introduction-probability.pdf

In other words the distribution function of the

Info icon This preview shows pages 43–47. Sign up to view the full content.

View Full Document Right Arrow Icon
In other words: the distribution-function of the random vector ( f, g ) has a density which is the product of the densities of f and g . Often one needs the existence of sequences of independent random variables f 1 , f 2 , ... : Ω R having a certain distribution. How to construct such sequences? First we let Ω := R N = { x = ( x 1 , x 2 , ... ) : x n R } . Then we define the projections π n : R N R given by π n ( x ) := x n , that means π n filters out the n -th coordinate. Now we take the smallest σ -algebra such that all these projections are random variables, that means we take B ( R N ) = σ ( π - 1 n ( B ) : n = 1 , 2 , ..., B ∈ B ( R ) ) ,
Image of page 43

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

View Full Document Right Arrow Icon
44 CHAPTER 2. RANDOM VARIABLES see Proposition 1.1.6. Finally, let P 1 , P 2 , ... be a sequence of probability measures on B ( R ). Using Carath´eodory ’s extension theorem (Proposition 1.2.17) we find an unique probability measure P on B ( R N ) such that P ( B 1 × B 2 × · · · × B n × R × R × · · · ) = P 1 ( B 1 ) · · · P n ( B n ) for all n = 1 , 2 , ... and B 1 , ..., B n ∈ B ( R ), where B 1 × B 2 × · · · × B n × R × R × · · · := x R N : x 1 B 1 , ..., x n B n . Proposition 2.3.7 [Realization of independent random variab- les] Let ( R N , B ( R N ) , P ) and π n : R N R be defined as above. Then ( π n ) n =1 is a sequence of independent random variables such that the law of π n is P n , that means P ( π n B ) = P n ( B ) for all B ∈ B ( R ) . Proof . Take Borel sets B 1 , ..., B n ∈ B ( R ). Then P ( { x : π 1 ( x ) B 1 , ..., π n ( x ) B n } ) = P ( B 1 × B 2 × · · · × B n × R × R × · · · ) = P 1 ( B 1 ) · · · P n ( B n ) = n k =1 P ( R × · · · × R × B k × R × · · · ) = n k =1 P ( { x : π k ( x ) B k } ) .
Image of page 44
Chapter 3 Integration Given a probability space (Ω , F , P ) and a random variable f : Ω R , we define the expectation or integral E f = Ω fd P = Ω f ( ω ) d P ( ω ) and investigate its basic properties. 3.1 Definition of the expected value The definition of the integral is done within three steps. Definition 3.1.1 [step one, f is a step-function] Given a probability space (Ω , F , P ) and an F -measurable g : Ω R with representation g = n i =1 α i 1I A i where α i R and A i ∈ F , we let E g = Ω gd P = Ω g ( ω ) d P ( ω ) := n i =1 α i P ( A i ) . We have to check that the definition is correct, since it might be that different representations give different expected values E g . However, this is not the case as shown by Lemma 3.1.2 Assuming measurable step-functions g = n i =1 α i 1I A i = m j =1 β j 1I B j , one has that n i =1 α i P ( A i ) = m j =1 β j P ( B j ) . 45
Image of page 45

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

View Full Document Right Arrow Icon
46 CHAPTER 3. INTEGRATION Proof . By subtracting in both equations the right-hand side from the left- hand one we only need to show that n i =1 α i 1I A i = 0 implies that n i =1 α i P ( A i ) = 0 . By taking all possible intersections of the sets A i and by adding appropriate complements we find a system of sets C 1 , ..., C N ∈ F such that (a) C j C k = if j = k , (b) N j =1 C j = Ω, (c) for all A i there is a set I i ⊆ { 1 , ..., N } such that A i = j I i C j .
Image of page 46
Image of page 47
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