184 in other words it follows from 183 and 184 that

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Unformatted text preview: − ￿)Y ≤ X } and therefore Ak,n,￿ ⊆ Ak,n+1,￿ ⊆ {(1 − ￿)Y ≤ X } ∩ Ak . (18.2) Since Y ≤ X by assumption, we know that the event {(1 − ￿)Y ≤ X } is equal to Ω. Thus, (18.2) implies Ak,n,￿ ⊆ Ak for all n and so ∞ ￿ n=1 Ak,n,￿ ⊆ Ak . (18.3) Conversely, let ω ∈ Ak = {(1 − ￿)Y ≤ X } ∩ Ak = {ω : (1 − ￿)Y (ω ) ≤ X (ω )} ∩ Ak so that (1 − ￿)ak ≤ X (ω ). Since Y ≤ X , which is to say that Y (ω ) ≤ X (ω ) = lim Xn (ω ) n→∞ for all ω , we know that if ω ∈ Ak , then there is some N such that Y (ω ) = ak ≤ Xn (ω ) whenever n > N . We therefore conclude that if ω ∈ Ak and n > N , then (1 − ￿)ak < Xn (ω ). (This requires that X is not identically 0.) That is, ω ∈ Ak,n,￿ which proves that ∞ ￿ n=1 Ak,n,￿ ⊇ Ak . (18.4) In other words, it follows from (18.3) and (18.4) that Ak,n,￿ increases to Ak , so by the continuity of probability theorem (Theorem 10.2), we conclude that lim P {Ak,n,￿ } = P {Ak } . n→∞ Hence, E(Yn,￿ ) = (1 − ￿) That...
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This document was uploaded on 04/07/2014.

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