851lecture04

# Note that the total length of 0 316 1516 1 is 316

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: r = 15/16, then [r, 1/4 + r] = [15/16, 19/16] which when “wrapped around” becomes [0, 3/16] ∪ [15/16, 1]. Note that the total length of [0, 3/16] ∪ [15/16, 1] is 3/16 + 1/16 = 1/4. That is, using ﬁnite additivity, P {[r, 1/4 + r]} = P {[0, 3/16] ∪ [15/16, 1]} = 4–1 1 = P {[0, 1/4]} . 4 We can write this (allowing for “wrapping around”) using the ⊕ symbol so that ￿ [r, 1/4 + r], if 0 < r ≤ 3/4, [0, 1/4] ⊕ r = [0, 1/4 + r − 1] ∪ [r, 1], if 3/4 < r < 1. Hence, if 0 < r ≤ 3/4, then P {[0, 1/4] ⊕ r} = P {[r, 1/4 + r]} = 1 1 +r−r = 4 4 while if 3/4 < r < 1, then P {[0, 1/4] ⊕ r} = P {[0, 1/4 + r − 1] ∪ [r, 1]} = P {[0, 1/4 + r − 1]} + P {[r, 1]} ￿ ￿ 1 1 = + r − 1 + (1 − r) = . 4 4 In general, if A ⊆ [0, 1] is any subset of [0, 1], then we can deﬁne the shift of A by r for any 0 < r < 1 as A ⊕ r = {a + r : a ∈ A, a + r ≤ 1} ∪ {a + r − 1 : a ∈ A, a + r > 1}. 0 1 ⊕r 0 1 A⊕r A And s...
View Full Document

## This document was uploaded on 04/07/2014.

Ask a homework question - tutors are online