STAT 830 Convergence in Distribution - People.stat.sfu.ca - Simon ...

Scaling mattersMultiply both X n and Y n by n 1/2 and let X ∼ N(0,1). Then√ nXn ∼ N(n −1/2 ,1) and √ nY n ∼ N(0,1).Use characteristic functions to prove that both √ nX n and √ nY nconverge to N(0,1) in distribution.If you now let X n ∼ N(n −1/2 ,1/n) and Y n ∼ N(0,1/n) then againboth X n and Y n converge to 0 in distribution.If you multiply X n and Y n in the previous point by n 1/2 thenn 1/2 X n ∼ N(1,1) and n 1/2 Y n ∼ N(0,1) so that n 1/2 X n and n 1/2 Y nare not close together in distribution.You can check that 2 −n → 0 in distribution.Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 — Fall 2011 10 / 31