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Monrad.410.09 - Chapter 2 Multivariate Distributions 2.1.2...

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Unformatted text preview: Chapter 2 Multivariate Distributions 2.1.2 2.1.5 2.1.6 2.1.7 2 \$2+\$2 7r x2+\$2] dIBdm fa [WI—b/Fj 1 2 II II Why is —10g 2 > 0? H [0: [OW/2i29(p)/7rp]pd9dp f0 9(p)dp= 1. P(X+Ysz)=/ / e_\$_ydydx 0 0 z / -- e‘(2-3)]€—\$ dx :21_ 6—2 _ 26-2. 0 2e” 0<z<oo G’(z) ={ O elsewhere 1 1 P(XY S 2) = 1 “ / / dydx 2 2/3 1-/:(1_; 0(2) = { 11 Z d\$=Z-zlogz ) “iogz 0 elsewhere. 12 Multivariate Distributions 2.1.8 13)(13 ( 26 ) m 13-—:—— x Z (Ly 2 0,1 + y g 13,31: and y integers f(\$ay) = 13 0 elsewhere. 2.1.10 1/2 1—121 P(X1+X2 Sl)=15/ z? 1E2d1122:ld.’131. 0 231 2.1.14 00 oo 1 i+j E[et1X1+t2X2] : ZZetii+t2j <_> i=1j=1 2 00 i 00 j 1 1 = z eh) 2e) 1:]. < 2 j==1 2 1 1 = [1 —— 2‘16‘1 — 1] [1 — 2‘16‘2 *1] ’ provided t,- < log 2, i = 1,2. 2.2.1 2 2-312 p(y1,y2)={ (gay (21:) :leslégil;(0’0)’(*1illi(1a1),“),2) 2.2.2 __ 3/1/36 3/1 = y2,2y2,3y2;y2 =1,2,3 p(yl’y2) - { 0 elsewhere. an m» was 2.2.4 The inverse transformation is given by \$1 = ylyz and x2 = 3/2 with Jacobian J = y2. By noting what the ‘boundaries of the space 8(X1, X2) map into, it follows that the space T(Y1, Y2): {(311,312) :0 < y.- < 1,2" = 1,2}. The pdf of (Y1, is fY1,Y2(3/1s 3/2) = 2.2.5 The inverse transformation is 11:1 = y1 — y2 and 2:2 = y; with Jacobian J = 1. The space of (Y1,Y2) is T = {(y1,y2) : —00 < y. < 00,71 = 1,2}. Thus the joint pdf of (Y1, Y2) is hag/2011,30) = fX;,X2(yl — 212,3!2), which leads to formula (2.2.1). 2.3.2 2.3.3 f2(\$2) f1|2(351l932) E(X1|\$2) (3(y) 9(y) E(Y) Vhr(ij 13LX1) Var(X1) 22 01/ 231/1173 (11121 = O 01 2 10m1m§,0 < m; < 932 < 1;zero elsewhere 1/2 f... 1/2 1/4 II II II 64 12 2:55 (21%) 2m1/‘(5/8)2 dm 3 1 1/2 / 10321323 (1:32de 2/ —x1(1— dxl 2 1‘1 1 /4 1132 f0 21x§x3dm1 = 7:33, 0 < 132 < 1. = 55 Zlmfmg/Z'Bg = 3zf/m‘3, 0 < :01 < :32. 3 —.’132. \$2 / m1(32§/mg)dm1 = 0 4 4y/3 P<§ngy)=/ 7\$3d\$2=< y 0 7QY¢ 0<y<§ 0 elsewhere. 7 3 __ 21 §Z‘§' 7 1024' g1 32' 553 7 “>— 15360 1024 ' 2.3.8 The marginal pdf of X is 00 fx(\$) = 2/ (fie—y dy = 26“”, 0 < :1: < 00. Hence, the conditional pdf of Y given X = a: is 26"”6'1’ _ _ fylxlylx) = 264\$ = 6 (y x), with conditional mean W E(YlX = 9:) =/ ye_(y“”) dy = a: + 1, a: > 0. 4 3 _—_-1=>c1=23ndc2=5. 0<x<y<oo, 13 ,W 15% QUIZ 1 SOLUTION (VEKSmN M Di: 7%, COMWWLA (Jam/{wk X Nara 700% j”) = x/é J 7cm Z<><<L{, H = ijewa z ska/Ma = [79?] = é? d; = EOﬂ ‘ﬂ; = 10” (“\$32 =, "E? but Y: VX v 77% dish‘buﬁom mam 01C Y is EM = P(Yég\ = PCXW/g‘) = 5; {wk T95 some TO quz 2 g 5%) X>0) (3)0}, ><+g<4 ’ 4/2 “X V PLXW) r. 3&6 {(xl3>d%>dx 72* ' .1 , : Si(g:wq><pa\d><)a(g + 18/2( O ‘*9( “RCA 2 _g: “1 SQ éxdé :: .. 00 \$093} w Ex (9% (820% icio‘) 6% g) ‘4 éxU—x) 0% ix 1 :32 (_X “X z 8 % mlﬂx 0W3 ’- -— —f104x<1 SOLUTION TO QUIZ 2 £993) '-= 60ng)2(><>J Vol x+g< 1 PORVB = \$1 {(xlﬂmé) 61X X 11.7. ' = 5 (quxvzﬁdxwg + [g/Xigjw‘wkwj O ...
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