STAT 514 - HW 2 Solution Key

# STAT 514 - HW 2 Solution Key - 541 a Xi ~ exponentialﬁ px...

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Unformatted text preview: 541 a.) Xi ~ exponentialﬁ). px = 1, Val-X z 1. From the CLT, in is approximately r-(l, 1 [{1} 55.55 So in «1 in - ' i/‘W andP(1/~m%< <x)-—»P(ng). b (1 I ~ .._ d __ _ 1 «2/2 ) EP(Z S X) — EFZ(X) — {2(3) ... We . M???“ %(,§S‘i‘“'+“) (Wagwmm) ‘ «_~ ﬁFW(xﬁ+n) = fw(xﬁ + n) «a = rleJmﬁ‘l‘ n)n“1e‘<“'ﬁ+m)ﬁ _ Therefore, (1/1100) (xii+n)n-1e“(m+nkﬁs 3;..3'1‘2/2 1r yields n! as nu+1l2€nw321n as n—rao. Substituting 1: == 0 " 5.43 (a) Let X ~ n(0,0'2). For Va > 0, limPQY— yl<g)=1imPQ\/Z(Ynwy)f<\/Ea)w—.PQX|<OO)=1, LIZ—”mu [Iv-“N70 Mam 5.44 (6083/2116 (3LT, ﬂu; w p)w‘L>n(0,p(1—p)) (b) Let g(x)= x(1—x) then g‘(x)=1—2x. For p ¢1/2,i.e., g'(p) i G by the Delta Method, J; (510’; )— g( p))—d> 11(0, 0‘2 [g‘ (p)]2 ), then «5020 — 1;)“ p(1 - p))—~"‘i-->n(0,(i *pr 19(1” p)) f5,24 a) ForU ~uniform(0,1),_EU :1/2, Veri=1/12. Then 355i 2 3—1/2 x— “Elm-6 1211— 5: «IE(—1~m) is in the form T((ﬁwEU)/U) with n = 12, so X is approximately n(0,1) by the Central Limit Theorem. b) The approximation does not have the same range as Z ~ n(0, 1) where -¢o < Z -< +¢o, since *6<X<'8. . 2 ' c) . EX=E(:E21U:;6)= 22EUi -6‘— —(§:1%)n6:6-6=0. i“1i=i i= 'VarX—T-V 22Ui -=6) Vat 2U—12VHU1=1 i=1 - i=1 EX3 2 0 since X is symmetric about 0. (In fact, all add moments of X are 0.) Thus, the ﬁrst three moments of X all agree with the first three moments of a. stamina-d normal. The fourth moment is not easy to get, one way to do it' is to get the mgf of X. Sinee EetU 2: (e t--1)/t, Eet( 2111;- 6)_¢_6;(§f_‘1)12=(et/213't/2)12 Computing the feurth derivative and evaluating it at t .—.. 0 gives us 3x4. This is a lengthy calculation, but is made easier with a symbolic nianipnlation computer package. The answer is EX4 2329/10, slightly smalles- than E Z4 a 3, Where Z ~ n(0,1). Problem from notes Contruct a 95% conﬁdence interval for p. Justify your method eompietely. X; N independent Bernouiﬁﬁp)‘ . 13 = 3;; Xi 1. B th . WWE—zv} “id y e CLT we know that W N(O,1) 2. By the WLLN we know that 33 —+p p 3. Deﬁne a function g(m)m \/\$(i «w :73) .Then 91(55) is a. )continuous function. We apply theorem 5.54a11dgetg(p)——>Pg(p)= 1/p(pi-—);Mp\/ (1—39 4.1381338 another continuous function Mas) 23/1113.) We apply theorem 5.5.4 again and get h(\/p(1wp))—>Ph(M)= V___:(:: 21. 5‘ ﬂow) 2 \fﬁfﬁ—p) \$41 13) x/ﬂﬁ p) d ‘ «ea—:5) v’pe—mx man 5) By SlutSkyS woo—p) N(0,1) So the 95% conﬁdence interval for p is: 13 :i: 1.96 x 23 1—29) n ...
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