Dr. Hackney STA Solutions pg 17

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## 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.

Ask a homework question - tutors are online