Re: How to numerically estimate an asymptotic equivalent?

*To*: mathgroup at smc.vnet.net*Subject*: [mg46358] Re: How to numerically estimate an asymptotic equivalent?*From*: Bill Rowe <readnewsciv at earthlink.net>*Date*: Mon, 16 Feb 2004 09:00:00 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

On 2/14/04 at 10:19 PM, Franck.Michel at wanadoo.fr (Franck Michel) wrote: >I have computed a sequence a(n) up to a level N. I know this >sequence as an asymptotic equivalent when n -> +infinity: a(n) ~ >(n!)^s*A^n*n^beta*M >but I ignore the values of s, A, beta and M, and I would like to >perform a numerical estimation of these 4 values. >Taking ln(a(n)), one gets a linear expression: s*ln(n!) + ln(A)*n + >beta*ln(n) + ln(M) >Therefore, I've used a linear least squares regression on a >subsequence of a(n) for identifying the four parameters s, ln(A), >beta and ln(M) with confidence intervals. I've used Mathematica >function Regress. All seems to work fine but I have a doubt on the >confidence intervals. >To check the validity if this method, I used another sequence a(n) >which has a an analogous asymptotic equivalent, and for which the 4 >parameters have been theoretically identified. The estimation of >the parameters s and A is good, but the confidence intervals are >false (below the results of the estimation). The basic problem is you are applying statistical methods and concepts to a non-statistical problem. For the usual linear regression problem, you are trying to find a relationship between a measured response and some set of predictors. The measured response is assumed to be the sum of the true response and an error term. The error term is assumed to come from a normal distribution with mean = zero. The confidence intervals Regress computes are based on this assumption about errors. But you don't have errors in the same sense in your problem. You know for several values of n the precise value of the sequence since you have computed it. There is no error term. Consequently, you cannot expect Regress to produce meaningful confidence limits. In fact, the normal meaning of confidence limits doesn't apply to your problem. >How can I manage to have the good confidence intervals? Or, simply, >to have a proof that some parameters are well estimated? Rather than confidence limits, I think you should be more interested in the maximum difference between the computed sequence values and the regression model. If that difference is suitably small, then the estimated parameters (and model) are sufficient. Note, since you are looking for the model that is asymptotically correct, it follows the difference between the computed sequence values and the model should decrease as n gets bigger. So, I would plot the residuals to see this is the case. Also, this points out another difference between the usual regression problem and the problem you are trying to solve. In the normal regression problem you expect the residuals to be independent of the predictor variables. This should not be the case for your problem. >Are there other possible methods to numerically estimate an asymptotic >equivalent like this one? There are a variety of techniques. I would probably use NMinimize to minimize ths sum of the absolute differences rather than Regress. -- To reply via email subtract one hundred and four