09_02_09_0 - 0 as n , and if > 4, P ( | X n 3 |...

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

View Full Document Right Arrow Icon
STAT 409 Examples for 09/02/2009 Fall 2009 Def Let U 1 , U 2 , … be an infinite sequence of random variables, and let W be another random variable. Then the sequence { U n } converges in probability to W, if for all ε > 0, ( 29 0 W U P lim ε = - n n , and write W U P n . Example 1 : Let X n have p.d.f. f n ( x ) = n x n – 1 , for 0 < x < 1, zero otherwise. Then 1 X P n , since if 0 < ε 1, P ( | X n – 1 | ε ) = P ( X n 1 – ε ) = ( 1 – ε ) n 0 as n , and if ε > 1, P ( | X n – 1 | ε ) = 0. Example 2 : Let X n have p.d.f. f n ( x ) = n e n x , for x > 0, zero otherwise. Then 0 X P n , since if ε > 0, P ( | X n – 0 | ε ) = P ( X n ε ) = e n ε 0 as n .
Background image of page 1

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

View Full DocumentRight Arrow Icon
Example 3 : Let X n have p.m.f. P ( X n = 3 ) = 1 – n 1 , P ( X n = 7 ) = n 1 . Then 3 X P n , since if 0 < ε 4, P ( | X n – 3 | ε ) = n 1
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 as n , and if &gt; 4, P ( | X n 3 | ) = 0. Example 4 : Suppose U ~ Uniform ( 0, 1 ). Let X n = - -+ + 1 , 1 3 2 U if 3 1 3 2 , 1 3 1 U if 2 1 3 1 , U if 1 n n n n X = 1 , 3 2 U if 3 3 2 , 3 1 U if 2 3 1 , U if 1 Then P ( | X n X | ) = n 2 , 0 &lt; &lt; 1, P ( | X n X | ) = 0, 1. Therefore, X X P n ....
View Full Document

Page1 / 2

09_02_09_0 - 0 as n , and if &amp;amp;gt; 4, P ( | X n 3 |...

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