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Assignment 4_Solution

Assignment 4_Solution - The University of Hong Kong...

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Unformatted text preview: The University of Hong Kong , Department of Statistics and Actuarial Science Stat 1301 Probability & Statistics Assignment 4 suggested solution Section A f I 1.ECX) = zixﬂx) = 3.3, ECXZ) = zgxzfix) = 13.5, VarCX) = EQXE) —- [59012 = 2.51 2. a)f0“‘f{x}dx = 1,j‘o“‘c(1+x)“3dx= 1. —§_—[(1 Mfg-Jig” = 1,; = LC: 2. b) m a: 1n» a: 13.3; 1 m 1 cc 1 Ea) 2"; xfﬂﬁax z!” (1+x33axz‘f0 [fléxfi‘s‘ (1+xjsdx‘2f0 (mama "211’ (1+3? {1+x)"1 a: {1+x}“2 a: =.—2-—:1—|G ~2“———_—§--—|G =—2{0-1)+{0-—1)=1 we) zfxfﬁzhu: r” 3 rim: 2”“? :1-— 1 — a 0 "9(1+u)5 -z 9 (1%}2 ' J 1 .1 ’1 1 .53» x=F“1{cz) 2(1—51.) 3-1,Medmn ofx:=F‘1(E) ~1=0AJA d) The inter quartile range ofX = F—1 — F”1 = 0.845 3. Let x=number of princes with expense \$2.0 before Prince Desire is selected xEmN—ﬂ Option1 x=O, Pr(Prince Desire is selected ] x=0)=1/N, expense .=\$1.0 x=1, Pr(Prince Desire is selected | x=1)=“:1 1 AH 1 expense = \$2;O+1.0 = \$3.0 -:l .i’l.‘ N-iN—z 1 _1 '—_.1 V AH ,‘M expense =2x \$2.0+\$1 = \$50 x=2, Pr(Prince Desire is selected | x=2)= x=N-1, Pr(Prince Desire is selected | x=N-1) = expense = \$2 (N-1) + \$1.0 I 1 E{Ex‘perzse optionl) = 31.0 + 53.04%?) + + [52(N - 1) + \$113] N~1 1 1 1 V 2w — 1);“: - 5)22“+1"(§)[——f_+”]-N k=0 Option2 ' ‘2“ k ECExpemse optionz) = Egg} + 1) (i), where k=number of princes failed 1 = ﬁekf‘wkﬁ)£%)=C‘%7Efsak(%)kn +:\$E?=o(%)k= 1=N-1+1=N Both options have the same expected expense. 4. a) £00 = Eixfcx) = 2.1, EC?) = Zia-ma = 2.33 b) 50:11) = zixzﬂx) = 4.9, ECYZ) = 232%!) = 6.63 ,VarOf) = 30(3) —— [E0012 = 0 .49,Var(¥) = 12011 C) Cov[X,Y]=E[(x-E(X))(y—E(Y))]=-0.063 Covth'Y) 9(3) Y) = = 41.0821 5. See example class 10 6. a) Max} = 2:5 = «2-2»ng 9:3" = e-M, see-2r) = 33; 5":ch 3 9.11m Varctx) = e—kt‘l _ 8—-21+2t}. b) Em = EWCW (1-22391‘5“ = 23615—321 :.~ 2234” =: (1 —- pl“? (i4- 53 = Z tgx px (1 ~' pju-x == {1 "9)“ (1 + 1pj229>ﬂ VarCtx’) = E’Ctz") ._ [Egg]: 3-" 0) Mt”) = 23‘ Ml - p) ‘29 : EC? - \$27)”? = g: t—t+1 max) = p p32 —— 11-2 + 1 d) E{tx) = txAE‘xlcix = fle'xa‘zﬂgf)dx= 1 ‘ l-iog (3; EL " fax : -——-———— ET ) 1’1 — 210g (t) e) Mix) zili fax: {2‘1 and gaze) : z“L1 '1 2 Ztlag{t} - 45: Egg: 7. a) The following graph shows there is a noticeable relationship between xiand 3’5. b) pg: .—. xi) = EgﬁiPCX = raj—J = yi) = + Q + + 01.ThusE(X) = 2&4 xiPﬂX = xi) = SEE : 6.214286 and similar to msz we have EGZY) = 22.64286. We can compute the ﬁn) = Zixiyjpcx = x0? = 3.3.) = 311.0993, and p = = = 0.9609897 It shows a strong relationship between X and Y. IE3(IXI) =1E(IXI,|XI Z 6) +1E(|X|,|X| < 6) 2 E(le,le Z 6) We know that lE(|X|,|X[ 2 e) 2 lE(e,|X| 2 e) = elP(]X[ _>_ e), which requested.1 2)First we deﬁne Xi as X. _ 1 if the ith couple repatriated 1 _ 0 0.10 we should calculate 1E(X1 + + XN). hog.) = 190;; =’ 1) = (2N—Mnglgl—M—1), S O M X1 +_ . . + XN) = NE<X1 = (2N—M)SVN—M—1)' We also deﬁne Yi as Y _ 1 if the ith couple seperated 1 _ O 0.11) Similar to previous part we should calulate lE(Y1 + - - ~ + YN). 1E(Y;) = Pa; = 1) = W80 1E(Y1+-- -+YN) = (ZN—jeﬁrf the goverment use a fair coin to decide which asylum would stay in Hong Kong, the probabilty of refugee is 0.5 for each person thus according to deﬁnition of Xi, Y; we have lE(X1 + ‘ - - + XN) = N/4 and lE(Yl +- - -+YN) = N / 2. To ﬁnd the range we should examine 1ng1>WﬂorN2_(2N-M)(2N—M—1)>o, which is polynomial of second degree and has the following roots: {% (4N—W—1)}-,{-;-(4N+m~1)} for the second part we examine Jiv- > W, or (M-—N)2 > 0. 1Since [Xl 2 e we can decrease the value by changing IX | with e 1 Since X be a non—negative random variable, X can be expressed as X = X I{X_>_0} = X Iu;1{i—1gx<i} = X Z I{z‘—1\$X<i} E XXI/1,- i=l i=1 Thus Z‘=1 i=1 i=1 Take expectation on both sides and X as well, we have _ 1) ’ S < ‘ E(IAi). Notice that 2a —- 1) -1E(IA,.) = 20‘ — 1) ~IP<Ai> i=1 20‘ ~ 1)-[1P(X 2 z"- 1) — M 2 2)] —_. Za—1)-]P(X2i~1)—Z(i—1)-P(X2i). Make a tranformation j = z?— 1 and notice that the summation is unrelated to footnote, then we have II M jg N v §> Similar argument we have i=1 = ii.p(X2i—1)~IP(X22'] 7,- = ;(j+1)IP(XZj)—;J IP(X>J) .._. Spurn) 2 1P(X20)+§:IP(XZZ’) z 1+§MX§2> If X follows exponential distribution of rate A, then lE(X) = X’1 and This implies (1 — A)? < 1 and e)‘ -—l)\ > 1. Or A + e‘)‘ > 1 and e"‘(1 + A) g 1. From Question B3(b), Assignment 3, we showed that X ~ N (0, 1) and Y ~ N(O, 1), thus Cov(X, Y) = IE(X, Y) +00 +00 = f / wyft’v, y)dwdy +°° +°° 1 (a2 - 2/1202 + (1 ~ ram/2} = :1: —————— e —-———-— dzd [-00 [—00 y27r\/1 - P2 xp{ m p2) y @953} {Km dw} dy Let z = (a: — py)/\/1_——p—§, then Cov(X, Y) 1 +00 +00 2 2 = ET. we /_m (x/l—p2yz+py2)exp{~%~%}dydz p 2 __ 2 +00 _z2 57; m y exp{-2—}dy/_m eXp{--§-}dz+ “ 12; p2 yeXp{:-g~2-}dy/+ 5p; - x/Z—va/ZF + 0 = p Let fX(:1:) and fy(y) denote the marginal distribution of X and Y respectively. 1! +00 Z J ‘W X and Y are independent :> ﬂan, y) = fX(m) - fy(y) <=> p = 0 <=> X and Y are uncorrelated ...
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Assignment 4_Solution - The University of Hong Kong...

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