{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

411_hw01_soln

# 411_hw01_soln - Stat 411 Homework 01 Solutions 1 Since the...

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

Stat 411 – Homework 01 Solutions 1. Since the X i ’s are independent, E ( Y n ) = E ( X 1 ) · · · E ( X n ) = c n . By Markov’s in- equality, P ( Y n > ε ) ε - 1 E ( Y n ) = ε - 1 c n 0 , n → ∞ , since c (0 , 1). Since ε > 0 is arbitrary, Y n p 0. 2. (a) The maximum of a list of numbers less than x if and only if all the numbers are less than x . Therefore, for any x (0 , 1), F M n ( x ) = P ( M n x ) = P ( X 1 x, . . . , X n x ) = P ( X 1 x ) n = x n ; if x 0, then F M n ( x ) = 0, and if x 1, then F M n ( x ) = 1. (b) Let Z n = n (1 - M n ). The CDF of Z n can be found as follows: F Z n ( z ) = P { n (1 - M n ) z } = P { M n 1 - z n } = 1 - (1 - z n ) n , z > 0 . A well-known fact from calculus about the exponential function gives lim n →∞ F Z n ( z ) = 1 - lim n →∞ (1 - z n ) n = 1 - e - z . This limit is the CDF of an exponential random variable Z with mean 1. 3. Write z ? in place of z ? 1 - α/ 2 . From the hint, P μ {C α contains μ } = P μ n X - z ? n μ X + z ? n o . Do some simple algebra inside {· · · } to get X in the middle: P μ n X - z ?

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 ]}

### Page1 / 2

411_hw01_soln - Stat 411 Homework 01 Solutions 1 Since the...

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

View Full Document
Ask a homework question - tutors are online