03_24 - X x =&&& ±&&& ² ³ ≥<...

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

View Full Document Right Arrow Icon
STAT 410 Examples for 03/24/2008 Spring 2008 Let { X n } be a sequence of random variables and let X be a random variable. Let F X n and F X be, respectively, the c.d.f.s of X n and X. Let C ( F X ) denote the set of all points where F X is continuous. We say that X n converges in distribution to X if ( ) ( ) x x n n X X F F lim = , for all x C ( F X ). We denote this convergence by X X D n . Example 1 : Consider { X n } with p.m.f.s P ( X n = 0 ) = 1 – n 1 , P ( X n = 1 ) = n 1 . Note that 0 X P n , since if 0 < ε 1, P ( | X n – 0 | ε ) = n 1 0 as n , and if ε > 1, P ( | X n – 0 | ε ) = 0. Consider X with p.m.f. P ( X = 0 ) = 1. F X n ( x ) = ± ² ³ < - < 1 1 1 0 1 1 0 0 x x n x F X ( x ) = ± ² ³ < 0 1 0 0 x x ( ) ( ) x x n n X X F F lim = , for all x R . ´ X X D n . Example 2 : Let { X n }, X be i.i.d. with p.m.f. P ( X n = n 1 ) = n 1 2 1 - , P ( X n = 1 ) = n 1 2 1 + . Then F X n ( x ) = ± ² ³ < - < 1 1 1 1 1 2 1 1 0 x x n n n x . ( ) ± ² ³ < < = 1 1 1 0 2 1 0 0 F X lim x x x x n 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
Consider X with p.m.f. P ( X = 0 ) = 2 1 , P ( X = 1 ) = 2 1 . Then F
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: X ( x ) = & & & ± & & & ² ³ ≥ < ≤ < 1 1 1 2 1 x x x ( ) ( ) x x n n X X F F lim = ∞ → , for all x ≠ 0. F X n ( ) = 0 for all n , but F X ( ) = 2 1 . ( ) ( ) F F X X lim ≠ ∞ → n n . Since 0 ∉ C ( F X ), X X D n → . Example 3 : Suppose P ( X n = i ) = 6 3 + + n i n , for i = 1, 2, 3. Then X X D n → , where P ( X = i ) = 3 1 , for i = 1, 2, 3. Example 4 : Let X n have p.d.f. f n ( x ) = n x n – 1 , for 0 < x < 1, zero elsewhere. Then F X n ( x ) = & & & ± & & & ² ³ ≥ < ≤ < 1 1 1 x x x x n . ( ) & ± & ² ³ ≥ < ∞ = → 1 1 1 F X lim x x x n n . Therefore, X X D n → , where P ( X = 1 ) = 1. Note that 1 X P n → , since if 0 < ε ≤ 1, P ( | X n – 1 | ≥ ε ) = ( 1 – ε ) n → 0 as n → ∞ , and if ε > 1, P ( | X n – 1 | ≥ ε ) = 0....
View Full Document

This note was uploaded on 02/11/2011 for the course STAT 410 taught by Professor Monrad during the Spring '08 term at University of Illinois, Urbana Champaign.

Page1 / 2

03_24 - X x =&&& ±&&& ² 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