STAT 514 - HW 5 Solution Key

# STAT 514 - HW 5 Solution Key - 7.1 7.2 a 1.7 7.8 a b For...

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Unformatted text preview: 7.1 7.2 a) - 1.7 7.8 ' a) b) For each value of x, the MLE 3 is the value of 9 that maximizes f(x|9). Thae values are in the foltowing table. 2: 9‘ 1 2 3 4 1 1' 20:3 3 ‘5) “I At x; 2, {(3:12) a {(2:13) ~.= 1/4 are both maxim, so both 3 2 2 or 3 a 3 m MLEa. . n __ ‘ ‘3 n “.1 _ . 3 mm) = E malwa’ﬁﬂ‘ =51 =W[Hxi] e 25’ . . a“: : iogLUBIx) = w 1081‘(G)”-Ml°sﬁ+1°8[ﬁxi] “"219: ' ix]. _ ' Set the partial denvative equal to 0 and solve for 5 to obtain 19— Eli/(nor). To check that this as a maximum, calculate _urx 22x; (mos 2M3 (an? ' —-‘-2'— T 1?." <0. LE 18*? 7216(qu (E ‘1) Since ﬂ is; the unique point where the derivative' is 0 and it as a local maximum, it is a. 32105 L 332 global maximum. That is, B is the MLE. L(0!x)=1,0<:§<1,andL(§lx)= 11f: 1/(2ﬁ),0<xi<l. ThmtheMLEhﬂﬂ 1_>_ Eli-1242; and the MLEis 1 in < I]. :13/(24'7’ E12... — VarX + 142 a 62. Therefore X2 is an unbiased estimator of 02. I ' .. L(a' |x)= mes-x 2/(262). LogLﬁr Ix): log(2w)ﬁ1l2-—loga~x2/(2v2). 31 L .§g~2—%+ﬁssee => and = 34:35:.” a . ._ 2 - 12%; 31602-5- :2, which is negative at 3 = M. 0’6 . Thus, 3’ a: Ix] as a local maximum. Sinoeit is the only piacewhere the'ﬁrst derivative is zero, it is also a. global maximum. ' “ 7.11 a.) {(1:19) 2-" ﬂax?! = 9n(HJri)9-l = L(8]x) - ' i=1 ' . . £31031; = ad? 111039 +(9-l)lognxi]a%+2103xi. Set the derivative equal to zero and solve for 0 to obtain 3 : (w %ZIGgXi)-1. The second ' derivative is — 11/92 < a, so this is the MLE. Tq calculate the variance of 3, note that Y} = ;— 103% ~ expouential(1/9), so —):]ogxi ~ gmn, 1/9). Thus 3 = n/T, where '1‘ ~ gamma(n, 1/8). We can either calculate the ﬁrst and second momenta directly or use the faqt that 3‘ is inn-cried gamma (Example 2.1.4). We have 1-9“ m1—14» _9“L(§1_-_§,1_, a ET—ﬁ-ETJIO It” 9 dt—man_1 __§:T' an “’1 1 4 .... 931131-21- 92 ".Jo 2724“” e mum—17 an‘i “(n-1)(n-2)' ...
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