Dr. Hackney STA Solutions pg 20

Dr. Hackney STA Solutions pg 20 - Second Edition 2-5...

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

View Full Document Right Arrow Icon
Second Edition 2-5 Therefore, F Y ( y ) = d dy F Y ( y ) = f X ( y ) + f X ( - y ) = 1 2 π e - y 2 + 1 2 π e - y 2 = r 2 π e - y 2 . Thus, E Y = Z 0 y r 2 π e - y 2 dy = r 2 π Z 0 e - u du = r 2 π ± - e - u ² ² 0 ³ = r 2 π , where u = y 2 2 . E Y 2 = Z 0 y 2 r 2 π e - y 2 dy = r 2 π ´ - ye - y 2 ² ² ² 0 + Z 0 e - y 2 dy µ = r 2 π r π 2 = 1 . This was done using integration by part with u = y and dv = ye - y 2 dy . Then Var( Y ) = 1 - 2 π . 2.12 We have tan x = y/d , therefore tan - 1 ( y/d ) = x and d dy tan - 1 ( y/d ) = 1 1+( y/d ) 2 1 d dy = dx. Thus, f Y ( y ) = 2 πd 1 1+( y/d ) 2 , 0 < y < . This is the Cauchy distribution restricted to (0 , ), and the mean is infinite. 2.13 P ( X = k ) = (1 - p ) k p + p k (1 - p ), k = 1 , 2 ,. .. . Therefore, E X = X k =1 k [(1 - p ) k p + p k (1 - p )] = (1 - p ) p " X k =1 k (1 - p ) k - 1 + X k =1 kp k - 1 # = (1 - p ) p ´ 1 p 2 + 1 (1 - p ) 2 µ = 1 - 2 p + 2 p 2 p (1 - p ) . 2.14 Z 0 (1 - F X ( x )) dx = Z 0 P ( X > x ) dx = Z 0 Z x f X ( y ) dydx = Z 0 Z y 0 dxf X ( y ) dy = Z 0 yf X ( y ) dy = E X, where the last equality follows from changing the order of integration.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the 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