Statistical independence

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I want to phrase this intuitively, put that statement formally, then show that's equiv to claim 1 below.

I also want to explore how suppose $\mathbb{P}[A\vert B]\eq\mathbb{P}[A]$ is all we know, what can we say about independence then? Alec (talk) 13:47, 17 October 2017 (UTC)

Main reference: Template:RMSADAR - Mathematical Statistics and Data Analysis - 2nd edition - John A. Rice Alec (talk) 13:47, 17 October 2017 (UTC)

$\newcommand{\P}[1]{\mathbb{P}\left[#1\right]}$

Definition

Let $(S,\Omega,\mathbb{P})$ be a probability space and let $A,B\in\Omega$ be events. We say "$A$ and $B$ are statistically independent if:

• The probability of $A$ is the same as the probability of $A$ given $B$ has occurred, and,

  the probability of $B$ is the same as the probability of $B$ given $A$ has occurred.

  • We may write this symbolically (conditional probability) as:
    1. $\mathbb{P}[A]\eq\mathbb{P}[A\vert B]$
    2. $\mathbb{P}[B]\eq\mathbb{P}[B\vert A]$

Formally:

• For $A,B\in\Omega$, $A$ and $B$ are independent if $\big[\big(\mathbb{P}[A]\eq\mathbb{P}[A\vert B]\big)\wedge \big(\mathbb{P}[B]\eq\mathbb{P}[B\vert A]\big)\big]$
  • Claim 1: $A$ and $B$ are statistically independent $\iff\big[\mathbb{P}[A\cap B]\eq\mathbb{P}[A]\cdot\mathbb{P}[B]\big]$
    • Formally: $\forall A,B\in\Omega\left[\big(\P{A\vert B}\eq\P{A}\wedge\P{B\vert A}\eq\P{B}\big)\iff\big(\P{A\cap B}\eq\P{A}\cdot\P{B}\big)\right]$

Proof of claims

Claim 1:

Claim: $\forall A,B\in\Omega\left[\big(\P{A\vert B}\eq\P{A}\wedge\P{B\vert A}\eq\P{B}\big)\iff\big(\P{A\cap B}\eq\P{A}\cdot\P{B}\big)\right]$