Assignment1_Solutions

Assignment1_Solutions - 1 Solutions of Assignment #1 - STAT...

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1 Solutions of Assignment #1 - STAT 330 Due in class: Thursday Jan. 21 Important Note: You need to print out this page as the cover page for your assignment. LAST NAME: FIRST NAME: ID. NO.: QUESTION 1. QUESTION 2. QUESTION 3. QUESTION 4. QUESTION 5. QUESTION 6. TOTAL:
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2 1. Sol’n: a) To be a density, f needs to satisfy 2 conditions: 1. f ( x ) 0: If - 1 x 1 then f ( x ) [0 , 1] so f ( x ) 0. 2. ± -∞ f ( x ) dx = 1: ² -∞ f ( x ) dx = ² - 1 -∞ 0 dx + ² 1 - 1 (1 + αx ) / 2 dx + ² 1 0 dx = 0 + 1 2 ³ x + αx 2 / 2 ´ µ µ µ µ 1 - 1 +0 = 1 2 ([1 + α/ 2] - [ - 1+ α/ 2]) = 1 2 (2) =1 Therefore, f is a density. b) The corresponding c.d.f. is, for any x R , F ( x )= ² x -∞ f ( t ) dt = ± x -∞ 0 dt, x < - 1 ± - 1 -∞ 0 dt + ± x - 1 (1 + αt ) / 2 dt, - 1 x 1 ± - 1 -∞ 0 dt + ± 1 - 1 (1 + αt ) / 2 dt + ± x 1 0 dt, x > 1 = 0 , x < - 1 1 2 x + αx 2 / 2 · - [ - α/ 2] ) , - 1 x 1 1 , x > 1 = 0 , x < - 1 αx 2 / 4+ x/ 2+1 / 2 - α/ 4 , - 1 x 1 1 , x > 1
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3 2. Sol’n: (a) To be a c.d.f., F needs to satisfy 4 conditions. 1. lim x →-∞ F ( x ) = 0: This holds as F ( x ) = 0 whenever x< 0. 2. lim x →∞ F ( x ) = 1: lim x →∞ F ( x ) = 1 - lim x →∞ e - αx β =1 - 0 = 1 (as α > 0 ,β > 0) 3. F is right-continuous: We need lim h 0 + F ( x + h )= F ( x ) for any x R . This condition holds as F is composed of two functions, both of which are (right) continuous.
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Assignment1_Solutions - 1 Solutions of Assignment #1 - STAT...

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