8_R

Independence

Two events are statistically independent if the occurrence of one does not affect the probability of occurrence of the other (equivalently, does not affect the odds) [1, 2]. In other words, two variables are independent if the realization of one does not affect the probability distribution of the other.

P(A | B) = P(A)

then A is statistically independent of B, since occurence of B makes no difference to how often A happens.

Also, based on conditional probabilities where

P(A | B) = P(A and B) / P(B)

=> P(A and B) = P(A | B) * P(B), if A is independent of B, then

=> P(A and B) = P(A) * P(B), if this is true then B is also independent of A

=> P(B | A) = P(A and B) / P(A) = (P(A) * P(B)) / P(A) = P(B)

so A and B are independent

For example, one way to test the independence

P(A and B) = P(A) * P(B) = > P(Male and Football) = P(Male ) * P(Football)

=> P(Male and Football) = 0.25 * 0.75 = 0.1875

=>P(Female and Football) = 0.75 * 0.75 = 0.5625, thus they are independent,

This relation is based on, property that for every cell Xij (joint distribution), its relation with marginal probability (X_j column and Xi_ row) is equal to relation of marginal probability of the row (if Xij/X_j) to total probability which is 1 (X). Note that, this applies not only for relational frequencies.

Xij/X_j = Xi_/X => X * Xij = Xi_ * X_j, which same for the row relation

Xij/X_i = X_j/X => X * Xij = Xi_ * X_j

For example,

P(Male and Football) / P(Male) = P(Football) / P(Total) = 0.75

P(Male and Football) / P(Male) = P(Football) / P(Total) =>

P(Male and Football) * P(Total) = P(Football) * P(Male) = 0.1875

For simple frequencies

F(Male and Football) / F(Male) = 15/20 = 0.75 <=> F(Football) / F(Total) = 60/80 = 0.75

F(Male and Football) * F(Total) = 15*80 = 1200 <=> F(Football) * F(Male) = 20*60 = 1200