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Say $d+1$ independent, equally reliable experts give you probabilities $p_0,\ldots,p_d$ of an event A occurring. What should you think the probability of $A$ is?

I'll give my solution (Edit: I retract my claim to having a solution. At the very least this is not, in general, normalized correctly. There are also other serious issues as pointed out in the comments and answers): $$ \begin{align} &P(A \mid p_0,\ldots,p_d) \\ &= \frac{P(p_0,\ldots,p_d \mid A) P(A)}{P(p_0,\ldots,p_d)}\\ &= P(A) \prod_{l=0}^{d}\frac{P(p_l \mid A)}{P(p_l)}\\ &= P(A) \prod_{l=0}^{d}\frac{P(A \mid p_l)}{P(A)}\\ &= \frac{1}{P(A)^d} \prod_{l=0}^{d} P(A \mid p_l)\\ &= \frac{1}{P(A)^d} \prod_{l=0}^{d} p_l\\ \end{align}$$

I think this is reasonable (let me know if you disagree) but I have a number of questions about it:

  1. I have couched this in an explicity Bayesian manner and I'm not sure if this is necessary. Can you instead talk about some estimates of $p_l$?

  2. Is line 3 ok? I'm dividing by zero... Can this be fixed by considering some density function?

  3. Why do we divide by $P(A)^d$ (and not to the power $d+1$ say)? It makes sense to me to multiply the probabilities together, I guess this is just normalizing?

  4. Does $P(A \mid p_l)=p_l$? It certainly seems like it should.

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    Could you justify $$P(p_0, p_1, \ldots, p_d) = \prod_{i=0}^{d} P(p_i)$$ a little? In many cases, conditionally independent events (e.g. the justification for $$P(p_0, p_1, \ldots, p_d|A) = \prod_{i=0}^{d} P(p_i|A)$$ are not necessarily independent in the absence of conditioning.2012-03-21
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    I don't think your combined estimate is justifiable. If even a single expert predicts $0$ then their estimate overwhelms the effect of all the other estimates!2012-03-21
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    Thanks for your answers and comments everyone. I had a feeling this might be controversial and you haven't disappointed. I retract my claim to having a solution.2012-03-22

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