This is an improved version of Backwards Bayesian argument for alien visitation?
It is said that extraordinary claims require extraordinary evidence and therefore this criterion is applied to every claim independently. But can extraordinary claims be legitimately combined before being weighed against our prior knowledge?
Let A = the proposition that aliens are visiting Earth
Let {C} = a set of witness testimonies of different close encounter events
I wish to use Bayes theorem backwards to deduce the prior odds P(A) / P(Not A) given that I end up with an evens posterior odds P(A | {C}) / P(Not A | {C}) = 1.
Bayes theorem can be written in terms of odds as:
P(A | {C}) / P(Not A | {C}) = P({C} | A) / P({C} | Not A) * P(A) / P(Not A)
If P(A | {C}) / P(Not A | {C}) = 1 we obtain the following expression for the prior odds given by
P(A) / P(Not A) = P({C} | Not A) / P({C} | A)
As the events are independent we can assume the witness testimonies are as well so that
P({C}) = P(C_1) * P(C_2) * P(C_3) * ...
Therefore we have
P(A) / P(Not A) = Product_i [ P(C_i | Not A) / P(C_i | A) ]
Let us assume there are 4 explanations for a typical close encounter witness testimony:
- They are lying.
- They were hallucinating.
- They were the victims of a hoax.
- It was aliens.
As we are assuming no prior knowledge at this stage we can assign equal probability to each alternative.
Therefore:
P(C_i | Not A) = P_1 + P_2 + P_3 = 3/4
P(C_i | A) = P_1 + P_2 + P_3 + P_4 = 1
Therefore the prior odds of alien visitation implied by even posterior odds is given by
P(A) / P(Not A) = (3/4)^N
where N is the number of close encounter testimonies.
Thus, given 100 testimonies of close encounters with aliens, even if our prior odds for alien visitation is only 1 in 10^13 we end up with a posterior 50% belief in alien visitation.