assi_soln - 1 Solutions of Assignment #1 - STAT 330 Due in...

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Unformatted text preview: 1 Solutions of Assignment #1 - STAT 330 Due in class: Thursday May 19 Important Note: You need to print out this page as the cover page for your assignment. LAST NAME: FIRST NAME: ID. NO.: QUESTION 1. /12 QUESTION 2. /8 QUESTION 3. /3 QUESTION 4. /2 QUESTION 5. /9 QUESTION 6. /6 TOTAL: /40 2 1. Solution: (a) (i) X x =0 f ( x ) = 1 X x =0 k (0 . 3) x = k 1 1- . 3 = 1 k = 0 . 7 (ii) F ( x ) = P ( X x ) = X r x f ( r ) = x X r =0 . 7(0 . 3) r = 0 . 7 1- . 3 x +1 1- . 3 = 1- . 3 x +1 , x = 0 , 1 , 2 ,... which gives a step function: F ( x ) = 1- . 3 n +1 , n x < n + 1 , otherwise where n = 0 , 1 , 2 ,... (iii) E ( X ) = X x =0 xf ( x ) = X x =0 x . 7 (0 . 3) x = X x =1 . 21 x (0 . 3) x- 1 = 0 . 21 X x =1 x (0 . 3) x- 1 = 0 . 21( 1 1- . 3 ) 2 = 3 7 where we used the technique: X n =1 ny n- 1 = X n =1 ( y n ) = ( X n =1 y n ) = ( y 1- y ) = 1 (1- y ) 2 , for | y | < 1 3 (b) (i) Z kx 2 e- x dx = 1 , let y = x, then Z k ( y ) 2 e- y 1 dy = k 3 Z y 3- 1 e- y dy = k 3 (3) = 2 k 3 k = 3 2 (ii) When x > 0: F ( x ) = Z x f ( t ) dt = Z x 3 2 t 2 e- t dt = Z x 1 2 y 2 e- y dy, let y = t = Z x 1 2 y 2 d (- e- y ) =- 1 2 y 2 e- y x + Z x e- y ydy =- 1 2 ( x ) 2 e- x + y (- e- y ) x + Z x e- y dy = 1- e- x 1 + x + 1 2 ( x ) 2 When x 0: F ( x ) = 0, therefore, F ( x ) = 1- e- x (1 + x + 1 2 ( x ) 2 ) , x > , x 4 (iii) E ( X ) = Z - xf ( x ) dx = Z x 3 2 x 2 e- x dx = 3 2 Z x 3 e- x dx = 1 2 Z y 4- 1 e- y dy, let y = x = 1 2 (4) = 3! 2 = 3 (c) (i) Z k (1 + x )- ( +1) dx = 1 - k (1 + x )- = 1 k = 1 k = (ii) when x > 0: F ( x ) = Z x (1 + t )- ( +1) dt = [- (1 + t )- | x ] = 1- (1 + x )- when x 0: F ( x ) = 0 Therefore, F ( x ) = 1- (1 + x )- , x > , x (iii) E ( X ) = Z x (1 + x )- ( +1) dx = Z 1 ( y- 1) y- ( +1) dy, let y = 1 + x = Z 1 y- dy- Z 1 y- ( +1) dy = 1- + 1 y- +1 1-...
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assi_soln - 1 Solutions of Assignment #1 - STAT 330 Due in...

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