A Probability Path.pdf

# The relationship between lim sup and lim inf is lim

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The relationship between lim sup and lim inf is lim inf An C lim sup An n->oo n->00

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8 1. Sets and Events since {w: wE An, for all n ;::: no(w)} C {w: wE An infinitely often} =lim sup An. 2. Connections via de Morgan's laws: (liminfAn)c = since applying de Morgan's laws twice yields =lim sup For a sequence of random variables IXn, n :::: 0}, suppose we need to show X n X o almost surely. As we will see, this means that we need to show P{w: lim Xn(w) = Xo(w)} = 1. We will show later that a criterion for this is that for all £ > 0 P{[IXn- Xol > s] i.o.} = 0. That is , with An = [IXn- Xol > s], we need to check P (limsupAn) = 0. 1.4 Monotone Sequences A sequence of sets {An} is monotone non-decreasing if A 1 C Az C · · · . The sequence {An} is monotone non-increasing if A1 :J Az :J A3 ···.To indicate a monotone sequence we will use the notation An /' or An t for non-decreasing sets and An "\. or An .J, for non-increasing sets. For a monotone sequence of sets, the limit always exists. Proposition 1.4.1 Suppose {An} is a monotone sequence of subsets . (1) If An /', then An = An.
1.4 Monotone Sequences 9 (2) If An \o then limn-+oo An = Consequently, since for any sequences Bm we have it follows that liminfBn = lim (inf Bk), n-+oo n-+oo k;::n lim sup Bn = lim (sup Bk) . n-+oo n-+oo k;::n Proof. (1) We need to show 00 liminfAn =lim sup An= U An. n-+oo n-+00 n=l Since A j c A j+l> and therefore Likewise lim sup An n-+00 Thus equality prevails and = liminfAn n-+oo (from (1.1)) C limsupAn. n-+oo lim sup An C U Ak C lim sup An; n-+oo k?:l n-+oo therefore 00 lim sup An= U Ak. n-+00 k=l This coupled with (1.1) yields (1 ). The proof of (2) is similar. (1.1) 0

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10 1. Sets and Events Example 1.4.1 As an easy exercise, check that you believe that lim [0, 1- 1/n] = [0, 1) n-+oo lim [0, 1- 1/n) = [0, 1) n-+00 lim [0, 1 + 1/n] = [0, 1] n-+oo lim [0, 1 + 1/n) = [0, 1]. n-+00 0 Here are relations that provide additional parallels between sets and functions and further illustrate the algebraic properties of sets. As usual let {An} be a se- quence of subsets of n. 1. We have 2. The following inequality holds : lunAn ::S L lAn n and if the sequence {A;} is mutually disjoint, then equality holds. 3. We have 4. Symmetric difference satisfies the relation = lA + ls (mod 2) . Note (3) follows from (1) since and from (1) this is inf 1supk Ak. n;::l Again using (1) we get inf sup lAk = lim sup lAn , n;::l k;::n n-+oo from the definition of the lim sup of a sequence of numbers . To prove (1), we must prove two functions are equal. But 1 inf An (w) = 1 iff n>k w E infn;::k An = An iff w E An for all n 2: k iff lAn (w) ;;;, 1 for all n 2: k iff infn;::k IAn (w) = 1. 0
1.5 Set Operations and Closure 11 1.5 Set Operations and Closure In order to think about what it means for a class to be closed under certain set operations, let us first consider some typical set operations. Suppose C C 'P(Q) is a collection of subsets of Q. (1) Arbitrary union: Let T be any arbitrary index set and assume for each t e T that A 1 e C. The word arbitrary here reminds us that T is not nec- essarily finite, countable or a subset of the real line. The arbitrary union is (2) Countable union: Let An. n > 1 be any sequence of subsets in C. The countable union is (3) Finite union: Let A1, ... , An be any finite collection of subsets in C.

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