Machine Learning

Logistic Regression

Now we are switching from regression problems to classification problems. Don’t be confused by the name “Logistic Regression”; it is named that way for historical reasons and is actually an approach to classification problems, not regression problems.

Binary Classification

Instead of our output vector y being a continuous range of values, it will only be 0 or 1.

\[y \in [0,1]\]

Where 0 is usually taken as the “negative class” and 1 as the “positive class”, but you are free to assign any representation to it.

We’re only doing two classes for now, called a “Binary Classification Problem.”

One method is to use linear regression and map all predictions greater than 0.5 as a 1 and all less than 0.5 as a 0. This method doesn’t work well because classification is not actually a linear function.

Hypothesis Representation

Our hypothesis should satisfy:

\[0 \leq h_\theta (x) \leq 1\]

Our new form uses the “Sigmoid Function,” also called the “Logistic Function”:

\[h_\theta (x) = g(\theta^T x)\]

\[z = \theta^T x\]

\[g(z) = \frac {1}{1 + e^{-z}}\]

The function g(z), shown here, maps any real number to the (0, 1) interval, making it useful for transforming an arbitrary-valued function into a function better suited for classification. Try playing with interactive plot of sigmoid function : (https://www.desmos.com/calculator/bgontvxotm).

We start with our old hypothesis (linear regression), except that we want to restrict the range to 0 and 1. This is accomplished by plugging \(\theta^T x\) into the Logistic Function.

\(h_\theta\) will give us the probability that our output is 1. For example, \(h_\theta (x) =0.7\) gives us the probability of 70% that our output is 1.

\[h_\theta(x) = P(y=1|x;\theta) = 1 - P(y = 0|x;\theta)\]

\[P(y = 0 | x; \theta) + P(y=1|x;\theta) = 1\]

Decision Boundary

In order to get our discrete 0 or 1 classification, we can translate the output of the hypothesis function as follows:

\[h_\theta(x) \geq 0.5 \rightarrow y = 1\]

\[h_\theta(x) < 0.5 \rightarrow y = 0 \]

The way our logistic function g behaves is that when its input is greater than or equal to zero, its output is greater than or equal to 0.5:

\[g(z) \geq 0.5 \]

\[when \ z \geq 0 \]

Remember.-

\[z = 0, e^0 = 1 \rightarrow g(z) = \frac{1}{2} \]

\[z \rightarrow \infty , e^{\neg \infty} \rightarrow 0 \rightarrow g(z) = 1 \]

\[z \rightarrow \neg \infty , e^{ \infty} \rightarrow \infty \rightarrow g(z) = 0 \]

So if our input to g is \(\theta^T X\), then that means:

\[h_\theta (x) = g(\theta^T x) \geq 0.5\]

\[when \ \theta^T x \geq 0\]

From these statements we can now say:

\[\theta^T x \geq 0 \rightarrow y = 1\]

\[\theta^T x \leq \rightarrow = 0\]

The decision boundary is the line that separates the area where y = 0 and where y = 1. It is created by our hypothesis function.

Example:

\[\left [\matrix {5 \\ \neg 1 \\ 0} \right ]\]

\[y = 1 \ if \ 5 + (-1)x_1 + 0x_2 \geq 0\]

\[5 - x_1 \geq 0 \]

\[-x_1 \geq -5\]

\[x_1 \geq 5 \]

In this case, our decision boundary is a straight vertical line placed on the graph where \(x_1 = 5\), and everything to the left of that denotes y = 1, while everything to the right denotes y = 0.

Again, the input to the sigmoid function g(z) (e.g. \(\theta^T X\)) doesn’t need to be linear, and could be a function that describes a circle (e.g. \(z = \theta_0 + \theta_1 x_1^2 + \theta_2 x_2^2\)) or any shape to fit our data.

Cost Function

We cannot use the same cost function that we use for linear regression because the Logistic Function will cause the output to be wavy, causing many local optima. In other words, it will not be a convex function.

Instead, our cost function for logistic regression looks like:

\[J (\theta) = \frac {1} {m} \sum_{i = 1}^{m} Cost(h_\theta(x^{(ij)}), y^{(ij)})\]

\[Cost(h_\theta (x), y) = -log(h_\theta (x)) \ if \ y = 1 \]

\[Cost(h_\theta (x), y) = -log(1 - h_\theta (x)) \ if \ y = 0 \]

The more our hypothesis is off from y, the larger the cost function output. If our hypothesis is equal to y, then our cost is 0:

\[Cost(h_\theta (x), y) = 0 \ if \ h_\theta (x) = y \]

\[Cost(h_\theta (x), y) \rightarrow \infty \ if \ y = 0 \ and \ h_\theta (x) \rightarrow 1 \]

\[Cost(h_\theta (x), y) \rightarrow \infty \ if \ y = 1 \ and \ h_\theta (x) \rightarrow 0 \]

if our correct answer ‘y’ is 0, then the cost function will be 0 if our hypothesis function also outputs 0. If our hypothesis approaches 1, then the cost function will approach infinity.

If our correct answer ‘y’ is 1, then the cost function will be 0 if our hypothesis function outputs 1. If our hypothesis approaches 0, then the cost function will approach infinity.

Note that writing the cost function in this way guarantees that \(J(\theta)\) is convex for logistic regression.

Simplified Cost Function and Gradient Descent

We can compress our cost function’s two conditional cases into one case:

\[Cost(h_\theta (x), y) = -y \ log(h_\theta (x)) - (1 - y) log(1-h_\theta(x))\]

Notice that when y is equal to 1, then the second term \((1-y)log(1 - h_\theta (x))\) will be zero and will not affect the result. If y is equal to 0, then the first term \(-ylog(h_\theta(x))\) will be zero and will not affect the result.

We can fully write out our entire cost function as follows:

\[J(\theta) = - \frac{1}{m} \sum_{i = 1}^{m} [y^{(i)} log(h_\theta(x^{(i)})) + (1 - y^{(i)})log(1 - h_\theta (x^{(i)})]) \]

A vectorized implementation is:

\[h = g(X\theta)\]

\[J(\theta) = \frac{1}{m} * (-y^T log(h) - (1 - y)^T log(1-h))\]

Gradient Descent

Remember that the general form of gradient descent is:

Repeat [

\[\theta_j := \theta_j - \alpha \frac{\partial}{\partial \theta_j} J(\theta) \]

]

We can work out the derivative part using calculus to get:

Repeat [

\[\theta_j := \theta_j - \frac{\alpha}{m} \sum_{i=1}^m (h_\theta (x^{(i)})) - y^{(i)}x_j^{(i)} \]

]

Notice that this algorithm is identical to the one we used in linear regression. We still have to simultaneously update all values in theta.

A vectorized implementation is:

\[\theta := \theta - \frac{\alpha}{m} X^T (g(X\theta) - \bar{y})\]

Partial derivative of \(J(\theta)\)

First calculate derivative of sigmoid function (it will be useful while finding partial derivative of \(J(\theta)\):

\[\begin{align*}\sigma(x)'&=\left(\frac{1}{1+e^{-x}}\right)'=\frac{-(1+e^{-x})'}{(1+e^{-x})^2}=\frac{-1'-(e^{-x})'}{(1+e^{-x})^2}=\frac{0-(-x)'(e^{-x})}{(1+e^{-x})^2}=\frac{-(-1)(e^{-x})}{(1+e^{-x})^2}=\frac{e^{-x}}{(1+e^{-x})^2} \newline &=\left(\frac{1}{1+e^{-x}}\right)\left(\frac{e^{-x}}{1+e^{-x}}\right)=\sigma(x)\left(\frac{+1-1 + e^{-x}}{1+e^{-x}}\right)=\sigma(x)\left(\frac{1 + e^{-x}}{1+e^{-x}} - \frac{1}{1+e^{-x}}\right)=\sigma(x)(1 - \sigma(x))\end{align*}\]

Now we are ready to find out resulting partial derivative:

\[\begin{align*}\frac{\partial}{\partial \theta_j} J(\theta) &= \frac{\partial}{\partial \theta_j} \frac{-1}{m}\sum_{i=1}^m \left [ y^{(i)} log (h_\theta(x^{(i)})) + (1-y^{(i)}) log (1 - h_\theta(x^{(i)})) \right ] \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ y^{(i)} \frac{\partial}{\partial \theta_j} log (h_\theta(x^{(i)})) + (1-y^{(i)}) \frac{\partial}{\partial \theta_j} log (1 - h_\theta(x^{(i)}))\right ] \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ \frac{y^{(i)} \frac{\partial}{\partial \theta_j} h_\theta(x^{(i)})}{h_\theta(x^{(i)})} + \frac{(1-y^{(i)})\frac{\partial}{\partial \theta_j} (1 - h_\theta(x^{(i)}))}{1 - h_\theta(x^{(i)})}\right ] \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ \frac{y^{(i)} \frac{\partial}{\partial \theta_j} \sigma(\theta^T x^{(i)})}{h_\theta(x^{(i)})} + \frac{(1-y^{(i)})\frac{\partial}{\partial \theta_j} (1 - \sigma(\theta^T x^{(i)}))}{1 - h_\theta(x^{(i)})}\right ] \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ \frac{y^{(i)} \sigma(\theta^T x^{(i)}) (1 - \sigma(\theta^T x^{(i)})) \frac{\partial}{\partial \theta_j} \theta^T x^{(i)}}{h_\theta(x^{(i)})} + \frac{- (1-y^{(i)}) \sigma(\theta^T x^{(i)}) (1 - \sigma(\theta^T x^{(i)})) \frac{\partial}{\partial \theta_j} \theta^T x^{(i)}}{1 - h_\theta(x^{(i)})}\right ] \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ \frac{y^{(i)} h_\theta(x^{(i)}) (1 - h_\theta(x^{(i)})) \frac{\partial}{\partial \theta_j} \theta^T x^{(i)}}{h_\theta(x^{(i)})} - \frac{(1-y^{(i)}) h_\theta(x^{(i)}) (1 - h_\theta(x^{(i)})) \frac{\partial}{\partial \theta_j} \theta^T x^{(i)}}{1 - h_\theta(x^{(i)})}\right ] \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ y^{(i)} (1 - h_\theta(x^{(i)})) x^{(i)}_j - (1-y^{(i)}) h_\theta(x^{(i)}) x^{(i)}_j\right ] \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ y^{(i)} (1 - h_\theta(x^{(i)})) - (1-y^{(i)}) h_\theta(x^{(i)}) \right ] x^{(i)}_j \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ y^{(i)} - y^{(i)} h_\theta(x^{(i)}) - h_\theta(x^{(i)}) + y^{(i)} h_\theta(x^{(i)}) \right ] x^{(i)}_j \newline&= - \frac{1}{m}\sum_{i=1}^m \left [ y^{(i)} - h_\theta(x^{(i)}) \right ] x^{(i)}_j \newline&= \frac{1}{m}\sum_{i=1}^m \left [ h_\theta(x^{(i)}) - y^{(i)} \right ] x^{(i)}_j\end{align*}\]

The vectorized version;

\[\nabla J (\theta) = \frac{1}{m} * X^T * (g(X * \theta) -\bar{y})\]

Advanced Optimization

“Conjugate gradient”, “BFGS”, and “L-BFGS” are more sophisticated, faster ways to optimize \(\theta\) that can be used instead of gradient descent. A. Ng suggests not to write these more sophisticated algorithms yourself (unless you are an expert in numerical computing) but use the libraries instead, as they’re already tested and highly optimized. Octave provides them.

We first need to provide a function that evaluates the following two functions for a given input value \(\theta\):

\[\begin{align*} & J(\theta) \newline & \dfrac{\partial}{\partial \theta_j}J(\theta)\end{align*}\]

# Function to calculate both cost function and gradient descent
function [jVal, gradient] = costFunction(theta)
  jVal = [...code to compute J(theta)...];
  gradient = [...code to compute derivative of J(theta)...];
end

Then we can use octave’s “fminunc()” optimization algorithm along with the “optimset()” function that creates an object containing the options we want to send to “fminunc()”.

# Use fminunc optimization algorithm
options = optimset('GradObj', 'on', 'MaxIter', 100);
      initialTheta = zeros(2,1);
      [optTheta, functionVal, exitFlag] = fminunc(@costFunction, initialTheta, options);

We give to the function “fminunc()” our cost function, our initial vector of theta values, and the “options” object that we created beforehand.

Multiclass Classification: One-vs-all

Now we will approach the classification of data into more than two categories. Instead of y = {0,1} we will expand our definition so that y = {0,1…n}.

In this case we divide our problem into n+1 (+1 because the index starts at 0) binary classification problems; in each one, we predict the probability that ‘y’ is a member of one of our classes.

\[\begin{align*}& y \in \lbrace0, 1 ... n\rbrace \newline& h_\theta^{(0)}(x) = P(y = 0 | x ; \theta) \newline& h_\theta^{(1)}(x) = P(y = 1 | x ; \theta) \newline& \cdots \newline& h_\theta^{(n)}(x) = P(y = n | x ; \theta) \newline& \mathrm{prediction} = \max_i( h_\theta ^{(i)}(x) )\newline\end{align*}\]

We are basically choosing one class and then lumping all the others into a single second class. We do this repeatedly, applying binary logistic regression to each case, and then use the hypothesis that returned the highest value as our prediction.

Regularization

The Problem of Overfitting

Regularization is designed to address the problem of overfitting.

High bias or underfitting is when the form of our hypothesis function h maps poorly to the trend of the data. It is usually caused by a function that is too simple or uses too few features. eg. if we take \(h_\theta (x) = \theta_0 + \theta_1 x_1 + \theta_2 x_2\) then we are making an initial assumption that a linear model will fit the training data well and will be able to generalize but that may not be the case.

At the other extreme, overfitting or high variance is caused by a hypothesis function that fits the available data but does not generalize well to predict new data. It is usually caused by a complicated function that creates a lot of unnecessary curves and angles unrelated to the data.

This terminology is applied to both linear and logistic regression. There are two main options to address the issue of overfitting:

  1. Reduce the number of features:

  1. Manually select which features to keep.

  2. Use a model selection algorithm (studied later in the course).

  1. Regularization

Keep all the features, but reduce the parameters \(\theta_j\)

Regularization works well when we have a lot of slightly useful features.

Cost Function

If we have overfitting from our hypothesis function, we can reduce the weight that some of the terms in our function carry by increasing their cost.

Say we wanted to make the following function more quadratic:

\[\theta_0 + \theta_1x + \theta_2 x^2 + \theta_3 x^3 + \theta_4 x^4\]

We’ll want to eliminate the influence of \(\theta_3 x^3\) and \(\theta_4 x^4\) . Without actually getting rid of these features or changing the form of our hypothesis, we can instead modify our cost function:

\[min_\theta \ \frac{1}{2m} \sum_{i =1} ^{m} (h_\theta(x^{(i)} ) - y^{(i)})^2 + 1000* \theta_3^2 + 1000* \theta_4^2\]

We’ve added two extra terms at the end to inflate the cost of \(\theta_3\) and \(\theta_4\). Now, in order for the cost function to get close to zero, we will have to reduce the values of \(\theta_3\) and \(\theta_4\) to near zero. This will in turn greatly reduce the values of \(\theta_3 x^3\) and \(\theta_4 x^4\) in our hypothesis function.

We could also regularize all of our theta parameters in a single summation:

\[min_\theta \ \frac{1}{2m} \left [\sum_{i =1} ^{m} (h_\theta(x^{(i)} ) - y^{(i)})^2 + \lambda \sum_{j=1}^n \theta_j^2 \right ] \]

The \(\lambda\), or lambda, is the regularization parameter. It determines how much the costs of our theta parameters are inflated. You can visualize the effect of regularization in this interactive plot : https://www.desmos.com/calculator/1hexc8ntqp

Using the above cost function with the extra summation, we can smooth the output of our hypothesis function to reduce overfitting. If lambda is chosen to be too large, it may smooth out the function too much and cause underfitting.

Regularized Linear Regression

We can apply regularization to both linear regression and logistic regression. We will approach linear regression first.

Gradient Descent

We will modify our gradient descent function to separate out \(\theta_0\) from the rest of the parameters because we do not want to penalize \(\theta_0\).

\[\begin{align*} & \text{Repeat}\ \lbrace \newline & \ \ \ \ \theta_0 := \theta_0 - \alpha\ \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_0^{(i)} \newline & \ \ \ \ \theta_j := \theta_j - \alpha\ \left[ \left( \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)} \right) + \frac{\lambda}{m}\theta_j \right] &\ \ \ \ \ \ \ \ \ \ j \in \lbrace 1,2...n\rbrace\newline & \rbrace \end{align*}\]

The term \(\frac{\lambda}{m}\theta_j \) performs our regularization.

With some manipulation our update rule can also be represented as:

\[\theta_j := \theta_j (1 - \alpha \frac{\lambda} {m}) - \alpha \frac{1}{m} \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)}) x_i^{(i)}\]

The first term in the above equation, \(1 - \alpha\frac{\lambda}{m}\) will always be less than 1. Intuitively you can see it as reducing the value of \(\theta_j\) by some amount on every update.

Notice that the second term is now exactly the same as it was before.

Normal Equation

Now let’s approach regularization using the alternate method of the non-iterative normal equation.

To add in regularization, the equation is the same as our original, except that we add another term inside the parentheses:

\[\begin{align*}& \theta = \left( X^TX + \lambda \cdot L \right)^{-1} X^Ty \newline& \text{where}\ \ L = \begin{bmatrix} 0 & & & & \newline & 1 & & & \newline & & 1 & & \newline & & & \ddots & \newline & & & & 1 \newline\end{bmatrix}\end{align*}\]

L is a matrix with 0 at the top left and 1’s down the diagonal, with 0’s everywhere else. It should have dimension (n+1)×(n+1). Intuitively, this is the identity matrix (though we are not including \(x_0\)), multiplied with a single real number \(\lambda\).

Recall that if \(m \leq n\), then \(X^T X\)is non-invertible. However, when we add the term \(\lambda - L\), then \(X^T X + ??*L\) becomes invertible.

Regularized Logistic Regression

We can regularize logistic regression in a similar way that we regularize linear regression. Let’s start with the cost function.

Cost Function

Recall that our cost function for logistic regression was:

\[J(\theta) = - \frac{1}{m} \sum_{i = 1}^{m} [y^{(i)} log(h_\theta(x^{(i)})) + (1 - y^{(i)})log(1 - h_\theta (x^{(i)})]) \]

We can regularize this equation by adding a term to the end:

\[J(\theta) = - \frac{1}{m} \sum_{i = 1}^{m} [y^{(i)} log(h_\theta(x^{(i)})) + (1 - y^{(i)})log(1 - h_\theta (x^{(i)})]) + \frac{\lambda}{2m} \sum_{j = 1}^n \theta_j^2 \]

Gradient Descent

Just like with linear regression, we will want to separately update \(\theta_0\) and the rest of the parameters because we do not want to regularize \(\theta_0\).

\[\begin{align*}& \text{Repeat}\ \lbrace \newline& \ \ \ \ \theta_0 := \theta_0 - \alpha\ \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_0^{(i)} \newline& \ \ \ \ \theta_j := \theta_j - \alpha\ \left[ \left( \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)} \right) + \frac{\lambda}{m}\theta_j \right] &\ \ \ \ \ \ \ \ \ \ j \in \lbrace 1,2...n\rbrace\newline& \rbrace\end{align*}\]

This is identical to the gradient descent function presented for linear regression.

Initial Ones Feature Vector

Constant Feature

As it turns out it is crucial to add a constant feature to your pool of features before starting any training of your machine. Normally that feature is just a set of ones for all your training examples.

Concretely, if X is your feature matrix then \(X_0\) is a vector with ones.

Below are some insights to explain the reason for this constant feature. The first part draws some analogies from electrical engineering concept, the second looks at understanding the ones vector by using a simple machine learning example.