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 [ilmath]\mathbb{P}[A\vert B]\eq\mathbb{P}[A][/ilmath] 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)

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

Definition

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

• The probability of [ilmath]A[/ilmath] is the same as the probability of [ilmath]A[/ilmath] given [ilmath]B[/ilmath] has occurred, and,

  the probability of [ilmath]B[/ilmath] is the same as the probability of [ilmath]B[/ilmath] given [ilmath]A[/ilmath] has occurred.

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

Formally:

• For [ilmath]A,B\in\Omega[/ilmath], [ilmath]A[/ilmath] and [ilmath]B[/ilmath] are independent if [ilmath]\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][/ilmath]
  • Claim 1: [ilmath]A[/ilmath] and [ilmath]B[/ilmath] are statistically independent [ilmath]\iff\big[\mathbb{P}[A\cap B]\eq\mathbb{P}[A]\cdot\mathbb{P}[B]\big][/ilmath]
    • Formally: [ilmath]\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][/ilmath]

Proof of claims

Claim 1:

Claim: [ilmath]\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][/ilmath]