> Consider the 97.5-th percentile to be a 2.5-th percentile > when you count from the other end. So when you take > rank-50 as a point estimate with CI of (37,65), you have > the symmetrical values at the other end of n values -- > rank (n-50) with CI (N-65, N-37).
Of course - silly me! In other words, it's like estimating a value of the inverse of R(y)=p(Y>=y)=1-p(Y<=y)=G(y), instead than a value of the inverse of G(y).
> > The Poisson is the distribution observed in random counts. > When a proportion is small, you can use Poisson as a good > approximation to what you would get for the Binomial, which > is the more general case of "counts out of a total". >
Understood - when the number of counts becomes very small compared with the total, i.e., when the proportion goes to 0, then the distribution of random counts and the distribution of "counts out of a total" are bound to become similar, because it's like letting the total go to infinity.
> The theory for the CI is that you can estimate the variance > of a transformation of X by taking the right derivative.
Hmmm, surely I'm missing something here. Right derivative? I thought that the theory of CI was to find the asymptotic distribution of an estimator.
This > works out as follows. From that estimate, the standard > deviation of the sqrt(Poisson) = 1/2 (approximately). > And the distribution of the sqrt(Poisson) is very close to > normal, once the counts are above a few.
Ok...intuitively this makes sense, because if either N is large or the proportion is closer to 0.5, then the normal approximation for the CI is good. Thus, I would expect this Poisson distribution to get close to the normal distribution.
> > Thus, the +/- 2SD range around the sqrt(Poisson) is the > 95% CI, to a pretty good approximation. Or, +/- 1. Then > you square that, to get the (slightly asymmetrical) CI for the > original distribution. You are typically going to round the > final results to integers, since this is the CI of "counts". > > When you want some CI other than 95%, two-tailed, you > multiple the SD=1/2 by some other multiplier than 2.0. >
Great! That's the result I needed. Rich, thanks a bunch for your help! I didn't know about this Poisson distribution, but you told me all I need to write down a simple formula for the estimate. Now I'm all set! Owe you a favour :)
> > -- > Rich Ulrich
ps any chances you can shed some light on my other question too? :)))) The link: