Editing Independence (probability theory) (section)

===Pairwise and mutual independence===
[[File:Pairwise independent.svg|thumb|Pairwise independent, but not mutually independent, events]]
[[File:Mutually independent.svg|thumb|Mutually independent events]]

Consider the two probability spaces shown. In both cases, <math>\mathrm{P}(A) = \mathrm{P}(B) = 1/2</math> and <math>\mathrm{P}(C) = 1/4</math>. The events in the first space are pairwise independent because <math>\mathrm{P}(A|B) = \mathrm{P}(A|C)=1/2=\mathrm{P}(A)</math>, <math>\mathrm{P}(B|A) = \mathrm{P}(B|C)=1/2=\mathrm{P}(B)</math>, and <math>\mathrm{P}(C|A) = \mathrm{P}(C|B)=1/4=\mathrm{P}(C)</math>; but the three events are not mutually independent. The events in the second space are both pairwise independent and mutually independent. To illustrate the difference, consider conditioning on two events. In the pairwise independent case, although any one event is independent of each of the other two individually, it is not independent of the intersection of the other two:

:<math>\mathrm{P}(A|BC) = \frac{\frac{4}{40}}{\frac{4}{40} + \frac{1}{40}} = \tfrac{4}{5} \ne \mathrm{P}(A)</math>
:<math>\mathrm{P}(B|AC) = \frac{\frac{4}{40}}{\frac{4}{40} + \frac{1}{40}} = \tfrac{4}{5} \ne \mathrm{P}(B)</math>
:<math>\mathrm{P}(C|AB) = \frac{\frac{4}{40}}{\frac{4}{40} + \frac{6}{40}} = \tfrac{2}{5} \ne \mathrm{P}(C)</math>

In the mutually independent case, however,
:<math>\mathrm{P}(A|BC) = \frac{\frac{1}{16}}{\frac{1}{16} + \frac{1}{16}} = \tfrac{1}{2} = \mathrm{P}(A)</math>
:<math>\mathrm{P}(B|AC) = \frac{\frac{1}{16}}{\frac{1}{16} + \frac{1}{16}} = \tfrac{1}{2} = \mathrm{P}(B)</math>
:<math>\mathrm{P}(C|AB) = \frac{\frac{1}{16}}{\frac{1}{16} + \frac{3}{16}} = \tfrac{1}{4} = \mathrm{P}(C)</math>