STAT 514 - HW 8 Solution Key

STAT 514 - HW 8 Solution Key - 6 By the Factorization...

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Unformatted text preview: 6.} By the Factorization Theorem, [X1 in sufficient because we can write the pdf 01' X as fX(xI52) = $UE-X2/2a2m mac {elite/26' _ SUXHV‘?) _ gig“ 50‘) 6.2 By the Factorization Theorem, T(X)... - mag/if m gumdgnt because we can Write the joint pdf of X1" ..., Xn as rel, New) ills-"”5 10,, “0, our ._. e‘" 1a,”, (Ten) if: - w ammo) “‘3 Notice, we use the fact that i > 9, and the fact that all xis > i8 if and only if mifig/i) > 0 3" 1 6.3 Let 1(1): minixi. Then We can write the joint pdf as a, a .1.. ~( xi'~#)/¢ eP/V / ‘ - 1 . {(x110°°)xni ”3 ) iImIlanl It”, m)(x’1iw-) - ("7—)1—Exl vim-1)) 112;) r 3(‘(])! mi 1 #10.) Thus, by the Factorization Theorem, (X0), 2:11)“ a sufficient stetistic for (11,6). 6,5 The sampie density is given by . ‘ =2"; 19] {(xi IE) =Hm1§li-§I(-i(5~l) < xi__ < i(9+1)) I .i-WI i: ll (21—9 n(i-I:]1~i1-)I(min¥ 2 ~{9»1))I(max§g 9+1) so (min'Xi/ i, max Xi/i) is sufficient for 9. 6.6 The sample density is given by “- 1 n .. - ‘lfi 1‘ a . “Sign? {on}... .=,x,,la,m Iii—"”5"" X1 1 e X: "’r_"“'""( mama) (gig) e . i: _. By the Factorization Theorem, (1—1:: Xi' £1_1Xi)m sufficient for (a, fi). 11 0""‘1 7.10 a) rode): ._iI_j[1 53:59!“ IE0 ,3 (m): _.§%( ) “Ml 1M, (3:01)) I[0 0°)(x(1))“- .. L(a,filx). By the Factorization Theorem, (HXi,X(n)) are sufficient. b) For any fixed a, L(c¢,fl Ix) z 0 if ,6! < 3:01), and L(a,fi fix) a. decreasing function of 19 if [3 2 x01). Thus, X01) is the MLE of [3. For the MLE of (I calculate 32—1-16ng 5—8 [nloga— nalogfl +(awUlOSHXi1: :37 "" n 3°85 +103 HIE: Set the derivative equal to zero and use 3 r...- Xm) to obtain . 1 3 ._ n "‘ was: 1:01) ioglIXi [HE (1°3x(n) ‘°3 19]“ - The second derivative is —n/ar2 < 6, so this' 13 the MLE. 6) x01) = 25.0,1eg Hxi a: Slog xi = 43.95 2:» 3 = 25.9, 5; ~_— 12_59_ ...
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