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1 n A 9 s d 1 Nj H f n DtC E J pJ x Y dh JCĖ clT s m P a wp R ݤ 2 䔛 T 5 3 F ΖE from MS 1 at Illinois Institute Of TechnologyBut then E(XjY =y)=åx xfXjY(xjy)=åx xfX(x)=E(X) 2 2 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 g''(x) = − e−x − (e−x − xe−x) ∴ g''(x) = − 2e−x xe−x Similarly the third derivative g(3)(x) = 2e−x e−x − xe−x ∴ g(3)(x) = 3e−x − xe−x So it looks like clear pattern is forming, but let us just check by looking at the fourth derivative; Prezentaciya Uprazhn Log Œ€ê"ŃEƒ‹ƒgƒ‰ƒ}ƒ"r/b ƒZƒŒƒNƒg ãJ‚̃NƒŠƒXƒ^ƒ‹ "zM

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