Tom K. answered • 07/10/20
Knowledgeable and Friendly Math and Statistics Tutor
P(A│B) = P(A│BC)
P(A|B) = P(AB)/P(B)
P(A│BC) = P(ABc)/P(Bc)
Let P(A│B) = k for some k. Then, as P(A│BC) = P(A│B), P(A│BC) = k
As P(A|B) = k, P(AB) = k P(B)
As P(A│BC) = k, P(ABc) = k P(Bc)
As P(AB) = k P(B) and P(ABc) = k P(Bc) ,
P(AB) + P(ABc) = k P(B) + k P(Bc)
P(AB) + P(ABc) = k(P(B) + P(Bc))
AB and ABc are disjoint, as BBc is the null set, and ABABc = ABBc is a subset of the null set, so it is also the null set.
As AB and ABc are disjoint, P(AB) + P(ABc ) = P(AB U ABc) = P(A (B U Bc)) = P(A)
As B and Bc are disjoint, P(B) + P(Bc) = P(B U Bc) = 1
Then, as P(AB) + P(ABc) = k(P(B) + P(Bc)) from above,
P(A) = k(1) = k
By assumption, P(AB) = k P(B)
Yet, k = P(A). Substituting for k,
P(AB) = P(A) P(B)
Yet, this is what we are trying to show for independence. Thus, A and B are independent.
Ysabelle C.
Sir Tom, for the second time, you had answered my question. Thank you again, sir! To more teachers like you, Sir!07/10/20