411 = W‘ - E Jig/9 - B H.“ .1. i=1 1 i=1 1 ii: 25°- Difi'erentiation show that in the numerator 30 = (21:1 + mi)/(n+m), while in the denominator 3 2 5E and it = i. i ' - e .‘ Therefore. Many) =_ + l = (n+m)n+m (mnwygm nfiEm + flyin-m‘ And the LRT is to reject H9 if Mr, y) S c. n m A _ (n+m)n+m Bari __ (n+m)n+m “ nfififi 2f,- + E E,- +Eyi " 115551 Timion A is a function of T. A is a unimadal function orrr which is maximized when T = min. Rejection for A S c is equivalent to rejection {or T 5 a or T 2 b, where a and b are constants that satisfy an(I-a)m a bun—b)”. Tn(1 —T)'n. When He is true, ZXi ~ gamma(n,9} and EYi ~ gamma.(m,9) and they are independent. So by an extension of Exercise 4.192), T ~ beta.(n,m). L?- w a.) The likelihood function is Mafia!) = #“(l'llq)”"19“(llyi)9“l- Lannimafiwnxinyga-l, and maximizing as above yields the restricted MLE, a 3 w n + m M 0 Slogan?» Slosyi' The LET statistic is , ' am“ a _‘- a -5- . A(IJ)=E-‘.’r§~a,a(llx;) 9 “(Evil 0 - n ‘ _ 1:) Substituting in the formulas for b.‘ :1, and 30 yields ([1:990” “(nyifo’ 9 A( ) _,,. £314.11 _ 131+“ m(m+tl)n T an I an! if‘nrem—Efigfi"? (-l - So rejecting if MI, 3!) 5 c is equivalent to rejecting if aloud This is a unimodal function of T. T 5 c1 or T 2 c2, where c1 and :2 are appropriately chosen constants. c) A simple transformation yields r—log Xi ~ exponentialfl/p) and -logYi ~ exponential(1/ 0). Therefore, T 2 W/(W+V) where W and V are independent, W ~ gmMn,1/p), and A V ~ gmmflmJ/B). Under 130, the scale parameters of W and V are equal. Then, a simple generalization of Exercise 4.1§b yields T we bote{n,m). The constants c; and c2 are (ier by the two equations ‘ . . y Pfiswpazem and (l-cl)“‘c‘£=(I-c2)‘“c‘2‘- a“? r A a —P lineal w1 P m‘6"l< z 1 “PA-fii-Eg i—oo 5%) , 7 7_ “ -ca/'~l'fi+30—3 ifig cor/«I’fi-i-Bg-E «1—P9( a/fi 011-13- 9 -9 0 —0 ._ m .. .0 ' 9 ~ #1 P( c+amszgc+alfi)_ (z mm) 90- 80-9 _ (Q is the standard normaledf) b) The size is .05 = 5096) = 1 + @(—e) w<1>(c) z: c = 1.96. The power (1 — type 11 error) is .75 g 13(90—1—0) = 1 + @(-c-~i§)-¢(c-—‘Tfi) 2 1 + owes—r) -—‘1’(1.96-\[fi). “NIH—I'd . #0 From Table 1, Q(-.675) to .25 => 136—411" m—.675'=> n = 6.943 $3 7. ...
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This note was uploaded on 03/17/2008 for the course STAT 514 taught by Professor Senturk,damla during the Spring '07 term at Pennsylvania State University, University Park.

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STAT 514 - HW 11 Solution Key - m'ieme'mmfimj sup WU my )...

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