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

[summary:
$$~$
\newcommand{\bP}{\mathbb{P}}
$~$$
We say that two [event_probability 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)$~$.
Another way to state independence is that $~$\bP(A,B) = \bP(A) \bP(B)$~$. ]

We say that two [event_probability 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_probability 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)$~$.

## Comments

Eric Rogstad

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?