851_lectures17_24

# That is take ak y ak let 0 1 and dene yn

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Unformatted text preview: ) Therefore, it follows from (18.1), the deﬁnition of E(X ), that a ≤ E(X ). We will now show a ≥ E(X ). As a result of (18.1), we only need to show that if Y is a simple random variable with 0 ≤ Y ≤ X , then E(Y ) ≤ a. That is, by deﬁnition, E(X ) = sup{E(Y ) : Y is simple with 0 ≤ Y ≤ X } and so if Y is an arbitrary simple random variable satisfying 0 ≤ Y ≤ X and E(Y ) ≤ a, then the deﬁnition of supremum implies E(X ) ≤ a. To this end, let Y be simple and write Y= m ￿ k=1 ak 1{Y = ak }. That is, take Ak = {ω : Y (ω ) = ak }. Let 0 &lt; ￿ ≤ 1 and deﬁne Yn,￿ = (1 − ￿)Y 1{(1−￿)Y ≤Xn } . Note that Yn,￿ = (1 − ￿)ak on the set {(1 − ￿)Y ≤ Xn } ∩ Ak = Ak,n,￿ and that Yn,￿ = 0 on the set {(1 − ￿)Y &gt; Xn }. Clearly Yn,￿ ≤ Xn and so E(Yn,￿ ) = (1 − ￿) m ￿ k=1 ak P {Ak,n,￿ } ≤ E(Xn ). 18–2 We will now show that Ak,n,￿ increases to Ak . Since Xn ≤ Xn+1 and Xn ↑ X we conclude that {(1 − ￿)Y ≤ Xn } ⊆ {(1 − ￿)Y ≤ Xn+1 } ⊆ {(1...
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