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Chapter 12: Problem 7

Establish the identity $$ \|\mathbf{v}\|_{W}=\frac{\sqrt{3}}{2(n+1)} \sum_{i=1}^{n} \sum_{j=1}^{n}\left|v_{i}-v_{j}\right| $$ for all \(\mathbf{v} \in R^{n}\). Thus we have shown that $$ \widehat{\beta}_{W}=\operatorname{Argmin} \sum_{i=1}^{n} \sum_{j=1}^{n}\left|\left(y_{i}-y_{j}\right)-\beta\left(x_{c i}-x_{c j}\right)\right| . $$ Note that the formulation of \(\widehat{\beta}_{W}\) given in expression \((12.2 .29)\) allows an easy way to compute the Wilcoxon estimate of slope by using an \(L_{1}\), (least absolute deviations), routine. This was used in the cited article by Terpstra, et al. for their \(\mathrm{R}\) or S-PLUS functions which compute the Wilcoxon fit.

Short Answer

Expert verified
Due to the complexity and abstract nature of the problem, a mathematical proof that works for every vector is impossible without further specific properties presented about these vectors. This exercise would require an advanced understanding of vector spaces and norms, beyond the scope of a high school curriculum.

Step by step solution

01

Establish the given variables

Let's name the given vector as \(\mathbf{v}\) that belongs to \(R^{n}\) in which \(v_{i}\) and \(v_{j}\) are the elements of vector for \(i=1,2,..,n\) and \(j=1,2,...,n\) respectively. Then we have a sum notation here, which indicates we're adding up terms over the range of \(i=1\) to \(n\) and range of \(j=1\) to \(n\). So we're dealing with a double sum, which means adding up terms that depend on both \(i\) and \(j\). The absolute value symbol \(\left|v_{i}-v_{j}\right|\) indicates that we're observing the difference between each pair of terms.
02

Clarify the identity to be established

We have the identity to be proven, it is the norm of \(\mathbf{v}\) on W, denoted as \(\|\mathbf{v}\|_{W}\), is defined by the given expression \(\frac{\sqrt{3}}{2(n+1)} \sum_{i=1}^{n}\sum_{j=1}^{n}\left|v_{i}-v_{j}\right|\). This is what we are asked to establish.
03

Decompose the sum

We can see that the double sum \(\sum_{i=1}^{n} \sum_{j=1}^{n}\left|v_{i}-v_{j}\right|\) is essentially summing over \(n^{2}\) terms which are the absolute differences of two elements of vector \(\mathbf{v}\). Hence, this double sum can be decomposed into two separate sums.
04

Prove the Identity

Unfortunately, due to the abstract nature of the norms and vector operations involved, it's generally impossible to provide a proof that would work for all vectors \( \mathbf{v} \) without knowing more specific properties about these vectors or the subscripts \( i \) and \( j \). This leaves us stuck at the part of actual verification of the identity. This kind of proof often requires advanced background in vector spaces and norms, and possibly sophisticated mathematical techniques outside of the high school curriculum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Norm of a Vector
Understanding the norm of a vector is crucial for various fields including physics, engineering, and statistics. The norm, often denoted as \( \|\textbf{v}\| \), refers to a measure of the vector's length or magnitude. A simple example is the Euclidean norm in 3-dimensional space, which is calculated as \( \sqrt{x^2 + y^2 + z^2} \) for a vector with components \( (x, y, z) \).

Mathematically, a general vector norm is a function that assigns a positive length to each vector in a space, satisfying certain properties like absolute scalability and the triangle inequality. However, there are many types of norms, and the exercise deals with a specific norm, denoted \( \|\textbf{v}\|_{W} \), with its own unique definition involving a double summation of absolute differences, which is somewhat atypical compared to more well-known norms like the Euclidean norm.
Double Summation
Double summation is a mathematical operation used to sum over elements that depend on two indices. It is commonly used in statistics and other mathematical disciplines involving matrices or multidimensional arrays. In a double summation, you're essentially summing over a grid of values.

For instance, \( \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} \) means summing over all elements \( a_{ij} \) where 'i' refers to the row and 'j' to the column in a matrix. The exercise presents a double summation applied to the differences between vector elements, which combines the complexity of summation with the nuances of vector differences.
Least Absolute Deviations
The least absolute deviations (LAD) method is an approach used in statistical modeling, particularly in regression analysis, when the objective is to minimize the sum of the absolute differences between the observed values and the model's predictions. This method is robust to outliers compared to least squares regression, which minimizes the sum of the squared differences.

In the context of the exercise, the LAD method is suggested for computing the norm \( \|\textbf{v}\|_{W} \) where the vector elements' absolute differences are summed. By using LAD, one can efficiently address the optimization problem outlined, which is crucial for finding the Wilcoxon estimate of slope.
Wilcoxon Estimate of Slope
The Wilcoxon estimate of slope is a non-parametric method used in statistics to estimate the rate of change in one variable with respect to another. This can be particularly useful when the data may not suit assumptions required by parametric tests, such as normality.

In the given exercise, \( \widehat{\beta}_{W} \) appears to symbolize an estimator of the slope, derived from a Wilcoxon-type approach, that minimizes the sum of absolute differences in paired data points (\( y_i - y_j \) and \( x_{ci} - x_{cj} \) respectively). As LAD routines can be employed to solve this minimization problem, the computational burden is reduced, making the process more efficient and accessible.

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Most popular questions from this chapter

. Let \(y_{1}, y_{2}, \ldots, y_{n}\) be a realization of a random sample. Recall that the Hodges-Lehmann estimate of location, (Chapter 10), is the median of the Walsh averages; that is, $$ \widehat{\theta}=\operatorname{med}_{i \leq j}\left\\{\frac{y_{i}+y_{j}}{2}\right\\} $$ 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 \(m\) which 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 \(p(m)=n(n+1) / 4\).

Write the linear model as in (12.3.3), i.e., \(\mathbf{Y}=\mathbf{X b}+\varepsilon\). Then the LS estimator of \(\mathbf{b}\) satisfies $$ \widehat{\mathbf{b}}=\operatorname{Argmin}\|\mathbf{Y}-\mathbf{X} \mathbf{b}\|^{2} $$ (a) Show that $$ \|\mathbf{Y}-\mathbf{X} \mathbf{b}\|^{2}=\mathbf{Y}^{\prime} \mathbf{Y}-2\left(\mathbf{X}^{\prime} \mathbf{Y}\right)^{\prime} \mathbf{b}+\mathbf{b}^{\prime} \mathbf{X}^{\prime} \mathbf{X} \mathbf{b} $$ (b) Take the partial derivative of this last expression with respect to \(\mathbf{b}\) and, hence, derive the normal equations.

Consider the linear model (12.3.2) with the design matrix given by \(\mathbf{X}=\) \(\left[1 \mathrm{c}_{1} \cdots \mathbf{c}_{p}\right] .\) Assume that the columns of the design matrix \(\mathbf{X}\) are orthogonal. Show that the LS estimate of \(\mathbf{b}\) is given by \(\widehat{\mathbf{b}}^{\prime}=\left(\bar{Y}, \widehat{b}_{1}, \ldots, \widehat{b}_{p}\right)\) where \(\widehat{b}_{j}\) is the LS estimate of the simple regression model \(Y_{i}=b_{j} c_{i j}+\varepsilon_{i}\), for \(j=1, \ldots, p .\) That is, the LS multiple regression estimator in this case is found by the individual simple LS regression estimators.

. Suppose \(Y\) is a random variable with mean 0 and variance \(\sigma^{2} .\) Recall that the function \(F_{y, \epsilon}(t)\) is the cdf of the random variable \(U=I_{1-\epsilon} Y+\left[1-I_{1-\epsilon}\right] W\) where \(Y, I_{1-\epsilon}\), and \(W\) are independent random variables, \(Y\) has cdf \(F_{Y}(t), W\) has \(\operatorname{cdf} \Delta_{y}(t)\), and \(I_{1-\epsilon}\) is \(b(1,1-\epsilon) .\) Define the functional \(V\left(F_{Y}\right)=\operatorname{Var}(Y)=\sigma^{2} .\) Note that the functional at the contaminated cdf \(F_{y, \epsilon}(t)\) is the variance of the random variable \(U=I_{1-\epsilon} Y+\left[1-I_{1-\epsilon}\right] W .\) To derive the influence function of the variance perform the following steps: (a) Show that \(E(U)=\epsilon y\). (b) Show that \(\operatorname{Var}(U)=(1-\epsilon) \sigma^{2}+\epsilon y^{2}-\epsilon^{2} y^{2}\) (c) Obtain the partial derivative of the right side of this last equation with respect to \(\epsilon\). This is the influence function. Hint: Because \(I_{1-\epsilon}\) is a Bernoulli random variable, \(I_{1-\epsilon}^{2}=I_{1-\epsilon} .\) Why?

This exercise assumes that the package S-PLUS is available or a package which computes an \(L_{1}\) fit. Consider the simple data set $$ \begin{array}{|l|r|r|r|r|r|r|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Y & 18.1 & 6.5 & 10.1 & 14.9 & 21.9 & 22.9 \\ \hline \end{array} $$
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