Independence Complement Theorem

If sets and are Independent, then so are and , where is the complement of (i.e., the set of all possible outcomes not contained in ). Let $\cup$ denote ``or'' and $\cap$ denote ``and.'' Then

$\displaystyle P(E)$	$\textstyle =$	$\displaystyle P(EF\cup EF')$	(1)
	$\textstyle =$	$\displaystyle P(EF)+P(EF')-P(EF\cap EF'),$	(2)

where

is an abbreviation for $A\cap B$ . But

and

are independent, so

$\begin{displaymath} P(EF)=P(E)P(F). \end{displaymath}$

(3)

Also, since

and

are complements, they contain no common elements, which means that

$\begin{displaymath} P(EF\cap EF')=0 \end{displaymath}$

(4)

for any

. Plugging (4) and (3) into (2) then gives

$\begin{displaymath} P(E)= P(E)P(F)+P(EF'). \end{displaymath}$

(5)

Rearranging,

$\begin{displaymath} P(EF') = P(E)[1-P(F)] = P(E)P(F'), \end{displaymath}$

(6)

Q.E.D.

See also Independent Statistics