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Decision Trees, Ensemble Methods and Boosting
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Dimensionality Reduction
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Exercises week 34
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Week 34: Introduction to the course, Logistics and Practicalities
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Exercises week 36
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Week 36: Linear Rgeression and Statistical interpretations
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<h1>Exercises week 36<a class="headerlink" href="#exercises-week-36" title="Permalink to this headline"></a></h1>
<p><strong>September 2-6, 2024</strong></p>
<p>Date: <strong>Deadline is Friday September 6 at midnight</strong></p>
<div class="section" id="overarching-aims-of-the-exercises-this-week">
<h2>Overarching aims of the exercises this week<a class="headerlink" href="#overarching-aims-of-the-exercises-this-week" title="Permalink to this headline"></a></h2>
<p>This set of exercises form an important part of the first project. The
analytical exercises deal with the material covered last week on the
mathematical interpretations of ordinary least squares and of Ridge
regression. The numerical exercises can be seen as a continuation of
exercise 3 from week 35, with the inclusion of Ridge regression. This
material enters also the discussions of the first project.</p>
</div>
<div class="section" id="exercise-1-analytical-exercises">
<h2>Exercise 1: Analytical exercises<a class="headerlink" href="#exercise-1-analytical-exercises" title="Permalink to this headline"></a></h2>
<p>The aim here is to derive the expression for the optimal parameters
using Ridge regression. Furthermore, using the singular value
decomposition, we will analyze the difference between the ordinary
least squares approach and Ridge regression.</p>
<p>The expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, was given by the
optimization problem</p>
<div class="math notranslate nohighlight">
\[
{\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
\]</div>
<p>which we can also write as</p>
<div class="math notranslate nohighlight">
\[
{\displaystyle \min_{\boldsymbol{\beta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2,
\]</div>
<p>where we have used the definition of a norm-2 vector, that is</p>
<div class="math notranslate nohighlight">
\[
\vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}.
\]</div>
<p>By minimizing the above equation with respect to the parameters
<span class="math notranslate nohighlight">\(\boldsymbol{\beta}\)</span> we could then obtain an analytical expression for the
parameters <span class="math notranslate nohighlight">\(\boldsymbol{\beta}\)</span>.</p>
<p>We can add a regularization parameter <span class="math notranslate nohighlight">\(\lambda\)</span> by
defining a new cost function to be optimized, that is</p>
<div class="math notranslate nohighlight">
\[
{\displaystyle \min_{\boldsymbol{\beta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2
\]</div>
<p>which leads to the Ridge regression minimization problem. One can require as part of the optimization problem
that <span class="math notranslate nohighlight">\(\vert\vert \boldsymbol{\beta}\vert\vert_2^2\le t\)</span>, where <span class="math notranslate nohighlight">\(t\)</span> is
a finite number larger than zero. We will not implement that here.</p>
<div class="section" id="a-expression-for-ridge-regression">
<h3>a) Expression for Ridge regression<a class="headerlink" href="#a-expression-for-ridge-regression" title="Permalink to this headline"></a></h3>
<p>Show that the optimal parameters</p>
<div class="math notranslate nohighlight">
\[
\hat{\boldsymbol{\beta}}_{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y},
\]</div>
<p>with <span class="math notranslate nohighlight">\(\boldsymbol{I}\)</span> being a <span class="math notranslate nohighlight">\(p\times p\)</span> identity matrix with the constraint that</p>
<div class="math notranslate nohighlight">
\[
\sum_{i=0}^{p-1} \beta_i^2 \leq t,
\]</div>
<p>with <span class="math notranslate nohighlight">\(t\)</span> a finite positive number. In the optimization, we will not require that the latter is satisfied.</p>
<p>The ordinary least squares result is</p>
<div class="math notranslate nohighlight">
\[
\hat{\boldsymbol{\beta}}_{\mathrm{OLS}} = \left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y},
\]</div>
</div>
<div class="section" id="b-the-singular-value-decomposition">
<h3>b) The singular value decomposition<a class="headerlink" href="#b-the-singular-value-decomposition" title="Permalink to this headline"></a></h3>
<p>Here we will use the singular value decomposition of an <span class="math notranslate nohighlight">\(n\times p\)</span> matrix <span class="math notranslate nohighlight">\(\boldsymbol{X}\)</span> (our design matrix)</p>
<div class="math notranslate nohighlight">
\[
\boldsymbol{X}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T,
\]</div>
<p>to study properties of Ridge regression and ordinary least squares regression.
Here <span class="math notranslate nohighlight">\(\boldsymbol{U}\)</span> and <span class="math notranslate nohighlight">\(\boldsymbol{V}\)</span> are orthogonal matrices of dimensions
<span class="math notranslate nohighlight">\(n\times n\)</span> and <span class="math notranslate nohighlight">\(p\times p\)</span>, respectively, and <span class="math notranslate nohighlight">\(\boldsymbol{\Sigma}\)</span> is an
<span class="math notranslate nohighlight">\(n\times p\)</span> matrix which contains the singular values only. This material was discussed during the lectures of week 35.</p>
<p>Show that you can write the
OLS solutions in terms of the eigenvectors (the columns) of the orthogonal matrix <span class="math notranslate nohighlight">\(\boldsymbol{U}\)</span> as</p>
<div class="math notranslate nohighlight">
\[
\tilde{\boldsymbol{y}}_{\mathrm{OLS}}=\boldsymbol{X}\boldsymbol{\beta} = \sum_{j=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\boldsymbol{y}.
\]</div>
<p>For Ridge regression, show that the corresponding equation is</p>
<div class="math notranslate nohighlight">
\[
\tilde{\boldsymbol{y}}_{\mathrm{Ridge}}=\boldsymbol{X}\boldsymbol{\beta}_{\mathrm{Ridge}} = \boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{\Sigma}^2\boldsymbol{V}^T+\lambda\boldsymbol{I} \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\sum_{j=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\boldsymbol{y},
\]</div>
<p>with the vectors <span class="math notranslate nohighlight">\(\boldsymbol{u}_j\)</span> being the columns of <span class="math notranslate nohighlight">\(\boldsymbol{U}\)</span> from the SVD of the matrix <span class="math notranslate nohighlight">\(\boldsymbol{X}\)</span>.</p>
<p>Give an interpretation of the results. <a class="reference external" href="https://link.springer.com/book/10.1007/978-0-387-84858-7">Section 3.4 of Hastie et als textbook gives a good discussion of the above results</a>.</p>
</div>
</div>
<div class="section" id="exercise-2-adding-ridge-regression">
<h2>Exercise 2: Adding Ridge Regression<a class="headerlink" href="#exercise-2-adding-ridge-regression" title="Permalink to this headline"></a></h2>
<p>This exercise is a continuation of exercise 3 from week 35, see <a class="reference external" href="https://compphysics.github.io/MachineLearning/doc/LectureNotes/_build/html/exercisesweek35.html">https://compphysics.github.io/MachineLearning/doc/LectureNotes/_build/html/exercisesweek35.html</a>. We will use the same function to
generate our data set, still staying with a simple function <span class="math notranslate nohighlight">\(y(x)\)</span>
which we want to fit using linear regression, but now extending the
analysis to include the Ridge regression method.</p>
<p>In this exercise you need to include the same elements from last week, that is</p>
<ol class="simple">
<li><p>scale your data by subtracting the mean value from each column in the design matrix.</p></li>
<li><p>perform a split of the data in a training set and a test set.</p></li>
</ol>
<p>The addition to the analysis this time is the introduction of the hyperparameter <span class="math notranslate nohighlight">\(\lambda\)</span> when introducing Ridge regression.</p>
<p>Extend the code from exercise 3 from <a class="reference external" href="https://compphysics.github.io/MachineLearning/doc/LectureNotes/_build/html/exercisesweek35.html">week 35</a> to include Ridge regression with the hyperparameter <span class="math notranslate nohighlight">\(\lambda\)</span>. The optimal parameters <span class="math notranslate nohighlight">\(\hat{\beta}\)</span> for Ridge regression can be obtained by matrix inversion in a similar way as done for ordinary least squares. You need to add to your code the following equations</p>
<div class="math notranslate nohighlight">
\[
\hat{\boldsymbol{\beta}}_{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}.
\]</div>
<p>The ordinary least squares result you encoded last week is given by</p>
<div class="math notranslate nohighlight">
\[
\hat{\boldsymbol{\beta}}_{\mathrm{OLS}} = \left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y},
\]</div>
<p>Use these results to compute the mean squared error for ordinary least
squares and Ridge regression first for a polynomial of degree five
with <span class="math notranslate nohighlight">\(n=100\)</span> data points and five selected values of
<span class="math notranslate nohighlight">\(\lambda=[0.0001,0.001, 0.01,0.1,1.0]\)</span>. Compute thereafter the mean
squared error for the same values of <span class="math notranslate nohighlight">\(\lambda\)</span> for polynomials of degree ten
and <span class="math notranslate nohighlight">\(15\)</span>. Discuss your results for the training MSE and test MSE with
Ridge regression and ordinary least squares.</p>
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