Documents
Resources
Learning Center
Upload
Plans & pricing Sign in
Sign Out

Bayes_Theorem

VIEWS: 17 PAGES: 2

									An economist believes that during periods of high economic growth, the U.S.
dollar appreciates with probability 0.70; in periods of moderate economic
growth, the dollar appreciates with probability 0.40; and during periods of
low economic growth, the dollar appreciates with probability 0.20. During
any period of time, the probability of high economic growth is 0.30, the
probability of moderate growth is 0.50, and the probability of low economic
growth is 0.20. Suppose the dollar has been appreciating during the present
period. What is the probability we are experiencing a period of high
economic growth?

                                            Prior Probabilities:
High economic growth:             H               P(H) = 0.30
Moderate economic growth:         M               P(M) = 0.50
Low economic growth:              L               P(L) = 0.20

Let A denote the event that the dollar appreciates.

     P(A|H) = 0.70; P(A|M) = 0.40; P(A|L) = 0.20

     Extended Bayes’ theorem:
                                      n
     P(B1|A) = [P(A | B1)P(B1)] / [ P(A | Bi) P(Bi)]
                                      i=1



     P(H|A) = [P(A|H)P(H)] /

                 [P(A|H)P(H) + P(A|M)P(M) + P(A|L)P(L)]

                 = [(0.70)(0.30)] /

                 [(0.70)(0.30) + (0.40)(0.50) + (0.20)(0.20)]

                 = 0.21 / 0.45 = 0.4667
Event   Prior         Conditional     Joint           Posterior
        Probability   Probability     Probability     Probability

 H      P(H) = 0.30   P(A|H) = 0.70   P(AH) = 0.21   P(H|A) = 0.467
                                                      (0.21 / 0.45)

 M      P(M) = 0.50   P(A|M) = 0.40   P(AM) = 0.20   P(M|A) = 0.444
                                                      (0.20 / 0.45)

 L      P(L) = 0.20   P(A|L) = 0.20   P(AL) = 0.04   P(L|A) = 0.089
                                                      (0.04 / 0.45)
        SUM: 1.00                     P(A) = 0.45        Sum: 1.000

								
To top