August 29, 2012
The Bayesian immortals
A hundred billion or so humans have ever lived, but only seven billion are alive now (which gives the human condition a 93% mortality rate)
While this is true given the sample, it reeks of frequentism. What should a Bayesian say?
Assume the true mortality of humans is p. We know that out of 100 billion trials 93 billion are dead ("success") and 7 billion have not died ("no success"). So assuming a uniform initial prior for p and that the above data is an outcome of a binomial distribution gives us a Beta(93 billion, 7 billion) distribution as the posterior for p. The mean is 0.93, agreeing with XKCD. This is also more or less the median or mode. But we also get the standard deviation, telling us how uncertain we should be: 8*10-7! So obviously we should be really confident in that 7% of people are immortal, right?
The mistake is of course using still-living people as evidence. Clearly some might, hypothetically, die. So this can be viewed as an experiment in lifetime analysis suffering from type I censoring.
One approach might be to say that there is a certain probability q that someone is born immortal. What is the probability distribution for q given that all immortals are included in the set of people born less than 120 years ago? Bayes tells us:
P(q|all immortals recent) = P(all immortals recent|q) P(q)/P(all immortals recent)
We can break up the first term:
P(all immortals recent|q) = sum P(N immortals recent|q) P(N immortals|q)
If the probability of being immortal is q, then P(N immortals|q)= (100 billion-N over N) qN (1-q)(100 billion-N). It is of course strongly peaked around 100 billion times q.
If we assume human history to be the last 200,000 years, then the chance that a random person is born in the last 120 years is r=0.0006. If there are N immortals the probability of them all being in this range is rN. So, again assuming a uniform prior for q (I doubt the Jeffreys prior changes things) and ignoring normalisation,
P(q|all immortals recent) = sum rN Binomial(100 billion-N, N) qN (1-q)(100 billion-N)
This is unfortunately a somewhat hard to evaluate sum because of the big numbers. If we make the assumption that N is going to be 100 billion times q with probability 1 and 0 otherwise, then it simplifies to P(q|all immortals recent) = r(100 billion * q). This is a rather sharp exponential: essentially all probability mass lies at q=0. So, sadly, Bayes doesn't believe in immortals very much.
Of course, what I am betting on is that we can change the probabilities.
Posted by Anders3 at August 29, 2012 12:57 PM