Dr. Hackney STA Solutions pg 36

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

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Unformatted text preview: Second Edition 3-9 The first two moments are EY EY 2 and so VarY = d. fx (x) = = = 1 (a - 1)ba-1 1 e-1/by = a (a)b 0 y (a)ba (a - 2)ba-2 1 = , (a)ba (a - 1)(a - 2)b2 a = 1 (a - 1)b 1 (a-1)2 (a-2)b2 . 1 x3/2-1 e-x/ , (3/2) 3/2 x > 0. For Y = (X/)1/2 , fY (y) = 2 0 calculate the moments we use integration-by-parts with u = y , dv = ye EY = 2 (3/2) 0 2 2 2 -y 2 , (3/2) y e -y 2 y > 0. To to obtain y 3 e-y dy = 2 2 (3/2) ye-y dy = 2 1 (3/2) and with u = y 3 , dv = ye-y to obtain EY 2 = 2 (3/2) 0 2 y 4 e-y dy = 2 3 (3/2) 0 y 2 e-y dy = 2 3 . (3/2) 1 Using the fact that 2 - y 2 e-y = 1, since it is the variance of a n(0, 2), symmetry yields 2 2 -y y e dy = . Thus, VarY = 6 - 4/, using (3/2) = 1 . 2 0 e. fx (x) = e-x , x > 0. For Y = - log X, fY (y) = e-e e of EY and EY 2 cannot be done in closed form. If we define -y -y 1 , - < y < . Calculation I1 = 0 log xe-x dx, I2 = 0 (log x)2 e-x dx, then EY = E( - log x) = - I1 , and EY 2 = E( - log x)2 = 2 - 2I1 + 2 I2 .The constant I1 = .5772157 is called Euler's constant. 3.25 Note that if T is continuous then, P (t T t+|t T ) = = = Therefore from the definition of derivative, hT (t) = Also, - 3.26 a. fT (t) = 1 -t/ e P (t T t+, t T ) P (t T ) P (t T t+) P (t T ) FT (t+) - F T (t) . 1-F T (t) 1 1 - FT (t) = 0 lim FT (t + ) - FT (t) = FT (t) 1 - FT (t) = fT (t) . 1-F T (t) d 1 (log[1 - F T (t)]) = - (-fT (t)) = hT (t). dt 1-F T (t) t 1 -x/ e dx 0 and FT (t) = = - e-x/ t 0 = 1 - e-t/ . Thus, hT (t) = fT (t) (1/)e-t/ 1 = . = -t/ 1-F T (t) 1-(1 - e ) ...
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