Let 1A be the indicator function of the event A so it is 1 w

Let 1_A be the indicator function of the event A (so it is 1 when the event happens, otherwise it is 0). Likewise 1_B is the indicator function for the event B. Show that the Cov(1_A, 1_B)= P(B)[(P(A|B)-P(A)].

Solution

Let 1A be the indicator function of the event A(so it is 1 when the event happens, otherwise it is 0). Likewise 1B is the indicator function for the event B.

According to the definition of indicator functions

Cov(1A,1B)=P(AB)-P(A)P(B)

From Probability we know that, P(A/B)=)=P(A/B)P(B)

Cov(1A,1B)= P(A/B)P(B)- P(A)P(B)

Cov(1A,1B)=P(B)[ P(A/B)- P(B)]