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doc/LectureNotes/.ipynb_checkpoints/exercisesweek37-checkpoint.ipynb
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{
"cells": [
{
"cell_type": "markdown",
"id": "7d56b2d5",
"metadata": {
"editable": true
},
"source": [
"<!-- HTML file automatically generated from DocOnce source (https://github.com/doconce/doconce/)\n",
"doconce format html exercisesweek37.do.txt -->\n",
"<!-- dom:TITLE: Exercises week 36 -->"
]
},
{
"cell_type": "markdown",
"id": "c7a8e9c7",
"metadata": {
"editable": true
},
"source": [
"# Exercises week 36\n",
"**Implementing gradient descent for Ridge and ordinary Least Squares Regression**\n",
"\n",
"Date: **September 8-12, 2025**"
]
},
{
"cell_type": "markdown",
"id": "cf8f0ecb",
"metadata": {
"editable": true
},
"source": [
"## Learning goals\n",
"\n",
"After having completed these exercises you will have:\n",
"1. Your own code for the implementation of the simplest gradient descent approach applied to ordinary least squares (OLS) and Ridge regression\n",
"\n",
"2. Be able to compare the analytical expressions for OLS and Rudge regression with the gradient descent approach\n",
"\n",
"3. Explore the role of the learning rate in the gradient descent approach and the hyperparameter $\\lambda$ in Ridge regression\n",
"\n",
"4. Scale the data properly"
]
},
{
"cell_type": "markdown",
"id": "a67ae548",
"metadata": {
"editable": true
},
"source": [
"## Ridge regression and a new Synthetic Dataset\n",
"\n",
"We create a synthetic linear regression dataset with a sparse\n",
"underlying relationship. This means we have many features but only a\n",
"few of them actually contribute to the target. In our example, we’ll\n",
"use 10 features with only 3 non-zero weights in the true model. This\n",
"way, the target is generated as a linear combination of a few features\n",
"(with known coefficients) plus some random noise. The steps we include are:\n",
"\n",
"Decide on the number of samples and features (e.g. 100 samples, 10 features).\n",
"Define the **true** coefficient vector with mostly zeros (for sparsity). For example, we set $\\hat{\\boldsymbol{\\theta}} = [5.0, -3.0, 0.0, 0.0, 0.0, 0.0, 2.0, 0.0, 0.0, 0.0]$, meaning only features 0, 1, and 6 have a real effect on y.\n",
"\n",
"Then we sample feature values for $\\boldsymbol{X}$ randomly (e.g. from a normal distribution). We use a normal distribution so features are roughly centered around 0.\n",
"Then we compute the target values $y$ using the linear combination $\\boldsymbol{X}\\hat{\\boldsymbol{\\theta}}$ and add some noise (to simulate measurement error or unexplained variance).\n",
"\n",
"Below is the code to generate the dataset:"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "f2d4a55d",
"metadata": {
"collapsed": false,
"editable": true
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"outputs": [],
"source": [
"import numpy as np\n",
"\n",
"# Set random seed for reproducibility\n",
"np.random.seed(0)\n",
"\n",
"# Define dataset size\n",
"n_samples = 100\n",
"n_features = 10\n",
"\n",
"# Define true coefficients (sparse linear relationship)\n",
"theta_true = np.array([5.0, -3.0, 0.0, 0.0, 0.0, 0.0, 2.0, 0.0, 0.0, 0.0])\n",
"\n",
"# Generate feature matrix X (n_samples x n_features) with random values\n",
"X = np.random.randn(n_samples, n_features) # standard normal distribution\n",
"\n",
"# Generate target values y with a linear combination of X and theta_true, plus noise\n",
"noise = 0.5 * np.random.randn(n_samples) # Gaussian noise\n",
"y = X.dot @ theta_true + noise"
]
},
{
"cell_type": "markdown",
"id": "a445583b",
"metadata": {
"editable": true
},
"source": [
"This code produces a dataset where only features 0, 1, and 6\n",
"significantly influence $\\boldsymbol{y}$. The rest of the features have zero true\n",
"coefficient, so they only contribute noise. For example, feature 0 has\n",
"a true weight of 5.0, feature 1 has -3.0, and feature 6 has 2.0, so\n",
"the expected relationship is:"
]
},
{
"cell_type": "markdown",
"id": "4a81ddf9",
"metadata": {
"editable": true
},
"source": [
"$$\n",
"y \\approx 5 \\times X_0 \\;-\\; 3 \\times X_1 \\;+\\; 2 \\times X_6 \\;+\\; \\text{noise}.\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "ae590275",
"metadata": {
"editable": true
},
"source": [
"## Exercise 1, scale your data\n",
"\n",
"Before fitting a regression model, it is good practice to normalize or\n",
"standardize the features. This ensures all features are on a\n",
"comparable scale, which is especially important when using\n",
"regularization. Here we will perform standardization, scaling each\n",
"feature to have mean 0 and standard deviation 1:\n",
"\n",
"Compute the mean and standard deviation of each column (feature) in $bm{X}$.\n",
"Subtract the mean and divide by the standard deviation for each feature.\n",
"\n",
"We will also center the target $\\boldsymbol{y}$ to mean $0$. Centering $\\boldsymbol{y}$\n",
"(and each feature) means the model won’t require a separate intercept\n",
"term – the data is shifted such that the intercept is effectively 0\n",
". (In practice, one could include an intercept in the model and not\n",
"penalize it, but here we simplify by centering.)"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "8b40c47a",
"metadata": {
"collapsed": false,
"editable": true
},
"outputs": [],
"source": [
"# Standardize features (zero mean, unit variance for each feature)\n",
"X_mean = X.mean(axis=0)\n",
"X_std = X.std(axis=0)\n",
"X_std[X_std == 0] = 1 # safeguard to avoid division by zero for constant features\n",
"X_norm = (X - X_mean) / X_std\n",
"\n",
"# Center the target to zero mean (optional, to simplify intercept handling)\n",
"y_mean = ?\n",
"y_centered = ?"
]
},
{
"cell_type": "markdown",
"id": "ff9c0c81",
"metadata": {
"editable": true
},
"source": [
"### 1a)\n",
"\n",
"Fill in the necessary details.\n",
"\n",
"After this preprocessing, each column of $\\boldsymbol{X}_norm$ has mean zero and standard deviation $1$\n",
"and $\\boldsymbol{y}_centered$ has mean 0. This makes the optimization landscape\n",
"nicer and ensures the regularization penalty $\\lambda \\sum_j\n",
"\\beta_j^2$ treats each coefficient fairly (since features are on the\n",
"same scale)."
]
},
{
"cell_type": "markdown",
"id": "d27c70e4",
"metadata": {
"editable": true
},
"source": [
"## Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\\boldsymbol{theta}$"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "9f1e5184",
"metadata": {
"collapsed": false,
"editable": true
},
"outputs": [],
"source": [
"# Set regularization parameter, either a single value or a vector of values\n",
"lambda = ?\n",
"\n",
"# Analytical form for OLS and Ridge solution: theta_Ridge = (X^T X + lambda * I)^{-1} X^T y and theta_OLS = (X^T X)^{-1} X^T y\n",
"I = np.eye(n_features)\n",
"theta_closed_formRidge = ?\n",
"theta_closed_formOLS = ?\n",
"\n",
"print(\"Closed-form Ridge coefficients:\", theta_closed_form)\n",
"print(\"Closed-form OLS coefficients:\", theta_closed_form)"
]
},
{
"cell_type": "markdown",
"id": "2ec556b9",
"metadata": {
"editable": true
},
"source": [
"This computes the ridge and OLS regression coefficients directly. The identity\n",
"matrix $I$ has the same size as $X^T X$ (which is n_features x\n",
"n_features), and lam * I adds $\\lambda$ to the diagonal of $X^T X. We\n",
"then invert this matrix and multiply by $X^T y. The result\n",
"for $\\boldsymbol{\\theta}$ is a NumPy array of shape (n_features,) containing the\n",
"fitted weights."
]
},
{
"cell_type": "markdown",
"id": "a821f0c5",
"metadata": {
"editable": true
},
"source": [
"### 2a)\n",
"\n",
"Finalize the OLS and Ridge regression determination of the optimal parameters $bm{\\theta}$."
]
},
{
"cell_type": "markdown",
"id": "d637130e",
"metadata": {
"editable": true
},
"source": [
"### 2b)\n",
"\n",
"Explore the results as function of different values of the hyperparameter $\\lambda$. See for example exercise 4 from week 36."
]
},
{
"cell_type": "markdown",
"id": "b455ce7e",
"metadata": {
"editable": true
},
"source": [
"## Implementing the simplest form for gradient descent\n",
"\n",
"Alternatively, we can fit the ridge regression model using gradient\n",
"descent. This is useful to visualize the iterative convergence and is\n",
"necessary if $n$ and $p$ are so large that the closed-form might be\n",
"too slow or memory-intensive. We derive the gradients from the cost\n",
"functions defined above. Use the gradients of the Ridge and OLS cost functions with respect to\n",
"the parameters $\\boldsymbol{\\theta}$ and set up (using the template below) your own gradient descent code for OLS and Ridge regression.\n",
"\n",
"Below is a template code for gradient descent implementation of ridge:"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "cfa1eb29",
"metadata": {
"collapsed": false,
"editable": true
},
"outputs": [],
"source": [
"# Gradient descent parameters, learning rate eta first\n",
"eta = 0.1\n",
"# Then number of iterations\n",
"num_iters = 1000\n",
"\n",
"# Initialize weights for gradient descent\n",
"theta = np.zeros(n_features)\n",
"\n",
"# Arrays to store history for plotting\n",
"cost_history = np.zeros(num_iters)\n",
"\n",
"# Gradient descent loop\n",
"m = n_samples # number of examples\n",
"for t in range(num_iters):\n",
" # Compute prediction error\n",
" error = X_norm.dot(theta) - y_centered \n",
" # Compute cost for OLS and Ridge (MSE + regularization for Ridge) for monitoring\n",
" cost_OLS = ?\n",
" cost_Ridge = ?\n",
" cost_history[t] = ?\n",
" # Compute gradients for OSL and Ridge\n",
" grad_OLS = ?\n",
" grad_Ridge = ?\n",
" # Update parameters theta\n",
" theta_gdOLS = ?\n",
" theta_gdRidge = ? \n",
"\n",
"# After the loop, theta contains the fitted coefficients\n",
"theta_gdOLS = ?\n",
"theta_gdRidge = ?\n",
"print(\"Gradient Descent OLS coefficients:\", theta_gdOLS)\n",
"print(\"Gradient Descent Ridge coefficients:\", theta_gdRidge)"
]
},
{
"cell_type": "markdown",
"id": "dc78d58d",
"metadata": {
"editable": true
},
"source": [
"### 3a)\n",
"\n",
"Discuss the results as function of the learning rate paramaters and the number of iterations."
]
},
{
"cell_type": "markdown",
"id": "15060acb",
"metadata": {
"editable": true
},
"source": [
"### 3b)\n",
"\n",
"Add a stopping parameter as function of the number iterations. \n",
"\n",
"If everything worked correctly, the learned coefficients should be\n",
"close to the true values [5.0, -3.0, 0.0, …, 2.0, …] that we used to\n",
"generate the data. Keep in mind that due to regularization and noise,\n",
"the learned values will not exactly equal the true ones, but they\n",
"should be in the same ballpark."
]
}
],
"metadata": {},
"nbformat": 4,
"nbformat_minor": 5
}
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doc/LectureNotes/_build/html/_sources/exercisesweek37.ipynb
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@@ -2,7 +2,7 @@
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@@ -14,7 +14,7 @@
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@@ -27,7 +27,7 @@
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@@ -46,7 +46,7 @@
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"id": "74fb184e",
"metadata": {
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@@ -72,7 +72,7 @@
{
"cell_type": "code",
"execution_count": 1,
"id": "f2d4a55d",
"id": "9e6acfef",
"metadata": {
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@@ -101,33 +101,43 @@
},
{
"cell_type": "markdown",
"id": "a445583b",
"id": "f2d03ca8",
"metadata": {
"editable": true
},
"source": [
"This code produces a dataset where only features 0, 1, and 6\n",
"significantly influence $\\boldsymbol{y}$. The rest of the features have zero true\n",
"coefficient, so they only contribute noise. For example, feature 0 has\n",
"coefficient. For example, feature 0 has\n",
"a true weight of 5.0, feature 1 has -3.0, and feature 6 has 2.0, so\n",
"the expected relationship is:"
]
},
{
"cell_type": "markdown",
"id": "4a81ddf9",
"id": "d2d64f9b",
"metadata": {
"editable": true
},
"source": [
"$$\n",
"y \\approx 5 \\times X_0 \\;-\\; 3 \\times X_1 \\;+\\; 2 \\times X_6 \\;+\\; \\text{noise}.\n",
"y \\approx 5 \\times x_0 \\;-\\; 3 \\times x_1 \\;+\\; 2 \\times x_6 \\;+\\; \\text{noise}.\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "ae590275",
"id": "b4248e9d",
"metadata": {
"editable": true
},
"source": [
"You can remove the noise if you wish to."
]
},
{
"cell_type": "markdown",
"id": "5fed181f",
"metadata": {
"editable": true
},
@@ -138,14 +148,24 @@
"standardize the features. This ensures all features are on a\n",
"comparable scale, which is especially important when using\n",
"regularization. Here we will perform standardization, scaling each\n",
"feature to have mean 0 and standard deviation 1:\n",
"feature to have mean 0 and standard deviation 1."
]
},
{
"cell_type": "markdown",
"id": "6ec0227c",
"metadata": {
"editable": true
},
"source": [
"### 1a)\n",
"\n",
"Compute the mean and standard deviation of each column (feature) in $bm{X}$.\n",
"Compute the mean and standard deviation of each column (feature) in $\\boldsymbol{X}$.\n",
"Subtract the mean and divide by the standard deviation for each feature.\n",
"\n",
"We will also center the target $\\boldsymbol{y}$ to mean $0$. Centering $\\boldsymbol{y}$\n",
"(and each feature) means the model won’t require a separate intercept\n",
"term – the data is shifted such that the intercept is effectively 0\n",
"(and each feature) means the model does not require a separate intercept\n",
"term, the data is shifted such that the intercept is effectively 0\n",
". (In practice, one could include an intercept in the model and not\n",
"penalize it, but here we simplify by centering.)"
]
@@ -153,7 +173,7 @@
{
"cell_type": "code",
"execution_count": 2,
"id": "8b40c47a",
"id": "a140aac7",
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@@ -173,36 +193,34 @@
},
{
"cell_type": "markdown",
"id": "ff9c0c81",
"id": "57ad18f5",
"metadata": {
"editable": true
},
"source": [
"### 1a)\n",
"\n",
"Fill in the necessary details.\n",
"\n",
"After this preprocessing, each column of $\\boldsymbol{X}_norm$ has mean zero and standard deviation $1$\n",
"and $\\boldsymbol{y}_centered$ has mean 0. This makes the optimization landscape\n",
"After this preprocessing, each column of $\\boldsymbol{X}_{\\mathrm{norm}}$ has mean zero and standard deviation $1$\n",
"and $\\boldsymbol{y}_{\\mathrm{centered}}$ has mean 0. This makes the optimization landscape\n",
"nicer and ensures the regularization penalty $\\lambda \\sum_j\n",
"\\beta_j^2$ treats each coefficient fairly (since features are on the\n",
"\\theta_j^2$ in Ridge regression treats each coefficient fairly (since features are on the\n",
"same scale)."
]
},
{
"cell_type": "markdown",
"id": "d27c70e4",
"id": "2886697d",
"metadata": {
"editable": true
},
"source": [
"## Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\\boldsymbol{theta}$"
"## Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\\boldsymbol{\\theta}$"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "9f1e5184",
"id": "97ac6cb6",
"metadata": {
"collapsed": false,
"editable": true
@@ -223,34 +241,33 @@
},
{
"cell_type": "markdown",
"id": "2ec556b9",
"id": "3efb067b",
"metadata": {
"editable": true
},
"source": [
"This computes the ridge and OLS regression coefficients directly. The identity\n",
"matrix $I$ has the same size as $X^T X$ (which is n_features x\n",
"n_features), and lam * I adds $\\lambda$ to the diagonal of $X^T X. We\n",
"then invert this matrix and multiply by $X^T y. The result\n",
"for $\\boldsymbol{\\theta}$ is a NumPy array of shape (n_features,) containing the\n",
"fitted weights."
"This computes the Ridge and OLS regression coefficients directly. The identity\n",
"matrix $I$ has the same size as $X^T X$. It adds $\\lambda$ to the diagonal of $X^T X for Ridge regression. We\n",
"then invert this matrix and multiply by $X^T y$. The result\n",
"for $\\boldsymbol{\\theta}$ is a NumPy array of shape (n$\\_$features,) containing the\n",
"fitted parameters $\\boldsymbol{\\theta}$.."
]
},
{
"cell_type": "markdown",
"id": "a821f0c5",
"id": "53be2bf8",
"metadata": {
"editable": true
},
"source": [
"### 2a)\n",
"\n",
"Finalize the OLS and Ridge regression determination of the optimal parameters $bm{\\theta}$."
"Finalize, in the above code, the OLS and Ridge regression determination of the optimal parameters $\\boldsymbol{\\theta}$."
]
},
{
"cell_type": "markdown",
"id": "d637130e",
"id": "e4126591",
"metadata": {
"editable": true
},
@@ -262,12 +279,12 @@
},
{
"cell_type": "markdown",
"id": "b455ce7e",
"id": "642d0850",
"metadata": {
"editable": true
},
"source": [
"## Implementing the simplest form for gradient descent\n",
"## Exercise 3, Implementing the simplest form for gradient descent\n",
"\n",
"Alternatively, we can fit the ridge regression model using gradient\n",
"descent. This is useful to visualize the iterative convergence and is\n",
@@ -282,7 +299,7 @@
{
"cell_type": "code",
"execution_count": 4,
"id": "cfa1eb29",
"id": "a67af634",
"metadata": {
"collapsed": false,
"editable": true
@@ -325,32 +342,32 @@
},
{
"cell_type": "markdown",
"id": "dc78d58d",
"id": "1c8c35dc",
"metadata": {
"editable": true
},
"source": [
"### 3a)\n",
"\n",
"Discuss the results as function of the learning rate paramaters and the number of iterations."
"Discuss the results as function of the learning rate parameters and the number of iterations."
]
},
{
"cell_type": "markdown",
"id": "15060acb",
"id": "899fec5c",
"metadata": {
"editable": true
},
"source": [
"### 3b)\n",
"\n",
"Add a stopping parameter as function of the number iterations. \n",
"Try to add a stopping parameter as function of the number iterations. How would you define a stopping criterion? \n",
"\n",
"If everything worked correctly, the learned coefficients should be\n",
"close to the true values [5.0, -3.0, 0.0, …, 2.0, …] that we used to\n",
"generate the data. Keep in mind that due to regularization and noise,\n",
"the learned values will not exactly equal the true ones, but they\n",
"should be in the same ballpark."
"should be in the same ballpark. Which method (OLS or Ridge) gives the best results?"
]
}
],
doc/LectureNotes/_build/html/exercisesweek37.html
+28 -28
View File
@@ -383,12 +383,12 @@ document.write(`
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#a">1a)</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercise-2-use-the-analytical-formulae-for-ols-and-ridge-regression-to-find-the-optimal-paramters-boldsymbol-theta">Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters <span class="math notranslate nohighlight">\(\boldsymbol{theta}\)</span></a><ul class="nav section-nav flex-column">
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercise-2-use-the-analytical-formulae-for-ols-and-ridge-regression-to-find-the-optimal-paramters-boldsymbol-theta">Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters <span class="math notranslate nohighlight">\(\boldsymbol{\theta}\)</span></a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#id1">2a)</a></li>
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#b">2b)</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#implementing-the-simplest-form-for-gradient-descent">Implementing the simplest form for gradient descent</a><ul class="nav section-nav flex-column">
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercise-3-implementing-the-simplest-form-for-gradient-descent">Exercise 3, Implementing the simplest form for gradient descent</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#id2">3a)</a></li>
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#id3">3b)</a></li>
</ul>
@@ -459,13 +459,14 @@ y = X.dot @ theta_true + noise
</div>
<p>This code produces a dataset where only features 0, 1, and 6
significantly influence <span class="math notranslate nohighlight">\(\boldsymbol{y}\)</span>. The rest of the features have zero true
coefficient, so they only contribute noise. For example, feature 0 has
coefficient. For example, feature 0 has
a true weight of 5.0, feature 1 has -3.0, and feature 6 has 2.0, so
the expected relationship is:</p>
<div class="math notranslate nohighlight">
\[
y \approx 5 \times X_0 \;-\; 3 \times X_1 \;+\; 2 \times X_6 \;+\; \text{noise}.
y \approx 5 \times x_0 \;-\; 3 \times x_1 \;+\; 2 \times x_6 \;+\; \text{noise}.
\]</div>
<p>You can remove the noise if you wish to.</p>
</section>
<section id="exercise-1-scale-your-data">
<h2>Exercise 1, scale your data<a class="headerlink" href="#exercise-1-scale-your-data" title="Link to this heading">#</a></h2>
@@ -473,12 +474,14 @@ y \approx 5 \times X_0 \;-\; 3 \times X_1 \;+\; 2 \times X_6 \;+\; \text{noise}.
standardize the features. This ensures all features are on a
comparable scale, which is especially important when using
regularization. Here we will perform standardization, scaling each
feature to have mean 0 and standard deviation 1:</p>
<p>Compute the mean and standard deviation of each column (feature) in <span class="math notranslate nohighlight">\(bm{X}\)</span>.
feature to have mean 0 and standard deviation 1.</p>
<section id="a">
<h3>1a)<a class="headerlink" href="#a" title="Link to this heading">#</a></h3>
<p>Compute the mean and standard deviation of each column (feature) in <span class="math notranslate nohighlight">\(\boldsymbol{X}\)</span>.
Subtract the mean and divide by the standard deviation for each feature.</p>
<p>We will also center the target <span class="math notranslate nohighlight">\(\boldsymbol{y}\)</span> to mean <span class="math notranslate nohighlight">\(0\)</span>. Centering <span class="math notranslate nohighlight">\(\boldsymbol{y}\)</span>
(and each feature) means the model won’t require a separate intercept
term – the data is shifted such that the intercept is effectively 0
(and each feature) means the model does not require a separate intercept
term, the data is shifted such that the intercept is effectively 0
. (In practice, one could include an intercept in the model and not
penalize it, but here we simplify by centering.)</p>
<div class="cell docutils container">
@@ -496,18 +499,16 @@ y_centered = ?
</div>
</div>
</div>
<section id="a">
<h3>1a)<a class="headerlink" href="#a" title="Link to this heading">#</a></h3>
<p>Fill in the necessary details.</p>
<p>After this preprocessing, each column of <span class="math notranslate nohighlight">\(\boldsymbol{X}_norm\)</span> has mean zero and standard deviation <span class="math notranslate nohighlight">\(1\)</span>
and <span class="math notranslate nohighlight">\(\boldsymbol{y}_centered\)</span> has mean 0. This makes the optimization landscape
<p>After this preprocessing, each column of <span class="math notranslate nohighlight">\(\boldsymbol{X}_{\mathrm{norm}}\)</span> has mean zero and standard deviation <span class="math notranslate nohighlight">\(1\)</span>
and <span class="math notranslate nohighlight">\(\boldsymbol{y}_{\mathrm{centered}}\)</span> has mean 0. This makes the optimization landscape
nicer and ensures the regularization penalty <span class="math notranslate nohighlight">\(\lambda \sum_j
\beta_j^2\)</span> treats each coefficient fairly (since features are on the
\theta_j^2\)</span> in Ridge regression treats each coefficient fairly (since features are on the
same scale).</p>
</section>
</section>
<section id="exercise-2-use-the-analytical-formulae-for-ols-and-ridge-regression-to-find-the-optimal-paramters-boldsymbol-theta">
<h2>Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters <span class="math notranslate nohighlight">\(\boldsymbol{theta}\)</span><a class="headerlink" href="#exercise-2-use-the-analytical-formulae-for-ols-and-ridge-regression-to-find-the-optimal-paramters-boldsymbol-theta" title="Link to this heading">#</a></h2>
<h2>Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters <span class="math notranslate nohighlight">\(\boldsymbol{\theta}\)</span><a class="headerlink" href="#exercise-2-use-the-analytical-formulae-for-ols-and-ridge-regression-to-find-the-optimal-paramters-boldsymbol-theta" title="Link to this heading">#</a></h2>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span># Set regularization parameter, either a single value or a vector of values
@@ -524,23 +525,22 @@ print(&quot;Closed-form OLS coefficients:&quot;, theta_closed_form)
</div>
</div>
</div>
<p>This computes the ridge and OLS regression coefficients directly. The identity
matrix <span class="math notranslate nohighlight">\(I\)</span> has the same size as <span class="math notranslate nohighlight">\(X^T X\)</span> (which is n_features x
n_features), and lam * I adds <span class="math notranslate nohighlight">\(\lambda\)</span> to the diagonal of <span class="math notranslate nohighlight">\(X^T X. We
then invert this matrix and multiply by \)</span>X^T y. The result
for <span class="math notranslate nohighlight">\(\boldsymbol{\theta}\)</span> is a NumPy array of shape (n_features,) containing the
fitted weights.</p>
<p>This computes the Ridge and OLS regression coefficients directly. The identity
matrix <span class="math notranslate nohighlight">\(I\)</span> has the same size as <span class="math notranslate nohighlight">\(X^T X\)</span>. It adds <span class="math notranslate nohighlight">\(\lambda\)</span> to the diagonal of <span class="math notranslate nohighlight">\(X^T X for Ridge regression. We
then invert this matrix and multiply by \)</span>X^T y<span class="math notranslate nohighlight">\(. The result
for \)</span>\boldsymbol{\theta}<span class="math notranslate nohighlight">\( is a NumPy array of shape (n\)</span>_<span class="math notranslate nohighlight">\(features,) containing the
fitted parameters \)</span>\boldsymbol{\theta}$..</p>
<section id="id1">
<h3>2a)<a class="headerlink" href="#id1" title="Link to this heading">#</a></h3>
<p>Finalize the OLS and Ridge regression determination of the optimal parameters <span class="math notranslate nohighlight">\(bm{\theta}\)</span>.</p>
<p>Finalize, in the above code, the OLS and Ridge regression determination of the optimal parameters <span class="math notranslate nohighlight">\(\boldsymbol{\theta}\)</span>.</p>
</section>
<section id="b">
<h3>2b)<a class="headerlink" href="#b" title="Link to this heading">#</a></h3>
<p>Explore the results as function of different values of the hyperparameter <span class="math notranslate nohighlight">\(\lambda\)</span>. See for example exercise 4 from week 36.</p>
</section>
</section>
<section id="implementing-the-simplest-form-for-gradient-descent">
<h2>Implementing the simplest form for gradient descent<a class="headerlink" href="#implementing-the-simplest-form-for-gradient-descent" title="Link to this heading">#</a></h2>
<section id="exercise-3-implementing-the-simplest-form-for-gradient-descent">
<h2>Exercise 3, Implementing the simplest form for gradient descent<a class="headerlink" href="#exercise-3-implementing-the-simplest-form-for-gradient-descent" title="Link to this heading">#</a></h2>
<p>Alternatively, we can fit the ridge regression model using gradient
descent. This is useful to visualize the iterative convergence and is
necessary if <span class="math notranslate nohighlight">\(n\)</span> and <span class="math notranslate nohighlight">\(p\)</span> are so large that the closed-form might be
@@ -588,16 +588,16 @@ print(&quot;Gradient Descent Ridge coefficients:&quot;, theta_gdRidge)
</div>
<section id="id2">
<h3>3a)<a class="headerlink" href="#id2" title="Link to this heading">#</a></h3>
<p>Discuss the results as function of the learning rate paramaters and the number of iterations.</p>
<p>Discuss the results as function of the learning rate parameters and the number of iterations.</p>
</section>
<section id="id3">
<h3>3b)<a class="headerlink" href="#id3" title="Link to this heading">#</a></h3>
<p>Add a stopping parameter as function of the number iterations.</p>
<p>Try to add a stopping parameter as function of the number iterations. How would you define a stopping criterion?</p>
<p>If everything worked correctly, the learned coefficients should be
close to the true values [5.0, -3.0, 0.0, …, 2.0, …] that we used to
generate the data. Keep in mind that due to regularization and noise,
the learned values will not exactly equal the true ones, but they
should be in the same ballpark.</p>
should be in the same ballpark. Which method (OLS or Ridge) gives the best results?</p>
</section>
</section>
</section>
@@ -663,12 +663,12 @@ should be in the same ballpark.</p>
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#a">1a)</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercise-2-use-the-analytical-formulae-for-ols-and-ridge-regression-to-find-the-optimal-paramters-boldsymbol-theta">Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters <span class="math notranslate nohighlight">\(\boldsymbol{theta}\)</span></a><ul class="nav section-nav flex-column">
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercise-2-use-the-analytical-formulae-for-ols-and-ridge-regression-to-find-the-optimal-paramters-boldsymbol-theta">Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters <span class="math notranslate nohighlight">\(\boldsymbol{\theta}\)</span></a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#id1">2a)</a></li>
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#b">2b)</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#implementing-the-simplest-form-for-gradient-descent">Implementing the simplest form for gradient descent</a><ul class="nav section-nav flex-column">
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exercise-3-implementing-the-simplest-form-for-gradient-descent">Exercise 3, Implementing the simplest form for gradient descent</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#id2">3a)</a></li>
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#id3">3b)</a></li>
</ul>
doc/LectureNotes/_build/html/searchindex.js
+1 -1
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doc/LectureNotes/_build/jupyter_execute/exercisesweek37.ipynb
+59 -42
View File
@@ -2,7 +2,7 @@
"cells": [
{
"cell_type": "markdown",
"id": "7d56b2d5",
"id": "d3aa801d",
"metadata": {
"editable": true
},
@@ -14,7 +14,7 @@
},
{
"cell_type": "markdown",
"id": "c7a8e9c7",
"id": "7c64e6da",
"metadata": {
"editable": true
},
@@ -27,7 +27,7 @@
},
{
"cell_type": "markdown",
"id": "cf8f0ecb",
"id": "51e35698",
"metadata": {
"editable": true
},
@@ -46,7 +46,7 @@
},
{
"cell_type": "markdown",
"id": "a67ae548",
"id": "74fb184e",
"metadata": {
"editable": true
},
@@ -72,7 +72,7 @@
{
"cell_type": "code",
"execution_count": 1,
"id": "f2d4a55d",
"id": "9e6acfef",
"metadata": {
"collapsed": false,
"editable": true
@@ -101,33 +101,43 @@
},
{
"cell_type": "markdown",
"id": "a445583b",
"id": "f2d03ca8",
"metadata": {
"editable": true
},
"source": [
"This code produces a dataset where only features 0, 1, and 6\n",
"significantly influence $\\boldsymbol{y}$. The rest of the features have zero true\n",
"coefficient, so they only contribute noise. For example, feature 0 has\n",
"coefficient. For example, feature 0 has\n",
"a true weight of 5.0, feature 1 has -3.0, and feature 6 has 2.0, so\n",
"the expected relationship is:"
]
},
{
"cell_type": "markdown",
"id": "4a81ddf9",
"id": "d2d64f9b",
"metadata": {
"editable": true
},
"source": [
"$$\n",
"y \\approx 5 \\times X_0 \\;-\\; 3 \\times X_1 \\;+\\; 2 \\times X_6 \\;+\\; \\text{noise}.\n",
"y \\approx 5 \\times x_0 \\;-\\; 3 \\times x_1 \\;+\\; 2 \\times x_6 \\;+\\; \\text{noise}.\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "ae590275",
"id": "b4248e9d",
"metadata": {
"editable": true
},
"source": [
"You can remove the noise if you wish to."
]
},
{
"cell_type": "markdown",
"id": "5fed181f",
"metadata": {
"editable": true
},
@@ -138,14 +148,24 @@
"standardize the features. This ensures all features are on a\n",
"comparable scale, which is especially important when using\n",
"regularization. Here we will perform standardization, scaling each\n",
"feature to have mean 0 and standard deviation 1:\n",
"feature to have mean 0 and standard deviation 1."
]
},
{
"cell_type": "markdown",
"id": "6ec0227c",
"metadata": {
"editable": true
},
"source": [
"### 1a)\n",
"\n",
"Compute the mean and standard deviation of each column (feature) in $bm{X}$.\n",
"Compute the mean and standard deviation of each column (feature) in $\\boldsymbol{X}$.\n",
"Subtract the mean and divide by the standard deviation for each feature.\n",
"\n",
"We will also center the target $\\boldsymbol{y}$ to mean $0$. Centering $\\boldsymbol{y}$\n",
"(and each feature) means the model won’t require a separate intercept\n",
"term – the data is shifted such that the intercept is effectively 0\n",
"(and each feature) means the model does not require a separate intercept\n",
"term, the data is shifted such that the intercept is effectively 0\n",
". (In practice, one could include an intercept in the model and not\n",
"penalize it, but here we simplify by centering.)"
]
@@ -153,7 +173,7 @@
{
"cell_type": "code",
"execution_count": 2,
"id": "8b40c47a",
"id": "a140aac7",
"metadata": {
"collapsed": false,
"editable": true
@@ -173,36 +193,34 @@
},
{
"cell_type": "markdown",
"id": "ff9c0c81",
"id": "57ad18f5",
"metadata": {
"editable": true
},
"source": [
"### 1a)\n",
"\n",
"Fill in the necessary details.\n",
"\n",
"After this preprocessing, each column of $\\boldsymbol{X}_norm$ has mean zero and standard deviation $1$\n",
"and $\\boldsymbol{y}_centered$ has mean 0. This makes the optimization landscape\n",
"After this preprocessing, each column of $\\boldsymbol{X}_{\\mathrm{norm}}$ has mean zero and standard deviation $1$\n",
"and $\\boldsymbol{y}_{\\mathrm{centered}}$ has mean 0. This makes the optimization landscape\n",
"nicer and ensures the regularization penalty $\\lambda \\sum_j\n",
"\\beta_j^2$ treats each coefficient fairly (since features are on the\n",
"\\theta_j^2$ in Ridge regression treats each coefficient fairly (since features are on the\n",
"same scale)."
]
},
{
"cell_type": "markdown",
"id": "d27c70e4",
"id": "2886697d",
"metadata": {
"editable": true
},
"source": [
"## Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\\boldsymbol{theta}$"
"## Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\\boldsymbol{\\theta}$"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "9f1e5184",
"id": "97ac6cb6",
"metadata": {
"collapsed": false,
"editable": true
@@ -223,34 +241,33 @@
},
{
"cell_type": "markdown",
"id": "2ec556b9",
"id": "3efb067b",
"metadata": {
"editable": true
},
"source": [
"This computes the ridge and OLS regression coefficients directly. The identity\n",
"matrix $I$ has the same size as $X^T X$ (which is n_features x\n",
"n_features), and lam * I adds $\\lambda$ to the diagonal of $X^T X. We\n",
"then invert this matrix and multiply by $X^T y. The result\n",
"for $\\boldsymbol{\\theta}$ is a NumPy array of shape (n_features,) containing the\n",
"fitted weights."
"This computes the Ridge and OLS regression coefficients directly. The identity\n",
"matrix $I$ has the same size as $X^T X$. It adds $\\lambda$ to the diagonal of $X^T X for Ridge regression. We\n",
"then invert this matrix and multiply by $X^T y$. The result\n",
"for $\\boldsymbol{\\theta}$ is a NumPy array of shape (n$\\_$features,) containing the\n",
"fitted parameters $\\boldsymbol{\\theta}$.."
]
},
{
"cell_type": "markdown",
"id": "a821f0c5",
"id": "53be2bf8",
"metadata": {
"editable": true
},
"source": [
"### 2a)\n",
"\n",
"Finalize the OLS and Ridge regression determination of the optimal parameters $bm{\\theta}$."
"Finalize, in the above code, the OLS and Ridge regression determination of the optimal parameters $\\boldsymbol{\\theta}$."
]
},
{
"cell_type": "markdown",
"id": "d637130e",
"id": "e4126591",
"metadata": {
"editable": true
},
@@ -262,12 +279,12 @@
},
{
"cell_type": "markdown",
"id": "b455ce7e",
"id": "642d0850",
"metadata": {
"editable": true
},
"source": [
"## Implementing the simplest form for gradient descent\n",
"## Exercise 3, Implementing the simplest form for gradient descent\n",
"\n",
"Alternatively, we can fit the ridge regression model using gradient\n",
"descent. This is useful to visualize the iterative convergence and is\n",
@@ -282,7 +299,7 @@
{
"cell_type": "code",
"execution_count": 4,
"id": "cfa1eb29",
"id": "a67af634",
"metadata": {
"collapsed": false,
"editable": true
@@ -325,32 +342,32 @@
},
{
"cell_type": "markdown",
"id": "dc78d58d",
"id": "1c8c35dc",
"metadata": {
"editable": true
},
"source": [
"### 3a)\n",
"\n",
"Discuss the results as function of the learning rate paramaters and the number of iterations."
"Discuss the results as function of the learning rate parameters and the number of iterations."
]
},
{
"cell_type": "markdown",
"id": "15060acb",
"id": "899fec5c",
"metadata": {
"editable": true
},
"source": [
"### 3b)\n",
"\n",
"Add a stopping parameter as function of the number iterations. \n",
"Try to add a stopping parameter as function of the number iterations. How would you define a stopping criterion? \n",
"\n",
"If everything worked correctly, the learned coefficients should be\n",
"close to the true values [5.0, -3.0, 0.0, …, 2.0, …] that we used to\n",
"generate the data. Keep in mind that due to regularization and noise,\n",
"the learned values will not exactly equal the true ones, but they\n",
"should be in the same ballpark."
"should be in the same ballpark. Which method (OLS or Ridge) gives the best results?"
]
}
],
doc/LectureNotes/exercisesweek37.ipynb
+59 -42
View File
@@ -2,7 +2,7 @@
"cells": [
{
"cell_type": "markdown",
"id": "7d56b2d5",
"id": "d3aa801d",
"metadata": {
"editable": true
},
@@ -14,7 +14,7 @@
},
{
"cell_type": "markdown",
"id": "c7a8e9c7",
"id": "7c64e6da",
"metadata": {
"editable": true
},
@@ -27,7 +27,7 @@
},
{
"cell_type": "markdown",
"id": "cf8f0ecb",
"id": "51e35698",
"metadata": {
"editable": true
},
@@ -46,7 +46,7 @@
},
{
"cell_type": "markdown",
"id": "a67ae548",
"id": "74fb184e",
"metadata": {
"editable": true
},
@@ -72,7 +72,7 @@
{
"cell_type": "code",
"execution_count": 1,
"id": "f2d4a55d",
"id": "9e6acfef",
"metadata": {
"collapsed": false,
"editable": true
@@ -101,33 +101,43 @@
},
{
"cell_type": "markdown",
"id": "a445583b",
"id": "f2d03ca8",
"metadata": {
"editable": true
},
"source": [
"This code produces a dataset where only features 0, 1, and 6\n",
"significantly influence $\\boldsymbol{y}$. The rest of the features have zero true\n",
"coefficient, so they only contribute noise. For example, feature 0 has\n",
"coefficient. For example, feature 0 has\n",
"a true weight of 5.0, feature 1 has -3.0, and feature 6 has 2.0, so\n",
"the expected relationship is:"
]
},
{
"cell_type": "markdown",
"id": "4a81ddf9",
"id": "d2d64f9b",
"metadata": {
"editable": true
},
"source": [
"$$\n",
"y \\approx 5 \\times X_0 \\;-\\; 3 \\times X_1 \\;+\\; 2 \\times X_6 \\;+\\; \\text{noise}.\n",
"y \\approx 5 \\times x_0 \\;-\\; 3 \\times x_1 \\;+\\; 2 \\times x_6 \\;+\\; \\text{noise}.\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "ae590275",
"id": "b4248e9d",
"metadata": {
"editable": true
},
"source": [
"You can remove the noise if you wish to."
]
},
{
"cell_type": "markdown",
"id": "5fed181f",
"metadata": {
"editable": true
},
@@ -138,14 +148,24 @@
"standardize the features. This ensures all features are on a\n",
"comparable scale, which is especially important when using\n",
"regularization. Here we will perform standardization, scaling each\n",
"feature to have mean 0 and standard deviation 1:\n",
"feature to have mean 0 and standard deviation 1."
]
},
{
"cell_type": "markdown",
"id": "6ec0227c",
"metadata": {
"editable": true
},
"source": [
"### 1a)\n",
"\n",
"Compute the mean and standard deviation of each column (feature) in $bm{X}$.\n",
"Compute the mean and standard deviation of each column (feature) in $\\boldsymbol{X}$.\n",
"Subtract the mean and divide by the standard deviation for each feature.\n",
"\n",
"We will also center the target $\\boldsymbol{y}$ to mean $0$. Centering $\\boldsymbol{y}$\n",
"(and each feature) means the model won’t require a separate intercept\n",
"term – the data is shifted such that the intercept is effectively 0\n",
"(and each feature) means the model does not require a separate intercept\n",
"term, the data is shifted such that the intercept is effectively 0\n",
". (In practice, one could include an intercept in the model and not\n",
"penalize it, but here we simplify by centering.)"
]
@@ -153,7 +173,7 @@
{
"cell_type": "code",
"execution_count": 2,
"id": "8b40c47a",
"id": "a140aac7",
"metadata": {
"collapsed": false,
"editable": true
@@ -173,36 +193,34 @@
},
{
"cell_type": "markdown",
"id": "ff9c0c81",
"id": "57ad18f5",
"metadata": {
"editable": true
},
"source": [
"### 1a)\n",
"\n",
"Fill in the necessary details.\n",
"\n",
"After this preprocessing, each column of $\\boldsymbol{X}_norm$ has mean zero and standard deviation $1$\n",
"and $\\boldsymbol{y}_centered$ has mean 0. This makes the optimization landscape\n",
"After this preprocessing, each column of $\\boldsymbol{X}_{\\mathrm{norm}}$ has mean zero and standard deviation $1$\n",
"and $\\boldsymbol{y}_{\\mathrm{centered}}$ has mean 0. This makes the optimization landscape\n",
"nicer and ensures the regularization penalty $\\lambda \\sum_j\n",
"\\beta_j^2$ treats each coefficient fairly (since features are on the\n",
"\\theta_j^2$ in Ridge regression treats each coefficient fairly (since features are on the\n",
"same scale)."
]
},
{
"cell_type": "markdown",
"id": "d27c70e4",
"id": "2886697d",
"metadata": {
"editable": true
},
"source": [
"## Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\\boldsymbol{theta}$"
"## Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\\boldsymbol{\\theta}$"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "9f1e5184",
"id": "97ac6cb6",
"metadata": {
"collapsed": false,
"editable": true
@@ -223,34 +241,33 @@
},
{
"cell_type": "markdown",
"id": "2ec556b9",
"id": "3efb067b",
"metadata": {
"editable": true
},
"source": [
"This computes the ridge and OLS regression coefficients directly. The identity\n",
"matrix $I$ has the same size as $X^T X$ (which is n_features x\n",
"n_features), and lam * I adds $\\lambda$ to the diagonal of $X^T X. We\n",
"then invert this matrix and multiply by $X^T y. The result\n",
"for $\\boldsymbol{\\theta}$ is a NumPy array of shape (n_features,) containing the\n",
"fitted weights."
"This computes the Ridge and OLS regression coefficients directly. The identity\n",
"matrix $I$ has the same size as $X^T X$. It adds $\\lambda$ to the diagonal of $X^T X for Ridge regression. We\n",
"then invert this matrix and multiply by $X^T y$. The result\n",
"for $\\boldsymbol{\\theta}$ is a NumPy array of shape (n$\\_$features,) containing the\n",
"fitted parameters $\\boldsymbol{\\theta}$.."
]
},
{
"cell_type": "markdown",
"id": "a821f0c5",
"id": "53be2bf8",
"metadata": {
"editable": true
},
"source": [
"### 2a)\n",
"\n",
"Finalize the OLS and Ridge regression determination of the optimal parameters $bm{\\theta}$."
"Finalize, in the above code, the OLS and Ridge regression determination of the optimal parameters $\\boldsymbol{\\theta}$."
]
},
{
"cell_type": "markdown",
"id": "d637130e",
"id": "e4126591",
"metadata": {
"editable": true
},
@@ -262,12 +279,12 @@
},
{
"cell_type": "markdown",
"id": "b455ce7e",
"id": "642d0850",
"metadata": {
"editable": true
},
"source": [
"## Implementing the simplest form for gradient descent\n",
"## Exercise 3, Implementing the simplest form for gradient descent\n",
"\n",
"Alternatively, we can fit the ridge regression model using gradient\n",
"descent. This is useful to visualize the iterative convergence and is\n",
@@ -282,7 +299,7 @@
{
"cell_type": "code",
"execution_count": 4,
"id": "cfa1eb29",
"id": "a67af634",
"metadata": {
"collapsed": false,
"editable": true
@@ -325,32 +342,32 @@
},
{
"cell_type": "markdown",
"id": "dc78d58d",
"id": "1c8c35dc",
"metadata": {
"editable": true
},
"source": [
"### 3a)\n",
"\n",
"Discuss the results as function of the learning rate paramaters and the number of iterations."
"Discuss the results as function of the learning rate parameters and the number of iterations."
]
},
{
"cell_type": "markdown",
"id": "15060acb",
"id": "899fec5c",
"metadata": {
"editable": true
},
"source": [
"### 3b)\n",
"\n",
"Add a stopping parameter as function of the number iterations. \n",
"Try to add a stopping parameter as function of the number iterations. How would you define a stopping criterion? \n",
"\n",
"If everything worked correctly, the learned coefficients should be\n",
"close to the true values [5.0, -3.0, 0.0, …, 2.0, …] that we used to\n",
"generate the data. Keep in mind that due to regularization and noise,\n",
"the learned values will not exactly equal the true ones, but they\n",
"should be in the same ballpark."
"should be in the same ballpark. Which method (OLS or Ridge) gives the best results?"
]
}
],
doc/src/week37/exercisesweek37.do.txt
+23 -23
View File
@@ -52,31 +52,33 @@ y = X.dot @ theta_true + noise
This code produces a dataset where only features 0, 1, and 6
significantly influence $\bm{y}$. The rest of the features have zero true
coefficient, so they only contribute noise. For example, feature 0 has
coefficient. For example, feature 0 has
a true weight of 5.0, feature 1 has -3.0, and feature 6 has 2.0, so
the expected relationship is:
!bt
\[
y \approx 5 \times X_0 \;-\; 3 \times X_1 \;+\; 2 \times X_6 \;+\; \text{noise}.
y \approx 5 \times x_0 \;-\; 3 \times x_1 \;+\; 2 \times x_6 \;+\; \text{noise}.
\]
!et
You can remove the noise if you wish to.
===== Exercise 1, scale your data =====
Before fitting a regression model, it is good practice to normalize or
standardize the features. This ensures all features are on a
comparable scale, which is especially important when using
regularization. Here we will perform standardization, scaling each
feature to have mean 0 and standard deviation 1:
feature to have mean 0 and standard deviation 1.
Compute the mean and standard deviation of each column (feature) in $bm{X}$.
=== 1a) ===
Compute the mean and standard deviation of each column (feature) in $\bm{X}$.
Subtract the mean and divide by the standard deviation for each feature.
We will also center the target $\bm{y}$ to mean $0$. Centering $\bm{y}$
(and each feature) means the model won’t require a separate intercept
term – the data is shifted such that the intercept is effectively 0
(and each feature) means the model does not require a separate intercept
term, the data is shifted such that the intercept is effectively 0
. (In practice, one could include an intercept in the model and not
penalize it, but here we simplify by centering.)
@@ -92,17 +94,16 @@ y_mean = ?
y_centered = ?
!ec
=== 1a) ===
Fill in the necessary details.
After this preprocessing, each column of $\bm{X}_norm$ has mean zero and standard deviation $1$
and $\bm{y}_centered$ has mean 0. This makes the optimization landscape
After this preprocessing, each column of $\bm{X}_{\mathrm{norm}}$ has mean zero and standard deviation $1$
and $\bm{y}_{\mathrm{centered}}$ has mean 0. This makes the optimization landscape
nicer and ensures the regularization penalty $\lambda \sum_j
\beta_j^2$ treats each coefficient fairly (since features are on the
\theta_j^2$ in Ridge regression treats each coefficient fairly (since features are on the
same scale).
===== Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\bm{theta}$ =====
===== Exercise 2, use the analytical formulae for OLS and Ridge regression to find the optimal paramters $\bm{\theta}$ =====
!bc pycod
# Set regularization parameter, either a single value or a vector of values
@@ -117,20 +118,19 @@ print("Closed-form Ridge coefficients:", theta_closed_form)
print("Closed-form OLS coefficients:", theta_closed_form)
!ec
This computes the ridge and OLS regression coefficients directly. The identity
matrix $I$ has the same size as $X^T X$ (which is n_features x
n_features), and lam * I adds $\lambda$ to the diagonal of $X^T X. We
then invert this matrix and multiply by $X^T y. The result
for $\bm{\theta}$ is a NumPy array of shape (n_features,) containing the
fitted weights.
This computes the Ridge and OLS regression coefficients directly. The identity
matrix $I$ has the same size as $X^T X$. It adds $\lambda$ to the diagonal of $X^T X for Ridge regression. We
then invert this matrix and multiply by $X^T y$. The result
for $\bm{\theta}$ is a NumPy array of shape (n$\_$features,) containing the
fitted parameters $\bm{\theta}$..
=== 2a) ===
Finalize the OLS and Ridge regression determination of the optimal parameters $bm{\theta}$.
Finalize, in the above code, the OLS and Ridge regression determination of the optimal parameters $\bm{\theta}$.
=== 2b) ===
Explore the results as function of different values of the hyperparameter $\lambda$. See for example exercise 4 from week 36.
===== Implementing the simplest form for gradient descent =====
===== Exercise 3, Implementing the simplest form for gradient descent =====
Alternatively, we can fit the ridge regression model using gradient
descent. This is useful to visualize the iterative convergence and is
@@ -178,10 +178,10 @@ print("Gradient Descent Ridge coefficients:", theta_gdRidge)
!ec
=== 3a) ===
Discuss the results as function of the learning rate paramaters and the number of iterations.
Discuss the results as function of the learning rate parameters and the number of iterations.
=== 3b) ===
Add a stopping parameter as function of the number iterations.
Try to add a stopping parameter as function of the number iterations. How would you define a stopping criterion?
@@ -189,5 +189,5 @@ If everything worked correctly, the learned coefficients should be
close to the true values [5.0, -3.0, 0.0, …, 2.0, …] that we used to
generate the data. Keep in mind that due to regularization and noise,
the learned values will not exactly equal the true ones, but they
should be in the same ballpark.
should be in the same ballpark. Which method (OLS or Ridge) gives the best results?