
Re: A "plausible range" for a random variable
Posted:
Jun 8, 2013 8:36 AM


> 1) > Selecting Lower and Upper limits, L and U, as you notice, is > usually done symmetrically (if not onetailed). That is done, > mainly, for lack of any other good reason to outweigh the > equal emphasis on each end. > > Statistical estimation theory occasionally looks at the > "narrowest" CI. That is *the* important characteristic > of onetailed tests, determining UMP (Uniformly Most > Powerful). Because tails can be asymmetrical, no twotailed > test is UMP. > > Decision theory would suggest that you apply a lossfunction > to determine what degree of asymmetry might apply  I > was intrigued, long ago, by the suggestion that the "power" of > standard research might be improved by splitting the conventional > 5% into 4% at the "expected" end and 1% at the other end, > for a gain in general power without losing the right to report > stronger effects in the opposite direction. I read that at least > 30 years ago, so you can see that the idea never caught on. > > 2) > A parametric approach to L and U for Extreme Values is not > going to be at all efficient. What is used for estimation is what > your bootstrapping would converge to: The CI based for L > (or U) based on rankorder in the original sample. > > Poisson consideration gives a good approximation for small > proportions. This is applied for your N=2000, 2 1/2%, as follows. > > Rank 50 is the point estimate of L. The +/ 2SD range for Poisson > can be estimated as ( Square(Sqrt(L)  1), Square(Sqrt(L) + 1) ) > > The square root of 50 is about 7; the square of 6 is 36, and the > square of 8 is 64. That gives (approximately) the CI for L=50 > is (37, 65). > >  > Rich Ulrich
Great! Thanks a lot, Rich, that's precisely what I needed. You mention that this Poisson approximation is valid for small proportions: what about U, i.e., the 97.5th percentile? This is a "big" proportion, I mean, it's close to 1. Can I still claim that the +/ 2SD CI for U=1950 is given by [(sqrt(U)1)^2, (sqrt(U)+1)^2]? Or do I need another formula? Also, which is the formula for a general C.I., like 90% or 99% or 99.9%, etc.? When using the normal approximation, one substitutes the value 1.96 with the corresponding percentile of the normal distribution, z_{1/2+alpha/2} where alpha={.90, .95, .99,...} etc. I'm not familiar with this Poisson approximation, however, so I don't know how to proceed in this case.
Thanks again,
Best Regards
Sergio Rossi

