introduction-probability.pdf

# Proposition 332 let f r r be a continuous function

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Proposition 3.3.2 Let f : R R be a continuous function such that -∞ | f ( x ) | p ( x ) dx < .

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3.3. CONNECTIONS TO THE RIEMANN-INTEGRAL 57 Then -∞ f ( x ) p ( 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 ( R , B ( R ) , P ) on the right-hand side. Let us consider two examples indicating the difference between the Riemann- integral and our expected value. Example 3.3.3 We give the standard example of a function which has an expected value, but which is not Riemann-integrable. Let f ( x ) := 1 , x [0 , 1] irrational 0 , x [0 , 1] rational . Then f is not Riemann integrable, but Lebesgue integrable with E f = 1 if we use the probability space ([0 , 1] , B ([0 , 1]) , λ ). Example 3.3.4 The expression lim t →∞ t 0 sin x x dx = π 2 is defined as limit in the Riemann sense although 0 sin x x + dx = and 0 sin x x - dx = . Transporting this into a probabilistic setting we take the exponential distri- bution with parameter λ > 0 from Section 1.3.6. Let f : R R be given by f ( x ) = 0 if x 0 and f ( x ) := sin x λx e λx if x > 0 and recall that the exponential distribution μ λ with parameter λ > 0 is given by the density p λ ( x ) = 1I [0 , ) ( x ) λe - λx . The above yields that lim t →∞ t 0 f ( x ) p λ ( x ) dx = π 2 but R f ( x ) + λ ( x ) = R f ( x ) - λ ( x ) = . Hence the expected value of f does not exists, but the Riemann-integral gives a way to define a value, which makes sense. The point of this example is that the Riemann-integral takes more information into the account than the rather abstract expected value.
58 CHAPTER 3. INTEGRATION 3.4 Change of variables in the expected value We want to prove a change of variable formula for the integrals Ω fd P . In many cases, only by this formula it is possible to compute explicitly expected values. Proposition 3.4.1 [Change of variables] Let , F , P ) be a probability space, ( E, E ) be a measurable space, ϕ : Ω E be a measurable map, and g : E R be a random variable. Assume that P ϕ is the image measure of P with respect to ϕ 1 , that means P ϕ ( A ) = P ( { ω : ϕ ( ω ) A } ) = P ( ϕ - 1 ( A )) for all A ∈ E . Then A g ( η ) d P ϕ ( η ) = ϕ - 1 ( A ) g ( ϕ ( ω )) d P ( ω ) for all A ∈ E in the sense that if one integral exists, the other exists as well, and their values are equal. Proof . (i) Letting g ( η ) := 1I A ( η ) g ( η ) we have g ( ϕ ( ω )) = 1I ϕ - 1 ( A ) ( ω ) g ( ϕ ( ω )) so that it is sufficient to consider the case A = E . Hence we have to show that E g ( η ) d P ϕ ( η ) = Ω g ( ϕ ( ω )) d P ( ω ) . (ii) Since, for f ( ω ) := g ( ϕ ( ω )) one has that f + = g + ϕ and f - = g - ϕ it is sufficient to consider the positive part of g and its negative part separately. In other words, we can assume that g ( η ) 0 for all η E . (iii) Assume now a sequence of measurable step-function 0 g n ( η ) g ( η ) for all η E which does exist according to Lemma 3.2.2 so that g n ( ϕ ( ω )) g ( ϕ ( ω )) for all ω Ω as well. If we can show that E g n ( η ) d P ϕ ( η ) = Ω g n ( ϕ ( ω )) d P ( ω ) then we are done. By additivity it is enough to check g n ( η ) = 1I B ( η ) for some B ∈ E (if this is true for this case, then one can multiply by real numbers and can take sums and the equality remains true). But now we get

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