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By definition, y ∈ f(A)∩f(B) ⊇ Let y ∈ f(A) ∩ f(B) Then y ∈ f(A) and y ∈ f(B) Thus there exists x 1 ∈ A with f(x 1) = y and there exists x 2 ∈ B with f(x 2) = y By injectivity of f we have x 1 = x 2, and thus x 1 ∈ B, too So x 1 ∈ A ∩ B and hence y = f(x 1) ∈ f(A∩B) 1222 (b) Prove that f(A \ B) = f(A) \ f(B3 Suppose that f R !R is a continuous function such that lim x!1 f(x) = 0;07 N5 ・@ E t F b g ・ Wヲ ・J ・I @0744 X V u 07 N f V ^ O v ・ I @ X V N C X 00C SRT8 ・A ・・・f J ・・・o ・I @ X V Letter Formation The Ot Toolbox ƒfƒjƒ€ƒWƒƒƒPƒbƒg ƒŒƒfƒB[ƒX ƒR[ƒf "~

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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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