introduction-probability.pdf

# The proof is an exercise proposition 215 properties

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The proof is an exercise. Proposition 2.1.5 [properties of random variables] Let , F ) be a measurable space and f, g : Ω R random variables and α, β R . Then the following is true: (1) ( αf + βg )( ω ) := αf ( ω ) + βg ( ω ) is a random variable. (2) ( fg )( ω ) := f ( ω ) g ( ω ) is a random-variable. (3) If g ( ω ) = 0 for all ω Ω , then f g ( ω ) := f ( ω ) g ( ω ) is a random variable. (4) | f | is a random variable. Proof . (2) We find measurable step-functions f n , g n : Ω R such that f ( ω ) = lim n →∞ f n ( ω ) and g ( ω ) = lim n →∞ g n ( ω ) . Hence ( fg )( ω ) = lim n →∞ f n ( ω ) g n ( ω ) . Finally, we remark, that f n ( ω ) g n ( ω ) is a measurable step-function. In fact, assuming that f n ( ω ) = k i =1 α i 1I A i ( ω ) and g n ( ω ) = l j =1 β j 1I B j ( ω ) , yields ( f n g n )( ω ) = k i =1 l j =1 α i β j 1I A i ( ω )1I B j ( ω ) = k i =1 l j =1 α i β j 1I A i B j ( ω ) and we again obtain a step-function, since A i B j ∈ F . Items (1), (3), and (4) are an exercise. 2.2 Measurable maps Now we extend the notion of random variables to the notion of measurable maps, which is necessary in many considerations and even more natural.

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38 CHAPTER 2. RANDOM VARIABLES Definition 2.2.1 [measurable map] Let (Ω , F ) and ( M, Σ) be measurable spaces. A map f : Ω M is called ( F , Σ) -measurable , provided that f - 1 ( B ) = { ω Ω : f ( ω ) B } ∈ F for all B Σ . The connection to the random variables is given by Proposition 2.2.2 Let , F ) be a measurable space and f : Ω R . Then the following assertions are equivalent: (1) The map f is a random variable. (2) The map f is ( F , B ( R )) -measurable. For the proof we need Lemma 2.2.3 Let , F ) and ( M, Σ) be measurable spaces and let f : Ω M . Assume that Σ 0 Σ is a system of subsets such that σ 0 ) = Σ . If f - 1 ( B ) ∈ F for all B Σ 0 , then f - 1 ( B ) ∈ F for all B Σ . Proof . Define A := B M : f - 1 ( B ) ∈ F . By assumption, Σ 0 ⊆ A . We show that A is a σ –algebra. (1) f - 1 ( M ) = Ω ∈ F implies that M ∈ A . (2) If B ∈ A , then f - 1 ( B c ) = { ω : f ( ω ) B c } = { ω : f ( ω ) / B } = Ω \ { ω : f ( ω ) B } = f - 1 ( B ) c ∈ F . (3) If B 1 , B 2 , · · · ∈ A , then f - 1 i =1 B i = i =1 f - 1 ( B i ) ∈ F . By definition of Σ = σ 0 ) this implies that Σ ⊆ A , which implies our lemma. Proof of Proposition 2.2.2. (2) = (1) follows from ( a, b ) ∈ B ( R ) for a < b which implies that f - 1 (( a, b )) ∈ F . (1) = (2) is a consequence of Lemma 2.2.3 since B ( R ) = σ (( a, b ) : -∞ < a < b < ).
2.2. MEASURABLE MAPS 39 Example 2.2.4 If f : R R is continuous, then f is ( B ( R ) , B ( R )) - measurable. Proof . Since f is continuous we know that f - 1 (( a, b )) is open for all -∞ < a < b < , so that f - 1 (( a, b )) ∈ B ( R ). Since the open intervals generate B ( R ) we can apply Lemma 2.2.3. Now we state some general properties of measurable maps. Proposition 2.2.5 Let 1 , F 1 ) , 2 , F 2 ) , 3 , F 3 ) be measurable spaces. Assume that f : Ω 1 Ω 2 is ( F 1 , F 2 ) -measurable and that g : Ω 2 Ω 3 is ( F 2 , F 3 ) -measurable. Then the following is satisfied: (1) g f : Ω 1 Ω 3 defined by ( g f )( ω 1 ) := g ( f ( ω 1 )) is ( F 1 , F 3 ) -measurable.

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