Derivation 2 sy 1 n1 1 n1 1 n1 b2 n yi y

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: )2 i =1 1 n−1 n ¯ (xi − x )2 i =1 Let yi = bxi , then 2 2 sy = b2 sx . Derivation. 2 sy = = = 1 n−1 1 n−1 1 n−1 = b2 n ¯ (yi − y )2 i =1 n ¯ (bxi − bx )2 i =1 n ¯ b2 (xi − x )2 i =1 1 n−1 2 = b 2 sx n ¯ (xi − x )2 i =1 Let yi = byi + a, then rxy = Let yi = byi + a, then rxy = rxy . Let yi = byi + a, then rxy = rxy . Derivation. rxy = 1 n−1 n i =1 (xi ¯ ¯ − x )(yi − y ) sx sy Let yi = byi + a, then rxy = rxy . Derivation. rxy = 1 n−1 = 1 n−1 n i =1 (xi ¯ ¯ − x )(yi − y ) sx sy n ¯ ¯ i =1 (xi − x )([byi + a] − b [y + a]) sx bsy Let yi = byi + a, then rxy = rxy . Derivation. rxy = 1 n−1 = 1 n−1 = 1 n−1 n i =1 (xi ¯ ¯ − x )(yi − y ) sx sy n ¯ ¯ i =1 (xi − x )([byi + a] − b [y + a]) sx bsy n ¯ ¯ i =1 (xi − x )(yi − y ) sx sy Let yi = byi + a, then rxy = rxy . Derivation. rxy = 1 n−1 = 1 n−1 = 1 n−1 = rxy n i =1 (xi ¯ ¯ − x )(yi − y ) sx sy n ¯ ¯ i =1 (xi − x )([byi + a] − b [y + a]) sx bsy n ¯ ¯ i =1 (xi − x )(yi − y ) sx sy Let yi = byi + a, then rxy = rxy . Derivation. rxy = 1 n−1 = 1 n−1 = 1 n−1 = rxy n i =1 (xi ¯ ¯ − x )(yi − y ) sx sy n ¯ ¯ i =1 (xi − x )([byi + a] − b [y + a]) sx bsy n ¯ ¯ i =1 (xi − x )(yi − y ) sx sy Let yi = bxi + a, then rxy = Let yi = bxi + a, then rxy = 1 (or -1, if b is negative). Let yi = bxi + a, then rxy = 1 (or -1, if b is negative). Derivation. rxy = 1 n−1 n i =1 (xi ¯ ¯ − x )(y − y ) sx sy Let yi = bxi + a, then rxy = 1 (or -1, if b is negative). Derivation. rxy = 1 n−1 n i =1 (xi = 1 n−1 n i =1 (xi ¯ ¯ − x )(y − y ) sx sy ¯ ¯ − x )([bxi + a] − [bx + a]) sx 2 b2 sx Let yi = bxi + a, then rxy = 1 (or -1, if b is negative). Derivation. rxy = 1 n−1 n i =1 (xi = 1 n−1 n i =1 (xi = 1 n−1 n i =1 (xi ¯ ¯ − x )(y − y ) sx sy ¯ ¯ − x )([bxi + a] − [bx + a]) sx 2 b2 sx ¯ ¯ − x )(xi − x ) sign(b) sx sx Let yi = bxi + a, then rxy = 1 (or -1, if b is negative). Derivation. rxy = 1 n−1 n i =1 (xi = 1 n−1 n i =1 (xi = 1 n−1 n i =1 (xi ¯ ¯ − x )(y − y ) sx sy ¯ ¯ − x )([bxi + a] − [bx + a]) sx 2 b2 sx ¯ ¯ − x )(xi − x ) sign(b) sx sx 2 = sx sign(b) /sx sx Let yi = bxi + a, then rxy = 1 (or -1, if b is negative). Derivation. rxy = 1 n−...
View Full Document

This document was uploaded on 03/02/2014 for the course STATS 4150 at Barnard College.

Ask a homework question - tutors are online