Dr. Hackney STA Solutions pg 17

Dr. Hackney STA Solutions pg 17 - 2-2 Solutions Manual for...

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Unformatted text preview: 2-2 Solutions Manual for Statistical Inference b. Let X be a random variable with density f (x). P (X < t) = 1 where, - 2 ex dx = Therefore, t 1 x t 2e - t - 0 - 1 x 2 e dx t 1 1 x -x dx 2 e dx+ 0 2 e t 1 e-x dx 0 2 if t < 0 if t 0 t 0 = 1 et and 2 = - 1 e-x 2 = - 1 e-t + 1 . 2 2 P (X < t) = c. P (|X| < t) = 0 for t < 0, and for t 0, 1 t 2e 1 - 1 e-t dx 2 if t < 0 if t 0 0 P (|X| < t) = P (-t < X < t) = -t 1 x e dx + 2 t 0 1 -x e dx 2 = 1 1 1 - e-t + -e-t +1 2 2 = 1 - e-t . 2.5 To apply Theorem 2.1.8. Let A0 = {0}, A1 = (0, ), A3 = (, 3 ) and A4 = ( 3 , 2). Then 2 2 2 -1 -1 gi (x) = sin2 (x) on Ai for i = 1, 2, 3, 4. Therefore g1 (y) = sin-1 ( y), g2 (y) = - sin-1 ( y), -1 -1 -1 -1 g3 (y) = sin ( y) + and g4 (y) = 2 - sin ( y). Thus fY (y) = = 1 1 1 1 1 1 1 1 1 1 1 1 - - + + + 2 2 2 2 1-y2 y 1-y2 y 1-y2 y 1-y2 y 1 , 0y1 y(1 - y) To use the cdf given in (2.1.6) we have that x1 = sin-1 ( y) and x2 = - sin-1 ( y). Then by differentiating (2.1.6) we obtain that fY (y) = = = d d 2fX (sin-1 ( y) (sin-1 ( y) - 2fX ( - sin-1 ( y) ( - sin-1 ( y) dy dy 1 1 1 1 -1 1 2( ) - 2( ) 2 1 - y 2 y 2 1 - y 2 y 1 y(1 - y) 2.6 Theorem 2.1.8 can be used for all three parts. a. Let A0 = {0}, A1 = (-, 0) and A2 = (0, ). Then g1 (x) = |x| = -x3 on A1 and 3 g2 (x) = |x| = x3 on A2 . Use Theorem 2.1.8 to obtain fY (y) = 1 -y1/3 -2/3 e y , 3 0<y< 3 . b. Let A0 = {0}, A1 = (-1, 0) and A2 = (0, 1). Then g1 (x) = 1 - x2 on A1 and g2 (x) = 1 - x2 on A2 . Use Theorem 2.1.8 to obtain fY (y) = . 3 3 (1 - y)-1/2 + (1 - y)1/2 , 8 8 0<y<1 ...
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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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