On Sun, Mar 8, 2020 at 6:14 PM Russell Standish <[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]> > 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). Bruce -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLTTikTege169WoO-yN-MpxWsT1JX5NY4VN3-0FH3b0ybg%40mail.gmail.com.
