Anonymous said... Thanks Lydia,
If you would, I have another question related to Bayes that would be very helpful for me and others that I speak with to get a handle on.
Just as the cumulative force of the likelihood ratios for multiple, independent eyewitness testimony can be mathematically modeled to show the Bayes Factor, or the direction the evidence is pointing, I was hoping to model the cumulative case for Christian theism from the arguments of Natural Theology in a similar manner.
If you would, I have another question related to Bayes that would be very helpful for me and others that I speak with to get a handle on.
Just as the cumulative force of the likelihood ratios for multiple, independent eyewitness testimony can be mathematically modeled to show the Bayes Factor, or the direction the evidence is pointing, I was hoping to model the cumulative case for Christian theism from the arguments of Natural Theology in a similar manner.
For example, if we have 5 independent arguments for the existence of God, and a person finds each of them to make the existence of God 80% likely, then how would that be modeled along a Bayesian line?
Would it be [p/q]^5 where p=.2 and q=.8? Using these numbers I get 0.0009765625. Does this mean that for the person in the epistemic situation under consideration that the arguments from Natural Theology make the evidence 10,000 more probable if H is true than if H is false? How low of a prior probability could a person in such an epistemic situation rationally claim that the evidence can overcome for them?
Thank you,
Kevin
5/24/2011 7:41 PM Tim said... Kevin,
This is a fascinating question. A full answer that chased every rabbit down its own separate hole would get very long indeed. So if you don't mind, I'll sketch an answer to a simplified version of the question.
First, consider the case of a single valid deductive argument, one where every premise is required for the inference to go through, for the conclusion "God exists." There may be some doubt about one or more of the premises, but if they are all true, then the conclusion must also be true. The sum of the uncertainties of the premises is the maximum uncertainty of the conclusion. So if the argument has two premises and each premise has a probability of, say, .7, then the conclusion that follows from those premises must have a probability of at least .4 -- perhaps more, though without further information we can't say how much more.
We can get a slightly higher lower bound if we know that the premises are probabilistically independent: then we can multiply, getting .7 x .7 = .49. Once again, this isn't the probability of the conclusion; it's the probability that both premises are true. But unless the conclusion is the logically strongest claim derivable from those premises, that is to say, unless it is logically equivalent to their conjunction, the probability of the conclusion may be higher than this.
Now, suppose that we have two valid deductive arguments like this for that same conclusion -- call them Argument A and Argument B -- each with two probabilistically independent premises, and (for simplicity) let's suppose that each premise of each argument has a probability of .7. Furthermore, let's suppose that across arguments, the premises are also independent: whether premise 1 of Argument B is true is probabilistically independent of whether premise 2 of Argument A is true, etc. How should we assess their combined strength?
[To be continued ...]
5/28/2011 11:15 PM Tim said... The key to this problem is that if both premises are true for either argument, then the conclusion must be true; after all, it follows deductively from the truth of those premises. So in order for the conclusion to be false, at least one premise in each argument must be false. (This is, of course, only a necessary condition for the conclusion to be false, not a sufficient one.) Out of the 16 possible combinations of truth values for the four premises in question, nine meet this criterion:
~A1 & A2 & ~B1 & B2
~A1 & A2 & B1 & ~B2
A1 & ~A2 & ~B1 & B2
A1 & ~A2 & B1 & ~B2
~A1 & ~A2 & ~B1 & B2
~A1 & ~A2 & B1 & ~B2
~A1 & A2 & ~B1 & ~B2
A1 & ~A2 & ~B1 & ~B2
~A1 & ~A2 & ~B1 & ~B2
I have grouped them for computational convenience. Since each premise has, ex hypothesi, a probability of .7, its negation has a probability of .3; since they are, again ex hypothesi, independent, we can calculation the probability of a conjunction of them by multiplying the respective individual probabilities. So in the first group, each line contains two assertions and two negations and therefore has probability .0441 (= .3 x .3 x .7 x .7), and there are four of them, for a total probability of .1764. In the second group, each line has probability .0189 (= .3 x .3 x .3 x .7), and their are once again four of them, for a total probability of .0756. The third group has only one line, with a probability .0081 (= .3 x .3 x .3 x .3). Summing the probabilities of all three groups -- all nine ways for there to be at least some failure in both Argument A and Argument B -- we get .2601. This is the maximum uncertainty for the common conclusion. Hence, the probability of the conclusion is greater than or equal to .7399 (= 1 - .2601). Thus, combining these two arguments, neither of which, by itself, guaranteed a conclusion with a probability greater than 0.49, yields a probability greater than .7399 for the conclusion.
Another way to get to the same result is to consider the possible combinations of truth values that do lead to at least one valid argument. I won't drag you through the calculation, but the partial sums are .0882 + .4116 + .2401 = .7399. This calculation serves as a cross check on our first one.
The key points in this calculation are
(1) The assumption that the arguments were simple deductively valid arguments,
(2) The assumption of a probability for each premise -- for simplicity's sake, we chose .7 for each,
(3) The assumption that within each argument, the premises were probabilistically independent, and
(4) The assumption that across arguments, the premises were probabilistically independent.
Without these assumptions in place, the calculations become more difficult. In theory, it is possible to have so much positive dependence across arguments (failure of 4) that the second argument adds nothing to the first (say, in case we knew that, if either premise of the first argument were true, both premises of the second would have to be true). That is, of course, an unrealistic scenario for many arguments of interest -- a mere limiting case.
I hope that this is the sort of answer you were looking for.
Would it be [p/q]^5 where p=.2 and q=.8? Using these numbers I get 0.0009765625. Does this mean that for the person in the epistemic situation under consideration that the arguments from Natural Theology make the evidence 10,000 more probable if H is true than if H is false? How low of a prior probability could a person in such an epistemic situation rationally claim that the evidence can overcome for them?
Thank you,
Kevin
5/24/2011 7:41 PM Tim said... Kevin,
This is a fascinating question. A full answer that chased every rabbit down its own separate hole would get very long indeed. So if you don't mind, I'll sketch an answer to a simplified version of the question.
First, consider the case of a single valid deductive argument, one where every premise is required for the inference to go through, for the conclusion "God exists." There may be some doubt about one or more of the premises, but if they are all true, then the conclusion must also be true. The sum of the uncertainties of the premises is the maximum uncertainty of the conclusion. So if the argument has two premises and each premise has a probability of, say, .7, then the conclusion that follows from those premises must have a probability of at least .4 -- perhaps more, though without further information we can't say how much more.
We can get a slightly higher lower bound if we know that the premises are probabilistically independent: then we can multiply, getting .7 x .7 = .49. Once again, this isn't the probability of the conclusion; it's the probability that both premises are true. But unless the conclusion is the logically strongest claim derivable from those premises, that is to say, unless it is logically equivalent to their conjunction, the probability of the conclusion may be higher than this.
Now, suppose that we have two valid deductive arguments like this for that same conclusion -- call them Argument A and Argument B -- each with two probabilistically independent premises, and (for simplicity) let's suppose that each premise of each argument has a probability of .7. Furthermore, let's suppose that across arguments, the premises are also independent: whether premise 1 of Argument B is true is probabilistically independent of whether premise 2 of Argument A is true, etc. How should we assess their combined strength?
[To be continued ...]
5/28/2011 11:15 PM Tim said... The key to this problem is that if both premises are true for either argument, then the conclusion must be true; after all, it follows deductively from the truth of those premises. So in order for the conclusion to be false, at least one premise in each argument must be false. (This is, of course, only a necessary condition for the conclusion to be false, not a sufficient one.) Out of the 16 possible combinations of truth values for the four premises in question, nine meet this criterion:
~A1 & A2 & ~B1 & B2
~A1 & A2 & B1 & ~B2
A1 & ~A2 & ~B1 & B2
A1 & ~A2 & B1 & ~B2
~A1 & ~A2 & ~B1 & B2
~A1 & ~A2 & B1 & ~B2
~A1 & A2 & ~B1 & ~B2
A1 & ~A2 & ~B1 & ~B2
~A1 & ~A2 & ~B1 & ~B2
I have grouped them for computational convenience. Since each premise has, ex hypothesi, a probability of .7, its negation has a probability of .3; since they are, again ex hypothesi, independent, we can calculation the probability of a conjunction of them by multiplying the respective individual probabilities. So in the first group, each line contains two assertions and two negations and therefore has probability .0441 (= .3 x .3 x .7 x .7), and there are four of them, for a total probability of .1764. In the second group, each line has probability .0189 (= .3 x .3 x .3 x .7), and their are once again four of them, for a total probability of .0756. The third group has only one line, with a probability .0081 (= .3 x .3 x .3 x .3). Summing the probabilities of all three groups -- all nine ways for there to be at least some failure in both Argument A and Argument B -- we get .2601. This is the maximum uncertainty for the common conclusion. Hence, the probability of the conclusion is greater than or equal to .7399 (= 1 - .2601). Thus, combining these two arguments, neither of which, by itself, guaranteed a conclusion with a probability greater than 0.49, yields a probability greater than .7399 for the conclusion.
Another way to get to the same result is to consider the possible combinations of truth values that do lead to at least one valid argument. I won't drag you through the calculation, but the partial sums are .0882 + .4116 + .2401 = .7399. This calculation serves as a cross check on our first one.
The key points in this calculation are
(1) The assumption that the arguments were simple deductively valid arguments,
(2) The assumption of a probability for each premise -- for simplicity's sake, we chose .7 for each,
(3) The assumption that within each argument, the premises were probabilistically independent, and
(4) The assumption that across arguments, the premises were probabilistically independent.
Without these assumptions in place, the calculations become more difficult. In theory, it is possible to have so much positive dependence across arguments (failure of 4) that the second argument adds nothing to the first (say, in case we knew that, if either premise of the first argument were true, both premises of the second would have to be true). That is, of course, an unrealistic scenario for many arguments of interest -- a mere limiting case.
I hope that this is the sort of answer you were looking for.
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