, If |r ? p(1 ? p)| ? ? 2 /4, then, by considering the off-diagonal term, we have: M (y) ? M 0 ? ? |? ? pp ? r|

, If |r ? p(1 ? p)| ? ? 2 /4, then, by considering the first term on the diagonal, we have M (y) ? M 0 ? ? ? 2 /4. If |r ? p(1 ? p)| ? ? 2 /4, then, by considering the off-diagonal term, we have: M (y) ? M 0 ? ? |? ? pp + r|. , we get

, Assume then that 1 ? p ? p < ??/2. For ? = r ? p(1 ? p) ?

, with f = W (y, ·), g = W (y , ·) and ? = 1 ? p ? p , we get that: ? ? (1 ? ?)(? ? ?)

, If |r ? p(1 ? p)| ? ? 2 /4, then, by considering the first term on the diagonal, we have M (y) ? M 0 ? ? ? 2 /4. If |r ? p(1 ? p)| ? ? 2 /4, then, by considering the off-diagonal term, we have: M (y) ? M 0 ? ? |? ? pp + r|

, 123) holds when M 0 ? F + , when M 0 ? F ? and either |1 ? p ? p | > ?/2 or

. Proof, Set f 1 = min(f, g) and g 1 = max(f, g) so 1 = [0,1] f g. Set h = min(f 1 , (1 ? ? ? g 1 )) as well as f 2 = f 1 ? h and

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