{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

09831HW5solutions

# 09831HW5solutions - HOMEWORK 5 SOLUTIONS Problem 1 a We can...

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

HOMEWORK 5 SOLUTIONS Problem 1 a) We can show this immediately using Corollary 6.4 from Chapter 1 of Durret. b)Let X = c and for any > 0 define the open interval A = ( c - , c + ). The boundary ∂A = { c - , c + } , therefore P { X ∂A } = 0 . By the Portmanteau Theorem 1 = P { X A } = lim n →∞ P { X n A } = lim n →∞ P {| X n - c | < } Problem 2: Observe that f n is always nonnegative. To show that is a density function we need to compute the integral: Z 1 0 1 - cos(2 πnx ) dx = x - sin(2 πnx ) 2 πn 1 x =0 = 1 . For x [0 , 1] , define F n ( x ) = x - sin(2 πnx ) 2 πn . Observe that for all x [0 , 1], (1) lim n →∞ F n ( x ) = F ( x ) = x which is the distribution function of a uniform [0,1] r.v. We actually proved that for any point of continuity of F ( x ) the convergence (1) holds, therefore the μ n converge weakly to the Lebesque measure. Problem 3: Let σ 2 = V ar ( X 1 ) . Classical CLT implies that S n σ n = X N (0 , 1) . Let M > 0 . The set ( σM, + ) is open and so by the Portmanteau Theorem, (2) lim inf n →∞ P S n σ n > M P { X > M } = 1 - Φ( M ) > 0 .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}