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

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Unformatted text preview: Homework 3 Solutions (a) H 2 ( P,Q ) = 1 2 Z ( √ p- √ q ) 2 dμ = 1 2 Z ( p + q- 2 √ pq ) dμ = 1 2 Z (2- 2 √ pq ) dμ = 1- Z √ pqdμ = 1- ρ ( 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 ) dμ | = Z p ≥ q ( p- q ) dμ ∨ Z p<q ( q- p ) dμ . However, from the equation Z p ≥ q ( p- q ) dμ- Z p<q ( q- p ) dμ = Z pdμ- Z qdμ = 0 , d TV ( P,Q ) = Z p ≥ q ( p- q ) dμ = 1- 1- Z p ≥ q ( p- q ) dμ = 1- Z p ≥ q pdμ- 1- Z p ≥ q qdμ = 1- Z p ≥ q pdμ- Z p<q qdμ = 1- Z p ∧ qdμ. 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 ∧ qdμ 2 ≤ 1- ρ 2 ( P,Q ) = 1- Z √ pqdμ 2 which is transformed to- 2 Z p ∧ qdμ + Z p ∧ qdμ 2 + Z √ pqdμ 2 ≤ . Now we have an equality pq = ( p ∧ q )( p ∨ q ). By Cauchy’s inequality (R √ pq ) 2 ≤ R ( p ∧ q ) R ( p ∨ q ). Plugging this into RHS of the above inequality,- 2 Z p ∧ qdμ + Z p ∧ qdμ 2 + Z √ pqdμ 2 ≤ - 2 Z p ∧ qdμ + Z p ∧ qdμ 2 + Z p ∧ qdμ Z p ∨ qdμ = Z p ∧ qdμ- 2 + Z p ∧ qdμ + Z p ∨ qdμ = 0 since p ∧ q + p ∨ q = p + q . The third inequality follows from Cauchy’s inequality....
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hw3w09sol - Homework 3 Solutions(a H 2 P,Q = 1 2 Z √ p...

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