introduction-probability.pdf

Proposition 326 lemma of fatou let ω f p be a

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Proposition 3.2.6 [Lemma of Fatou] Let , F , P ) be a probability space and g, f 1 , f 2 , ... : Ω R be random variables with | f n ( ω ) | ≤ g ( ω ) a.s. Assume that g is integrable. Then lim sup f n and lim inf f n are integrable and one has that E lim inf n →∞ f n lim inf n →∞ E f n lim sup n →∞ E f n E lim sup n →∞ f n . Proof . We only prove the first inequality. The second one follows from the definition of lim sup and lim inf, the third one can be proved like the first one. So we let Z k := inf n k f n so that Z k lim inf n f n and, a.s., | Z k | ≤ g and | lim inf n f n | ≤ g.
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3.2. BASIC PROPERTIES OF THE EXPECTED VALUE 55 Applying monotone convergence in the form of Corollary 3.2.5 gives that E lim inf n f n = lim k E Z k = lim k E inf n k f n lim k inf n k E f n = lim inf n E f n . Proposition 3.2.7 [Lebesgue’s Theorem, dominated convergence] Let , F , P ) be a probability space and g, f, f 1 , f 2 , ... : Ω R be random variables with | f n ( ω ) | ≤ g ( ω ) a.s. Assume that g is integrable and that f ( ω ) = lim n →∞ f n ( ω ) a.s. Then f is integrable and one has that E f = lim n E f n . Proof . Applying Fatou’s Lemma gives E f = E lim inf n →∞ f n lim inf n →∞ E f n lim sup n →∞ E f n E lim sup n →∞ f n = E f. Finally, we state a useful formula for independent random variables. Proposition 3.2.8 If f and g are independent and E | f | < and E | g | < , then E | fg | < and E fg = E f E g. The proof is an exercise. Concerning the variance of the sum of independent random variables we get the fundamental Proposition 3.2.9 Let f 1 , ..., f n be independent random variables with finite second moment. Then one has that var( f 1 + · · · + f n ) = var( f 1 ) + · · · + var( f n ) . Proof . The formula follows from var( f 1 + · · · + f n ) = E (( f 1 + · · · + f n ) - E ( f 1 + · · · + f n )) 2 = E n i =1 ( f i - E f i ) 2 = E n i,j =1 ( f i - E f i )( f j - E f j ) = n i,j =1 E (( f i - E f i )( f j - E f j ))
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56 CHAPTER 3. INTEGRATION = n i =1 E ( f i - E f i ) 2 + i = j E (( f i - E f i )( f j - E f j )) = n i =1 var( f i ) + i = j E ( f i - E f i ) E ( f j - E f j ) = n i =1 var( f i ) because E ( f i - E f i ) = E f i - E f i = 0. 3.3 Connections to the Riemann-integral In two typical situations we formulate (without proof) how our expected value connects to the Riemann-integral. For this purpose we use the Lebesgue measure defined in Section 1.3.4. Proposition 3.3.1 Let f : [0 , 1] R be a continuous function. Then 1 0 f ( x ) dx = E f with the Riemann-integral on the left-hand side and the expectation of the random variable f with respect to the probability space ([0 , 1] , B ([0 , 1]) , λ ) , where λ is the Lebesgue measure, on the right-hand side. Now we consider a continuous function p : R [0 , ) such that -∞ p ( x ) dx = 1 and define a measure P on B ( R ) by P (( a 1 , b 1 ] ∩ · · · ∩ ( a n , b n ]) := n i =1 b i a i p ( x ) dx for -∞ ≤ a 1 b 1 ≤ · · · ≤ a n b n ≤ ∞ (again with the convention that ( a, ] = ( a, )) via Carath´ eodory’s Theorem (Proposition 1.2.17). The function p is called density of the measure P .
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