On 3/8/2020 12:08 AM, Bruce Kellett wrote:
On Sun, Mar 8, 2020 at 6:14 PM Russell Standish <[email protected] <mailto:[email protected]>> wrote:On Thu, Mar 05, 2020 at 09:45:38PM +1100, Bruce Kellett wrote: > On Thu, Mar 5, 2020 at 5:26 PM Russell Standish <[email protected] <mailto:[email protected]>> wrote: > > But a very large proportion of them (→1 as N→∞) will report being > within ε (called a confidence interval) of 50% for any given ε>0 > chosen at the outset of the experiment. This is simply the law of > large numbers theorem. You can't focus on the vanishingly small > population that lie outside the confidence interval. > > > This is wrong. Them's fighting words. Prove it! I have, in other posts and below. > In the binary situation where both outcomes occur for every > trial, there are 2^N binary sequences for N repetitions of the experiment. This > set of binary sequences exhausts the possibilities, so the same sequence is > obtained for any two-component initial state -- regardless of the amplitudes. > You appear to assume that the natural probability in this situation is p = 0.5 > and, what is more, your appeal to the law of large numbers applies only for > single-world probabilities, in which there is only one outcome on each trial. I didn't mention proability once in the above paragraph, not even implicitly. I used the term "proportion". That the proportion will be equal to the probability in a single universe case is a frequentist assumption, and should be uncontroversial, but goes beyond what I stated above.Sure. But the proportion of the 2^N sequences that exhibit any particular p value (proportion of 1's) decreases with N.> 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. I wasn't talking about that. I was talking about the proportion of sequences whose ratio of 0 bits to 1 bits lie within ε of 0.5, rather than the proportion of sequences that have exactly equal 0 or 1 bits. That proportion grows as sqrt N.No, it falls as 1/sqrt(N). Remember, the confidence interval depends on the standard deviation, and that falls as 1/sqrt(n). Consequently deviations from equal numbers of zeros and ones for p to be within the CI of 0.5 must decline as n becomes large> Now sequences with small departures from equal numbers will still give > probabilities within the confidence interval of p = 0.5. But this confidence > interval also shrinks as 1/sqrt(N) as N increases, so these additional > sequences do not contribute a growing number of cases giving p ~ 0.5 as N > increases. The confidence interval ε is fixed.No, it is not. The width of, say the 95% CI, decreases with N since the standard deviation falls as 1/sqrt(N).
Right. But that's just a different way of saying the density of results concentrates around the expected value. The CI interval in constructed to contain a certain fraction, but it's width contracts as 1/sqrt(N). Or if you take a fixed deviation interval around the expected value, e.g. 0.333_+_0.01 then the proportion within that interval goes to 1 as N->oo.
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