From: *Brent Meeker* <[email protected] <mailto:[email protected]>>
On 2/20/2020 5:52 PM, Bruce Kellett wrote:
On Fri, Feb 21, 2020 at 12:08 PM 'Brent Meeker' via Everything List
<[email protected]
<mailto:[email protected]>> wrote:
On 2/20/2020 4:26 PM, Bruce Kellett wrote:
This argument has worried me, so I thought that some
serious calculations were in order. If you have an even
number of trials, N = 2M, the the number of binary strings
that have equal numbers of ones and zeros (both equal to M)
is given, from the binomial distribution, as N!/M!*M!.
Using the Stirling expansion for the factorial, as N gets
large,
N! ~ sqrt(N) N^N.
So (2M)!/M!^2 ~ 1/sqrt(N).
I think you misplaced a 2. I get sqrt(pi*M) 2^2M. Anyway the
number obviously goes to infinity. And you can see from the
Gaussian approx that the distribution of 1s becomes more
concentrated near N/2 as N->oo.
I was a little too quick with saying this was the number of
bitstrings with equal zeros and ones. What I was actually calculating
was the proportion of such strings in the 2^N strings from N trials.
Sure, the number of strings with equal zeros and ones increases as
2^N, but so does the total number of strings, so the proportion goes
as 1/sqrt(N) as advertised. I think you have you factors of
sqrt(pi*M) upside down.
Right. Stirling approximation M! ~ sqrt(2pi*M) (M/e)^M
Number with M out of 2M = (2M)!/(M!)^2 =
[sqrt(2pi*2M)(2M/e)^2M]/[sqrt(2pi*M)(M/e)^M]^2
= [1/sqrt(p*M)] 2^2M
(M/e)^2M / (M/e)^2M = 2^2M/sqrt(pi*M)
See below, I say "the proportion of trials with equal numbers of
zeros and ones decreases as 1/sqrt(N) as N becomes large." Equal
numbers of zeros and ones do not dominate as N increases, so it is
not the case that the majority of observers see a probability of 0.5.
Of course that's true. But the more relevant value is the fraction of
sequences with the proportion of 1s within some narrow range of 0.5.
For large N, the distribution is Gaussian with std deviation ~sqrt(N)
so almost equal numbers of 1s and 0s do predominate.
I was aware of that, but they only dominate in a narrow range when p =
0.5. My thinking was that since the confidence interval around the
estimated probability shrinks as 1/sqrt(N) for large N, outside a small
range of small deviations from equal numbers of zeros and ones, the
confidence interval on the probability estimates would no longer capture
p = 0.5. Also, looking at numbers of zeros within +- a small number of
N/2 would give results for the asymptotic proportion similar to those
for N/2 zeros. Since my calculation systematically ignores factors of
the order of one, I doubt that including such bit strings with close to
equal numbers of zeros and ones would make any significant difference to
the conclusion that such strings do not dominate in the limit. In other
words, I think my conclusion that the majority of the 2^N observers
would not estimate probabilities close to 0.5 is secure. (Ignoring
factors of order one in the calculation!)
Bruce
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