introduction-probability.pdf

# In other words the distribution function of the

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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 ) ) ,

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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 } ) .
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

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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 .
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