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How does one assign a probability to statements that are themselves probabilistic? For example, how would one assign a probability to the statement, "There is an 80% chance that it will rain on September 16, 2017"? And how would one assign a probability of THAT statement? etc. Is there a paper where someone talks about this problem?

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    If you accept past performance as measuring future probability (not true), you could make statements like: "it has rained on 60% of the days when Bob the weatherman has said there is an 80% chance of rain". This is something you may do if you were comparing different predictions. Note that "probability of rain" is a bad example, because it could mean it will rain over 80% of the forecast area, not that there's a 20% chance there will be no rain at all.
    – user935
    Sep 13, 2017 at 18:39
  • One interpretation is that the probability in the sentence is statistical probability of rain, whereas probability of the sentence concerns the credibility of its source. Since estimating credibility may include considerations beyond forecasts of rain such second order statements can not be accomodated into first order statements. This is similar to two-dimensional modal semantics, which interprets sentences like "it is possible that stable wormholes are impossible" as expressing two different types of possible, see e.g. Chalmers
    – Conifold
    Sep 13, 2017 at 23:52
  • Probabilities multiply... Sep 14, 2017 at 4:52
  • Fuzzy logic involves the notion of meta-fuzzy rules: Rules about your measures of certainty that may themselves need to be tuned to the situation as you find out your predictions of uncertainties are wrong in a systematic way.
    – user9166
    Sep 14, 2017 at 22:42

3 Answers 3

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There is much more to so called higher-order probabilities than taking the mean of a random variable.

Yes, there are a bunch of papers which talk about this problem. A good starting point is:

Do We Need Higher-Order Probabilities and, If So, What Do They Mean? https://arxiv.org/abs/1304.2716

The takeaway by Pearl and others is that yes you can talk about higher-order probabilities (people do it all the time), but that typically they are derived from an existing causal model.

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  • Who are the others besides Pearl? How does the causal model fit in with this? Jul 8, 2018 at 3:00
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My analysis would be that such statements represent propositions that are about other propositions. The example you gave proposes that the proposition [it will rain on September 16, 2017] has an 80% likelihood of being true. I would claim that such propositions will always be probabilistic in nature because there is nothing to be said about a proposition except its truth value.

If I were to make a 'higher order' proposition about the likelihood of the probabilistic assertion, I claim that the same meaning, with respect to the non-probabilistic proposition, can be achieved by revising the lower order claim. For instance, if I were to say [There is a 20% chance the exemplary proposition is true], all I have to do is multiply .2 and .8 to get the revised probability that [it will rain on September 16, 2017], by the multiplicative rule of probabilities. I have no real conclusion about this kind of proposition, but I've described what they are and how I think they work. Check out Bayes' Rule if you are interested in reading more about revising probabilities.

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You're talking about the mean (first moment), as opposed to the variance (second moment), etc, and perhaps more directly about quantiles. See https://en.wikipedia.org/wiki/Moment_%28mathematics%29#Significance_of_the_moments   and   https://en.wikipedia.org/wiki/Quantile_function   or just google   "statistics moments quantiles"   for detailed discussions.

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    The standard deviation does not cover the odds the mean is wrong, it handles the error that is not accounted for by the mean. Say you do a t-test. The p value is the chance your mean is wrong. What is the chance your p-value is wrong? It is not the p-value on the F distribution on the standard deviation. That answers a totally different question. The original p-value already has the standard deviation baked into it.
    – user9166
    Sep 14, 2017 at 22:37
  • @jobermark thanks for contributing the additional statistics info.
    – user19423
    Sep 15, 2017 at 5:54

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