Re: Simple question about a tape and a few strings.

In article <slrnhhcgi4.n6v.tim@xxxxxxxxxxxxxxxxxxxxxxxxxx>,
Tim Little <tim@xxxxxxxxxxxxxxxxxx> wrote:

I believe, as T approaches infinity, the ratio #S_k / sum_t (#S_t)
converges to fixed number or each k.

Your proposition is slightly ambiguous: you mention the ratio
(#S_k)/sum_t(#S_t), but do not mention whether this should be
interpreted as a ratio of expected values, or an expected value of the
ratio.

For example, suppose |S_1| = 1, |S_2| = 5, T = 5. Then E(#S_2) = 1/2
(it appears iff it appears first), while E(#S_1) = 31/32. So then
E(#S_1) / E(sum_t(#S_t)) = 31/47 and E(#S_2) / E(sum_t(#S_t)) = 16/47.

However, E(#S_1 / sum_t(#S_t)) = E(#S_2 / sum_t(#S_t)) = 1/2.


On 2009-12-02, Barb Knox <see@xxxxxxxxx> wrote:
Does anyone reading this have a good intuitive combinatorial (i.e.,
not probabilistic) argument showing that considering all possible
completed tapes, every S_k occurs with equal frequency regardless of
the tape length?

As the above example indicates, they don't. Short strings do occur
more frequently - in that example, the shorter string appears 31/16
times more frequently overall.

My bad. I misread the original problem as saying #S_i = i. Since Paul
Black used an example isomorphic to this he too got equal frequencies.

I expect there is a simple argument why when #S_i = i the frequencies
must be equal, but I can't see it. Any ideas?


If you disregard the probabilities of each completed tape and just
look at the combinatorial possibilities, then the disparity is even
greater: the possibilities are 1, 11, 111, 1111, 11111, and 2.

It is true that the final strings are uniformly distributed, by
reversal symmetry and the condition that the initial strings are.


- Tim



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