An Introduction to Probability Theory by Geiss

By Geiss

Show description

Read or Download An Introduction to Probability Theory PDF

Best probability books

Applied Adaptive Statistical Methods: Tests of Significance and Confidence Intervals

ASA-SIAM sequence on facts and utilized likelihood 12 Adaptive statistical checks, built over the past 30 years, are usually extra strong than conventional exams of value, yet haven't been standard. thus far, discussions of adaptive statistical equipment were scattered around the literature and usually don't comprise the pc courses essential to make those adaptive equipment a realistic replacement to standard statistical equipment.

Additional resources for An Introduction to Probability Theory

Example text

1) we assume random variables ϕ, ψ : Ω → ❘ such that ϕ ≥ 0 and ψ ≥ 0. We find ∞ measurable step functions (ϕn )∞ n=1 and (ψn )n=1 with 0 ≤ ϕn (ω) ↑ ϕ(ω) and 0 ≤ ψn (ω) ↑ ψ(ω) for all ω ∈ Ω. 3, and ϕn (ω) + ψn (ω) ↑ ϕ(ω) + ψ(ω) give that ❊ϕ + ❊ψ = lim ❊ϕn + lim ❊ψn = lim ❊(ϕn + ψn) = ❊(ϕ + ψ). n n n (2) is an exercise. (3) If ❊f − = ∞ or ❊g + = ∞, then ❊f = −∞ or ❊g = ∞ so that nothing is to prove. Hence assume that ❊f − < ∞ and ❊g + < ∞. The inequality f ≤ g gives 0 ≤ f + ≤ g + and 0 ≤ g − ≤ f − so that f and g are integrable and ❊f = ❊f + − ❊f − ≤ ❊g+ − ❊g− = ❊g.

K=1 Fubini’s Theorem In this section we consider iterated integrals, as they appear very often in applications, and show in Fubini’s Theorem that integrals with respect to product measures can be written as iterated integrals and that one can change the order of integration in these iterated integrals. In many cases this provides an appropriate tool for the computation of integrals. Before we start with Fubini’s Theorem we need some preparations. First we recall the notion of a vector space. 1 [vector space] A set L equipped with operations + : L × L → L and · : ❘ × L → L is called vector space over ❘ if the following conditions are satisfied: (1) x + y = y + x for all x, y ∈ L.

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. It ¨ lder-inequality. 5 [Ho (Ω, F, P) and random variables f, g : Ω → ❘. If 1 < p, q < ∞ with p1 + 1q = 1, then ❊|f g| ≤ (❊|f |p) p1 (❊|g|q ) 1q . Proof. We can assume that ❊|f |p > 0 and ❊|g|q > 0. s. s. and ❊|f g| = 0.

Download PDF sample

Rated 4.30 of 5 – based on 43 votes