# Linear Regression with Gradient Descent

Solving a linear regression problem with Gradient Descent optimization algorithm to better understand learning

## 1. Notations and Definitions

Assume the training set below: (a tibble in R)

library(dplyr)
as_tibble(cars)
## # A tibble: 50 x 2
##    speed  dist
##    <dbl> <dbl>
##  1     4     2
##  2     4    10
##  3     7     4
##  4     7    22
##  5     8    16
##  6     9    10
##  7    10    18
##  8    10    26
##  9    10    34
## 10    11    17
## # … with 40 more rows

m : Number of training examples (or data points). In this example it is 50.
x : Input variable speed
y : Output variable dist
$$(x,y)$$ denotes one sample from our training set.
$$(x^i, y^i)$$ is the $$i^{th}$$ training example. In our example, $$x^1$$ = 4, $$y^1$$ = 2

The hypothesis $$h$$ maps $$x$$’s to $$y's$$.
$$y$$ will be the estimated value of dist.

### 1.1 How to represent $$h$$?

$h_\theta(x) = \theta_0+\theta_1 x$

Because there is one variable, it is also called as univariate linear regression or linear regression with one variable.

## 2. Cost Function

How to choose $$\theta_0$$ and $$\theta_1$$ for the best fit?

### 2.1 Idea

Chose $$\theta_0$$ and $$\theta_1$$ so that $$h_\theta(x)$$ is close to our training examples $$y$$.

Mathematically speaking, find $$\theta_0$$ and $$\theta_1$$ that minimizes: $J(\theta_0, \theta_1) = \frac{1}{2m}\sum\limits_1^m(h_\theta(x^i)-y^i)^2$

$$J(\theta_0, \theta_1)$$ is the cost function. It is also called Squared Error Function.

### 2.2 Application

An illustration with cars datasets in R:
Plot speed on x axis and dist (distance to stop) on y axis.

library(ggplot2)
ggplot(cars, aes(x = speed, y = dist))+
geom_point()

Add some models to represent the relationship between speed and dist.
For simplicity, set $$\theta_0 = 0$$.

library(ggplot2)

# simulate data
set.seed(1010)
sim1 <- data_frame(
x = rep(cars$speed, 9), y = rep(cars$dist, 9),
model_no = paste("Model", sort(rep(1:9, 50))),
theta1 = sort(rep(seq(from = 1, to = 4.5, length.out = 9), 50)),
theta0 = 0)
## Warning: data_frame() is deprecated as of tibble 1.1.0.
## Please use tibble() instead.
## This warning is displayed once every 8 hours.
## Call lifecycle::last_warnings() to see where this warning was generated.
# add predictions
sim1 <- sim1 %>% mutate(prediction = theta0 + x*theta1)

# visualize
sim1 %>% ggplot(aes(x = x, y = y))+
geom_point(size = 0.5)+
geom_abline(aes(slope = theta1, intercept = theta0), color = "red") +
geom_segment(aes(x = x, xend = x,
y = prediction, yend = y), alpha = 0.3) +
facet_grid(.~model_no)

Now visualize cost function:

# sum of squared errors
sse <- sim1 %>%
mutate(squared_error = (prediction-y)^2) %>%
group_by(model_no, theta1) %>%
summarize(m = n(),
cost = (1/(2*m))*sum(squared_error))
## summarise() regrouping output by 'model_no' (override with .groups argument)
sse %>% ggplot(aes(x = theta1, y = cost)) +
geom_point(color = "blue") +
geom_text(aes(label = model_no), angle = 0, nudge_y = 20, size = 3)

It looks like a parabola. The model 5 is looks better comparing to other alternatives.

How to find the best one? There might be other solutions but the objective of this example is understanding learning. So let’s go with Gradient Descent.

Outline:
1. Start with some $$\theta_0, \theta_1$$
2. Keep changing $$\theta_0, \theta_1$$ to reduce $$J(\theta_0, \theta_1)$$ until we hopefully find a minimum.

It is an itterative optimization algorithm. It is not only useful for linear regression but many machine learning problems.

repeat until convergence: $\theta_j := \theta_j-\alpha\frac{\partial}{\partial_j}J(\theta_0,\theta_1)$ for i = 0 and i = 1

• $$:=$$ is the assigment operator
• $$\alpha$$ is the learning rate.
• $$\theta_0$$ and $$\theta_1$$ should simultaneously be updated.

Visualization with ggplot2:

# cost function
cost_function <- function(x) (x-2)^2

# derivative function
library(Deriv)
derivative <- Deriv(cost_function)

df <- data_frame(x = 0,
y = cost_function(x),
new_x = x,
new_y = y)

# learning rate
learning_rate = 0.2

# create dataframe
for (i in 1:20) {
x = df$new_x[i] y = df$new_y[i]
step = derivative(x)*learning_rate
new_x = x-step
new_y = cost_function(new_x)
new_df = data_frame(x = x, y = y, new_x = new_x, new_y = new_y)
df <- bind_rows(df, new_df)
rm(x, y, new_x, step, new_y, new_df)
}

# plot
ggplot(df, aes(x, y)) +
geom_point() +
geom_segment(aes(x = x, xend = new_x,
y =y, yend = new_y), color = "blue", linetype = "dotted") +
stat_function(fun = cost_function, alpha = 0.5) +
xlim(c(-1, 5)) +
ggtitle(label = "Gradient Descent Visualization", subtitle = "Learning rate : 0.2")

In this example the learning rate was 0.2.

Another example with a higher learning rate: 0.9 :

## 3.3 Finding Parameters with Gradient Descent

Note that the derivative of the cost function:
$\frac{1}{m}\sum_{i}^{m}(h_\theta(x^{(i)}-y^{(i)})x^{(i)}$

Find the parameters:

x <- cars$speed y <- cars$dist
theta1 <- 6
alpha <- 0.001
m <- nrow(cars)
yhat <- theta1*x
df <- data_frame(theta1 = as.double(),
cost = NA_real_,
iteration = 1)
for (i in 1:20){
theta1 <- theta1 - alpha * ((1 / m) * (sum((yhat - y) * x)))
yhat <- theta1*x
cost <- (1/m)*sum((yhat-y)^2)
df[i, 1] = theta1
df[i, 2] <- cost
df[i, 3] <- i
}
theta1
## [1] 2.915755

Visualize the linear regression line:

cars %>% ggplot(aes(speed, dist))+
geom_point()+
geom_abline(slope = theta1, intercept = 0)

Check the cost function:

df %>% ggplot(aes(x = iteration, y = cost))+
geom_line()+
geom_point()