xj xj 1 and we can get the rst j terms in the sum by

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Unformatted text preview: roc univariate data=prob2_7 plots; histogram fatalities; run; 2.7c The problem asks us to verify the formula, xj +1 = xj + ¯ ¯ xj +1 − xj ¯ . j+1 Note that this is a formula for xj +1 in terms of xj and xj +1 . To see why ¯ ¯ the formula should be correct, let’s imagine that we had only xj and xj +1 ¯ and that we wanted to compute xj +1 : how would we proceed? ¯ Well, to compute xj +1 , we would want to start with the sum ¯ x1 + x2 + . . . + xj + xj +1 . And we can get the first j terms in the sum by multiplying xj by j . That is, ¯ we can get the whole sum as j xj + xj +1 , ¯ so that we could get xj +1 as ¯ j xj + xj +1 ¯ . j+1 So the question comes down to whether xj +1 − xj ¯ j xj + xj +1 ¯ = xj + ¯ . j+1 j+1 To see make the comparison easier, let’s write the right hand side as a fraction with denominator j + 1, so that our question becomes whether j xj + xj +1 ¯ (j + 1)¯j + xj +1 − xj x ¯ = . j+1 j+1 And a tiny bit of algebra in the numerator on the right hand s...
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This document was uploaded on 03/02/2014 for the course STATS 4150 at Barnard College.

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