Dr. Hackney STA Solutions pg 24

# Dr. Hackney STA Solutions pg 24 - Second Edition 2-9 2.28...

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Second Edition 2-9 2.28 a. μ 3 = Z -∞ ( x - a ) 3 f ( x ) dx = Z a -∞ ( x - a ) 3 f ( x ) dx + Z a ( x - a ) 3 f ( x ) dx = Z 0 -∞ y 3 f ( y + a ) dy + Z 0 y 3 f ( y + a ) dy (change variable y = x - a ) = Z 0 - y 3 f ( - y + a ) dy + Z 0 y 3 f ( y + a ) dy = 0 . ( f ( - y + a ) = f ( y + a )) b. For f ( x ) = e - x , μ 1 = μ 2 = 1, therefore α 3 = μ 3 . μ 3 = Z 0 ( x - 1) 3 e - x dx = Z 0 ( x 3 - 3 x 2 + 3 x - 1) e - x dx = Γ(4) - 3Γ(3) + 3Γ(2) - Γ(1) = 3! - 3 × 2! + 3 × 1 - 1 = 3 . c. Each distribution has μ 1 = 0, therefore we must calculate μ 2 = E X 2 and μ 4 = E X 4 . (i) f ( x ) = 1 2 π e - x 2 / 2 , μ 2 = 1 , μ 4 = 3 , α 4 = 3 . (ii) f ( x ) = 1 2 , - 1 < x < 1, μ 2 = 1 3 , μ 4 = 1 5 , α 4 = 9 5 . (iii) f ( x ) = 1 2 e -| x | , -∞ < x < , μ 2 = 2 , μ 4 = 24 , α 4 = 6 . As a graph will show, (iii) is most peaked, (i) is next, and (ii) is least peaked. 2.29 a. For the binomial E X ( X - 1) = n X x =2 x ( x - 1) ± n x ² p x (1 - p ) n - x = n ( n - 1) p 2 n X x =2 ± n - 2 x ² p x - 2 (1 - p ) n - x = n ( n - 1) p 2 n - 2 X y =0 ± n - 2 y ² p y (1 - p ) n - 2 - y = n ( n - 1) p 2 , where we use the identity x ( x - 1) ( n x ) = n ( n - 1) ( n - 2 x ) , substitute y = x - 2 and recognize that the new sum is equal to 1. Similarly, for the Poisson
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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