By Dorogovtsev A.Ya., et al.
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Extra info for Probability theory: collection of problems
Sample text
Bn = An \ An−1 . Because the sequence is increasing, we have that the Bi ’s are disjoint: n P(An ) = P(B1 ∪ B2 ∪ · · · ∪ Bn ) = P(Bi ). i=1 Thus using countable additivity we obtain An P =P n≥1 Bn n≥1 ∞ = n P(Bi ) = lim i=1 n→∞ P(Bi ) i=1 = lim P(An ) n→∞ 2. Note that An ↓ A ⇔ An c ↑ Ac , and from part 1 this means 1 − P(An ) ↑ 1 − P(A). 3. Let B1 = A1 , B2 = A1 ∪ A2 , . . , Bn = A1 ∪ · · · ∪ An , . . 2. Applying that result, we obtain P(Bn ) = P(A1 ∪ · · · ∪ An ) ≤ P(A1 ) + · · · + P(An ).
25 Let (Fi )i∈I be an independent family of -fields and let (Ij )j∈J be a partition of the set I (that is, Ik ∪ Il = ∅ if j = / l and j∈J Ij = I ). For every j ∈ J we define ⎞ ⎛ Fi ⎠ . ⎝ Gj = i∈Ij Then the family of -fields (Gj )j∈J is independent. Thus no matter how we associate (or group) -algebras which are independent the groups still are independent as long as there is no overlap. Proof: To prove the theorem, define for every j ∈ J Mj := {A1 ∩ . . ∩ Anj | Ai ∈ Fi , i ∈ Ij }, where nj is the number of algebras Fi indexed by Ij .
2. Calculate the probabilities of the following events: A ∪ B, Ac , Bc , B ∩ Ac , A ∪ Bc , and Ac ∪ B c . 6. 2. 8. 2 A coin example Two fair coins are tossed; find the probability that two heads are obtained. Solution: Each coin has two possible outcomes let us denote them H (heads) and T (tails). The sample space is given by = {(H, T ), (H, H ), (T, H ), (T, T )}. Since we know that the coin is fair, then it is equally likely to land on either outcome H or T . ’’ Then E = {(H, H )}. Since all are equally likely, we have P(E ) = |E | 1 = .