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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 3E g(X) ≥ g( EX ) Source Wolfram MathWorld a) Proof Let g(X) be −√$, for all positive real numbers g(x) is a convex function as seen by the graph to the right Since g(X) satisfies the requirements for Jensen's inequality, we can use it to directly prove the inequalityLinks with this icon indicate that you are leaving the CDC website The Centers for Disease Control and Prevention (CDC) cannot attest to the accuracy of a nonfederal website Linking to a nonfederal website does not constitute an endorsement by CDC or any of its employees of the sponsors or the information and products presented on the website
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We see in the above pictures that (W ⊥) ⊥ = W Example The orthogonal complement of R n is {0}, since the zero vector is the only vector that is orthogonal to all of the vectors in R n For the same reason, we have {0} ⊥ = R n Subsection 622 Computing Orthogonal Complements Since any subspace is a span, the following proposition gives a recipe for computing the orthogonalPlayer 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 SelfPartnering Total Pos'n A A v2 TXAB vi 21 ZQCA v14 DAJM vii 3 EAVI v4 AFNO iii 19 RBGAThen E(y g(X)) 2 is minimized when g(X) = EYjX Lecture 26 Examples I Toss 100 coins What's the conditional expectation of the number of heads given the number of heads among the rst fty tosses?






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