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Dr. Hackney STA Solutions pg 36

# Dr. Hackney STA Solutions pg 36 - Second Edition 3-9 The...

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Second Edition 3-9 The first two moments are E Y = 1 Γ( a ) b a 0 1 y a e - 1 /by = Γ( a - 1) b a - 1 Γ( a ) b a = 1 ( a - 1) b E Y 2 = Γ( a - 2) b a - 2 Γ( a ) b a = 1 ( a - 1)( a - 2) b 2 , and so Var Y = 1 ( a - 1) 2 ( a - 2) b 2 . d. f x ( x ) = 1 Γ(3 / 2) β 3 / 2 x 3 / 2 - 1 e - x/β , x > 0. For Y = ( X/β ) 1 / 2 , f Y ( y ) = 2 Γ(3 / 2) y 2 e - y 2 , y > 0. To calculate the moments we use integration-by-parts with u = y 2 , dv = ye - y 2 to obtain E Y = 2 Γ(3 / 2) 0 y 3 e - y 2 dy = 2 Γ(3 / 2) 0 ye - y 2 dy = 1 Γ(3 / 2) and with u = y 3 , dv = ye - y 2 to obtain E Y 2 = 2 Γ(3 / 2) 0 y 4 e - y 2 dy = 3 Γ(3 / 2) 0 y 2 e - y 2 dy = 3 Γ(3 / 2) π. Using the fact that 1 2 π -∞ y 2 e - y 2 = 1, since it is the variance of a n(0 , 2), symmetry yields 0 y 2 e - y 2 dy = π . Thus, Var Y = 6 - 4 , using Γ(3 / 2) = 1 2 π . e. f x ( x ) = e - x , x > 0. For Y = α - γ log X , f Y ( y ) = e - e α - y γ e α - y γ 1 γ , -∞ < y < . Calculation of E Y and E Y 2 cannot be done in closed form. If we define I 1 = 0 log xe - x dx, I 2 = 0 (log x ) 2 e - x dx, then E Y = E( α - γ log x ) = α - γI 1 , and E Y 2 = E( α - γ log x ) 2 = α 2 - 2 αγI 1 + γ 2 I 2 .The constant I 1 = . 5772157 is called Euler’s constant. 3.25 Note that if T is continuous then,
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