Backwards Bayesian argument for alien visitation?

Question

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

$$P(A|E)=\frac{P(E|A)P(A)}{P(E|A)P(A)+P(E|\lnot A)P(\lnot A)}.\tag{1}$$

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

$$P(A|E)=\frac{1}{2}.\tag{2}$$

Substituting Eqn.$(2)$ into Eqn.$(1)$ we find

$$\frac{P(A)}{P(\lnot A)}=\frac{P(E|\lnot A)}{P(E|A)}.\tag{3}$$

Now let us assume that the evidence is made of cases that have natural explanations, $N$, and cases that don't have natural explanations $\lnot N$.

$$P(E)=P(E)P(N) + P(E)P(\lnot N).\tag{4}$$

On the hypothesis that aliens are visiting Earth both terms on the RHS of Eqn.$(4)$ are present so that we have

$$P(E|A)=P(E).\tag{5}$$

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

$$P(E|\lnot A)=P(E)P(N).\tag{6}$$

Substituting Eqn.$(5)$ and Eqn.$(6)$ into Eqn.$(3)$ we obtain

$$\frac{P(A)}{P(\lnot A)}=P(N).\tag{7}$$

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

https://youtu.be/x0Ec5NJLWOw

Let us assume that the probability of a natural explanation for the Winchester 1976 case (lying, hallucinating or victims of a hoax) is

$$P(N)=1/10.\tag{8}$$

Let us now generalize $P(N)$ in Eqn.$(7)$ so that it becomes 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?

Answer 1

The first part I don't understand how you get is (5). Can you rewrite it using conditional probabilities explicitly?

Anyway, I guess it will be much more simpler to rewrite Bayes theorem in odds form: $$\frac{P(A|E)}{P(\neg A|E)} =\frac{P(A)}{P(\neg A)}\cdot \frac{P(E|A)}{P(E|\neg A)}$$ - posterior odds are prior odds multiplied by how much likely it is to see evidence given $A$ then given $\neg A$.

It looks to me like you are assuming $P(E|A) = 1$ and $P(E | \neg A) = \frac{1}{10}$, in this case your arithmetic is correct - 10 such evidences are enough to get odds from $1:10^{10}$ to $1:1$.

What are correct priors here and how you estimate evidence strength is beyond scope of this site, however I would like to note that if your event is "Winchester said they saw an alien", then your assumptions (if I got them correct) imply, for example, that in every world where $A$ is true Winchester would say it.