Kireinacqiv

B e h a v i o u r P o l i c y ( i n c l u di n g E x c l u s i o n s ) P r e par e d by Se n i o r P ast o r al T e am L ast R e vi e we d Se pt e m be r 21 N e x t r e vi e w J u l y 22 St at ut or y Pol i cy C ont ent s Page A i m s of our pol i cy 1 W r i t t en s t at em ent of behavi our pr i nci pl es 1 T he 6 F HS val ues 2 Suppor t i ng our pos i t i ve behavi our cul t ur eSimilarly we can show that H(Z) H(X) (c) Consider the following joint distribution for Xand Y Let X= Y = ˆ 1 with probability 1=2 0 with probability 1=2 Then H(X) = H(Y) = 1, but Z= 0 with prob 1 and hence H(Z) = 0 (d) We have H(Z) H(X;Y) H(X) H(Y) because Zis a function of (X;Y) and H(X;Y) = H(X)H(YjX) H(X)H(Y) We have equality i (X;Y) is a function of Zand H(Y) = H(YjX),P V c a E R a Occ a a I H a a S C c W Z a b a a T a a ac c b R Bac , S a R 440 3 Geometry Topology Cohomology Homework 7 Solution Math 440 3 Studocu ƒƒ"ƒY "¯Œ^ ƒƒCƒ‹ƒh ƒxƒŠ[ƒVƒ‡[ƒg