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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 < ￿ ≤ 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 > 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...
View Full Document

Ask a homework question - tutors are online