[net.math] Rational expectation of survival

lew@ihuxr.UUCP (12/01/83)

Bayes' Theorem gives a very simple answer when applied to survival functions.
If S(t) is the probability of still being alive at time t, the probability,
S'(t'), of being alive at time t0 + t', given that we were alive at time t0,
is just S(t0 + t')/S(t0) .

If S(t) is an exponential decay, representing a constant probability
per unit time of being killed, S'(t') = S(t').  This means that if we feel
certain for some reason that there is a certain probability per unit time
of being killed (call it p), this certainty is rationally maintainable
in the face of any run of good luck, however long. This is equivalent to
maintaining a 50/50 expectation of heads/tails after a run of heads.

If we are uncertain as to the exact value of p, we can give our estimation
of the likelihood of various values in the form of a distribution
function, f(p). This means that we think the probability that p has
value between p' and p' + dp' is given by f(p') * dp'.

The survival function is then given by:

	S(t) = int 0 to inf of f(p) * exp( -p*t) * dp

This is just the weighted average of the survival functions for all p.
Note that the Bayesian result ( S'(t') = S(t'+t0)/S(t0) ) just redefines
our distribution function:

	S'(t') = int 0 to inf of f'(p) * exp( -p*t) * dp

	f'(p) = f(p) * exp( -p*t0 ) / S(t0)

This is just another form of Bayes' Theorem. Note that the model
based on a distribution over p is equivalent to simply specifying
the appropriate survival function. A nice choice of f(p) is:

	f(p) = (1/p0) * exp( - p/p0 )

This gives S(t) = 1/(1+p0*t). The reader can verify that

	S'(t') = S(t'+t0)/S(t0) = 1/(1+p1*t') ; p1 = p0/(1+p0*t0)

... so that our survival function retains the same form but with a
time constant which is optimistically revised the longer we survive.

	Lew Mammel, Jr. ihnp4!ihuxr!lew

P.S. This choice of f(p) gives something very close to Herb Norton's
answer. If we start with p0 = 1/yr, after 35 years we have p1 = 1/36 yrs.
I would emphasize, though that this isn't the only answer. The answer
always depends on our a priori expectations.