introduction-probability.pdf

Proposition 332 let f r r be a continuous function

Info icon This preview shows pages 56–59. Sign up to view the full content.

View Full Document Right Arrow Icon
Proposition 3.3.2 Let f : R R be a continuous function such that -∞ | f ( x ) | p ( x ) dx < .
Image of page 56

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 57
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
Image of page 58

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 59
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern