{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stat447F08HW3Sol

# Stat447F08HW3Sol - Solution Set to Homework 3 1 CB 3.44...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

• Spring '08
• Staff
• Probability distribution, Probability theory, probability density function, Cumulative distribution function

{[ snackBarMessage ]}

Ask a homework question - tutors are online