By Geiss C., Geiss S.

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**Example text**

Fn1 +n2 +n3 (ω)), ... are independent. The proof is an exercise. 3. 6 [independence and product of laws] Assume that (Ω, F, P) is a probability space and that f, g : Ω → ❘ are random variables with laws Pf and Pg and distribution-functions Ff and Fg , respectively. Then the following assertions are equivalent: (1) f and g are independent. (2) (3) P ((f, g) ∈ B) = (Pf × Pg )(B) for all B ∈ B(❘2). P(f ≤ x, g ≤ y) = Ff (x)Ff (y) for all x, y ∈ ❘. 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).

SOME INEQUALITIES 59 Proof. Let x0 = ❊f . Since g is convex we find a “supporting line”, that means a, b ∈ ❘ such that ax0 + b = g(x0 ) and ax + b ≤ g(x) for all x ∈ ❘. It follows af (ω) + b ≤ g(f (ω)) for all ω ∈ Ω and g(❊f ) = a❊f + b = ❊(af + b) ≤ ❊g(f ). 4 (1) The function g(x) := |x| is convex so that, for any integrable f , |❊f | ≤ ❊|f |. (2) For 1 ≤ p < ∞ the function g(x) := |x|p is convex, so that Jensen’s inequality applied to |f | gives that (❊|f |)p ≤ ❊|f |p . For the second case in the example above there is another way we can go.

3. 2 Let (Ω, F, P) be a probability space and f : Ω → random variable. ❘ be a (1) Then there exists a sequence of measurable step-functions fn : Ω → such that, for all n = 1, 2, . . and for all ω ∈ Ω, |fn (ω)| ≤ |fn+1 (ω)| ≤ |f (ω)| and ❘ f (ω) = lim fn (ω). n→∞ If f (ω) ≥ 0 for all ω ∈ Ω, then one can arrange fn (ω) ≥ 0 for all ω ∈ Ω. (2) If f ≥ 0 and if (fn )∞ n=1 is a sequence of measurable step-functions with 0 ≤ fn (ω) ↑ f (ω) for all ω ∈ Ω as n → ∞, then ❊f = n→∞ lim ❊fn . Proof. (1) It is easy to verify that the staircase-functions 4n −1 fn (ω) := k=−4n k 1I k k+1 (ω).