# Ecole d'Ete de Probabilites de Saint-Flour XIX - 1989 by Donald L. Burkholder, Etienne Pardoux, Alain-Sol Sznitman, PDF By Donald L. Burkholder, Etienne Pardoux, Alain-Sol Sznitman, Paul-Louis Hennequin

ISBN-10: 3540538410

ISBN-13: 9783540538417

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Extra info for Ecole d'Ete de Probabilites de Saint-Flour XIX - 1989

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This suggests that we deﬁne P by P(A) := ∑ p(ω ), ω ∈A where p(ω ) = p(e, d) is given by p(1, 1) := 1/4, p(0, 1) := 1/16, p(1, 0) := 1/2, and p(0, 0) := 3/16. 24) holds. 6: Independence Note 6. Here is an example of three events that are pairwise independent, but not mutually independent. Let Ω := {1, 2, 3, 4, 5, 6, 7}, and put P({ω }) := 1/8 for ω = 7, and P({7}) := 1/4. Take A := {1, 2, 7}, B := {3, 4, 7}, and C := {5, 6, 7}. Then P(A) = P(B) = P(C) = 1/2. and P(A ∩ B) = P(A ∩ C) = P(B ∩ C) = P({7}) = 1/4.

31) similarly, we have P(S2 ) = 3(1 − λ )λ 2 . , λ = 1/2, then P(S2 ) = 3/8. 26. If A1 , A2 , . . are mutually independent, show that P ∞ An ∞ = ∏ P(An ). n=1 n=1 Solution. 14), n=1 by independence, n=1 = ∞ ∏ P(An ), n=1 where the last step is just the deﬁnition of the inﬁnite product. h In working this example, we again do not explicitly specify the sample space Ω or the probability measure P. The interested reader can ﬁnd one possible choice for Ω and P in Note 8. 27. Consider the transmission of an unending sequence of bits over a noisy channel.

Similarly, if A consists of countably many 24 Introduction to probability sample points, say A = {ω1 , ω2 , . }, then directly from axiom (iii), P(A) = ∑∞ n=1 P({ωn }). Probability of a complement. Given an event A, we can always write Ω = A ∪ A c , which is a ﬁnite disjoint union. Hence, P(Ω) = P(A) + P(A c ). Since P(Ω) = 1, we ﬁnd that P(A c ) = 1 − P(A). 10) Monotonicity. If A and B are events, then A ⊂ B implies P(A) ≤ P(B). 11) To see this, ﬁrst note that A ⊂ B implies B = A ∪ (B ∩ A c ).