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Theorem 1.1 Let N be a countable set. Let p → [0, 1] be any function satisfying ΣωΩΡ(ω) = 1. Then there is a valid probability triple (N, F, P) where F = 22 and P(A) ΣWEAP(w). =

Theorem 1.1 Let N be a countable set. Let p → [0, 1] be any function satisfying ΣωΩΡ(ω) = 1. Then there is a valid probability triple (N, F, P) where F = 22 and P(A) ΣWEAP(w). =

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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Please if able explain the following theorem and example in more detail, I'm struggling with understanding them both. I'm quite new to Statistics

Theorem 1.1 Let be a countable set. Let p[0, 1] be any function satisfying Ewen P(w) = 1. Then there is a valid probability triple (N, F, P) where F = 2º and P(A) = ΣWEAP(W). - Example: Let us consider the geometric distribution with parameter p € (0, 1). Then 2 = = N₁ F = 22 and P(A) = ΣkɛA P(1 − p)k-1. The power set F is clearly a sigma-algebra since it contains every subset of N. We will show that P satisfies Kolmogorov's axioms. First of all P is a map from F to [0, 1]. Secondly ∞ P(N) = Σp(1 − p)k−¹ = pΣ(1 − p)k = 1 k=1 k=0 Р (1 - p) and third let us consider a sequence of disjoint events A₁, A2,.... Then ³ (ŨAn) - P = 1 ∞ Σ p(1-p)k-1 A; disjoint Σ Σ p(1 - p)k-1 = ΣP(An). KEU1 An n=1 k€ An n=1
Transcribed Image Text:Theorem 1.1 Let be a countable set. Let p[0, 1] be any function satisfying Ewen P(w) = 1. Then there is a valid probability triple (N, F, P) where F = 2º and P(A) = ΣWEAP(W). - Example: Let us consider the geometric distribution with parameter p € (0, 1). Then 2 = = N₁ F = 22 and P(A) = ΣkɛA P(1 − p)k-1. The power set F is clearly a sigma-algebra since it contains every subset of N. We will show that P satisfies Kolmogorov's axioms. First of all P is a map from F to [0, 1]. Secondly ∞ P(N) = Σp(1 − p)k−¹ = pΣ(1 − p)k = 1 k=1 k=0 Р (1 - p) and third let us consider a sequence of disjoint events A₁, A2,.... Then ³ (ŨAn) - P = 1 ∞ Σ p(1-p)k-1 A; disjoint Σ Σ p(1 - p)k-1 = ΣP(An). KEU1 An n=1 k€ An n=1
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