hw7sol - Math 280A, Fall 2011 Homework 7 Solutions 1....

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 280A, Fall 2011 Homework 7 Solutions 1. Problem 1 (page 260). Clearly P [ X k x ] = braceleftbigg 1 exp( ( x a )) , if x a ; , if x < a . Consequently, for x a , P [ Z n > z ] = P [ X 1 > x, X 2 > x, . . . , X n > x ] = n productdisplay k =1 P [ X k > x ] = exp( n ( x a )) , while P [ Z n > x ] = 1 if x < a . Therefore, given > 0, we have P [ | Z n a | > ] = P [ Z n > a + ] = exp( n ) , n , which means that Z n P a as required. 2. Let x 1 , x 2 , . . . , x n be strictly positive numbers. Show that 1 n n summationdisplay k =1 x k parenleftBigg n productdisplay k =1 x k parenrightBigg 1 /n , with equality only if x 1 = x 2 = = x n . [Hint: The natural logarithm log is a strictly concave function.] Solution. Let X be a random variable that is uniformly distributed over the finite set { x 1 , x 2 , . . . , x n } ; that is, P [ X = x k ] = 1 n , k = 1 , 2 , . . ., n. Clearly X is integrable and E [ X ] = 1 n n summationdisplay k =1 x k . The function log is strictly concave on (0 , ), and so by Jensens inequality (restated for concave functions) (2 . 1) E [log( X )] log( E [ X ]) ....
View Full Document

This note was uploaded on 12/11/2011 for the course MATH 280a taught by Professor Driver,b during the Fall '08 term at UCSD.

Page1 / 3

hw7sol - Math 280A, Fall 2011 Homework 7 Solutions 1....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online