# Machine Learning Coursera’s Notes – Week 3

01 September 2020

My notes from the Machine Learning Course provided by Andrew Ng on Coursera

# Logistic Regression

Example of classification problem:

• Email: Spam/Not spam?
• Online transaction: Fraudulent (Yes/No)?
• Tumour: Malingnat/Benign?

$y \in \{0,1\}$ where $0$ - "Ngative class" $1$ - "Positive class"

## Multiclass classification problem $y \in \{0,1,2,3,...\}$

### How to develop a classification algorithm?

Example of a training set for classification task:

For this data set applying linear regression algorithm: $h_{\theta}(x)=\theta^Tx$ and it could finish like below:

For prediction, it could be useful to try to do the threshold:

Threshold classifier output $h_{\theta}(x)$ at $0.5$: (vertical axis value)

• If $h_{\theta}(x) \geqslant 0.5$, predict "$y=1$"
• If $h_{\theta}(x) < 0.5$, predict "$y=0$"

### What happens if a problem will change a bit?

By adding one positive example cause linear regression to change, it's a straight line to fit the new data. It causes (in this example) a worse hypothesis.

### Conclusion: Applying linear regression to classification problem often isn't a good idea

Funny bit of using linear regression on classification problem: Classification: $y=0$ or $y=1$ but $h_{\theta}(x)$ can be $>1$ or $<0$ So even when if we know that label should be $0$ or $1$, the linear regression could back with output values much larger than $1$ or much smaller than $0$.

# Hypothesis Representation

### Logistic Regression Model

Want $0 \leqslant h_{\theta}(x) \leqslant 1$

$h_{\theta}(x) = g(\theta^Tx)$

$g(z) = {1 \over {1 + e^{-z}}}$ (two different names for the same function: sigmoid function and logistic function )

$\begin{rcases} h_{\theta}(x) = g(\theta^Tx) \\ g(z) = {1 \over {1 + e^{-z}}} \end{rcases}⇒h_{\theta}(x)={1 \over {1 + e^{-\theta^Tx}}}$

$z$ - real number

### Sigmoid function

Parameters: $\theta$

## Interpretation of Hypothesis Output

$h_{\theta}(x)$ - estimated probability that $y=1$ on input $x$

Example:

$\text{if } x = \begin{bmatrix} x_0 \\ x_1 \end{bmatrix} = \begin{bmatrix} 1 \\ \text{tumor size} \end{bmatrix}$

$h_{\theta}(x) = 0.7$ Tell patient that 7-% chance of tumor being malignant

$h_{\theta}(x) = P(y=1|x;\theta)$ - probability that $y=1$, give $x$, parameterized by $\theta$

Because this is a classification problem, then $y$ can get only two values: $0$ or $1$

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

$P(y=0|x;\theta) = 1 - P(y=1|x;\theta)$

# Decision Boundary

Logistic Regresion:

$h_{\theta}(x) = g(\theta^Tx)$

$g(z) = {1 \over {1 + e^{-z}}}$

Predict $"y=1"$ if $h_{\theta}(x) \geqslant 0.5$

Predict $"y=0"$ if $h_{\theta}(x) < 0.5$

$g(z) \geqslant 0.5$ when $z \geqslant 0$

$h_{\theta}(x) = g(\theta^Tx) \geqslant 0.5$ where $\theta^Tx \geqslant 0$

### Training set

$h(_{\theta}(x) = q(\theta_0 + \theta_1x_1 + \theta_2x_2)$

Based on the plot: $h(_{\theta}(x) = q(-3 + 1x_1 + 1x_2)$

so: $\theta = \begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix}$

Predict: $"y=1"$ if $-3 + x_1 + x_2 \geqslant 0$

$-3 + x_1 + x_2 \geqslant 0$

$x_1 + x_2 \geqslant 3$ - will create a sraight line

### Excercise:

$\theta_0=5; \theta_1=-1; \theta_2=0;$

$h_{\theta}(x) = g(5-x_1)$

so: $x_1 \geqslant 5$

### Non-linear decision boundaries

$h_{\theta}(x) = g(\theta_0 +\theta_2x_2 + \theta_3x_1^2 + \theta_4x_2^2)$

$\theta = \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}$

Predict $"y=1"$ if $-1 + x_1^2 + x_2^2 \geqslant 0$

$x_1^2 + x_2^2 = 1$

#### More complex decision boundaries

$h_{\theta}(x) = g(\theta_0 +\theta_2x_2 + \theta_3x_1^2 + \theta_4x_2^2 + \theta_5x_1^2x_2^2 + \theta_6x_1^3x_2 + ...)$

# Cost Function

## Supervised learning problem:

Training set: $\{(x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)}), ... , (x^{(m)}, y^{(m)})\}$

$m$ examples:

$x \in \begin{bmatrix} x_0 \\ x_1 \\ \vdots \\ x_n \end{bmatrix}$ $x_0 =1, y \in \{0,1\}$

$h_{\theta}(x) = {1 \over 1 + e - \theta^Tx}$

## How to choose parameter $\theta$?

### Cost function

Linear regression: $J(θ)=\frac{1}{m} \displaystyle\sum_{i=1}^m \frac{1}{2} (h_θ (x^{(i)})-y^{(i)})^2$

An alternative way of writing this function:

$J(θ)=\frac{1}{m} \displaystyle\sum_{i=1}^m cost (h_θ (x^{(i)}), y)$

so: $Cost (h_θ (x^{(i)}), y) = \frac{1}{2} (h_θ (x^{(i)})-y^{(i)})$

This function is fine for linear regression.

For logistic regression would also work fine, but as well would be a non-convex function of the parameters $\theta$

Example:

## Logistic regression cost function

$Cost (h_θ (x^{(i)}) = \begin{array}{cc} \log(h_{\theta}(x)) \text{if } y = 1 \\ \log(1 - h_{\theta}(x)) \text{if } y = 0 \end{array}$

$Cost = 0$ if $y = 1, h_{\theta}(x) = 1$

But as $h_{\theta}(x) \to 0$ and $Cost \to \infty$

Capture intuition that if $h_{\theta}(x) = 0$ (predict $P(y=1|x;\theta)=0$), but $y = 1$, we will penalize learning algorithm by a very large cost

## Simplified cost function and gradient descent

### Logistic regression cost function

$J(θ)=\frac{1}{m} \displaystyle\sum_{i=1}^m Cost (h_θ (x^{(i)}), y^{(i)})$

$Cost (h_θ (x^{(i)}) = \begin{array}{cc} \log(h_{\theta}(x)) \text{if } y = 1 \\ \log(1 - h_{\theta}(x)) \text{if } y = 0 \end{array}$

Always $y=0 \text{ or } y=1$

$Cost (h_θ (x),y) = -y\log(h_{\theta}(x)) - ((-y)\log(1-h_{\theta}(x))$ (compact way to write in one line)

if $y=1$: $Cost (h_θ (x),y) = -\log(h_{\theta}(x))$

if $y=0$: $Cost (h_θ (x),y) = -\log(1 - h_{\theta}(x))$

#### Logistic regression cost function

$J(θ)=\frac{1}{m} \displaystyle\sum_{i=1}^m Cost (h_θ (x^{(i)}), y^{(i)}) = -\frac{1}{m} \lbrack\displaystyle\sum_{i=1}^m y^{(i)} \log h_{\theta}(x^{(i)})+(1-y^{(i)})\log(1-h_{\theta}(x^{(i)})) \rbrack$$J(θ)=\frac{1}{m} \displaystyle\sum_{i=1}^m Cost (h_θ (x^{(i)}), y^{(i)})$

To fit parameter $\theta$: $\begin{matrix} min \\ \theta \end{matrix} J(\theta)$

To make a prediction given new $x$:

Output $h_{\theta}(x) = \frac{1}{1 + e^{-\theta T} x}$

$J(θ)=-\frac{1}{m} \lbrack\displaystyle\sum_{i=1}^m y^{(i)} \log h_{\theta}(x^{(i)})+(1-y^{(i)})\log(1-h_{\theta}(x^{(i)})) \rbrack$

Want $min_{\theta}J(\theta)$:

Repeat {
$θ_j := θ_j - \alpha \frac{∂}{∂θ_j} J(θ)$
}
(simultanously update for every $\theta_j$)

$\alpha \frac{∂}{∂θ_j} J(θ) = \frac{1}{m} \displaystyle\sum_{i=1}^m (h_θ (x^{(i)}) - y^{(i)})x_j^{(i)}$

so:

Repeat {
$θ_j := θ_j - \alpha \displaystyle\sum_{i=1}^m (h_θ (x^{(i)}) - y^{(i)})x_j^{(i)}$
}
(simultanously update all $\theta_j$)

$\theta$ = $\begin{bmatrix} \theta_0 \\ \theta_1 \\ \theta_2 \\ \vdots \\ \theta_n \end{bmatrix} h_{\theta}(x)=\frac{1}{1 + e^{\theta^Tx}}$

Algorithms look identical like linear regression but have a different $h_{\theta}(x)$ definition.

## Optimalisation algotithms

Cost funtion $J(\theta)$. Want $min_{\theta}J(\theta)$.

Given $\theta$, we already have a code that can compute:

• $J(\theta)$
• $\frac{∂}{∂θ_j} J(θ)$ (for $j=0,1,...,n$)

Repeat {
$θ_j := θ_j - \alpha \frac{∂}{∂θ_j} J(θ)$
}

### Other (than Gradient descent) optimazation algorithms:

• BFGS
• L-BFGS

• No need to manually peek $\alpha$

• have inline search algorithm that automatically tries different values for the learning rate $\alpha$, and automatically picks a good learning rate $\alpha$
• Often faster than gradient descent

Example: $\theta = \begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix}$

$\begin{matrix} min \\ \theta \end{matrix} J(\theta) \implies \theta_1=5; \theta_2=5$

$J(\theta) = (\theta_1 - 5)^2 + (\theta_2 - 5)^2$ (cost function)

$\frac{∂}{∂θ_1}J(\theta) = 2(\theta_1-5)$

$\frac{∂}{∂θ_2}J(\theta) = 2(\theta_2-5)$

### Prove in Octave:

>> options = optimset('GradObj', 'On', 'MaxIter', 100);
>> initialTheta = zeros(2,1);
>> function [jVal, gradient] = costFunction(theta)

jVal = (theta(1)-5)^2 + (theta(2)-5)^2;
endfunction
>> [optTheta, functionValm exitFlag] = fminunc(@costFunction, initialTheta, options)
optTheta =

5.0000
5.0000

functionValm =    1.5777e-30
exitFlag =  1
>>

fminunc - Advance Optimalization function in Octave

# Multi-class classification: ONE-VS-ALL

### Multiclass classification

Examples:

• Email foldering/tagging: Work $(y=1)$, Friends $(y=2)$, Family $(y=3)$, Hobby $(y=4)$
• Medical diagrams: Not ill $(y=1)$, Cold $(y=2)$, Flu $(y=3)$
• Weather: Sunny $(y=1)$, Cloudy $(y=2)$, Rain $(y=3)$, Snow $(y=4)$

## One-vs-all (one-vs-rest)

Step 1: Createing a new training set to fit the classifier $h_{\theta}^{(1)}(x)$

Step 2: Createing a new training set to fit the classifier $h_{\theta}^{(2)}(x)$

Step 3: Createing a new training set to fit the classifier $h_{\theta}^{(3)}(x)$

We fit three classifiers what is a probablity of one of the three classes:

$h_{\theta}^{(i)}(x) = P(y=i|x;\theta)$

## One-vs-all

Train a logistoc regression classifier $h_{\theta}^{(i)}(x)$ for each class $i$ to predict the probability that $y=i$ On a new input $x$, to make a prediction, pick the class $i$ ythat maximaizes: $\begin{matrix} max\\ i \end{matrix} h_{\theta}^{(i)}(x)$