Dr. Hackney STA Solutions pg 71

Dr. Hackney STA Solutions pg 71 - y will do but this one...

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

View Full Document Right Arrow Icon
Second Edition 5-5 5.15 a. ¯ X n +1 = n +1 i =1 X i n + 1 = X n +1 + n i =1 X i n + 1 = X n +1 + n ¯ X n n + 1 . b. nS 2 n +1 = n ( n + 1) - 1 n +1 X i =1 ( X i - ¯ X n +1 ) 2 = n +1 X i =1 ± X i - X n +1 + n ¯ X n n + 1 ² 2 (use (a)) = n +1 X i =1 ± X i - X n +1 n + 1 - n ¯ X n n + 1 ² 2 = n +1 X i =1 ³ ( X i - ¯ X n ) - ± X n +1 n + 1 - ¯ X n n + 1 ²´ 2 ( ± ¯ X n ) = n +1 X i =1 " ( X i - ¯ X n ) 2 - 2 ( X i - ¯ X n ) ± X n +1 - ¯ X n n + 1 ² + 1 ( n + 1) 2 ( X n +1 - ¯ X n ) 2 # = n X i =1 ( X i - ¯ X n ) 2 + ( X n +1 - ¯ X n ) 2 - 2 ( X n +1 - ¯ X n ) 2 n + 1 + n + 1 ( n + 1) 2 ( X n +1 - ¯ X n ) 2 µ since n X 1 ( X i - ¯ X n ) = 0 ! = ( n - 1) S 2 n + n n + 1 ( X n +1 - ¯ X n ) 2 . 5.16 a. 3 i =1 ( X i - i i ) 2 χ 2 3 b. ( X i - 1 i ) , v u u t 3 i =2 ( X i - i i ) 2 , 2 t 2 c. Square the random variable in part b). 5.17 a. Let U χ 2 p and V χ 2 q , independent. Their joint pdf is 1 Γ ( p 2 ) Γ ( q 2 ) 2 ( p + q ) / 2 u p 2 - 1 v q 2 - 1 e - ( u + v ) 2 . From Definition 5.3.6, the random variable X = ( U/p ) / ( V/q ) has an F distribution, so we make the transformation x = ( u/p ) / ( v/q ) and y = u + v . (Of course, many choices of
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: y will do, but this one makes calculations easy. The choice is prompted by the exponential term in the pdf.) Solving for u and v yields u = p q xy 1 + q p x , v = y 1 + q p x , and | J | = q p y ¶ 1 + q p x · 2 . We then substitute into f U,V ( u,v ) to obtain f X,Y ( x,y ) = 1 Γ ( p 2 ) Γ ( q 2 ) 2 ( p + q ) / 2 µ p q xy 1 + q p x ! p 2-1 µ y 1 + q p x ! q 2-1 e-y 2 q p y ¶ 1 + q p x · 2 ....
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