# ch5 - 5-1 CHAPTER 5 Section 5-15-1 First f(x,y ≥0 Let R...

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Unformatted text preview: 5-1 CHAPTER 5 Section 5-15-1. First, f(x,y) ≥0. Let R denote the range of (X,Y). Then, &=++++=Ryxf1),(81414181415-2 a) P(X < 2.5, Y < 3) = f(1.5,2)+f(1,1) = 1/8+1/4=3/8 b) P(X < 2.5) = f (1.5, 2) + f (1.5, 3) +f(1,1)= 1/8 + 1/4+1/4 = 5/8 c) P(Y < 3) = f (1.5, 2)+f(1,1) = 1/8+1/4=3/8 d) P(X > 1.8, Y > 4.7) = f ( 3, 5) = 1/8 5-3. E(X) = 1(1/4)+ 1.5(3/8) + 2.5(1/4) + 3(1/8) = 1.8125 E(Y) = 1(1/4)+2(1/8) + 3(1/4) + 4(1/4) + 5(1/8) = 2.875 5-4 a) marginal distribution of Xx f(x) 1 1/4 1.5 3/8 2.5 1/4 3 1/8 b) )5.1(),5.1()(5.1XXYYfyfyf=and )5.1(Xf= 3/8. Then, y )(5.1yfY2 (1/8)/(3/8)=1/3 3 (1/4)/(3/8)=2/3 c) )2()2,()(2YXYXfxfxf=and )2(Yf= 1/8. Then, x )(2yfX1.5 (1/8)/(1/8)=1 d) E(Y|X=1.5) = 2(1/3)+3(2/3) =2 1/3 e) Since fY|1.5(y)≠fY(y), X and Y are not independent 5-5 Let R denote the range of (X,Y). Because 1)654543432(),(=++++++++=&cyxfR, 36c = 1, and c = 1/36 5-6. a) 4/1)432()3,1()2,1()1,1()4,1(361=++=++=<=XYXYXYfffYXPb) P(X = 1) is the same as part a. = 1/4 c) 3/1)543()2,3()2,2()2,1()2(361=++=++==XYXYXYfffYPd) 18/1)2()1,1()2,2(361===<<XYfYXP5-2 5-7. [ ] [ ][ ]( )( )( )167.26/13321)3,3()2,3()1,3(3)3,2()2,2()1,2(2)3,1()2,1()1,1(1)(36153612369==×+×+×=++++++++=XYXYXYXYXYXYXYXYXYfffffffffXE639.)3()2()1()(36152613361226133692613=-+-+-=XV639.)(167.2)(==YVYE5-8 a) marginal distribution of Xx )3,()2,()1,()(xfxfxfxfXYXYXYX++=1 1/4 2 1/3 3 5/12 b) )1(),1()(XXYXYfyfyf=y fyY X()1 (2/36)/(1/4)=2/9 2 (3/36)/(1/4)=1/3 3 (4/36)/(1/4)=4/9 c) )2()2,()(YXYYXfxfxf=and 3/1)2,3()2,2()2,1()2(3612==++=XYXYXYYffffx fxX Y( )1 (3/36)/(1/3)=1/4 2 (4/36)/(1/3)=1/3 3 (5/36)/(1/3)=5/12 d) E(Y|X=1) = 1(2/9)+2(1/3)+3(4/9) = 20/9 e) Since fXY(x,y) ≠fX(x)fY(y), X and Y are not independent. 5-9.&=≥Ryxfandyxf1),(),(5-10. a) 834181)1,5.()2,1()5.1,5.(=+=--+--=<<XYXYffYXPb) 83)1,5.()2,1()5.(=--+--=<XYXYffXPc) 87)1,5.()1,5.()2,1()5.1(=+--+--=<XYXYXYfffYPd) 85)2,1()1,5.()5.4,25.(=+=<>XYXYffYXP5-3 5-11. 41812141818181214181)(2)(1)(1)(2)()(1)(5.)(5.)(1)(=++--==++--=YEXE5-12 a) marginal distribution of Xx )(xfX-1 1/8 -0.5 ¼ 0.5 ½ 1 1/8 b) )1(),1()(XXYXYfyfyf=y fyY X()2 1/8/(1/8)=1 c) )1()1,()(YXYYXfxfxf=x fxX Y( )0.5 ½/(1/2)=1 d) E(Y|X=1) = 0.5 e) no, Xand Yare not independent 5-13. Because X and Y denote the number of printers in each category, 4,=+≥≥YXandYX5-14. a) The range of (X,Y) is 321yx123Let H = 3, M = 2, and L = 1 denote the events that a bit has high, moderate, and low distortion, respectively. Then, 5-4 x,y fxy(x,y) 0,0 0.85738 0,1 0.1083 0,2 0.00456 0,3 0.000064 1,0 0.027075 1,1 0.00228 1,2 0.000048 2,0 0.000285 2,1 0.000012 3,0 0.000001 b) x fx(x) 0 0.970299 1 0.029835 2 0.000297 3 0.000001 c) )1(),1()(1XXYYfyfyf=, fx(1) = 0.29835 y fY|1(x) 0 0.09075 1 0.00764 2 0.000161 E(X)=0(0.970299)+1(0.029403)+2(0.000297)+3*(0.000001)=0.03 (or np=3*.01)....
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ch5 - 5-1 CHAPTER 5 Section 5-15-1 First f(x,y ≥0 Let R...

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