hw3w09sol - Homework 3 Solutions (a) H 2 ( P,Q ) = 1 2 Z (...

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Unformatted text preview: Homework 3 Solutions (a) H 2 ( P,Q ) = 1 2 Z ( p- q ) 2 d = 1 2 Z ( p + q- 2 pq ) d = 1 2 Z (2- 2 pq ) d = 1- Z pqd = 1- ( P,W ) . (b) Since { p q } and { p < q } are measurable, d TV ( P,Q ) = sup A A | P ( A )- Q ( A ) | = sup A A | Z A ( p- q ) d | = Z p q ( p- q ) d Z p<q ( q- p ) d . However, from the equation Z p q ( p- q ) d- Z p<q ( q- p ) d = Z pd- Z qd = 0 , d TV ( P,Q ) = Z p q ( p- q ) d = 1- 1- Z p q ( p- q ) d = 1- Z p q pd- 1- Z p q qd = 1- Z p q pd- Z p<q qd = 1- Z p qd. 1 (c) First inequality comes from (a), (b) and the fact pq p q . For the second inequality, by taking the square or both sides the inequality we have to prove forms 1- Z p qd 2 1- 2 ( P,Q ) = 1- Z pqd 2 which is transformed to- 2 Z p qd + Z p qd 2 + Z pqd 2 . Now we have an equality pq = ( p q )( p q ). By Cauchys inequality (R pq ) 2 R ( p q ) R ( p q ). Plugging this into RHS of the above inequality,- 2 Z p qd + Z p qd 2 + Z pqd 2 - 2 Z p qd + Z p qd 2 + Z p qd Z p qd = Z p qd- 2 + Z p qd + Z p qd = 0 since p q + p q = p + q . The third inequality follows from Cauchys inequality....
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hw3w09sol - Homework 3 Solutions (a) H 2 ( P,Q ) = 1 2 Z (...

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