Let A = the hypothesis that aliens are visiting Earth
Let E = evidence that aliens are visiting Earth
The posterior probability that aliens are visiting Earth, given some evidence, P(A|E), can be derived from the likelihood of the evidence P(E|A) and the prior P(A), using Bayes' theorem
P(A|E) = P(E|A) * P(A) / [ P(E|A) * P(A) + P(E|Not[A]) * P(Not[A]) ]. (1)
I wish to argue backwards from the assumption of a significant posterior probability in order to find an expression for the prior probability given the likelihoods
P(A|E) = 1/2. (2)
Substituting Eqn.2 into Eqn.1 we find
P(A) / P(Not[A]) = P(E|Not[A]) / P(E|A). (3)
Now let us assume that the evidence is made of cases that have natural explanations, N, and cases that don't have natural explanations Not[N].
P(E) = P(E) P(N) + P(E) P(Not[N]) (4)
On the hypothesis that aliens are visiting Earth both terms on the RHS of Eqn.4 are present so that we have
P(E|A) = P(E) (5)
On the hypothesis that aliens are not visiting Earth the second term on the RHS of Eqn.4 is zero (no non-natural explanations) so that we have
P(E|Not[A]) = P(E) P(N) (6)
Substituting Eqn.5 and Eqn.6 into Eqn.3 we obtain
P(A) / P(Not[A] = P(N) (7)
Let us assume that we have a report of a close encounter where the witnesses seem to be of good character, in a normal state of consciousness and were unlikely to be the victims of a hoax. For example see the following Winchester,UK encounter of 1976 with witnesses Joyce Bowles and Ted Pratt
Let us assume that the probability of a natural explanation for the Winchester 1976 case (lying, hallucinating or victims of a hoax) is
P(N) = 1/10. (8)
Thus, if we take P(N) in Eqn.7 to refer to the probability that a group of close encounter cases can be explained by natural causes, we only need 10 good independent cases like the one above in order to deduce that even if the prior odds of alien visitation is only 1:10^10 we will still end up with significant posterior odds of 1:1.
Is this backwards argument from likelihood to prior valid?