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EE(g(X)jY)=E(g(X)) Proof Set Z = g(X) Statement (i) of Theorem 1 applies to any two rv's Hence, applying it to Z and Y we obtain EE(ZjY)= E(Z) which is the same as EE(g(X)jY)= E(g(X)) 2 This property may seem to be more general statement than (i) in Theorem 1 The proof above shows that in fact these are equivalent statements 3REPS O v T C g Ѓ b v X ^ T t B b v X ^ T t B N ^ { N T p c l b g ^ Y E { N T p c X O W A X ^ Y E { N T p c X O W A X y V s X ^ Y A _ E G A W p ^ h b O@ l b g7 t 񓹂ɂ ƁA C 傪 炩 ɂ Ƃ ɂ ƁA ŋ߃ e g ` b v X ̃A ~ j E ܗL ʂ 𒴂 邱 Ƃ p Ă B Ƃ͐ Y ߒ ɃA ~ j E ܂܂ x L O p E _ Y 邾 炾 B A ~ j E Ȃ x L O p E _ ̉ i ́A ʂ̂ ̂Ɣ ׂ 3 `4 { ɂȂ B A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Badcode ƒ|ƒeƒgƒ`ƒbƒv ƒ|ƒeƒg ƒ`ƒbƒvƒX ƒCƒ‰ƒXƒg

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