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Usefulness of Bayes Rule to application of mental models
Hi, is the following Bayesian formulation generally well-known, when it comes to applying ideas/mental models to a given Context? "The probability that 'an Idea is applicable' to a Context, is equal to: the probability of how often this Context shows up within that Idea's applications, multiplied by the general applicability of the Idea and divided by the general probability of that Context."
P(Idea's applicability∣Context) = P(Context showing up∣Idea is applied)∗P(Idea applied) / P (Context)
Apologies if that sounds a bit abstract, but it's necessarily so because I'm thinking at the level of Ideas and their applicability in general, and Contexts also in general.
Sure, let me try with two different examples.
Let's say you come across a mathematical problem. And you wonder whether it can be solved using Calculus? My statement above implies that this would be a function of: how often problems that "look like this" (yeah this is slightly handwavy) show up within Calculus, how often problems like this show up at all(!), and how often Calculus is useful/applicable in general (!).
The quantitative eqn would be:
Probability (Calculus is applicable | Problem) = Probability (Problems look like this | Calculus has been successfully applied) * Probability (Calculus's applicability in general) / Probability (Problems look like this in general).
The "look like this" parts above are unwieldy but I don't know how else to characterize problems that look similar. There is also messing around with tenses as @JBlack has pointed out. Perhaps that's a fatal error. I haven't thought through that yet.
Applying this to say Cognitive Biases:
Let's say you are in a decision-making situation, and you're wondering if Loss Aversion might be at play here (although you don't know for sure yet). Applying the same principle, the quantitative eqn would be:
Probability (Loss Aversion is applicable | Situation) = Probability (Situation looks like this | Loss Aversion has been successfully applied) * Probability (Loss Aversion's applicability in general) / Probability (Situations like this arise).
To make this valid we need two things.
- Phrase this only in terms of events/propositions, not in terms of single words like "context", "situation" or "problem". Probability only applies to events which occur or don't occur, or to propositions that are true or false. Otherwise it is unclear what e.g. P(x | Situation) means.
- Bayes theorem involves exactly two events (or propositions), so we must make sure we express merely similar sounding events (Situation looks like this / Situation like this arise) with one and the same event, in order to not exceed the total amount of two events overall.
Before you read on, you may want to first try the above approach yourself. The following is my attempt at a solution.
Formalization attempt
Here is a possible formalization which uses only two propositions (albeit with two "free variables" and the indexical "current", which arguably is another variable):
- Idea A applies to the current context.
- The current context is of type B.
P(Idea A applies to the current context | The current context is of type B) = P(The current context is of type B | Idea A applies to the current context) * P(The idea A applies to the current context) / P(The current context is of type B)
Or perhaps more abstractly:
Not sure whether this accurately captures what you had in mind. It probably needs to be refined.
Thanks for the formalization attempt. After thinking and reading some more, I feel I've only restated in a vague manner the Hypothesis and Evidence version of Bayes' Theorem - https://en.wikipedia.org/wiki/Bayesian_inference. Quoting from that page: "
"Idea A applies" would be the Hypothesis in my case, and "current context is of type B" is the Evidence. To restate:
P(Idea A applies | current context is of type B) = P (current context is of type B | Idea A applies) * P (Idea A applies) / P (current context is of type B).
If the above version is correct but straightforward, I'm still impressed by two things. Even after you see some evidence the following two things matter: how likely/probable your Hypothesis is true *in general*, and also how likely/probable the Evidence can show up in cases where your Hypothesis is false (as captured in the denominator)!
Absent further explanation, that looks like a misapplication of the Bayes formula. The event "idea is applicable" is very definitely not the same event as "idea is applied". An idea can be applicable without being applied, and can be applied when it's not actually applicable (as I believe you have done here). Furthermore there is a tense difference which diverges the events even further.
So to answer your question: no this is not "well-known", for the reason that it is incorrect.
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