Bayes-and-Certainty.md

@@ -34,4 +34,54 @@ We can confirm that this is equivalent to the Bayes eqn as follows:
 
 0\.99(0.1) / ((0.99(0.1) + (1-0.99)(1-0.1) )
 
-We recognize this result as the Bayes eqn.
+We recognize this result as the Bayes eqn. for this case.
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+
+Now let's do the same with a test that is 99.9% accurate. We need 1000 subjects in this case:
+
+1000 healthy subjects ==\> Test 1 is 99.9% accurate ==\> 1 tests positive (false positive) ==\> Test 2 is 10% accurate ==\> .9 test positive again.
+
+1000 sick subjects ==\> Test 1 is 99% accurate ==\> 999 test positive ==\> Test 2 is 10% accurate ==\> 99.9 test positive again
+
+The number of people who tested positive twice = 0.9 + 99.9 = 100.8
+
+Of these, the number of people who are actually sick: 99.9
+
+P = Number of sick who tested positive twice / Number of people who tested positive twice = 99.9/(99.9 + 0.9) = 0.991
+
+Let's juxtapose the two relevant calculations:
+
+9\.9/(9.9 + 0.9) = 0.917
+
+99\.9/(99.9 + 0.9) = 0.991
+
+We see here that the numerator increases by a factor of 10, corresponding to the 10 times increase in sample size, and representing the number of people who test positive twice who are actually sick. The denominator is composed of this same number plus a constant, the number of people who falsely test positive twice. This constant remains so because as the test becomes more accurate, ever fewer results are bad (in percentage terms). For every 9 added to the decimal place, it becomes about 10 times harder to have a bad result. This is the basic reason why certainty pulls so hard in its favor. 
+
+Indeed, to counteract the effect of certainty we would need an equal and opposite level of uncertainty which, in this case, is simply certainty for the opposite opinion. If 99% represents our probability of a True result, then a 1% opinion for True is it's opposite (99% for False). When combined these two yield, via Bayes,
+
+0\.99(0.1) / (0.99(0.1) + 0.1(0.99)) = 0.5
+
+and thus cancel themselves out. 99.9% would require a 0.1% counterbalance to cancel itself, and so on. In terms of the evidentiary weight argument we've been using, the 99.9% requires 1000 experiments to establish but the 0.1% also requires 1000 experiments. They are "equal" in that sense.
+
+We can understand this numerically in terms of order of magnitude. The counterbalance against certainty (eg 99.9%) is a small number (eg 0.1%) which causes the failures to be ever smaller compared to the successes, as we've seen. For this case, 0.1% equates to 0.001 which has order of magnitude of -3. 99.99% (0.01%) would have order of magnitude of -4 and have 10 times the evidence in favor of it. Any time we see an order of magnitude difference, we know we are dealing with experiments which have large discrepancies in their evidence and their Bayesian combination will favor greatly the experiment with the more negative OOM. 
+
+To an extent we can also understand this in terms of decimal places. The more decimal places an opinion has, the more tests were done to confirm it, with each decimal place adding an order of magnitude to the the number of tests.
+
+This only works at the certainty (99%, 99.9%, 99.99%, etc.) and uncertainty end (1%, 0.1%, 0.01%, etc.) of the probability spectrum, which is why the OOM view is the more accurate way to look at it. If someone says they are 10.001% certain of something they must have done at least 100,000 experiments to confirm that (10001 succeeded, 89999 failed). But in Bayes, this is not very different than someone saying 10% where only 10 experiments were conducted for confirmation (9 succeeded, 1 failed). When combined with a 99.999%, which has the same number of decimal places, the result will be 99.991%, very close to 99.999%. The "decimal place heuristic" clearly only works at the ends of the spectrum (where the smaller opinions exhibit a difference in OOM).
+
+Nevertheless, this idea asks us to consider what happens when two opinions differ in the evidentiary weight behind them. Clearly an experiment with 1000 tests is not equivalent to one with 10. The Bayes equation, however, has no mechanism to judge the quality of the probabilities inserted into it except when those probabilities clearly differ in terms of their OOM.  
+
+So one idea to handle this is to weigh the opinions as follows:
+
+1) Use the regular Bayes equation to combine the two probabilities. This is the upper limit.
+
+2) Use the experiment with the most evidence for it (most decimal places) to establish the lower limit. 
+
+3) Decide on a factor for the difference in evidence between the two experiments using the decimal place of the reported probabilities. An 80.3% probability vs. 73% has one more decimal place and so gives the 80.3% opinion 10 times the weight of the 73% opinion. Of course, if the probability comes with the sample size used to calculate it we can just divide the sample sizes to obtain the factor.
+
+4) Use the factor just calculated as a weighting factor in establishing the combined opinion between the lower and upper limit.
+
+For example, using 80.3% and 73%,
+
+1) $P\_{bay} = 0.803(0.73)\\over {0.803(0.73) + 0.197(0.27)} = 0.917$
+
+2)     
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