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Unformatted text preview: Solution Set to Homework 3 1. CB 3.44 Define = 1,0,1,3 , = 1,2,3 . According to Definition 3.6 in the text book we know that: 1= 2 + 2 = 2 + 2 = 2 + 2 = 2 + 2 = 3 2 + 2 = 3 2 + 2 = 3 2 + 4 2 = 3 1 + 1 + 9 + 4 1 + 4 + 9 Thus = 89 2. CB 3.45 (a) 1, > 2 = ,1 , >2 1 = 89 1 2 1 + 2 = 89 89 9 = 1 89 2 + 9 ,1 = = (b) 1 89 29 89 2 + ,1 ,1 1+1 +39 = 0, 2 =
=0 , 2 1 2 + 2 = 89 2 = , 2 = , 2 1 2 1 = 89 89 , 2 =0 1 2 + 2 89 1 5 1+4 = 89 89 (c) + > 2 = 1 = 89 = 1 89 ,+ >2 1 2 + 2 89 2 + 2 ,+ >2 2 + 2 +
=0, =3 ,1 , 2 2 + 2 = 1 9+ 89 1 9+ 89 2 + ,1 ,2 , 2 2 = 2 2 + ,1 ,2 2 = = 3. CB 3.46 Proof: 1 9+2 89 2 + 2 ,1 ,2 2 = 55 89 1 9+2 1+9+4+9 89 0,1 = 2 1  0 = 2 3,1 = 2 1  3 =  If , can serve as the joint probability distribution of two random variables, then according to Theorem 3.8 part (1) in the text book, we know that: 0,1 = 2 0 => 0 3,1 =  0 => 0 we gain that k=0, thus , = 0. So, (, ) = 0 = 0 which conflict with Theorem 3.8 part (2). So we know that , cannot serve as the joint probability distribution of two random variables. 4. CB 3.50 Define = { , : 0 < < 1,0 < < 1, + < 1}. 1 + < = 2 1 16 (, ) =
1 + < 2 (, ) 1 + < 2 (, ) 24 = 1 2 0 1  2 0 24 = 5. CB 3.56 2 2 2     , = , = 1   + = =   =  , > 0, > 0. 6. CB 3.58 < , < = , <  , < = ,  ,  ,  , = ,  ,  ,  (, ) = , + ,  ,  (, ) 7. CB 3.70 (a) According to Definition 3.10, the marginal distribution of X is = (, )
1 120 7 15 7 15 So, 0 = 1 = (0, ) = 1 12 1 6 1 + + +
1 4 4 8 1 1 20 1 1 = ; (1, ) = + + =
1 ; 1 = (1, ) = + = . 24 40 15 (b) According to Definition 3.10, the marginal distribution of X is = (, )
7 So, 0 = 1 = 2 = (, 0) = 1 12 1 4 1 8 + +
1 6 4 1 1 1 24 1 = ; (, 1) = + + (, 2) = +
20 = 40 7 = 24 21 40 ; ; 40 3 = (, 3) = . 120 (c) According to Definition 3.12, the conditional distribution of X given Y=1 is, (, 1) 1 = (1) So, 0 1 = 1 1 = 2 1 =
(0,1) (1) (1,1) (1) (2,1) (1) 1 = = = 1 4 21 40 1 4 21 40 1 40 21 40 = = = 10 21 10 21 1 21 ; ; . (d) According to Definition 3.12, the conditional distribution of Y given X=0 is, (0, ) 0 = (0) So, 0 0 = 1 0 = 2 0 = 3 0 =
(0,0) (0) (0,1) (0) (0,2) (0) (0,3) (0) = = = = 1 12 7 15 1 4 7 15 1 8 7 15 1 120 7 15 = = = 5 28 15 28 15 56 ; ; ; . = 1 56 8. CB 3.79 Proof: Assume that f(x, y) is the joint probability density function of X and Y at (x, y), g(x) is the marginal density of X then
+ = +  , + =
 =
  + , = lim ,   = lim (, ) = (, ) So we know that the marginal distribution function of X is, = , = lim (, )
+ = lim 1   + 2 1   2 = 1   2 , > 0. 9. CB 3.92 Assume that F(x) is the cumulative distribution function of f(x) then, 1 = = = 36  2  6 6 288 1 1 = 144 + 36  3 288 3 1 1 3 1 = 36  + , 6 < < 6. 288 3 2 (a) 1 1 1 7 2 = 36 2 + 23 + = 288 3 2 27 (b) 1 1 1 539 325 1  1 = 1  36  + =1 = 288 3 2 864 864 (c) 1 1 1 1 95 1  3 = 36 +  36 3 + 33 = 288 3 288 3 432 (d) 5  lim () = 0
5 10. CB 3.95 Assume that F(x) is the cumulative distribution function of f(x) then, 1  = = = 3 9  0 0  = 1  1 + 3 , > 0. 3 (a) 6 6 = 1  1 + 2 = 1  3 2 = 0.5940 3 (b) 9 1  9 = 1 + 3 = 4 3 = 0.1991 3 11. CB 3.98 Y 0 0 X 1 2
3 28 9 28 3 28 1
3 14 3 14 2
1 28 3 2 3 2 3 3 1 0,0 = 2 = , 0,1 = 1 1 = , 0,2 = 2 = 8 8 8 28 14 28 2 2 2 3 3 3 2 9 3 1,0 = 1 1 = , 1,1 = 1 1 = 8 8 28 14 2 2 3 3 2,0 = 2 = 8 28 2 12. CB 3.106 (a) According to Definition 3.11 in the text book, the marginal density of P is,
+ + =
 , =
0 5  = 5, 0.2 < < 0.4. (b) According to Definition 3.13 in the text book, the conditional density of S given P=p is (, ) 5  = = =  , 0.2 < < 0.4, > 0. () 5 (c) 3 3 3 1 1 1 1 < 3 = = ( ) = 4 = 1  4 = 0.5276 4 4  0 4 13. CB 4.2 Proof: Using Theorem 4.1 with = + , we get + = ( + ) () = () + () = () + () = () + 14. CB 4.6 = 3 1 1 1 () = 1 + + 3 = 7 7 7 7
3 15. CB 4.10 = 1 2 = ( 1.8205) (3) 3  1 + 3 1 4 2 2 = = 2 = ( 3.6410) (3) 3  1 + 3 1 26 3 3 = = 3 = ( 7.8887) (3) 3(3)  1 16. CB 4.11 = + + 1 =
 =
0 2 2 1 2  5 + 3 +  5 + 3 2 2 1 3 3  2 +  5 + 3 2 2 1 13 19 11 =   = 24 12 24 6 ...
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This note was uploaded on 08/26/2009 for the course STAT 447 taught by Professor Staff during the Spring '08 term at Iowa State.
 Spring '08
 Staff

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