solpracfinal

# solpracfinal - 7 x x ~ Hbv|{k | t x b}\$ 5m t x |{zyxswA5m t...

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Unformatted text preview: 7 x x ~ Hbv|{k | t# x b}\$ 5m t# x @ |{zyxswA5m t# vom t# @ vosuE wostk @ m @ m # womsr pwom @ @ g womnlx#Gq ( g wonlxkj` % ihg f @ p @ m # e d # e1C . Then, . # # e'8ex6 9 6 9 i e'I ( i # [email protected] B UyGq 6 B s r 7 9 i e5 ( i # F 9 i Bu ( A!yxq B u B @ s r F eGB B HGEf! F @CB @ u ( A!yxq u B @ s r F eGB B B HFGEf! @CB @ u ( A!yxq u [email protected] v8thq B u s r B @ s r SQ 9 i HhgfpA TRP F @CB @ i F @ a Y B @ a Y cG b`X ! b`X W eRd SQ TRP F @CB @ IHGED!A 9 SQ P F @CB @ Hhgf! F @CB @ VGEU! , conclude that . Hence, (1) The cdf is 7 4 6 8' 4 5" & & '" # #\$" ! # 2301 ()% % and hence Since, so that (2) We know that (3) Let Solutions to Practice Final Exam 1 Q V x C x . x y B [email protected] PB r H |Ixx @ @ 9 8 A7Q 7 FD B GECa % B & [email protected] (B Y 6 yx 5% 1 6 3 x 4|v 1 2\$ x B {[email protected] x \$ )[email protected] (B 'Y B & x | B [email protected] x [email protected] p B "B @ bx #! E| ~ tkcbv ~ W x # x f f t# 7 |x x | f # i g i s r f tj f u % % v zI % 8thq 7 % I f b`X aY 8txq i s r f 7 t [email protected] @ k b t# ~ (3b) The estimated standard error of is . Now use the delta method: and the estimated standard error of is . The approximate confidence interval is f e# Gbx x E @ |x bx I f x x bv s|{k |f x @ k | [email protected] Ibx d f ~ # Now, (4) Note that is 0 if 0 if this happens for any . So . This is the same as . So, (3c) The variance of (3a) The mle is is or . The likelihood will be is non-zero only if for all . On the other hand, if |x 8e# {bx 0p "B @ x B [email protected] x { ! % ! (4c) First, x B [email protected] (4b) . For , 2 x m W T R USQ Q x { B [email protected] . The ARE is . E I . B B P B B B ` B B y( B B B B B y( B B `gAB v B B Q u E hg E E P i g "HG E g g B B B AB B n E i . D AB E F S ( B v B i F Y D E # #tqC . # 7tk ( bY a e A B g e # 7tk6C & # IEg 58e1C 3 4 7 # 2 q # ) 1 ! !Q 0(zI & ) ' g y0 % # #tk %I \$" n ( !Q wG i i Rh B [email protected] B i . . V x ` 0 1 q # % 0 Q 1 1 h Q C 7 1 0 I x Qx Q ||x " ! Q B [email protected] x B 0 B B B % Q B [email protected] Q kom om Q B )[email protected] Y B & m The density is for T R USQ (5a) (5b) Let ! 2# # \$ q !Q "hi 1 (5c) (5d) (6a) Let D 9 @8 (6b) as long as . Setting i.e. as long as . Setting where yields . The Fisher information is 3 for some and gives . B Q [email protected] m %t# 6 Q % [email protected] &D B ost# 6 % m Q % [email protected] % B om Q &D x om , V 'x C x x x x @ % 'x x 1 %5 % @ 7 @ FD B GECa @ 9 8 A7Q x # \$"! x 3 1 @ 1 yx p @ % @ yx % @ x @ hg i s I @ v g @ s i s i @ @ 7 # 6 for . | y0 ( % % w i m 7% 0 # x i m wh # tj| y0 % s ! A8 i @ s 0 % 0 0 q ( % t# # @ # kom ( vod i m i # kom ( # 6 wh # e# @ m oI vozI d vo i m i m 9 @8 g E E B B B B 8 B B i B B 8gAB vthq B B IuE `X i s r aY then and then . B B 8 i For (7a) First, (6c) The variance is . If , . For . A8s # So, ht y0 and % | y0 s for i hg (7b) Write (8a) Let unless (7c) Given . Hence, and and , where be the smallest and largest values. The likelihood is 0 i.e. . Thus, % @ 1 @ This is a decreasing function so x |v 1 2\$ (8b) First, . For . is a point mass at 4 W Q T R USQ . , . So s EC Ri T1H ek ( #C # i 1 7 % 0 1 0 1 0 # 7 0 @ W 0 0 f 0 @ W b @ W W 5 W e# & 7 Q Q 1 1 0 I I i m om Q V x C 3x 7 x x % 0 1 E| d cbb I 1 Q 1Q ! x x ' t# ||x " B x %t# x B %t# 5 6 The density is (9) Parts (i)-(iii) should be straightforward. For part (iv), note that . Let . Then, if is a nonnegative integer, i Q Q 7 Q & x t4 Q Q Q 5 Q 5 6 Q Q x g|b d c| j` x for T R USQ This is Gamma . So the posterior mean is B [email protected] B ji I f Q % For part (v), note that Q F F )GD Q (8c) For . So (10) Let the density is . The posterior density is . The mean of a Gamma ( . 5 E . The confidence interval is . . The mean is ) is (in this version) . Also, @ ! B [email protected] B ...
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