- No, I don't.
- I first estimated the prior probabilities of the different hypotheses included under ~H in 'chapter' II on 3/27/16, #3007
...
11.3. Re P(E|~H):
11.3.1. The probability (“likelihood”) of E given ~H, involves several specific hypothetical possibilities.
11.3.1.1. That only some of us have but one finite life.
11.3.1.2. That we each have numerous finite lives.
11.3.1.3. That only some of us have numerous finite lives.
11.3.1.4. That we each have an infinity of finite lives.
11.3.1.5. That only some of us have an infinity of finite lives.
11.3.1.6. That we each have an infinite life.
11.3.1.7. That only some of us have an infinite life.
11.3.1.8. That time isn’t what we think it is (to be explained).
11.3.1.9. Some other explanation.
11.3.2. Now I must estimate (roughly) the prior probability (rounded off to three decimal places) of each more specific possibility (hypothesis), given ~H.
11.3.2.1. That only some of us have but one finite life: .000
11.3.2.2. That we each have numerous finite lives: .002.
11.3.2.3. That only some of us have numerous finite lives: .000.
11.3.2.4. That we each have an infinity of finite lives; .002
11.3.2.5. That only some of us have an infinity of finite lives: 000.
11.3.2.6. That we each have an infinite life: .002
11.3.2.7. That only some of us have an infinite life: .000
11.3.2.8. That time isn’t what we think it is (to be explained): .002
11.3.2.9. Some other explanation: .002
- Also on 3/27/16, in #3014, I added the likelihoods of my current existence -- given each included hypothesis -- and multiplied them by the prior probabilities of each included hypothesis.
11.3.3. And now, I must estimate the likelihood of my own current existence given the different specific hypotheses under ~Hs.
11.3.3.1. That only some of us have but one finite life: .10.
11.3.3.2. That we each have numerous finite lives: .10.
11.3.3.3. That only some of us have numerous finite lives: .25.
11.3.3.4. That we each have an infinity of finite lives; 1.00
11.3.3.5. That only some of us have an infinity of finite lives: .50.
11.3.3.6. That we each have an infinite life: 1.00
11.3.3.7. That only some of us have an infinite life: .50
11.3.3.8. That time isn’t what we think it is (to be explained): .50
11.3.3.9. Some other explanation: .50
11.3.4. And now, I must multiply each of the probabilities of ~H above by the likelihoods of my current existence, given each specific hypothesis, and add up their products. And, the total likelihood of my current existence given ~H:
11.3.4.1. P(E|~H) = (0*.5) + (.002*.10) + (0*.25) + (.002*1.0) + (0*.5) + (1*..002) + (0*.5) + (.002*.5) + (.002*.5), or
11.3.4.2. P(E|~H) = 0 + .0002 + 0 +.002 + 0 + .002 + 0 + .001 + .001, or
11.3.4.3. P(E|~H) = .0052. And,
11.3.5. P(H|E) = 0*.99/(0*.99 + .0052*.01) = (0/.000052) = 0.
11.3.6. P(H|E) = 0.
11.3.7. IOW, given my current existence, the posterior probability that I will have one, finite life is zero…
- On 7/21/16, in chapterIII, #2115, jt says
BTW, Jabba, the form of Bayes' Theorem that you are using requires H and ~H to be complementary hypotheses, that is, they must be mutually exclusive and exhaustive. If ~H is a specific hypothesis, and there is more than one alternative to ~H, then H must comprise all the alternatives. The problem with that is how do you come up with a single P(E|H), if H represents a collection of alternative models, each implying a different value of P(E|H)?
- Then,on 7/23/16, in chapter III, #2149, jt says
Alas, after stating that I didn't want to know how you calculated P(E|~H), I read your explanation anyway. Although the following correction doesn't matter, because (1) none of the numbers enters into the calculation of the posterior, since your P(E|H)=0, and thus the 0 probability of your posterior is a foregone conclusion, and (2) all your numbers are just made up, I feel pathologically pedagogically compelled to point out that the prior probabilities 11.3.2.1–11.3.2.9 need to be normalized to sum to 1. After doing so, P(E|~H)=.62 (not .0052). Thus under your pet hypothesis, or ensemble of hypotheses, you had a 62% chance (strangely enough) of having come into existence.
Alternatively, if H and ~H are both specific hypotheses, then they must be the only possible hypotheses in the universe. If, on the other hand, there are more than two possible hypotheses, then you cannot use the form of Bayes' Theorem that you've been using; you must use the odds form of Bayes Theorem instead.
- Then, on 7/23/16, in chapter III, #2156, I accept jt's correction.
jt,
- Showing my ignorance, and lack of recent statistical study, I now agree with .62 rather than .0052.
- I'll have to think more about the need to normalize -- but for now, I agree with you on that also...
Jay,
- As you can see above, I was doing that before jt "instructed" me, but I did accept his correction about normalization.
- Also, having realized that the simple formula doesn't work, I've returned to my earlier, more complicated, formula -- that Dr Horel(sp?) approved.