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Consider a linear regression problem where \(p \gg N\), and assume the rank of \(\mathbf{X}\) is \(N\). Let the SVD of \(\mathbf{X}=\mathbf{U D V}^{T}=\mathbf{R V}^{T}\), where \(\mathbf{R}\) is \(N \times N\) nonsingular, and \(\mathbf{V}\) is \(p \times N\) with orthonormal columns. (a) Show that there are infinitely many least-squares solutions all with zero residuals. (b) Show that the ridge-regression estimate for \(\beta\) can be written $$ \hat{\beta}_{\lambda}=\mathbf{V}\left(\mathbf{R}^{T} \mathbf{R}+\lambda \mathbf{I}\right)^{-1} \mathbf{R}^{T} \mathbf{y} $$ (c) Show that when \(\lambda=0\), the solution \(\hat{\beta}_{0}=\mathbf{V D}^{-1} \mathbf{U}^{T} \mathbf{y}\) has residuals all equal to zero, and is unique in that it has the smallest Euclidean norm amongst all zero-residual solutions.

Short Answer

Expert verified
Question: Show how the given conditions on SVD and linear regression problem with more features than samples lead to: (a) Infinitely many least-squares solutions all with zero residuals. (b) The Ridge-regression estimate for \(\boldsymbol{\beta}\) is $$ \hat{\boldsymbol{\beta}}_{\lambda}=\mathbf{V}\left(\mathbf{R}^{T} \mathbf{R}+\lambda \mathbf{I}\right)^{-1} \mathbf{R}^{T} \mathbf{y} $$ (c) When \(\lambda=0\), the solution \(\hat{\boldsymbol{\beta}}_{0}\) has residuals all equal to zero, and is unique in that it has the smallest Euclidean norm amongst all zero-residual solutions.

Step by step solution

01

Least-squares solutions

In the case of a linear regression problem with \(p \gg N\), we have \(\operatorname{rank}(\mathbf{X})=N\). The rank-nullity theorem states that the dimension of the null space of \(\mathbf{X}\) is \(p-N > 0\). Since the null space has a non-zero dimension, there exists a non-zero vector \(\boldsymbol{\delta}\) in the null space of \(\mathbf{X}\). This means that $$\mathbf{X}\boldsymbol{\delta}=\mathbf{0}.$$
02

Infinite least-squares solutions

Consider a least-squares solution \(\boldsymbol{\beta}^*\), then the residual vector is \(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}^*\). If we add \(\boldsymbol{\delta}\) to \(\boldsymbol{\beta}^*\), we get $$\mathbf{X}(\boldsymbol{\beta}^* + \boldsymbol{\delta}) = \mathbf{X}\boldsymbol{\beta}^* + \mathbf{X}\boldsymbol{\delta} = \mathbf{X}\boldsymbol{\beta}^*.$$ Thus, \(\boldsymbol{\beta}^* + \boldsymbol{\delta}\) also has the same residual vector as \(\boldsymbol{\beta}^*\). Since \(\boldsymbol{\delta}\) is non-zero, infinitely many solutions exist for the least-squares problem, all with zero residuals.
03

Ridge-regression estimate for \(\boldsymbol{\beta}\)

The Ridge Regression solution of the problem is given as $$\hat{\boldsymbol{\beta}}_{\lambda}=(\mathbf{X}^{T} \mathbf{X}+\lambda \mathbf{I})^{-1} \mathbf{X}^{T} \mathbf{y}.$$ Using the given SVD decomposition, \(\mathbf{X}=\mathbf{R}\mathbf{V}^{T}\), we can simplify this as $$\hat{\boldsymbol{\beta}}_{\lambda}=(\mathbf{V}\mathbf{R}^{T} \mathbf{R}\mathbf{V}^{T}+\lambda \mathbf{I})^{-1} \mathbf{V}\mathbf{R}^{T} \mathbf{y}.$$ Since \(\mathbf{R}\) is nonsingular, we can simplify further to get $$\hat{\boldsymbol{\beta}}_{\lambda}=\mathbf{V}(\mathbf{R}^{T} \mathbf{R}+\lambda \mathbf{I})^{-1} \mathbf{R}^{T} \mathbf{y}.$$
04

\(\lambda = 0\) case

When \(\lambda = 0\), the Ridge Regression solution becomes $$\hat{\boldsymbol{\beta}}_{0}=\mathbf{V}(\mathbf{R}^{T} \mathbf{R})^{-1} \mathbf{R}^{T} \mathbf{y}=\mathbf{V}\mathbf{D}^{-2}\mathbf{U}^{T}\mathbf{y}.$$ Using the given SVD decomposition, the residual vector is $$\mathbf{y}-\mathbf{X}\hat{\boldsymbol{\beta}}_{0}=\mathbf{y}-\mathbf{U}\mathbf{D} \mathbf{V}^{T}\mathbf{V}\mathbf{D}^{-2}\mathbf{U}^{T}\mathbf{y}=\mathbf{y}-\mathbf{U}\mathbf{U}^{T}\mathbf{y},$$ which is equal to the null space of \(\mathbf{U}\mathbf{U}^{T}\). So, the residual is equal to zero. To show its uniqueness, we can express the Euclidean norm as $$||\hat{\boldsymbol{\beta}}_{0}||^2=\hat{\boldsymbol{\beta}}_{0}^{T}\hat{\boldsymbol{\beta}}_{0}=(\mathbf{V}\mathbf{D}^{-2}\mathbf{U}^{T}\mathbf{y})^{T}(\mathbf{V}\mathbf{D}^{-2}\mathbf{U}^{T}\mathbf{y})=\mathbf{y}^{T}\mathbf{U D}^{-2} \mathbf{U}^{T}\mathbf{y}.$$ Thus, the solution \(\hat{\boldsymbol{\beta}}_{0}\) is unique and has the smallest Euclidean norm amongst all zero-residual solutions.

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Key Concepts

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

Least-Squares Solutions
Least-squares solutions serve as a cornerstone of linear regression, especially valuable when dealing with over-determined systems where there are more equations (data points) than unknowns. This approach aims to minimize the residuals, which are the differences between the observed values and the values predicted by the model.

When the number of predictors, denoted as p, is larger than the number of observations, N, you have a situation where p ≫ N. These conditions can result in more than one solution to the least-squares problem, leading to the conundrum of determining which one is the 'best'. The exercise highlights that due to the rank-nullity theorem, the existence of non-zero vectors in the null space of the design matrix X implies infinite solutions with zero residuals.

Another interesting aspect of least-squares solutions is highlighted when singular value decomposition or ridge regression is applied, altering the solution's properties, such as uniqueness or stability, which is elucidated in the subsequent sections.
Singular Value Decomposition (SVD)
SVD is a powerful factorization technique in linear algebra, commonly used in signal processing and statistics. In the context of linear regression, SVD decomposes the original design matrix X into three components: a matrix U holding the left singular vectors, a diagonal matrix D with singular values, and the transpose of a matrix V that contains the right singular vectors. These components are crucial for understanding the geometry and stability of the solutions to the regression problem.

The singular vectors encapsulate important properties such as the directions of maximum variance. The singular values, found along the diagonal of D, are pivotal in determining the sensitivity of the output to the input and can be instrumental when addressing issues like multicollinearity or the presence of near-zero singular values that can cause numerical problems in calculations. By restructuring the regression in terms of U, D, and V, we gather insights into the nature and behavior of the regression equations, facilitating an enhanced grasp of the least-squares solutions.
Ridge Regression
Ridge regression is a variant of linear regression that incorporates a regularization parameter, λ, to impose a penalty on the size of coefficients. This technique combats overfitting by shrinking the coefficients, thereby making the model less sensitive to the individual data points.

When the design matrix X has a greater number of variables than observations, it is prone to instability and overfitting. Ridge regression ameliorates this by introducing bias into the estimates through λ, which helps manage multicollinearity (when independent variables are highly correlated) by adding a degree of bias to the regression estimates.

The exercise distinctly shows how the estimate for β changes with different values of λ. When the regularization parameter is set to zero, the method simplifies back to least-squares, as captured in the formula provided for βλ. However, nonzero values of λ alter the solution, offering a method to balance between the bias and variance, an essential concept in the bias-variance tradeoff, leading to more robust models in practical scenarios.

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Most popular questions from this chapter

Equivalence between Benjamini-Hochberg and plug-in methods. (a) In the notation of Algorithm 16.2, show that for rejection threshold \(p_{0}=p_{(L)}\), a proportion of at most \(p_{0}\) of the permuted values \(t_{j}^{k}\) exceed \(|T|_{(L)}\) where \(|T|_{(L)}\) is the \(L\) th largest value among the \(\left|t_{j}\right|\). Hence show that the plug-in FDR estimate \(\widehat{\mathrm{FDR}}\) is less than or equal to \(p_{0} \cdot M / L=\alpha\). (b) Show that the cut-point \(|T|_{(L+1)}\) produces a test with estimated FDR greater than \(\alpha\).

Nearest shrunken centroids and the lasso. Consider a (naive Bayes) Gaussian model for classification in which the features \(j=1,2, \ldots, p\) are assumed to be independent within each class \(k=1,2, \ldots, K .\) With observations \(i=1,2, \ldots, N\) and \(C_{k}\) equal to the set of indices of the \(N_{k}\) observations in class \(k\), we observe \(x_{i j} \sim N\left(\mu_{j}+\mu_{j k}, \sigma_{j}^{2}\right)\) for \(i \in C_{k}\) with \(\sum_{k=1}^{K} \mu_{j k}=0 .\) Set \(\hat{\sigma}_{j}^{2}=s_{j}^{2}\), the pooled within-class variance for feature \(j\), and consider the lasso-style minimization problem $$ \min _{\left\\{\mu_{j}, \mu_{j k}\right\\}}\left\\{\frac{1}{2} \sum_{j=1}^{p} \sum_{k=1}^{K} \sum_{i \in C_{k}} \frac{\left(x_{i j}-\mu_{j}-\mu_{j k}\right)^{2}}{s_{j}^{2}}+\lambda \sqrt{N_{k}} \sum_{j=1}^{p} \sum_{k=1}^{K} \mid \frac{\mu_{j k} \mid}{s_{j}} .\right\\}(16.54) $$ Show that the solution is equivalent to the nearest shrunken centroid estimator \((16.5)\), with \(s_{0}\) set to zero, and \(M_{k}\) equal to \(1 / N_{k}\) instead of \(1 / N_{k}-1 / N\) as before.

Bonferroni method for multiple comparisons. Suppose we are in a multiple- testing scenario with null hypotheses \(H_{0 j}, j=1,2, \ldots, M\), and corresponding \(p\)-values \(p_{j}, i=1,2, \ldots, M\). Let \(A\) be the event that at least one null hypothesis is falsely rejected, and let \(A_{j}\) be the event that the \(j\) th null hypothesis is falsely rejected. Suppose that we use the Bonferroni method, rejecting the \(j\) th null hypothesis if \(p_{j}<\alpha / M\). (a) Show that \(\operatorname{Pr}(A) \leq \alpha\). [Hint: \(\operatorname{Pr}\left(A_{j} \cup A_{j^{\prime}}\right)=\operatorname{Pr}\left(A_{j}\right)+\operatorname{Pr}\left(A_{j^{\prime}}\right)-\) \(\left.\operatorname{Pr}\left(A_{j} \cap A_{j^{\prime}}\right)\right]\) (b) If the hypotheses \(H_{0 j}, j=1,2, \ldots, M\), are independent, then \(\operatorname{Pr}(A)=\) \(1-\operatorname{Pr}\left(A^{C}\right)=1-\prod_{j=1}^{M} \operatorname{Pr}\left(A_{j}^{C}\right)=1-(1-\alpha / M)^{M}\). Use this to show that \(\operatorname{Pr}(A) \approx \alpha\) in this case.

Proof of result 16.20. Write $$ \begin{aligned} \mathrm{pFDR} &=\mathrm{E}\left(\frac{V}{R} \mid R>0\right) \\ &=\sum_{k=1}^{M} \mathrm{E}\left[\frac{V}{R} \mid R=k\right] \operatorname{Pr}(R=k \mid R>0) \end{aligned} $$ Use the fact that given \(R=k, V\) is a binomial random variable, with \(k\) trials and probability of success \(\operatorname{Pr}(H=0 \mid T \in \Gamma)\), to complete the proof.

Suppose we wish to select the ridge parameter \(\lambda\) by 10 -fold crossvalidation in a \(p \gg N\) situation (for any linear model). We wish to use the computational shortcuts described in Section 16.3.5. Show that we need only to reduce the \(N \times p\) matrix \(\mathbf{X}\) to the \(N \times N\) matrix \(\mathbf{R}\) once, and can use it in all the cross-validation runs.

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