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I want to forecast $P(A)$ where $A$ is a messy real-world event, for which I have no analytical expression or statistical model.

Assume, however, that for $b$ events $B_i$ I have forecasts for $P(B_i)$ and estimates for $P(A|B_i)$ and $P(A|\neg B_i)$. For each event I can use Jeffrey's rule to get: \begin{equation} P_i(A)=P(A|B_i)*P(B_i)+P(A|\neg B_i)*(1-P(B_i)) \end{equation} (See e.g. Diaconis & Zabell, 1983.)

However, this will give $b$ different estimates $P_i(A)$.

Is there any way to combine all the information I have to get a single estimate $P(A)$?

The $B_i$ are not assumed to be either mutually exclusive or jointly exhaustive.

If there is not a straightforward theoretical way like the Jeffrey's rule, I would appreciate reasonably motivated practical heuristics. My use case is getting a good practical estimate, not to prove anything.

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  • $\begingroup$ I think that the "Jeffrey's rule" might be otherwise known as the "law of total probability" $\endgroup$ – NofP Oct 22 '18 at 22:51
  • $\begingroup$ I'm assuming that the $b$ events $B_i$ are a complete disjoint set of events, i.e. that $\sum_i P(B_i) = 1$? $\endgroup$ – jwimberley Oct 23 '18 at 12:07
  • $\begingroup$ @jwimberley no, they are not. For each single $B_i$ you can take its complement, but nothing beyond that. $\endgroup$ – Jacob Lagerros Oct 23 '18 at 15:59
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You can use the $B_i$ to construct a partition, but I'm not sure how useful this is as it might be difficult to estimate the probabilities of the relevant events: $¬(B_1\cup B_2\dots\cup B_b),\ B_1\cap¬(B_2\cup\dots\cup B_b), \ B_2\cap¬(B_1\cup B_3\cup\dots\cup B_b),\ \dots \ , \ B_b\cap¬(B_1\cup B_2\cup\dots\cup B_{b-1}), \ B_1\cap B_2\cap ¬(B_3\cup\dots\cup B_b),\ \dots \ , \ B_1\cap B_2\cap\dots\cap B_b$

The easiest way to see how this works is to draw a venn diagram and see that each event corresponds to a single region. The idea is to pick out each region individually to form a partition.

Then you can apply the law of total probability to get a single estimate of $P(A)$.

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    $\begingroup$ That's progress, but still not very practically useful as it requires estimating $>2^b$ quantities if we're not assuming independence. $\endgroup$ – Jacob Lagerros Oct 23 '18 at 16:52
  • $\begingroup$ If you're not willing to assume independence, or anything else about the relationship it's the best you can do. You either need some assumption about the relationship, or you can estimate that relationship itself using the data, making an assumption about the form of the relationship in order to estimate it. $\endgroup$ – David Manheim Nov 18 '18 at 13:13

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