*This preview shows
page 1. Sign up
to
view the full content.*

**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...

View
Full
Document