Given S, as we observe an initial segment
of some string,
which is the most likely continuation?
Consider x's finite continuations
.
According to Bayes (compare Equation (15)),

(48)

where
is the measure of z^{2}, given z^{1}.
Having observed x we will predict those y that maximize
.
Which are those? In what follows,
we will confirm the intuition that
for
the
only probable continuations of x_{n} are those with fast
programs. The sheer number of ``slowly''
computable strings cannot balance the speed advantage
of ``more quickly'' computable strings with equal beginnings.

Definition 6.4 (tex2html_wrap_inline$p < k_i x$ etc.)
Write
if
finite program p ()
computes x within
less than k steps, and
if it does so within PHASE i of FAST.
Similarly for
and
(at most k steps),
,
(exactly k steps),
,
(at least k steps),
(more than k steps).

Theorem 6.1
Suppose
is S-describable, and
outputs x_{n} within at most f(n) steps for all n,
and
g(n) > O(f(n)).
Then

Proof. Since no program that requires at least g(n) steps for producing
x_{n} can compute x_{n} in a phase with number
< logg(n), we have

Here we have used the Kraft inequality [#!Kraft:49!#]
to obtain a rough upper bound for the enumerator:
when no p is prefix of another one, then

Hence, if we know a rather fast
finite program p^{x} for x, then Theorem 6.1 allows for predicting:
if we observe some x_{n} (n sufficiently large) then
it is very unlikely that it was produced by an x-computing algorithm
much slower than p^{x}.

Among the fastest algorithms for x is FAST itself, which is
at least as fast as p^{x}, save for a constant factor. It outputs
x_{n} after
O(2^{Kt(xn)}) steps. Therefore Theorem 6.1 tells us:

Corollary 6.1
Let
be S-describable.
For
,
with probability 1 the
continuation of x_{n}
is computable within
O(2^{Kt(xn)}) steps.

Given observation x with
,
we
predict a continuation y with minimal Kt(xy).

Example 6.1
Consider
Example 1.2 and Equation (1).
According to the weak anthropic principle,
the conditional probability of a particular observer finding herself
in one of the universes compatible with her existence equals 1.
Given S, we predict a universe with minimal Kt.
Short futures are more likely than long ones: the probability
that the universe's history so far will extend beyond the one computable
in the current phase of FAST (that is, it will be prolongated into
the next phase) is at most 50 %. Infinite futures have measure zero.