Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13
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Two positive-probability events are independent exactly when conditioning on either leaves the other's probability unchanged

Statement

Let A and B satisfy P(A)>0 and P(B)>0. The following are equivalent:

  1. A and B are independent;
  2. P(A∣B)=P(A);
  3. P(B∣A)=P(B).

Facts & Assumptions

Given: Positive-probability events A and B.

[L1]

Conditional probability is P(A∣B)=P(A∩B)/P(B) for P(B)>0 (Conditional probability P(A∣B) for P(B)>0).

[L2]

Independence means P(A∩B)=P(A)P(B) (Independent events, pairwise independence, and mutual independence of a finite family).

Proof

technique · direct
1.1

If A and B are independent, divide the identity in [L2] by P(B)>0 to obtain P(A∣B)=P(A).

L1L2algebra
1.2

The same calculation with A and B interchanged gives P(B∣A)=P(B).

L1L2algebra
1.3

Conversely, multiplying either conditional identity by its positive conditioning probability gives the product identity in [L2].

L1L2algebra
2.1

Thus each of conditions 2 and 3 is equivalent to condition 1, proving all three equivalent.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources