# Two independent events

$$\newcommand{\bP}{\mathbb{P}}$$

We say that two events, $$A$$ and $$B$$, are in­de­pen­dent when learn­ing that $$A$$ has oc­curred does not change your prob­a­bil­ity that $$B$$ oc­curs. That is, $$\bP(B \mid A) = \bP(B)$$. Equiv­a­lently, $$A$$ and $$B$$ are in­de­pen­dent if $$\bP(A)$$ doesn’t change if you con­di­tion on $$B$$: $$\bP(A \mid B) = \bP(A)$$.

Another way to state in­de­pen­dence is that $$\bP(A,B) = \bP(A) \bP(B)$$.

All these defi­ni­tions are equiv­a­lent:

$$\bP(A,B) = \bP(A)\; \bP(B \mid A)$$

by the chain rule, so

$$\bP(A,B) = \bP(A)\; \bP(B)\;\; \Leftrightarrow \;\; \bP(A)\; \bP(B \mid A) = \bP(A)\; \bP(B) \ ,$$

and similarly for $$\bP(B)\; \bP(A \mid B)$$.

Children:

Parents:

• Probability theory

The logic of sci­ence; co­her­ence re­la­tions on quan­ti­ta­tive de­grees of be­lief.

• I’m not sure what this equa­tion is try­ing to tell me. Some parts of it are only true if A and B are in­de­pen­dent, but some parts are true in gen­eral, right?

• Yeah this is maybe a place­holder to sup­port the lens, added stub tag.