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Oil Example

Dry (0.60). -100k. 52.50k. Low (0.25). Dome. 150k. (0.908). High (0.15). EMV=10k. 500k. Dry (0.85). -100k. -53.75k. Low (0.125). No Dome. 150k. Site 1. (0.092). High (0.025). 500k. Site 2. Dry (0.2). -200k. Low (0.8). 50k. EMV=0. Oil Example.

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Oil Example

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  1. Dry (0.60) -100k 52.50k Low (0.25) Dome 150k (0.908) High (0.15) EMV=10k 500k Dry (0.85) -100k -53.75k Low (0.125) No Dome 150k Site 1 (0.092) High (0.025) 500k Site 2 Dry (0.2) -200k Low (0.8) 50k EMV=0 Oil Example An oil company is considering a site for an exploratory well. If the rock strata underlying the site are characterized by what geologists call a “dome” structure, the chances of finding oil are somewhat greater than if no dome structure exists. The probability of a dome structure is Pr(Dome)=0.6 . The conditional probabilities of finding oil in this site are as follows. Pr(High|Dome) = 0.15 Pr(Dry|Dome) = 0.6 Pr(Low|Dome) = 0.25 Pr(High|No Dome) = 0.025 Pr(Low|No Dome) = 0.125 Pr(Dry|No Dome) = 0.85

  2. 7.32 If the company could collect information from a drilling core sample and analyze it to determine whether a dome structure exists at site 1. A positive result would indicate the presence of dome, and a negative result would indicate the absence of a dome. The test is not perfect, however. Particularly, Pr(+|Dome)=0.99, and Pr(-|No Dome)=0.85. Use these probabilities and information given in the example, and Bayes’ theorem to find the posterior probabilities Pr(Dome|+) and Pr(Dome|-). If the test gives a positive result, which site should be selected? If the test result is negative, which site should be selected? Pr(Dome |+) = Pr(Dome |– ) =

  3. Dry (0.60) -100k 52.50k Low (0.25) Dome 150k EMV(Site 1|+)=? High (0.15) (0.908) 500k Positive Dry (0.85) -100k -53.75k No Dome Low (0.125) (0.654) 150k (0.092) High (0.025) 500k Dry (0.60) -100k 52.50k Site 1 Low (0.25) Dome 150k High (0.15) (0.017) 500k Negative Dry (0.85) -100k -53.75k (0.346) Low (0.125) No Dome 150k Site 2 (0.983) High (0.025) 500k Dry (0.2) -200k Low (0.8) 50k EMV=0 If the test gives a positive result, what is the expected money value of Site 1? EMV(Site 1|+) = EMV (Dome) P(Dome | +) + EMV (No dome) P(No dome | +) = (52.50 K) 0.908 + (-53.75 K) 0.092 = $42.725 K Because EMV(Site 1|+) > 0, site 1 should be selected

  4. Dry (0.60) -100k 52.50k Low (0.25) Dome 150k High (0.15) (0.908) 500k Positive Dry (0.85) -100k -53.75k (0.654) No Dome Low (0.125) 150k High (0.025) (0.092) 500k Dry (0.60) -100k 52.50k Site 1 Low (0.25) EMV(Site 1|–)=? Dome 150k High (0.15) (0.017) 500k Negative Dry (0.85) -100k -53.75k (0.346) Low (0.125) No Dome 150k Site 2 High (0.025) (0.983) 500k Dry (0.2) -200k Low (0.8) 50k If the test gives a negative result, what is the expected money value of Site 1? EMV(Site 1|–) = EMV (Dome) P(Dome | –) + EMV (No dome) P(No dome | –) = (52.50 K) 0.017 + (-53.75 K) 0.983 = -$51.944 K Because EMV(Site 1| – ) < 0, site 2 should be selected

  5. Dry (0.60) -100k Low (0.25) Dome 150k (0.908) High (0.15) 500k Positive Dry (0.85) -100k (0.654) No Dome Low (0.125) 150k (0.092) High (0.025) 500k Dry (0.60) -100k Site 1 Low (0.25) Dome 150k (0.017) High (0.15) 500k Negative Dry (0.85) -100k (0.346) Low (0.125) No Dome 150k Site 2 (0.983) High (0.025) 500k Dry (0.2) -200k Low (0.8) 50k Pr(Dry|Dome), Pr(Low|Dome), Pr(High|Dome), Pr(Dry|No Dome), Pr(Low|No Dome) and Pr(High|No Dome) are not affected by the test result. Thus, we can conclude that the test result and the amount oil in the well are conditionally independent given the information about the dome structure of the well.

  6. Pr(Dome | + AND Dry) = 7.33 Pr(+ and Dome)=? Pr(+ and Dome + Dry)=? Pr(Dome|+ and Dry)=? Pr(+ AND Dome) = Pr(+| Dome) Pr(Dome) = 0.99(0.60) = 0.594 Pr(+ AND Dome and Dry) = Pr(Dry and + and Dome) = Pr(Dry | + AND Dome) Pr(+ AND Dome) Because the amount of oil and test result are conditionally independent given the dome structure, Pr(Dry | + AND Dome) =Pr(Dry| Dome) =0.6 Therefore, Pr(+ AND Dome AND Dry) = 0.60 (0.594) = 0.356 Pr(+ AND Dry) = Pr(+ AND Dry AND Dome) + Pr(+ AND Dry AND No Dome) Pr(+ AND Dry AND No Dome) = Pr(Dry AND + and No Dome) = Pr(Dry | + AND No Dome) Pr(+ AND No Dome) = Pr(Dry | No Dome) Pr(+ and No Dome) Pr(+ AND No Dome) = Pr(+|No Dome) Pr(No Dome) =(1-0.85)(0.4) = 0.06 So Pr (+AND Dry AND No Dome) = 0.85(0.06) = 0.051 Pr(+AND Dry) = 0.356 + 0.051=0.407 Pr(Dome | + AND Dry) = 0.356/0.407 = 0.875

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