Theorem

2
Queens College, Economics 249, Geordan Hull, 1 BAYES’ THEOREM Provides a means for making probability calculations after revising probabilities when obtaining new information in an important phase of probability analysis. When given P(A) and P(AB), one can calculate P(B/A) by manipulating the information in the Multiplication Rule. However, one could not calculate P(A/B). Similarly, when given P(B) and P(AB), one can calculate P(A/B) by manipulating the information in the Multiplication Rule. However, one could not calculate P(B/A). There is where one can now apply Bayes’ Theorem. BAYES’ THEOREM (TWO-EVENT CASE) P(A 1 /B) = P(A 1 )P(B/A 1 ) P(A 1 )P(B/A 1 ) + P(A 2 )P(B/A 2 ) P(A 2 /B) = P(A 2 )P(B/A 2 ) P(A 1 )P(B/A 1 ) + P(A 2 )P(B/A 2 ) BAYES’ THEOREM (MULTI-EVENT CASE) P(A i /B) = P(A i )P(B/A i ) P(A 1 )P(B/A 1 ) + P(A 2 )P(B/A 2 ) + … + P(A n )P(B/A n ) EXAMPLE Given: G = Having good seats, B = Having bad seats, A 1 = Seeing Wicked, A 2 = Seeing Mama Mia!, the probability of having good Prior Probabil ities New Informat ion Applicat ion of Baye’s Theorem Posterio r Probabil ities

description

the

Transcript of Theorem

BAYES THEOREM

Queens College, Economics 249, Geordan Hull, 1

BAYES THEOREM

Provides a means for making probability calculations after revising probabilities when obtaining new information in an important phase of probability analysis.

When given P(A) and P(A(B), one can calculate P(B/A) by manipulating the information in the Multiplication Rule. However, one could not calculate P(A/B). Similarly, when given P(B) and P(A(B), one can calculate P(A/B) by manipulating the information in the Multiplication Rule. However, one could not calculate P(B/A). There is where one can now apply Bayes Theorem.

BAYES THEOREM (TWO-EVENT CASE)

P(A1/B) = P(A1)P(B/A1)

P(A1)P(B/A1) + P(A2)P(B/A2)

P(A2/B) = P(A2)P(B/A2)

P(A1)P(B/A1) + P(A2)P(B/A2)

BAYES THEOREM (MULTI-EVENT CASE)

P(Ai/B) = P(Ai)P(B/Ai)

P(A1)P(B/A1) + P(A2)P(B/A2) + + P(An)P(B/An)

EXAMPLE

Given: G = Having good seats, B = Having bad seats, A1 = Seeing Wicked, A2 = Seeing Mama Mia!, the probability of having good seats given that you are seeing Wicked [P(G/A1)] = 0.98, the probability of having good seats given that you are seeing Mama Mia! [P(G/A2)] = 0.95, P(A1) = 0.65, and P(A2) = 0.35 Please find the probability of a) Seeing Wicked given you have bad seats, and b) Seeing Mama Mia! given you have bad seats.

P(B/A1) = 1 - P(G/A1) = 1 - 0.98 = 0.02; P(B/A2) = 1 - P(G/A2) = 1 - 0.95 = 0.05

a) P(A1/B) = [(0.65)(0.02)]/[(0.65)(0.02) + (0.35)(0.05)] = 0.0130/(0.0130 + 0.0175) = 0.0130/0.0305 = 0.4262

b) P(A2/B) = [(0.35)(0.05)]/[(0.65)(0.02) + (0.35)(0.05)] = 0.0175/(0.0130 + 0.0175) = 0.0175/0.0305 = 0.5738

Prior Probabilities

New Information

Application of Bayes Theorem

Posterior Probabilities