On 3/9/2020 3:08 PM, Bruce Kellett wrote:
On Tue, Mar 10, 2020 at 8:54 AM Russell Standish
<[email protected] <mailto:[email protected]>> wrote:
On Sun, Mar 08, 2020 at 10:10:23PM +1100, Bruce Kellett wrote:
>
> > > In order to infer a probability of p = 0.5, your
branch data must
> have
> > > approximately equal numbers of zeros and ones. The
number of
> branches
> > with
> > > equal numbers of zeros and ones is given by the binomial
> coefficient. For
> > large
> > > even N = 2M trials, this coefficient is N!/M!*M!.
Using the
> Stirling
> > > approximation to the factorial for large N, this
goes as 2^N/sqrt
> (N)
> > (within
> > > factors of order one). Since there are 2^N
sequences, the
> proportion with
> > n_0 =
> > > n_1 vanishes as 1/sqrt(N) for N large.
>
>
>
> This is the nub of the proof you wanted.
No - it is simply irrelevant. The statement I made was about the
proportion of strings whose bit ratio lies within certain percentage
of the expected value.
After all when making a measurement, you are are interested in the
value and its error bounds, eg 10mm +/- 0.1%, or 10mm +/- 0.01mm. We
can never know its exact value.
If you are using experimental data to estimate a quantity (and a p
value is a quantity in the required sense), then you are interested in
the confidence interval, not an absolute or percentage error. And the
confidence interval for a given probability of including the true
value decreases with the number of trials (since the standard error
decreases with N).
Right. So that's because the density of results concentrates around the
true value as N->oo. In statistician talk, the mean is a consistent
estimator.
Brent
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