Due Date

Due Mon Nov 11, 2013.

First Problem

The data set LakeHuron included in R (on-line help) gives annual measurements of the level, in feet, of Lake Huron 1875-1972.

The name of the time series is LakeHuron, for example,


does a time series plot.

The average water level over this period was

> mean(LakeHuron)
[1] 579.0041

Obtain a standard error for this estimate (i. e., mean(LakeHuron)) using the subsampling bootstrap with subsample length b = 10. Assume that the sample mean obeys the square root law (that is, the rate is square root of n), and assume the time series is stationary.

Second Problem

The documentation for the R function lmsreg says

There seems no reason other than historical to use the lms and lqs options. LMS estimation is of low efficiency (converging at rate n− 1 ⁄ 3) whereas LTS has the same asymptotic efficiency as an M estimator with trimming at the quartiles (Marazzi, 1993, p. 201). LQS and LTS have the same maximal breakdown value of (floor((n-p)/2) + 1)/n attained if floor((n+p)/2) <= quantile <= floor((n+p+1)/2). The only drawback mentioned of LTS is greater computation, as a sort was thought to be required (Marazzi, 1993, p. 201) but this is not true as a partial sort can be used (and is used in this implementation).

Thus it seems that LMS regression is the Wrong Thing (with a capital W and a capital T), something that survives only for historico-sociological reasons: it was invented first and most people that have heard of robust regression at all have only heard of it. To be fair to Efron and Tibshirani, the literature cited in the documentation for lmsreg is the same vintage as their book. So maybe they thought they were using the latest and greatest.

Anyway, the problem is to redo the LMS examples using LTS.

What changes? Does LTS really work better here than LMS? Describe the differences you see and why you think these differences indicate LTS is better (or worse, if that is what you think) than LMS.

Third Problem

The file


contains a vector x of data from a heavy tailed distribution such that the sample mean has rate of convergence n1 ⁄ 3, that is

n^(1 / 3) * (theta.hat - theta)

has nontrivial asymptotics (nontrivial here meaning it doesn't converge to zero in probability and also is bounded in probability, so n1 ⁄ 3 is the right rate) where theta.hat is the sample mean and theta is the true unknown population mean.

When the sample mean behaves as badly as this, the sample variance behaves even worse (it converges in probability to infinity), but robust measures of scale make sense, for example, the interquartile range (calculated by the IQR function in R, note that the capital letters are not a mistake).

  1. Using subsample size b = 20, do a subsampling bootstrap to estimate the distribution of the sample mean.

  2. Make a histogram of theta.star marking the point theta.hat.

  3. Make a histogram of b^(1 / 3) * (theta.star - theta.hat) which is the analog in the bootstrap world of the distribution of the quantity having nontrivial asymptotics displayed above.

  4. Calculate the subsampling bootstrap estimate of the IQR of theta.hat, rescaling by the ratio of rates in the appropriate fashion.


Answers in the back of the book are here.