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The proof is an exercise. 7 Assume that there are Riemann-integrable functions pf , pg : ❘ → [0, ∞) such that ❘ pf (x)dx = ❘ pg (x)dx = 1, x x pf (y)dy, Ff (x) = pg (y)dy and Fg (x) = −∞ −∞ for all x ∈ ❘ (one says that the distribution-functions Ff and Fg are absolutely continuous with densities pf and pg , respectively). Then the independence of f and g is also equivalent to x y −∞ −∞ pf (u)pg (v)d(u)d(v). F(f,g) (x, y) = 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.

1 Definition of the expected value The definition of the integral is done within three steps. 1 [step one, f is a step-function] Given a probability space (Ω, F, P) and an F-measurable g : Ω → ❘ with representation n g= αi 1IAi i=1 where αi ∈ ❘ and Ai ∈ F, we let ❊g = Ω gdP = Ω g(ω)dP(ω) := n αi P(Ai ). i=1 We have to check that the definition is correct, since it might be that different representations give different expected values ❊g. 2 Assuming measurable step-functions n g= m αi 1IAi = i=1 one has that n i=1 αi P(Ai ) = βj 1IBj , j=1 m j=1 βj P(Bj ).

2 [Convergence in distribution] Let (Ωn , Fn , Pn ) and (Ω, F, P) be probability spaces and let fn : Ωn → ❘ and f : Ω → ❘ be random variables. Then the sequence (fn )∞ n=1 converges in distribution d to f (fn → f ) if and only if ❊ψ(fn) → ψ(f ) as n → ∞ for all bounded and continuous functions ψ : ❘ → ❘. 63 64 CHAPTER 4. MODES OF CONVERGENCE We have the following relations between the above types of convergence. 3 Let (Ω, F, P) be a probability space and f, f1 , f2 , · · · : Ω → ❘ be random variables.

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An Introduction to Probability Theory by Geiss


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