Dealing with TMI in Statistics
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For instance, consider a random sample, i.i.d., from a Gaussian distribution. Then, a confidence interval for the mean is
We call it a cost since the new confidence interval is now larger (the Student distribution has higher upper-quantiles than the Gaussian distribution).
So usually, if we substitute an estimation to the true value, there is a price to pay.
A few years ago, with Jean David Fermanian and Olivier Scaillet, we were writing a survey on copula density estimation (using kernels, here). At the end, we wanted to add a small paragraph on the fact that we assumed that we wanted to fit a copula on a sample i.i.d. with distribution , a copula, but in practice, we start from a sample with joint distribution (assumed to have continuous margins, and - unique - copula ). But since margins are usually unknown, there should be a price for not observing them.
To be more formal, in a perfect wold, we would consider
My point is that when I ran simulations for the survey (the idea was more to give illustrations of several techniques of estimation, rather than proofs of technical theorems) we observed that the price to pay... was negative ! I.e. the variance of the estimator of the density (wherever on the unit square) was smaller on the pseudo sample than on perfect sample .
By that time, we could not understand why we got that counter-intuitive result: even if we do know the true distribution, it is better not to use it, and to use instead a nonparametric estimator. Our interpretation was based on the discrepancy concept and was related to the latin hypercube construction:
With ranks, the data are more regular, and marginal distributions are exactly uniform on the unit interval. So there is less variance.
This was our heuristic interpretation.
A couple of weeks ago, Christian Genest and Johan Segers proved that intuition in an article published in JMVA,
Well, we observed something for finite , but Christian and Johan obtained an analytical result. Hence, if we denote
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