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Morten Hjorth-Jensen 970ebe0921 updating week 38
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<h1>Data Analysis and Machine Learning: Logistic Regression</h1>
</center> <!-- document title -->
<!-- author(s): Morten Hjorth-Jensen -->
<center>
<b>Morten Hjorth-Jensen</b> [1, 2]
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<!-- institution(s) -->
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[1] <b>Department of Physics and Center for Computing in Science Education, University of Oslo</b>
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[2] <b>Department of Physics and Astronomy and Facility for Rare Isotope Beams, Michigan State University</b>
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<br>
<center>
<h4>September 22 and 23</h4>
</center> <!-- date -->
<br>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="plans-for-week-38">Plans for week 38 </h2>
<ul>
<li> Lab Wednesday and Thursday: work on project 1</li>
<li> Thursday: Summary of regression methods, cross-validation and discussion of project 1. Start Logistic Regression</li>
<li> Friday: Classification problems and Logistic Regression, from binary cases to several categories. Start optimization methods</li>
<li> Reading recommendations:
<ol type="a"></li>
<li> See lecture notes for week 37 on cross-validation and week 38 at <a href="https://compphysics.github.io/MachineLearning/doc/web/course.html." target="_blank"><tt>https://compphysics.github.io/MachineLearning/doc/web/course.html.</tt></a></li>
<li> Bishop 4.1, 4.2 and 4.3. Not all the material is relevant or will be covered. Section 4.3 is the most relevant, but 4.1 and 4.2 give interesting background readings for logistic regression</li>
<li> Hastie et al 4.1, 4.2 and 4.3 on logistic regression</li>
<li> For a good discussion on gradient methods, see Goodfellow et al section 4.3-4.5 and chapter 8. We will come back to the latter chapter in our discussion of Neural networks as well.</li>
</ol>
</ul>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="ridge-and-lasso-regression-reminder">Ridge and LASSO Regression, reminder </h2>
<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, that is
our optimization problem is
</p>
$$
{\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
$$
<p>or we can state it as</p>
$$
{\displaystyle \min_{\boldsymbol{\beta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2,
$$
<p>where we have used the definition of a norm-2 vector, that is</p>
$$
\vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}.
$$
<p>By minimizing the above equation with respect to the parameters
\( \boldsymbol{\beta} \) we could then obtain an analytical expression for the
parameters \( \boldsymbol{\beta} \). We can add a regularization parameter \( \lambda \) by
defining a new cost function to be optimized, that is
</p>
$$
{\displaystyle \min_{\boldsymbol{\beta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2
$$
<p>which leads to the Ridge regression minimization problem where we
require that \( \vert\vert \boldsymbol{\beta}\vert\vert_2^2\le t \), where \( t \) is
a finite number larger than zero. By defining
</p>
$$
C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1,
$$
<p>we have a new optimization equation</p>
$$
{\displaystyle \min_{\boldsymbol{\beta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1
$$
<p>which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator. </p>
<p>Here we have defined the norm-1 as </p>
$$
\vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert.
$$
<!-- !split -->
<h2 id="various-steps-in-cross-validation">Various steps in cross-validation </h2>
<p>When the repetitive splitting of the data set is done randomly,
samples may accidently end up in a fast majority of the splits in
either training or test set. Such samples may have an unbalanced
influence on either model building or prediction evaluation. To avoid
this \( k \)-fold cross-validation structures the data splitting. The
samples are divided into \( k \) more or less equally sized exhaustive and
mutually exclusive subsets. In turn (at each split) one of these
subsets plays the role of the test set while the union of the
remaining subsets constitutes the training set. Such a splitting
warrants a balanced representation of each sample in both training and
test set over the splits. Still the division into the \( k \) subsets
involves a degree of randomness. This may be fully excluded when
choosing \( k=n \). This particular case is referred to as leave-one-out
cross-validation (LOOCV).
</p>
<!-- !split -->
<h2 id="how-to-set-up-the-cross-validation-for-ridge-and-or-lasso">How to set up the cross-validation for Ridge and/or Lasso </h2>
<ul>
<li> Define a range of interest for the penalty parameter.</li>
<li> Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.</li>
<li> Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \boldsymbol{\sigma}_{-i}^2(\lambda) \), as</li>
</ul>
$$
\begin{align*}
\boldsymbol{\beta}_{-i}(\lambda) & = ( \boldsymbol{X}_{-i, \ast}^{T}
\boldsymbol{X}_{-i, \ast} + \lambda \boldsymbol{I}_{pp})^{-1}
\boldsymbol{X}_{-i, \ast}^{T} \boldsymbol{y}_{-i}
\end{align*}
$$
<ul>
<li> Evaluate the prediction performance of these models on the test set by \( C[y_i, \boldsymbol{X}_{i, \ast}; \boldsymbol{\beta}_{-i}(\lambda), \boldsymbol{\sigma}_{-i}^2(\lambda)] \). Or, by the prediction error \( |y_i - \boldsymbol{X}_{i, \ast} \boldsymbol{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.</li>
<li> Repeat the first three steps such that each sample plays the role of the test set once.</li>
<li> Average the prediction performances of the test sets at each grid point of the penalty bias/parameter. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.</li>
</ul>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="cross-validation-in-brief">Cross-validation in brief </h2>
<p>For the various values of \( k \)</p>
<ol>
<li> shuffle the dataset randomly.</li>
<li> Split the dataset into \( k \) groups.</li>
<li> For each unique group:
<ol type="a"></li>
<li> Decide which group to use as set for test data</li>
<li> Take the remaining groups as a training data set</li>
<li> Fit a model on the training set and evaluate it on the test set</li>
<li> Retain the evaluation score and discard the model</li>
</ol>
<li> Summarize the model using the sample of model evaluation scores</li>
</ol>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="code-example-for-cross-validation-and-k-fold-cross-validation">Code Example for Cross-validation and \( k \)-fold Cross-validation </h2>
<p>The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial. </p>
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<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.model_selection</span> <span style="color: #008000; font-weight: bold">import</span> KFold
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.linear_model</span> <span style="color: #008000; font-weight: bold">import</span> Ridge
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.model_selection</span> <span style="color: #008000; font-weight: bold">import</span> cross_val_score
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.preprocessing</span> <span style="color: #008000; font-weight: bold">import</span> PolynomialFeatures
<span style="color: #408080; font-style: italic"># A seed just to ensure that the random numbers are the same for every run.</span>
<span style="color: #408080; font-style: italic"># Useful for eventual debugging.</span>
np<span style="color: #666666">.</span>random<span style="color: #666666">.</span>seed(<span style="color: #666666">3155</span>)
<span style="color: #408080; font-style: italic"># Generate the data.</span>
nsamples <span style="color: #666666">=</span> <span style="color: #666666">100</span>
x <span style="color: #666666">=</span> np<span style="color: #666666">.</span>random<span style="color: #666666">.</span>randn(nsamples)
y <span style="color: #666666">=</span> <span style="color: #666666">3*</span>x<span style="color: #666666">**2</span> <span style="color: #666666">+</span> np<span style="color: #666666">.</span>random<span style="color: #666666">.</span>randn(nsamples)
<span style="color: #408080; font-style: italic">## Cross-validation on Ridge regression using KFold only</span>
<span style="color: #408080; font-style: italic"># Decide degree on polynomial to fit</span>
poly <span style="color: #666666">=</span> PolynomialFeatures(degree <span style="color: #666666">=</span> <span style="color: #666666">6</span>)
<span style="color: #408080; font-style: italic"># Decide which values of lambda to use</span>
nlambdas <span style="color: #666666">=</span> <span style="color: #666666">500</span>
lambdas <span style="color: #666666">=</span> np<span style="color: #666666">.</span>logspace(<span style="color: #666666">-3</span>, <span style="color: #666666">5</span>, nlambdas)
<span style="color: #408080; font-style: italic"># Initialize a KFold instance</span>
k <span style="color: #666666">=</span> <span style="color: #666666">5</span>
kfold <span style="color: #666666">=</span> KFold(n_splits <span style="color: #666666">=</span> k)
<span style="color: #408080; font-style: italic"># Perform the cross-validation to estimate MSE</span>
scores_KFold <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros((nlambdas, k))
i <span style="color: #666666">=</span> <span style="color: #666666">0</span>
<span style="color: #008000; font-weight: bold">for</span> lmb <span style="color: #AA22FF; font-weight: bold">in</span> lambdas:
ridge <span style="color: #666666">=</span> Ridge(alpha <span style="color: #666666">=</span> lmb)
j <span style="color: #666666">=</span> <span style="color: #666666">0</span>
<span style="color: #008000; font-weight: bold">for</span> train_inds, test_inds <span style="color: #AA22FF; font-weight: bold">in</span> kfold<span style="color: #666666">.</span>split(x):
xtrain <span style="color: #666666">=</span> x[train_inds]
ytrain <span style="color: #666666">=</span> y[train_inds]
xtest <span style="color: #666666">=</span> x[test_inds]
ytest <span style="color: #666666">=</span> y[test_inds]
Xtrain <span style="color: #666666">=</span> poly<span style="color: #666666">.</span>fit_transform(xtrain[:, np<span style="color: #666666">.</span>newaxis])
ridge<span style="color: #666666">.</span>fit(Xtrain, ytrain[:, np<span style="color: #666666">.</span>newaxis])
Xtest <span style="color: #666666">=</span> poly<span style="color: #666666">.</span>fit_transform(xtest[:, np<span style="color: #666666">.</span>newaxis])
ypred <span style="color: #666666">=</span> ridge<span style="color: #666666">.</span>predict(Xtest)
scores_KFold[i,j] <span style="color: #666666">=</span> np<span style="color: #666666">.</span>sum((ypred <span style="color: #666666">-</span> ytest[:, np<span style="color: #666666">.</span>newaxis])<span style="color: #666666">**2</span>)<span style="color: #666666">/</span>np<span style="color: #666666">.</span>size(ypred)
j <span style="color: #666666">+=</span> <span style="color: #666666">1</span>
i <span style="color: #666666">+=</span> <span style="color: #666666">1</span>
estimated_mse_KFold <span style="color: #666666">=</span> np<span style="color: #666666">.</span>mean(scores_KFold, axis <span style="color: #666666">=</span> <span style="color: #666666">1</span>)
<span style="color: #408080; font-style: italic">## Cross-validation using cross_val_score from sklearn along with KFold</span>
<span style="color: #408080; font-style: italic"># kfold is an instance initialized above as:</span>
<span style="color: #408080; font-style: italic"># kfold = KFold(n_splits = k)</span>
estimated_mse_sklearn <span style="color: #666666">=</span> np<span style="color: #666666">.</span>zeros(nlambdas)
i <span style="color: #666666">=</span> <span style="color: #666666">0</span>
<span style="color: #008000; font-weight: bold">for</span> lmb <span style="color: #AA22FF; font-weight: bold">in</span> lambdas:
ridge <span style="color: #666666">=</span> Ridge(alpha <span style="color: #666666">=</span> lmb)
X <span style="color: #666666">=</span> poly<span style="color: #666666">.</span>fit_transform(x[:, np<span style="color: #666666">.</span>newaxis])
estimated_mse_folds <span style="color: #666666">=</span> cross_val_score(ridge, X, y[:, np<span style="color: #666666">.</span>newaxis], scoring<span style="color: #666666">=</span><span style="color: #BA2121">&#39;neg_mean_squared_error&#39;</span>, cv<span style="color: #666666">=</span>kfold)
<span style="color: #408080; font-style: italic"># cross_val_score return an array containing the estimated negative mse for every fold.</span>
<span style="color: #408080; font-style: italic"># we have to the the mean of every array in order to get an estimate of the mse of the model</span>
estimated_mse_sklearn[i] <span style="color: #666666">=</span> np<span style="color: #666666">.</span>mean(<span style="color: #666666">-</span>estimated_mse_folds)
i <span style="color: #666666">+=</span> <span style="color: #666666">1</span>
<span style="color: #408080; font-style: italic">## Plot and compare the slightly different ways to perform cross-validation</span>
plt<span style="color: #666666">.</span>figure()
plt<span style="color: #666666">.</span>plot(np<span style="color: #666666">.</span>log10(lambdas), estimated_mse_sklearn, label <span style="color: #666666">=</span> <span style="color: #BA2121">&#39;cross_val_score&#39;</span>)
plt<span style="color: #666666">.</span>plot(np<span style="color: #666666">.</span>log10(lambdas), estimated_mse_KFold, <span style="color: #BA2121">&#39;r--&#39;</span>, label <span style="color: #666666">=</span> <span style="color: #BA2121">&#39;KFold&#39;</span>)
plt<span style="color: #666666">.</span>xlabel(<span style="color: #BA2121">&#39;log10(lambda)&#39;</span>)
plt<span style="color: #666666">.</span>ylabel(<span style="color: #BA2121">&#39;mse&#39;</span>)
plt<span style="color: #666666">.</span>legend()
plt<span style="color: #666666">.</span>show()
</pre>
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<h2 id="logistic-regression">Logistic Regression </h2>
<p>In linear regression our main interest was centered on learning the
coefficients of a functional fit (say a polynomial) in order to be
able to predict the response of a continuous variable on some unseen
data. The fit to the continuous variable \( y_i \) is based on some
independent variables \( \boldsymbol{x}_i \). Linear regression resulted in
analytical expressions for standard ordinary Least Squares or Ridge
regression (in terms of matrices to invert) for several quantities,
ranging from the variance and thereby the confidence intervals of the
parameters \( \boldsymbol{\beta} \) to the mean squared error. If we can invert
the product of the design matrices, linear regression gives then a
simple recipe for fitting our data.
</p>
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<h2 id="classification-problems">Classification problems </h2>
<p>Classification problems, however, are concerned with outcomes taking
the form of discrete variables (i.e. categories). We may for example,
on the basis of DNA sequencing for a number of patients, like to find
out which mutations are important for a certain disease; or based on
scans of various patients' brains, figure out if there is a tumor or
not; or given a specific physical system, we'd like to identify its
state, say whether it is an ordered or disordered system (typical
situation in solid state physics); or classify the status of a
patient, whether she/he has a stroke or not and many other similar
situations.
</p>
<p>The most common situation we encounter when we apply logistic
regression is that of two possible outcomes, normally denoted as a
binary outcome, true or false, positive or negative, success or
failure etc.
</p>
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<h2 id="optimization-and-deep-learning">Optimization and Deep learning </h2>
<p>Logistic regression will also serve as our stepping stone towards
neural network algorithms and supervised deep learning. For logistic
learning, the minimization of the cost function leads to a non-linear
equation in the parameters \( \boldsymbol{\beta} \). The optimization of the
problem calls therefore for minimization algorithms. This forms the
bottle neck of all machine learning algorithms, namely how to find
reliable minima of a multi-variable function. This leads us to the
family of gradient descent methods. The latter are the working horses
of basically all modern machine learning algorithms.
</p>
<p>We note also that many of the topics discussed here on logistic
regression are also commonly used in modern supervised Deep Learning
models, as we will see later.
</p>
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<h2 id="basics">Basics </h2>
<p>We consider the case where the dependent variables, also called the
responses or the outcomes, \( y_i \) are discrete and only take values
from \( k=0,\dots,K-1 \) (i.e. \( K \) classes).
</p>
<p>The goal is to predict the
output classes from the design matrix \( \boldsymbol{X}\in\mathbb{R}^{n\times p} \)
made of \( n \) samples, each of which carries \( p \) features or predictors. The
primary goal is to identify the classes to which new unseen samples
belong.
</p>
<p>Let us specialize to the case of two classes only, with outputs
\( y_i=0 \) and \( y_i=1 \). Our outcomes could represent the status of a
credit card user that could default or not on her/his credit card
debt. That is
</p>
$$
y_i = \begin{bmatrix} 0 & \mathrm{no}\\ 1 & \mathrm{yes} \end{bmatrix}.
$$
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<h2 id="linear-classifier">Linear classifier </h2>
<p>Before moving to the logistic model, let us try to use our linear
regression model to classify these two outcomes. We could for example
fit a linear model to the default case if \( y_i > 0.5 \) and the no
default case \( y_i \leq 0.5 \).
</p>
<p>We would then have our
weighted linear combination, namely
</p>
$$
\begin{equation}
\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{\beta} + \boldsymbol{\epsilon},
\label{_auto1}
\end{equation}
$$
<p>where \( \boldsymbol{y} \) is a vector representing the possible outcomes, \( \boldsymbol{X} \) is our
\( n\times p \) design matrix and \( \boldsymbol{\beta} \) represents our estimators/predictors.
</p>
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<h2 id="some-selected-properties">Some selected properties </h2>
<p>The main problem with our function is that it takes values on the
entire real axis. In the case of logistic regression, however, the
labels \( y_i \) are discrete variables. A typical example is the credit
card data discussed below here, where we can set the state of
defaulting the debt to \( y_i=1 \) and not to \( y_i=0 \) for one the persons
in the data set (see the full example below).
</p>
<p>One simple way to get a discrete output is to have sign
functions that map the output of a linear regressor to values \( \{0,1\} \),
\( f(s_i)=sign(s_i)=1 \) if \( s_i\ge 0 \) and 0 if otherwise.
We will encounter this model in our first demonstration of neural networks. Historically it is called the ``perceptron" model in the machine learning
literature. This model is extremely simple. However, in many cases it is more
favorable to use a ``soft" classifier that outputs
the probability of a given category. This leads us to the logistic function.
</p>
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<h2 id="simple-example">Simple example </h2>
<p>The following example on data for coronary heart disease (CHD) as function of age may serve as an illustration. In the code here we read and plot whether a person has had CHD (output = 1) or not (output = 0). This ouput is plotted the person's against age. Clearly, the figure shows that attempting to make a standard linear regression fit may not be very meaningful.</p>
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<pre style="line-height: 125%;"><span style="color: #408080; font-style: italic"># Common imports</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">os</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">pandas</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">pd</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.linear_model</span> <span style="color: #008000; font-weight: bold">import</span> LinearRegression, Ridge, Lasso
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.model_selection</span> <span style="color: #008000; font-weight: bold">import</span> train_test_split
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.utils</span> <span style="color: #008000; font-weight: bold">import</span> resample
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.metrics</span> <span style="color: #008000; font-weight: bold">import</span> mean_squared_error
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">IPython.display</span> <span style="color: #008000; font-weight: bold">import</span> display
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">pylab</span> <span style="color: #008000; font-weight: bold">import</span> plt, mpl
plt<span style="color: #666666">.</span>style<span style="color: #666666">.</span>use(<span style="color: #BA2121">&#39;seaborn&#39;</span>)
mpl<span style="color: #666666">.</span>rcParams[<span style="color: #BA2121">&#39;font.family&#39;</span>] <span style="color: #666666">=</span> <span style="color: #BA2121">&#39;serif&#39;</span>
<span style="color: #408080; font-style: italic"># Where to save the figures and data files</span>
PROJECT_ROOT_DIR <span style="color: #666666">=</span> <span style="color: #BA2121">&quot;Results&quot;</span>
FIGURE_ID <span style="color: #666666">=</span> <span style="color: #BA2121">&quot;Results/FigureFiles&quot;</span>
DATA_ID <span style="color: #666666">=</span> <span style="color: #BA2121">&quot;DataFiles/&quot;</span>
<span style="color: #008000; font-weight: bold">if</span> <span style="color: #AA22FF; font-weight: bold">not</span> os<span style="color: #666666">.</span>path<span style="color: #666666">.</span>exists(PROJECT_ROOT_DIR):
os<span style="color: #666666">.</span>mkdir(PROJECT_ROOT_DIR)
<span style="color: #008000; font-weight: bold">if</span> <span style="color: #AA22FF; font-weight: bold">not</span> os<span style="color: #666666">.</span>path<span style="color: #666666">.</span>exists(FIGURE_ID):
os<span style="color: #666666">.</span>makedirs(FIGURE_ID)
<span style="color: #008000; font-weight: bold">if</span> <span style="color: #AA22FF; font-weight: bold">not</span> os<span style="color: #666666">.</span>path<span style="color: #666666">.</span>exists(DATA_ID):
os<span style="color: #666666">.</span>makedirs(DATA_ID)
<span style="color: #008000; font-weight: bold">def</span> <span style="color: #0000FF">image_path</span>(fig_id):
<span style="color: #008000; font-weight: bold">return</span> os<span style="color: #666666">.</span>path<span style="color: #666666">.</span>join(FIGURE_ID, fig_id)
<span style="color: #008000; font-weight: bold">def</span> <span style="color: #0000FF">data_path</span>(dat_id):
<span style="color: #008000; font-weight: bold">return</span> os<span style="color: #666666">.</span>path<span style="color: #666666">.</span>join(DATA_ID, dat_id)
<span style="color: #008000; font-weight: bold">def</span> <span style="color: #0000FF">save_fig</span>(fig_id):
plt<span style="color: #666666">.</span>savefig(image_path(fig_id) <span style="color: #666666">+</span> <span style="color: #BA2121">&quot;.png&quot;</span>, <span style="color: #008000">format</span><span style="color: #666666">=</span><span style="color: #BA2121">&#39;png&#39;</span>)
infile <span style="color: #666666">=</span> <span style="color: #008000">open</span>(data_path(<span style="color: #BA2121">&quot;chddata.csv&quot;</span>),<span style="color: #BA2121">&#39;r&#39;</span>)
<span style="color: #408080; font-style: italic"># Read the chd data as csv file and organize the data into arrays with age group, age, and chd</span>
chd <span style="color: #666666">=</span> pd<span style="color: #666666">.</span>read_csv(infile, names<span style="color: #666666">=</span>(<span style="color: #BA2121">&#39;ID&#39;</span>, <span style="color: #BA2121">&#39;Age&#39;</span>, <span style="color: #BA2121">&#39;Agegroup&#39;</span>, <span style="color: #BA2121">&#39;CHD&#39;</span>))
chd<span style="color: #666666">.</span>columns <span style="color: #666666">=</span> [<span style="color: #BA2121">&#39;ID&#39;</span>, <span style="color: #BA2121">&#39;Age&#39;</span>, <span style="color: #BA2121">&#39;Agegroup&#39;</span>, <span style="color: #BA2121">&#39;CHD&#39;</span>]
output <span style="color: #666666">=</span> chd[<span style="color: #BA2121">&#39;CHD&#39;</span>]
age <span style="color: #666666">=</span> chd[<span style="color: #BA2121">&#39;Age&#39;</span>]
agegroup <span style="color: #666666">=</span> chd[<span style="color: #BA2121">&#39;Agegroup&#39;</span>]
numberID <span style="color: #666666">=</span> chd[<span style="color: #BA2121">&#39;ID&#39;</span>]
display(chd)
plt<span style="color: #666666">.</span>scatter(age, output, marker<span style="color: #666666">=</span><span style="color: #BA2121">&#39;o&#39;</span>)
plt<span style="color: #666666">.</span>axis([<span style="color: #666666">18</span>,<span style="color: #666666">70.0</span>,<span style="color: #666666">-0.1</span>, <span style="color: #666666">1.2</span>])
plt<span style="color: #666666">.</span>xlabel(<span style="color: #BA2121">r&#39;Age&#39;</span>)
plt<span style="color: #666666">.</span>ylabel(<span style="color: #BA2121">r&#39;CHD&#39;</span>)
plt<span style="color: #666666">.</span>title(<span style="color: #BA2121">r&#39;Age distribution and Coronary heart disease&#39;</span>)
plt<span style="color: #666666">.</span>show()
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<h2 id="plotting-the-mean-value-for-each-group">Plotting the mean value for each group </h2>
<p>What we could attempt however is to plot the mean value for each group.</p>
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<pre style="line-height: 125%;">agegroupmean <span style="color: #666666">=</span> np<span style="color: #666666">.</span>array([<span style="color: #666666">0.1</span>, <span style="color: #666666">0.133</span>, <span style="color: #666666">0.250</span>, <span style="color: #666666">0.333</span>, <span style="color: #666666">0.462</span>, <span style="color: #666666">0.625</span>, <span style="color: #666666">0.765</span>, <span style="color: #666666">0.800</span>])
group <span style="color: #666666">=</span> np<span style="color: #666666">.</span>array([<span style="color: #666666">1</span>, <span style="color: #666666">2</span>, <span style="color: #666666">3</span>, <span style="color: #666666">4</span>, <span style="color: #666666">5</span>, <span style="color: #666666">6</span>, <span style="color: #666666">7</span>, <span style="color: #666666">8</span>])
plt<span style="color: #666666">.</span>plot(group, agegroupmean, <span style="color: #BA2121">&quot;r-&quot;</span>)
plt<span style="color: #666666">.</span>axis([<span style="color: #666666">0</span>,<span style="color: #666666">9</span>,<span style="color: #666666">0</span>, <span style="color: #666666">1.0</span>])
plt<span style="color: #666666">.</span>xlabel(<span style="color: #BA2121">r&#39;Age group&#39;</span>)
plt<span style="color: #666666">.</span>ylabel(<span style="color: #BA2121">r&#39;CHD mean values&#39;</span>)
plt<span style="color: #666666">.</span>title(<span style="color: #BA2121">r&#39;Mean values for each age group&#39;</span>)
plt<span style="color: #666666">.</span>show()
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<p>We are now trying to find a function \( f(y\vert x) \), that is a function which gives us an expected value for the output \( y \) with a given input \( x \).
In standard linear regression with a linear dependence on \( x \), we would write this in terms of our model
</p>
$$
f(y_i\vert x_i)=\beta_0+\beta_1 x_i.
$$
<p>This expression implies however that \( f(y_i\vert x_i) \) could take any
value from minus infinity to plus infinity. If we however let
\( f(y\vert y) \) be represented by the mean value, the above example
shows us that we can constrain the function to take values between
zero and one, that is we have \( 0 \le f(y_i\vert x_i) \le 1 \). Looking
at our last curve we see also that it has an S-shaped form. This leads
us to a very popular model for the function \( f \), namely the so-called
Sigmoid function or logistic model. We will consider this function as
representing the probability for finding a value of \( y_i \) with a given
\( x_i \).
</p>
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<h2 id="the-logistic-function">The logistic function </h2>
<p>Another widely studied model, is the so-called
perceptron model, which is an example of a &quot;hard classification&quot; model. We
will encounter this model when we discuss neural networks as
well. Each datapoint is deterministically assigned to a category (i.e
\( y_i=0 \) or \( y_i=1 \)). In many cases, and the coronary heart disease data forms one of many such examples, it is favorable to have a &quot;soft&quot;
classifier that outputs the probability of a given category rather
than a single value. For example, given \( x_i \), the classifier
outputs the probability of being in a category \( k \). Logistic regression
is the most common example of a so-called soft classifier. In logistic
regression, the probability that a data point \( x_i \)
belongs to a category \( y_i=\{0,1\} \) is given by the so-called logit function (or Sigmoid) which is meant to represent the likelihood for a given event,
</p>
$$
p(t) = \frac{1}{1+\mathrm \exp{-t}}=\frac{\exp{t}}{1+\mathrm \exp{t}}.
$$
<p>Note that \( 1-p(t)= p(-t) \).</p>
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<h2 id="examples-of-likelihood-functions-used-in-logistic-regression-and-nueral-networks">Examples of likelihood functions used in logistic regression and nueral networks </h2>
<p>The following code plots the logistic function, the step function and other functions we will encounter from here and on.</p>
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<pre style="line-height: 125%;"><span style="color: #BA2121; font-style: italic">&quot;&quot;&quot;The sigmoid function (or the logistic curve) is a</span>
<span style="color: #BA2121; font-style: italic">function that takes any real number, z, and outputs a number (0,1).</span>
<span style="color: #BA2121; font-style: italic">It is useful in neural networks for assigning weights on a relative scale.</span>
<span style="color: #BA2121; font-style: italic">The value z is the weighted sum of parameters involved in the learning algorithm.&quot;&quot;&quot;</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">math</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">mt</span>
z <span style="color: #666666">=</span> numpy<span style="color: #666666">.</span>arange(<span style="color: #666666">-5</span>, <span style="color: #666666">5</span>, <span style="color: #666666">.1</span>)
sigma_fn <span style="color: #666666">=</span> numpy<span style="color: #666666">.</span>vectorize(<span style="color: #008000; font-weight: bold">lambda</span> z: <span style="color: #666666">1/</span>(<span style="color: #666666">1+</span>numpy<span style="color: #666666">.</span>exp(<span style="color: #666666">-</span>z)))
sigma <span style="color: #666666">=</span> sigma_fn(z)
fig <span style="color: #666666">=</span> plt<span style="color: #666666">.</span>figure()
ax <span style="color: #666666">=</span> fig<span style="color: #666666">.</span>add_subplot(<span style="color: #666666">111</span>)
ax<span style="color: #666666">.</span>plot(z, sigma)
ax<span style="color: #666666">.</span>set_ylim([<span style="color: #666666">-0.1</span>, <span style="color: #666666">1.1</span>])
ax<span style="color: #666666">.</span>set_xlim([<span style="color: #666666">-5</span>,<span style="color: #666666">5</span>])
ax<span style="color: #666666">.</span>grid(<span style="color: #008000; font-weight: bold">True</span>)
ax<span style="color: #666666">.</span>set_xlabel(<span style="color: #BA2121">&#39;z&#39;</span>)
ax<span style="color: #666666">.</span>set_title(<span style="color: #BA2121">&#39;sigmoid function&#39;</span>)
plt<span style="color: #666666">.</span>show()
<span style="color: #BA2121; font-style: italic">&quot;&quot;&quot;Step Function&quot;&quot;&quot;</span>
z <span style="color: #666666">=</span> numpy<span style="color: #666666">.</span>arange(<span style="color: #666666">-5</span>, <span style="color: #666666">5</span>, <span style="color: #666666">.02</span>)
step_fn <span style="color: #666666">=</span> numpy<span style="color: #666666">.</span>vectorize(<span style="color: #008000; font-weight: bold">lambda</span> z: <span style="color: #666666">1.0</span> <span style="color: #008000; font-weight: bold">if</span> z <span style="color: #666666">&gt;=</span> <span style="color: #666666">0.0</span> <span style="color: #008000; font-weight: bold">else</span> <span style="color: #666666">0.0</span>)
step <span style="color: #666666">=</span> step_fn(z)
fig <span style="color: #666666">=</span> plt<span style="color: #666666">.</span>figure()
ax <span style="color: #666666">=</span> fig<span style="color: #666666">.</span>add_subplot(<span style="color: #666666">111</span>)
ax<span style="color: #666666">.</span>plot(z, step)
ax<span style="color: #666666">.</span>set_ylim([<span style="color: #666666">-0.5</span>, <span style="color: #666666">1.5</span>])
ax<span style="color: #666666">.</span>set_xlim([<span style="color: #666666">-5</span>,<span style="color: #666666">5</span>])
ax<span style="color: #666666">.</span>grid(<span style="color: #008000; font-weight: bold">True</span>)
ax<span style="color: #666666">.</span>set_xlabel(<span style="color: #BA2121">&#39;z&#39;</span>)
ax<span style="color: #666666">.</span>set_title(<span style="color: #BA2121">&#39;step function&#39;</span>)
plt<span style="color: #666666">.</span>show()
<span style="color: #BA2121; font-style: italic">&quot;&quot;&quot;tanh Function&quot;&quot;&quot;</span>
z <span style="color: #666666">=</span> numpy<span style="color: #666666">.</span>arange(<span style="color: #666666">-2*</span>mt<span style="color: #666666">.</span>pi, <span style="color: #666666">2*</span>mt<span style="color: #666666">.</span>pi, <span style="color: #666666">0.1</span>)
t <span style="color: #666666">=</span> numpy<span style="color: #666666">.</span>tanh(z)
fig <span style="color: #666666">=</span> plt<span style="color: #666666">.</span>figure()
ax <span style="color: #666666">=</span> fig<span style="color: #666666">.</span>add_subplot(<span style="color: #666666">111</span>)
ax<span style="color: #666666">.</span>plot(z, t)
ax<span style="color: #666666">.</span>set_ylim([<span style="color: #666666">-1.0</span>, <span style="color: #666666">1.0</span>])
ax<span style="color: #666666">.</span>set_xlim([<span style="color: #666666">-2*</span>mt<span style="color: #666666">.</span>pi,<span style="color: #666666">2*</span>mt<span style="color: #666666">.</span>pi])
ax<span style="color: #666666">.</span>grid(<span style="color: #008000; font-weight: bold">True</span>)
ax<span style="color: #666666">.</span>set_xlabel(<span style="color: #BA2121">&#39;z&#39;</span>)
ax<span style="color: #666666">.</span>set_title(<span style="color: #BA2121">&#39;tanh function&#39;</span>)
plt<span style="color: #666666">.</span>show()
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<h2 id="two-parameters">Two parameters </h2>
<p>We assume now that we have two classes with \( y_i \) either \( 0 \) or \( 1 \). Furthermore we assume also that we have only two parameters \( \beta \) in our fitting of the Sigmoid function, that is we define probabilities </p>
$$
\begin{align*}
p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}),
\end{align*}
$$
<p>where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \). </p>
<p>Note that we used</p>
$$
p(y_i=0\vert x_i, \boldsymbol{\beta}) = 1-p(y_i=1\vert x_i, \boldsymbol{\beta}).
$$
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<h2 id="maximum-likelihood">Maximum likelihood </h2>
<p>In order to define the total likelihood for all possible outcomes from a
dataset \( \mathcal{D}=\{(y_i,x_i)\} \), with the binary labels
\( y_i\in\{0,1\} \) and where the data points are drawn independently, we use the so-called <a href="https://en.wikipedia.org/wiki/Maximum_likelihood_estimation" target="_blank">Maximum Likelihood Estimation</a> (MLE) principle.
We aim thus at maximizing
the probability of seeing the observed data. We can then approximate the
likelihood in terms of the product of the individual probabilities of a specific outcome \( y_i \), that is
</p>
$$
\begin{align*}
P(\mathcal{D}|\boldsymbol{\beta})& = \prod_{i=1}^n \left[p(y_i=1|x_i,\boldsymbol{\beta})\right]^{y_i}\left[1-p(y_i=1|x_i,\boldsymbol{\beta}))\right]^{1-y_i}\nonumber \\
\end{align*}
$$
<p>from which we obtain the log-likelihood and our <b>cost/loss</b> function</p>
$$
\mathcal{C}(\boldsymbol{\beta}) = \sum_{i=1}^n \left( y_i\log{p(y_i=1|x_i,\boldsymbol{\beta})} + (1-y_i)\log\left[1-p(y_i=1|x_i,\boldsymbol{\beta}))\right]\right).
$$
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<h2 id="the-cost-function-rewritten">The cost function rewritten </h2>
<p>Reordering the logarithms, we can rewrite the <b>cost/loss</b> function as</p>
$$
\mathcal{C}(\boldsymbol{\beta}) = \sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
$$
<p>The maximum likelihood estimator is defined as the set of parameters that maximize the log-likelihood where we maximize with respect to \( \beta \).
Since the cost (error) function is just the negative log-likelihood, for logistic regression we have that
</p>
$$
\mathcal{C}(\boldsymbol{\beta})=-\sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right).
$$
<p>This equation is known in statistics as the <b>cross entropy</b>. Finally, we note that just as in linear regression,
in practice we often supplement the cross-entropy with additional regularization terms, usually \( L_1 \) and \( L_2 \) regularization as we did for Ridge and Lasso regression.
</p>
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<h2 id="minimizing-the-cross-entropy">Minimizing the cross entropy </h2>
<p>The cross entropy is a convex function of the weights \( \boldsymbol{\beta} \) and,
therefore, any local minimizer is a global minimizer.
</p>
<p>Minimizing this
cost function with respect to the two parameters \( \beta_0 \) and \( \beta_1 \) we obtain
</p>
$$
\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \beta_0} = -\sum_{i=1}^n \left(y_i -\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right),
$$
<p>and </p>
$$
\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \beta_1} = -\sum_{i=1}^n \left(y_ix_i -x_i\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right).
$$
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<h2 id="a-more-compact-expression">A more compact expression </h2>
<p>Let us now define a vector \( \boldsymbol{y} \) with \( n \) elements \( y_i \), an
\( n\times p \) matrix \( \boldsymbol{X} \) which contains the \( x_i \) values and a
vector \( \boldsymbol{p} \) of fitted probabilities \( p(y_i\vert x_i,\boldsymbol{\beta}) \). We can rewrite in a more compact form the first
derivative of cost function as
</p>
$$
\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right).
$$
<p>If we in addition define a diagonal matrix \( \boldsymbol{W} \) with elements
\( p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}) \), we can obtain a compact expression of the second derivative as
</p>
$$
\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}.
$$
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<h2 id="extending-to-more-predictors">Extending to more predictors </h2>
<p>Within a binary classification problem, we can easily expand our model to include multiple predictors. Our ratio between likelihoods is then with \( p \) predictors</p>
$$
\log{ \frac{p(\boldsymbol{\beta}\boldsymbol{x})}{1-p(\boldsymbol{\beta}\boldsymbol{x})}} = \beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p.
$$
<p>Here we defined \( \boldsymbol{x}=[1,x_1,x_2,\dots,x_p] \) and \( \boldsymbol{\beta}=[\beta_0, \beta_1, \dots, \beta_p] \) leading to</p>
$$
p(\boldsymbol{\beta}\boldsymbol{x})=\frac{ \exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}{1+\exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}.
$$
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<h2 id="including-more-classes">Including more classes </h2>
<p>Till now we have mainly focused on two classes, the so-called binary
system. Suppose we wish to extend to \( K \) classes. Let us for the sake
of simplicity assume we have only two predictors. We have then following model
</p>
$$
\log{\frac{p(C=1\vert x)}{p(K\vert x)}} = \beta_{10}+\beta_{11}x_1,
$$
<p>and </p>
$$
\log{\frac{p(C=2\vert x)}{p(K\vert x)}} = \beta_{20}+\beta_{21}x_1,
$$
<p>and so on till the class \( C=K-1 \) class</p>
$$
\log{\frac{p(C=K-1\vert x)}{p(K\vert x)}} = \beta_{(K-1)0}+\beta_{(K-1)1}x_1,
$$
<p>and the model is specified in term of \( K-1 \) so-called log-odds or
<b>logit</b> transformations.
</p>
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<h2 id="more-classes">More classes </h2>
<p>In our discussion of neural networks we will encounter the above again
in terms of a slightly modified function, the so-called <b>Softmax</b> function.
</p>
<p>The softmax function is used in various multiclass classification
methods, such as multinomial logistic regression (also known as
softmax regression), multiclass linear discriminant analysis, naive
Bayes classifiers, and artificial neural networks. Specifically, in
multinomial logistic regression and linear discriminant analysis, the
input to the function is the result of \( K \) distinct linear functions,
and the predicted probability for the \( k \)-th class given a sample
vector \( \boldsymbol{x} \) and a weighting vector \( \boldsymbol{\beta} \) is (with two
predictors):
</p>
$$
p(C=k\vert \mathbf {x} )=\frac{\exp{(\beta_{k0}+\beta_{k1}x_1)}}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}.
$$
<p>It is easy to extend to more predictors. The final class is </p>
$$
p(C=K\vert \mathbf {x} )=\frac{1}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}},
$$
<p>and they sum to one. Our earlier discussions were all specialized to
the case with two classes only. It is easy to see from the above that
what we derived earlier is compatible with these equations.
</p>
<p>To find the optimal parameters we would typically use a gradient
descent method. Newton's method and gradient descent methods are
discussed in the material on <a href="https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html" target="_blank">optimization
methods</a>.
</p>
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<h2 id="friday-september-23">Friday September 23 </h2>
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<h2 id="wisconsin-cancer-data">Wisconsin Cancer Data </h2>
<p>We show here how we can use a simple regression case on the breast
cancer data using Logistic regression as our algorithm for
classification.
</p>
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<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.model_selection</span> <span style="color: #008000; font-weight: bold">import</span> train_test_split
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.datasets</span> <span style="color: #008000; font-weight: bold">import</span> load_breast_cancer
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.linear_model</span> <span style="color: #008000; font-weight: bold">import</span> LogisticRegression
<span style="color: #408080; font-style: italic"># Load the data</span>
cancer <span style="color: #666666">=</span> load_breast_cancer()
X_train, X_test, y_train, y_test <span style="color: #666666">=</span> train_test_split(cancer<span style="color: #666666">.</span>data,cancer<span style="color: #666666">.</span>target,random_state<span style="color: #666666">=0</span>)
<span style="color: #008000">print</span>(X_train<span style="color: #666666">.</span>shape)
<span style="color: #008000">print</span>(X_test<span style="color: #666666">.</span>shape)
<span style="color: #408080; font-style: italic"># Logistic Regression</span>
logreg <span style="color: #666666">=</span> LogisticRegression(solver<span style="color: #666666">=</span><span style="color: #BA2121">&#39;lbfgs&#39;</span>)
logreg<span style="color: #666666">.</span>fit(X_train, y_train)
<span style="color: #008000">print</span>(<span style="color: #BA2121">&quot;Test set accuracy with Logistic Regression: </span><span style="color: #BB6688; font-weight: bold">{:.2f}</span><span style="color: #BA2121">&quot;</span><span style="color: #666666">.</span>format(logreg<span style="color: #666666">.</span>score(X_test,y_test)))
<span style="color: #408080; font-style: italic">#now scale the data</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.preprocessing</span> <span style="color: #008000; font-weight: bold">import</span> StandardScaler
scaler <span style="color: #666666">=</span> StandardScaler()
scaler<span style="color: #666666">.</span>fit(X_train)
X_train_scaled <span style="color: #666666">=</span> scaler<span style="color: #666666">.</span>transform(X_train)
X_test_scaled <span style="color: #666666">=</span> scaler<span style="color: #666666">.</span>transform(X_test)
<span style="color: #408080; font-style: italic"># Logistic Regression</span>
logreg<span style="color: #666666">.</span>fit(X_train_scaled, y_train)
<span style="color: #008000">print</span>(<span style="color: #BA2121">&quot;Test set accuracy Logistic Regression with scaled data: </span><span style="color: #BB6688; font-weight: bold">{:.2f}</span><span style="color: #BA2121">&quot;</span><span style="color: #666666">.</span>format(logreg<span style="color: #666666">.</span>score(X_test_scaled,y_test)))
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<h2 id="using-the-correlation-matrix">Using the correlation matrix </h2>
<p>In addition to the above scores, we could also study the covariance (and the correlation matrix).
We use <b>Pandas</b> to compute the correlation matrix.
</p>
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<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.model_selection</span> <span style="color: #008000; font-weight: bold">import</span> train_test_split
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.datasets</span> <span style="color: #008000; font-weight: bold">import</span> load_breast_cancer
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.linear_model</span> <span style="color: #008000; font-weight: bold">import</span> LogisticRegression
cancer <span style="color: #666666">=</span> load_breast_cancer()
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">pandas</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">pd</span>
<span style="color: #408080; font-style: italic"># Making a data frame</span>
cancerpd <span style="color: #666666">=</span> pd<span style="color: #666666">.</span>DataFrame(cancer<span style="color: #666666">.</span>data, columns<span style="color: #666666">=</span>cancer<span style="color: #666666">.</span>feature_names)
fig, axes <span style="color: #666666">=</span> plt<span style="color: #666666">.</span>subplots(<span style="color: #666666">15</span>,<span style="color: #666666">2</span>,figsize<span style="color: #666666">=</span>(<span style="color: #666666">10</span>,<span style="color: #666666">20</span>))
malignant <span style="color: #666666">=</span> cancer<span style="color: #666666">.</span>data[cancer<span style="color: #666666">.</span>target <span style="color: #666666">==</span> <span style="color: #666666">0</span>]
benign <span style="color: #666666">=</span> cancer<span style="color: #666666">.</span>data[cancer<span style="color: #666666">.</span>target <span style="color: #666666">==</span> <span style="color: #666666">1</span>]
ax <span style="color: #666666">=</span> axes<span style="color: #666666">.</span>ravel()
<span style="color: #008000; font-weight: bold">for</span> i <span style="color: #AA22FF; font-weight: bold">in</span> <span style="color: #008000">range</span>(<span style="color: #666666">30</span>):
_, bins <span style="color: #666666">=</span> np<span style="color: #666666">.</span>histogram(cancer<span style="color: #666666">.</span>data[:,i], bins <span style="color: #666666">=50</span>)
ax[i]<span style="color: #666666">.</span>hist(malignant[:,i], bins <span style="color: #666666">=</span> bins, alpha <span style="color: #666666">=</span> <span style="color: #666666">0.5</span>)
ax[i]<span style="color: #666666">.</span>hist(benign[:,i], bins <span style="color: #666666">=</span> bins, alpha <span style="color: #666666">=</span> <span style="color: #666666">0.5</span>)
ax[i]<span style="color: #666666">.</span>set_title(cancer<span style="color: #666666">.</span>feature_names[i])
ax[i]<span style="color: #666666">.</span>set_yticks(())
ax[<span style="color: #666666">0</span>]<span style="color: #666666">.</span>set_xlabel(<span style="color: #BA2121">&quot;Feature magnitude&quot;</span>)
ax[<span style="color: #666666">0</span>]<span style="color: #666666">.</span>set_ylabel(<span style="color: #BA2121">&quot;Frequency&quot;</span>)
ax[<span style="color: #666666">0</span>]<span style="color: #666666">.</span>legend([<span style="color: #BA2121">&quot;Malignant&quot;</span>, <span style="color: #BA2121">&quot;Benign&quot;</span>], loc <span style="color: #666666">=</span><span style="color: #BA2121">&quot;best&quot;</span>)
fig<span style="color: #666666">.</span>tight_layout()
plt<span style="color: #666666">.</span>show()
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">seaborn</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">sns</span>
correlation_matrix <span style="color: #666666">=</span> cancerpd<span style="color: #666666">.</span>corr()<span style="color: #666666">.</span>round(<span style="color: #666666">1</span>)
<span style="color: #408080; font-style: italic"># use the heatmap function from seaborn to plot the correlation matrix</span>
<span style="color: #408080; font-style: italic"># annot = True to print the values inside the square</span>
plt<span style="color: #666666">.</span>figure(figsize<span style="color: #666666">=</span>(<span style="color: #666666">15</span>,<span style="color: #666666">8</span>))
sns<span style="color: #666666">.</span>heatmap(data<span style="color: #666666">=</span>correlation_matrix, annot<span style="color: #666666">=</span><span style="color: #008000; font-weight: bold">True</span>)
plt<span style="color: #666666">.</span>show()
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<h2 id="discussing-the-correlation-data">Discussing the correlation data </h2>
<p>In the above example we note two things. In the first plot we display
the overlap of benign and malignant tumors as functions of the various
features in the Wisconsing breast cancer data set. We see that for
some of the features we can distinguish clearly the benign and
malignant cases while for other features we cannot. This can point to
us which features may be of greater interest when we wish to classify
a benign or not benign tumour.
</p>
<p>In the second figure we have computed the so-called correlation
matrix, which in our case with thirty features becomes a \( 30\times 30 \)
matrix.
</p>
<p>We constructed this matrix using <b>pandas</b> via the statements</p>
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<pre style="line-height: 125%;">cancerpd <span style="color: #666666">=</span> pd<span style="color: #666666">.</span>DataFrame(cancer<span style="color: #666666">.</span>data, columns<span style="color: #666666">=</span>cancer<span style="color: #666666">.</span>feature_names)
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<p>and then</p>
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<pre style="line-height: 125%;">correlation_matrix <span style="color: #666666">=</span> cancerpd<span style="color: #666666">.</span>corr()<span style="color: #666666">.</span>round(<span style="color: #666666">1</span>)
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<p>Diagonalizing this matrix we can in turn say something about which
features are of relevance and which are not. This leads us to
the classical Principal Component Analysis (PCA) theorem with
applications. This will be discussed later this semester (<a href="https://compphysics.github.io/MachineLearning/doc/pub/week43/html/week43-bs.html" target="_blank">week 43</a>).
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<h2 id="other-measures-in-classification-studies-cancer-data-again">Other measures in classification studies: Cancer Data again </h2>
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<pre style="line-height: 125%;"><span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">matplotlib.pyplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">plt</span>
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">numpy</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">np</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.model_selection</span> <span style="color: #008000; font-weight: bold">import</span> train_test_split
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.datasets</span> <span style="color: #008000; font-weight: bold">import</span> load_breast_cancer
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.linear_model</span> <span style="color: #008000; font-weight: bold">import</span> LogisticRegression
<span style="color: #408080; font-style: italic"># Load the data</span>
cancer <span style="color: #666666">=</span> load_breast_cancer()
X_train, X_test, y_train, y_test <span style="color: #666666">=</span> train_test_split(cancer<span style="color: #666666">.</span>data,cancer<span style="color: #666666">.</span>target,random_state<span style="color: #666666">=0</span>)
<span style="color: #008000">print</span>(X_train<span style="color: #666666">.</span>shape)
<span style="color: #008000">print</span>(X_test<span style="color: #666666">.</span>shape)
<span style="color: #408080; font-style: italic"># Logistic Regression</span>
logreg <span style="color: #666666">=</span> LogisticRegression(solver<span style="color: #666666">=</span><span style="color: #BA2121">&#39;lbfgs&#39;</span>)
logreg<span style="color: #666666">.</span>fit(X_train, y_train)
<span style="color: #008000">print</span>(<span style="color: #BA2121">&quot;Test set accuracy with Logistic Regression: </span><span style="color: #BB6688; font-weight: bold">{:.2f}</span><span style="color: #BA2121">&quot;</span><span style="color: #666666">.</span>format(logreg<span style="color: #666666">.</span>score(X_test,y_test)))
<span style="color: #408080; font-style: italic">#now scale the data</span>
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.preprocessing</span> <span style="color: #008000; font-weight: bold">import</span> StandardScaler
scaler <span style="color: #666666">=</span> StandardScaler()
scaler<span style="color: #666666">.</span>fit(X_train)
X_train_scaled <span style="color: #666666">=</span> scaler<span style="color: #666666">.</span>transform(X_train)
X_test_scaled <span style="color: #666666">=</span> scaler<span style="color: #666666">.</span>transform(X_test)
<span style="color: #408080; font-style: italic"># Logistic Regression</span>
logreg<span style="color: #666666">.</span>fit(X_train_scaled, y_train)
<span style="color: #008000">print</span>(<span style="color: #BA2121">&quot;Test set accuracy Logistic Regression with scaled data: </span><span style="color: #BB6688; font-weight: bold">{:.2f}</span><span style="color: #BA2121">&quot;</span><span style="color: #666666">.</span>format(logreg<span style="color: #666666">.</span>score(X_test_scaled,y_test)))
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.preprocessing</span> <span style="color: #008000; font-weight: bold">import</span> LabelEncoder
<span style="color: #008000; font-weight: bold">from</span> <span style="color: #0000FF; font-weight: bold">sklearn.model_selection</span> <span style="color: #008000; font-weight: bold">import</span> cross_validate
<span style="color: #408080; font-style: italic">#Cross validation</span>
accuracy <span style="color: #666666">=</span> cross_validate(logreg,X_test_scaled,y_test,cv<span style="color: #666666">=10</span>)[<span style="color: #BA2121">&#39;test_score&#39;</span>]
<span style="color: #008000">print</span>(accuracy)
<span style="color: #008000">print</span>(<span style="color: #BA2121">&quot;Test set accuracy with Logistic Regression and scaled data: </span><span style="color: #BB6688; font-weight: bold">{:.2f}</span><span style="color: #BA2121">&quot;</span><span style="color: #666666">.</span>format(logreg<span style="color: #666666">.</span>score(X_test_scaled,y_test)))
<span style="color: #008000; font-weight: bold">import</span> <span style="color: #0000FF; font-weight: bold">scikitplot</span> <span style="color: #008000; font-weight: bold">as</span> <span style="color: #0000FF; font-weight: bold">skplt</span>
y_pred <span style="color: #666666">=</span> logreg<span style="color: #666666">.</span>predict(X_test_scaled)
skplt<span style="color: #666666">.</span>metrics<span style="color: #666666">.</span>plot_confusion_matrix(y_test, y_pred, normalize<span style="color: #666666">=</span><span style="color: #008000; font-weight: bold">True</span>)
plt<span style="color: #666666">.</span>show()
y_probas <span style="color: #666666">=</span> logreg<span style="color: #666666">.</span>predict_proba(X_test_scaled)
skplt<span style="color: #666666">.</span>metrics<span style="color: #666666">.</span>plot_roc(y_test, y_probas)
plt<span style="color: #666666">.</span>show()
skplt<span style="color: #666666">.</span>metrics<span style="color: #666666">.</span>plot_cumulative_gain(y_test, y_probas)
plt<span style="color: #666666">.</span>show()
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<h2 id="friday-september-25">Friday September 25 </h2>
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<h2 id="optimization-the-central-part-of-any-machine-learning-algortithm">Optimization, the central part of any Machine Learning algortithm </h2>
<a href="https://www.uio.no/studier/emner/matnat/fys/FYS-STK3155/h20/forelesningsvideoer/OverarchingAimsWeek39.mp4?vrtx=view-as-webpage" target="_blank">Overview Video, why do we care about gradient methods?</a>
<p>Almost every problem in machine learning and data science starts with
a dataset \( X \), a model \( g(\beta) \), which is a function of the
parameters \( \beta \) and a cost function \( C(X, g(\beta)) \) that allows
us to judge how well the model \( g(\beta) \) explains the observations
\( X \). The model is fit by finding the values of \( \beta \) that minimize
the cost function. Ideally we would be able to solve for \( \beta \)
analytically, however this is not possible in general and we must use
some approximative/numerical method to compute the minimum.
</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="revisiting-our-logistic-regression-case">Revisiting our Logistic Regression case </h2>
<p>In our discussion on Logistic Regression we studied the
case of
two classes, with \( y_i \) either
\( 0 \) or \( 1 \). Furthermore we assumed also that we have only two
parameters \( \beta \) in our fitting, that is we
defined probabilities
</p>
$$
\begin{align*}
p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\
p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}),
\end{align*}
$$
<p>where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \). </p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="the-equations-to-solve">The equations to solve </h2>
<p>Our compact equations used a definition of a vector \( \boldsymbol{y} \) with \( n \)
elements \( y_i \), an \( n\times p \) matrix \( \boldsymbol{X} \) which contains the
\( x_i \) values and a vector \( \boldsymbol{p} \) of fitted probabilities
\( p(y_i\vert x_i,\boldsymbol{\beta}) \). We rewrote in a more compact form
the first derivative of the cost function as
</p>
$$
\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right).
$$
<p>If we in addition define a diagonal matrix \( \boldsymbol{W} \) with elements
\( p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}) \), we can obtain a compact expression of the second derivative as
</p>
$$
\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}.
$$
<p>This defines what is called the Hessian matrix.</p>
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<h2 id="solving-using-newton-raphson-s-method">Solving using Newton-Raphson's method </h2>
<p>If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives. </p>
<p>Our iterative scheme is then given by</p>
$$
\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T}\right)^{-1}_{\boldsymbol{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}\right)_{\boldsymbol{\beta}^{\mathrm{old}}},
$$
<p>or in matrix form as</p>
$$
\boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X} \right)^{-1}\times \left(-\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{p}) \right)_{\boldsymbol{\beta}^{\mathrm{old}}}.
$$
<p>The right-hand side is computed with the old values of \( \beta \). </p>
<p>If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement. </p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="brief-reminder-on-newton-raphson-s-method">Brief reminder on Newton-Raphson's method </h2>
<p>Let us quickly remind ourselves how we derive the above method.</p>
<p>Perhaps the most celebrated of all one-dimensional root-finding
routines is Newton's method, also called the Newton-Raphson
method. This method requires the evaluation of both the
function \( f \) and its derivative \( f' \) at arbitrary points.
If you can only calculate the derivative
numerically and/or your function is not of the smooth type, we
normally discourage the use of this method.
</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="the-equations">The equations </h2>
<p>The Newton-Raphson formula consists geometrically of extending the
tangent line at a current point until it crosses zero, then setting
the next guess to the abscissa of that zero-crossing. The mathematics
behind this method is rather simple. Employing a Taylor expansion for
\( x \) sufficiently close to the solution \( s \), we have
</p>
$$
f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots.
\label{eq:taylornr}
$$
<p>For small enough values of the function and for well-behaved
functions, the terms beyond linear are unimportant, hence we obtain
</p>
$$
f(x)+(s-x)f'(x)\approx 0,
$$
<p>yielding</p>
$$
s\approx x-\frac{f(x)}{f'(x)}.
$$
<p>Having in mind an iterative procedure, it is natural to start iterating with</p>
$$
x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
$$
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<h2 id="simple-geometric-interpretation">Simple geometric interpretation </h2>
<p>The above is Newton-Raphson's method. It has a simple geometric
interpretation, namely \( x_{n+1} \) is the point where the tangent from
\( (x_n,f(x_n)) \) crosses the \( x \)-axis. Close to the solution,
Newton-Raphson converges fast to the desired result. However, if we
are far from a root, where the higher-order terms in the series are
important, the Newton-Raphson formula can give grossly inaccurate
results. For instance, the initial guess for the root might be so far
from the true root as to let the search interval include a local
maximum or minimum of the function. If an iteration places a trial
guess near such a local extremum, so that the first derivative nearly
vanishes, then Newton-Raphson may fail totally
</p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="extending-to-more-than-one-variable">Extending to more than one variable </h2>
<p>Newton's method can be generalized to systems of several non-linear equations
and variables. Consider the case with two equations
</p>
$$
\begin{array}{cc} f_1(x_1,x_2) &=0\\
f_2(x_1,x_2) &=0,\end{array}
$$
<p>which we Taylor expand to obtain</p>
$$
\begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1
\partial f_1/\partial x_1+h_2
\partial f_1/\partial x_2+\dots\\
0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1
\partial f_2/\partial x_1+h_2
\partial f_2/\partial x_2+\dots
\end{array}.
$$
<p>Defining the Jacobian matrix \( {\bf \boldsymbol{J}} \) we have</p>
$$
{\bf \boldsymbol{J}}=\left( \begin{array}{cc}
\partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\
\partial f_2/\partial x_1 &\partial f_2/\partial x_2
\end{array} \right),
$$
<p>we can rephrase Newton's method as</p>
$$
\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)=
\left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+
\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right),
$$
<p>where we have defined</p>
$$
\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)=
-{\bf \boldsymbol{J}}^{-1}
\left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right).
$$
<p>We need thus to compute the inverse of the Jacobian matrix and it
is to understand that difficulties may
arise in case \( {\bf \boldsymbol{J}} \) is nearly singular.
</p>
<p>It is rather straightforward to extend the above scheme to systems of
more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
</p>
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<h2 id="steepest-descent">Steepest descent </h2>
<p>The basic idea of gradient descent is
that a function \( F(\mathbf{x}) \),
\( \mathbf{x} \equiv (x_1,\cdots,x_n) \), decreases fastest if one goes from \( \bf {x} \) in the
direction of the negative gradient \( -\nabla F(\mathbf{x}) \).
</p>
<p>It can be shown that if </p>
$$
\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k),
$$
<p>with \( \gamma_k > 0 \).</p>
<p>For \( \gamma_k \) small enough, then \( F(\mathbf{x}_{k+1}) \leq
F(\mathbf{x}_k) \). This means that for a sufficiently small \( \gamma_k \)
we are always moving towards smaller function values, i.e a minimum.
</p>
<!-- !split -->
<h2 id="more-on-steepest-descent">More on Steepest descent </h2>
<p>The previous observation is the basis of the method of steepest
descent, which is also referred to as just gradient descent (GD). One
starts with an initial guess \( \mathbf{x}_0 \) for a minimum of \( F \) and
computes new approximations according to
</p>
$$
\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0.
$$
<p>The parameter \( \gamma_k \) is often referred to as the step length or
the learning rate within the context of Machine Learning.
</p>
<!-- !split -->
<h2 id="the-ideal">The ideal </h2>
<p>Ideally the sequence \( \{\mathbf{x}_k \}_{k=0} \) converges to a global
minimum of the function \( F \). In general we do not know if we are in a
global or local minimum. In the special case when \( F \) is a convex
function, all local minima are also global minima, so in this case
gradient descent can converge to the global solution. The advantage of
this scheme is that it is conceptually simple and straightforward to
implement. However the method in this form has some severe
limitations:
</p>
<p>In machine learing we are often faced with non-convex high dimensional
cost functions with many local minima. Since GD is deterministic we
will get stuck in a local minimum, if the method converges, unless we
have a very good intial guess. This also implies that the scheme is
sensitive to the chosen initial condition.
</p>
<p>Note that the gradient is a function of \( \mathbf{x} =
(x_1,\cdots,x_n) \) which makes it expensive to compute numerically.
</p>
<!-- !split -->
<h2 id="the-sensitiveness-of-the-gradient-descent">The sensitiveness of the gradient descent </h2>
<p>The gradient descent method
is sensitive to the choice of learning rate \( \gamma_k \). This is due
to the fact that we are only guaranteed that \( F(\mathbf{x}_{k+1}) \leq
F(\mathbf{x}_k) \) for sufficiently small \( \gamma_k \). The problem is to
determine an optimal learning rate. If the learning rate is chosen too
small the method will take a long time to converge and if it is too
large we can experience erratic behavior.
</p>
<p>Many of these shortcomings can be alleviated by introducing
randomness. One such method is that of Stochastic Gradient Descent
(SGD), see below.
</p>
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