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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. (4) |f | is a random variable. Proof. (2) We find measurable step-functions fn , gn : Ω → ❘ such that f (ω) = lim fn (ω) and g(ω) = lim gn (ω). n→∞ n→∞ Hence (f g)(ω) = lim fn (ω)gn (ω). n→∞ Finally, we remark, that fn (ω)gn (ω) is a measurable step-function. In fact, assuming that k fn (ω) = l αi 1IAi (ω) and gn (ω) = i=1 βj 1IBj (ω), j=1 yields k k l (fn gn )(ω) = l αi βj 1IAi ∩Bj (ω) αi βj 1IAi (ω)1IBj (ω) = i=1 j=1 i=1 j=1 and we again obtain a step-function, since Ai ∩ Bj ∈ F.
INTEGRATION On the other hand, using polar coordinates we get 1 2π |f (x, y)|d(µ × µ)(x, y) ≥ 4 [−1,1]×[−1,1] 0 0 1 = 2 0 | sin ϕ cos ϕ| dϕdr r 1 dr = ∞. r The inequality holds because on the right hand side we integrate only over the area {(x, y) : x2 + y 2 ≤ 1} which is a subset of [−1, 1] × [−1, 1] and 2π π/2 | sin ϕ cos ϕ|dϕ = 4 0 sin ϕ cos ϕdϕ = 2 0 follows by a symmetry argument. 6 Some inequalities In this section we prove some basic inequalities. 1 [Chebyshev’s inequality] Let f be a non-negative integrable random variable defined on a probability space (Ω, F, P).
How to construct such sequences? ) : xn ∈ ❘} . Then we define the projections Πn : ❘◆ → ❘ given by Πn (x) := xn , that means Πn filters out the n-th coordinate. , B ∈ B(❘) , 38 CHAPTER 2. 5. Finally, let P1 , P2 , ... 14) we B(❘). Using Carathe find an unique probability measure P on B(❘◆ ) such that P(B1 × B2 × · · · × Bn × ❘ × ❘ · · · ) = P1(B1) · · · Pn(Bn) for all n = 1, 2, ... , xn ∈ Bn . 8 [Realization of independent random variables] Let (❘◆ , B(❘◆ ), P) and πn : Ω → ❘ be defined as above.