By Geiss

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4) has as law the normal ❊(f − ❊f )2 = σ2. 5) Proof. By the change of variable x → m + σx it is sufficient √ to show the statements for m = 0 and σ = 1. 7 so that ❘ p0,1 (x)dx = 1. Secondly, ❘ xp0,1 (x)dx = 0 follows from the symmetry of the density p0,1 (x) = p0,1 (−x). Finally, by partial integration (use (x exp(−x2 /2)) = exp(−x2 /2) − x2 exp(−x2 /2)) one can also compute that 1 √ 2π ∞ x2 1 x2 e− 2 dx = √ 2π −∞ ∞ x2 e− 2 dx = 1. −∞ We close this section with a “counterexample” to Fubini’s Theorem.

We only prove the first inequality. The second one follows from the definition of lim sup and lim inf, the third one can be proved like the first one. , |Zk | ≤ g and | lim inf fn | ≤ g. 5 gives that ❊ limninf fn = lim ❊Zk = lim ❊ n≥k inf fn k k ≤ lim inf k n≥k ❊fn = lim inf ❊fn . 7 [Lebesgue’s Theorem, dominated convergence] Let (Ω, F, P) be a probability space and g, f, f1 , f2 , ... s. s. Then f is integrable and one has that ❊f = lim ❊fn. n Proof. Applying Fatou’s Lemma gives ❊f = ❊ lim inf fn n→∞ ≤ lim inf ❊fn ≤ lim sup ❊fn ≤ n→∞ n→∞ ❊ lim sup fn = ❊f.

4 Assume a measurable space (Ω, F) and a sequence of random variables fn : Ω → ❘ such that f (ω) := limn fn (ω) exists for all ω ∈ Ω. Then f : Ω → ❘ is a random variable. The proof is an exercise. 5 [properties of random variables] Let (Ω, F) be a measurable space and f, g : Ω → ❘ random variables and α, β ∈ ❘. Then the following is true: (1) (αf + βg)(ω) := αf (ω) + βg(ω) is a random variable. (2) (f g)(ω) := f (ω)g(ω) is a random-variable. (3) If g(ω) = 0 for all ω ∈ Ω, then f g (ω) := f (ω) g(ω) is a random variable.