Assignment_2-Sol

Assignment_2-Sol - l5) §C%3)={ Kq JP Hg “:32: i if} 0...

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Unformatted text preview: l5) §C%3)={ Kq JP Hg “:32: i if} 0 Oil) 0) t K w wgw KNEE K 325:: f3 (IF—Ff} :KqéybéquO7ILK K; Bomb .7: A 10,)”; z ) “rt/L 93mg) g 7% I” “If; 0 K CHDX \g "120,1 9.} 30;) 51X .2: 0' Qku/ (3% at)me a :8: I, «5‘ {Azall/l/ 0 J] .(U “TIL j‘mfie 3mm self 0;; (WW 5‘3 RR“ :{ (QC/WW } {:{1/1:0/‘7wr-">X{8}yfli‘~? omé gavmfl (1,9)é'RxY) '§‘(x,3),:§g¢(q)jivw) Tm ch >4 & x m «WNW/r. 15 C) 1>C><.:ac( ><+\(=Jc) ; W HMY rfl =W? 95) a) 19) ‘er xivgaw ) r- 31 QI/WICX/[j] EN): E[E[‘Y[r‘fl «rid/“J: “(5‘ vmm: E[VW('Y//»)]+VM[E(Y//x)] : E[ H + MP] : ©(fl “$— : jLM/N I: (9M) 3' Cf) W W _ I —fk Lg I 91-] '- NW —-l _f*(nl+ r M fire/Mu * 4% 3H. // fire/52‘ , \f :E N5(o<j;6L) .. 2.. 8 @Note that since 2 is a symetric matrix 2T : 2 so that 2—12 2 I (2 X 2 identity matrix) and STE” 2 I. Also uTt : tTu since MTt is a scalar and asTt : tTZII since th is a scalar. = [m —l (M + 20? :3"1 [w — (,u + Et)] 7 2ft — tTEt : — [.L)T — tTZT] 2‘1 — ,u) — 2t] 4 2,LLTt — tTEt = (ac — MT :3” (m — ,u) — (w — MT 2‘12t — tTETE‘Km — u) +tTZTE*1Zt # 2ft — tht z (m — MT E_1(x — p) — — mTt — tT(:13 v p) + tTEt — 2/,LTt v tTEt : (a —;L)TE_1($—u) *th+uTt—tTw+tT;L—2th * ($*#)TE_1($“H) -2tT1 — L.S as required; Now Zl/[(t1,t2) 00 (X) 1 : /400 /_00 ———27r|;|1/2 eXp(th)Cxp [—5 (m _ MT 2—1 _ m] [m1de _ oo oo 1 l _ T _1 I— _ T ‘ ' /_m/.m2w|2|1/2CXP{—2[(I M) ‘8 (l n) 22: 17]}d11d332 1 : fie Warp {'5 — (,u + 21%)]T53’1 [m _ (it + Em _ ZHTt _ tTEt)}dmld$2 1 1 1 i T __ T __ _ T #1 ‘ v t 7 1 : exp (th + EtTEt) , t E §R2 since 1 1 W exp {#5 ~ (lb-1' Et)]TE’1;L‘ — (it + 20]} is a BVNQL + 2t, 2) p.d.f. and therefore the integral is equal to one. 23 resume 1\/le(1€) : ]\/I(t,0) : exp (#115 + 7520?), t6 9? which is the n1.g.f. of a N(M1,a%) random variable7 then by the Uniqueness Theorem for in.g.fi’s X1 m N(;L1,a§)i By a similar argument X2 m NW2, 02). 19 28(crlsmce 8t18t2M(t1’t2) _ 02 T 1 T — atlatg exp (/1 15+ 5t 2t) 62 : exp ,ult1 + ltth —|— 115202 +t1t2p0102 +1302 atlatg 2 1 1 2 2 2 (9 672 [(#1 H10? + t2p0102) Aal(t1,t2)] : p0102A/[(t1,t2) + (#1 +t10’? +t2p0’102) (/12 +t203+t1p0102)A/[(t1,t2) therefore 62 E(XY) 8t16t2Al(t1)t2)l(t1,t2):(0,0) : p010'2 +u1p2 ll From (1)) we know E(X1) : p1 and E(X2) : #2. Therefore Cov(X,Y) : E(XY) — E(X)E(Y) : p0102 +ILL1/.L2 —,LL1;L2 : {20102. 2 By Theorem 38.2 X1 and X2 are independent random variables if and only if [l/I(t1, t2) = A/le (t1)]\/1X2(t2) then X1 and X2 are independent random variables if and only if 1 ex t r 1152 2 t —1t2 2 p M 1+,u2,2+ 2 101+ 1t2p0102+ 2 202 1 1 : exp (ultl + #0?) exp (Hth + E15303) for all (t1,t2) E 332 or 1 l 1 1 [11231 + #th + Effie? + t1t2p010’2 + 5&0; Z #1151 + 5&0”? + #th + 515303 01‘ t1t2p0’102 Z 0 for all (t1, t2) 6 §R2 which is true if and only if p : 0. Therefore X1 and X2 are independent random variables if and only if p : 0. 20 Since E[exp(tTY)] : E{exp[tT(AX+b)]} : E{cxp[(ATt)TX+th]} = exp(th)E{exp[(ATt)TX]} : eXp(th) exp (,LLT(ATt) + $(ATt)TE(ATt)) = exp <th + (AmTt + étTMEATfi) : exp [(Ah —- b)T t + étUAEATfi] , 156 $9 which is the n1.g.f. of a BVN(A;L vv b,AEAT) random variable, then by the Uniqueness Theorem for n1.g.f.’s, Y : AX + b m BVN(A;L + b, AZAT). First note that 1 2 1 is —~“—’— Z#1 : i: 02 '002102 :| _ (71‘ 01102 7 0%03 (1—p2) "P0102 01 (1—102) 4319372 33 1/2 0% P0102 1/2 2 2 2 1/2 2 izl :H palm 03 J * [0102—(P0102)] :0102 1—!) and (06 7 MT 3‘1 (1‘ e M) : Therefore 2 — U§(11_p2)[(m2 #2)? 2ij (961 mm u2>+p§2m m2] 2 2 1 [<a2—uzi—pflm—wr 02(1-102) ‘71 fl 1 {3:2 [;L2+p02($1 #1)]}2 owl—p2) 01 The conditional p.d.f. of X2 given X1 : 9:1 is f(351,$2) f1(331) W 0xp(tTm) 9X13 — ILL)T 2—1 ([17 A. 2 1 1 ‘ ~ 1 2M} exp [_5 (7216111 ) ] 0'1 —1 T C131 71L 2 — exp a: 1L 2'1 (:1: 4 1L) — ( 1) 271'0'10'2x/1 — p2 { 2 ) 01 2 1 1 1 p02 2 fl exp :L’g u + $1 ,LL , $69? V27T02V1’P2 [ 203(1#P2){ [2 01 ( 1) ] which is the p.d.f. of a N<u2 + “27—12 (9:1 — 1L1), 0% (1 — p2)) random variable. H 22 __ "“ I i ,— Q(/~fi) I. ‘——'t.:)_ _ 1.;6‘..%9_ ] z: 2. 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Assignment_2-Sol - l5) §C%3)={ Kq JP Hg “:32: i if} 0...

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