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Chapter 10: Problem 10

Let \(x_{1}, x_{2}, \ldots, x_{n}\) be a realization of a random sample. Consider the Hodges-Lehmann estimate of location given in expression (10.9.4). Show that the breakdown point of this estimate is \(0.29 .\) Hint: Suppose we corrupt \(m\) data points. We need to determine the value of \(m\) that results in corruption of one-half of the Walsh averages. Show that the corruption of \(m\) data points leads to $$ p(m)=m+\left(\begin{array}{c} m \\ 2 \end{array}\right)+m(n-m) $$ corrupted Walsh averages. Hence the finite sample breakdown point is the "correct" solution of the quadratic equation \(p(m)=n(n+1) / 4\).

Short Answer

Expert verified
The breakdown point of the Hodges-Lehmann estimate of location is approximately 0.29.

Step by step solution

01

Understand the Walsh averages and their corruption

Walsh averages are a method for measuring the central tendency of a data set, especially robust to outliers. By corrupting \(m\) data points, the number of corrupted Walsh averages \(p(m)\) will be \(m\) (for the corrupted points themselves), plus \(m \choose 2\) (for the corrupted averages among the corrupted points) and \(m(n-m)\) (for the corrupted averages between corrupted and non-corrupted points). This leads to the formula \(p(m) = m+\left(\begin{array}{c} m \ 2\end{array}\right)+m(n-m)\).
02

Solve for finite sample breakdown point

The finite sample breakdown point is the point at which one-half of the Walsh averages are corrupted. This is expressed mathematically as \(p(m) = \frac{n(n+1)}{4}\). To find this point, equate the formula for \(p(m)\) from step 1 to \(\frac{n(n+1)}{4}\) and solve for \(m\).
03

Compute the Breakdown Point

After solving the quadratic equation, it results to \(m\) being approximately 0.29 of \(n\). Therefore, the breakdown point is approximately 0.29, meaning that about 29% of the data can be corrupted without inducing more than half the Walsh averages to be corrupted.

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Most popular questions from this chapter

Let \(X\) be a random variable with cdf \(F(x)\) and let \(T(F)\) be a functional. We say that \(T(F)\) is a scale functional if it satisfies the three properties $$ \text { (i) } T\left(F_{a X}\right)=a T\left(F_{X}\right), \text { for } a>0 $$ (ii) \(T\left(F_{X+b}\right)=T\left(F_{X}\right), \quad\) for all \(b\) $$ \text { (iii) } T\left(F_{-X}\right)=T\left(F_{X}\right) \text { . } $$ Show that the following functionals are scale functionals. (a) The standard deviation, \(T\left(F_{X}\right)=(\operatorname{Var}(X))^{1 / 2}\). (b) The interquartile range, \(T\left(F_{X}\right)=F_{X}^{-1}(3 / 4)-F_{X}^{-1}(1 / 4)\).

Often influence functions are derived by differentiating implicitly the defining equation for the functional at the contaminated cdf \(F_{x, e}(t),(10.9 .13) .\) Consider the mean functional with the defining equation (10.9.10). Using the linearity of the differential, first show that the defining equation at the cdf \(F_{x, \epsilon}(t)\) can be expressed as $$ \begin{aligned} 0=\int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d F_{x, \epsilon}(t)=&(1-\epsilon) \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] f_{X}(t) d t \\ &+\epsilon \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d \Delta(t) \end{aligned} $$ Recall that we want \(\partial T\left(F_{x, \epsilon}\right) / \partial \epsilon .\) Obtain this by implicitly differentiating the above equation with respect to \(\epsilon\).

Consider the general score rank correlation coefficient \(r_{a}\) defined in Exercise 10.8.5. Consider the null hypothesis \(H_{0}: X\) and \(Y\) are independent. (a) Show that \(E_{H_{0}}\left(r_{a}\right)=0\). (b) Based on part (a) and \(H_{0}\), as a first step in obtaining the variance under \(H_{0}\), show that the following expression is true: $$ \operatorname{Var}_{H_{0}}\left(r_{a}\right)=\frac{1}{s_{a}^{4}} \sum_{i=1}^{n} \sum_{j=1}^{n} E_{H_{0}}\left[a\left(R\left(X_{i}\right)\right) a\left(R\left(X_{j}\right)\right)\right] E_{H_{0}}\left[a\left(R\left(Y_{i}\right)\right) a\left(R\left(Y_{j}\right)\right)\right] $$ (c) To determine the expectation in the last expression, consider the two cases \(i=j\) and \(i \neq j\). Then using uniformity of the distribution of the ranks, show that $$ \operatorname{Var}_{H_{0}}\left(r_{a}\right)=\frac{1}{s_{a}^{4}} \frac{1}{n-1} s_{a}^{4}=\frac{1}{n-1} $$

Obtain the sensitivity curves for the sample mean, the sample median and the Hodges-Lehmann estimator for the following data set. Evaluate the curves at the values \(-300\) to 300 in increments of 10 and graph the curves on the same plot. Compare the sensitivity curves. $$ \begin{array}{rrrrrrrr} -9 & 58 & 12 & -1 & -37 & 0 & 11 & 21 \\ 18 & -24 & -4 & -53 & -9 & 9 & 8 & \end{array} $$ Note that the \(\mathrm{R}\) command wilcox.test \((\mathrm{x}\), conf . int \(=\mathrm{T}\) ) \$est computes the Hodges Lehmann estimate for the \(\mathrm{R}\) vector \(\mathrm{x}\).

Let \(\widehat{F}_{n}(x)\) denote the empirical cdf of the sample \(X_{1}, X_{2}, \ldots, X_{n} .\) The distribution of \(\hat{F}_{n}(x)\) puts mass \(1 / n\) at each sample item \(X_{i} .\) Show that its mean is \(\bar{X}\). If \(T(F)=F^{-1}(1 / 2)\) is the median, show that \(T\left(\widehat{F}_{n}\right)=Q_{2}\), the sample median.
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