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CS229 Lecture notes
Andrew Ng
Supervised learning
Lets start by talking about a few examples of supervised learning problems.
Suppose we have a dataset giving the living areas and prices of 47 houses
from Portland, Oregon:
Living area (feet
2
) Price (1000$s)
2104 400
1600
330
2400
369
1416
232
3000
540
.
.
.
.
.
.
We can plot this data:
500 1000 1500 2000 2500 3000 3500 4000 4500 5000
0
100
200
300
400
500
600
700
800
900
1000
housing prices
square feet
price (in $1000)
Given data like this, how can we learn to predict the prices of other houses
in Portland, as a function of the size of their living areas?
1
CS229 Winter 2003 2
To establish notation for future use, we’ll use x
(i)
to denote the “inp ut”
variables (living area in this example), also called input features, and y
(i)
to denote the “output” or target variable that we are trying to predict
(price). A pair (x
(i)
, y
(i)
) is called a training example, and the dataset
that we’ll be using to learn—a list of m training examples {(x
(i)
, y
(i)
); i =
1, . . . , m}—is called a training set. Note that the superscript “(i)” in the
notation is simply an index into the training set, and has nothing to do with
exponentiation. We will also use X denote the space of input values, and Y
the space of output values. In this example, X = Y = R.
To describe the supervised learning problem slightly more formally, our
goal is, given a training set, to learn a function h : X 7→ Y so that h(x) is a
“good” predictor for the corresponding value of y. For historical reasons, th is
function h is called a hypothesis. Seen pictorially, the process is therefore
like this:
Training
set
house.)
(living area of
Learning
algorithm
h
predicted yx
(predicted price)
of house)
When the target variable that we’re trying to predict is continuous, such
as in our housing example, we call the learning problem a regression prob-
lem. When y can take on only a small number of d iscrete values (such as
if, given the living area, we wanted to predict if a dwelling is a house or an
apartment, say), we call it a classification problem.
3
Part I
Linear Regression
To make our housing example more interesting, lets consider a slightly richer
dataset in which we also know the number of bedrooms in each house:
Living area (feet
2
)
#bedrooms Price (1000$s)
2104 3 400
1600
3 330
2400
3 369
1416
2 232
3000
4 540
.
.
.
.
.
.
.
.
.
Here, the x’s are two-dimensional vectors in R
2
. For instance, x
(i)
1
is the
living area of the i-th house in the training set, and x
(i)
2
is its number of
bedrooms. (In general, when designing a learning problem, it will be up to
you to d ecide what features to choose, so if you are out in Portland gathering
housing data, you might also decide to include other features such as whether
each house has a fireplace, the number of bathrooms, and so on. We’ll say
more about feature selection later, but for now lets take the features as given.)
To perform super vised learning, we must decide how we’re going to rep-
resent functions/hypotheses h in a computer. As an initial choice, lets say
we decide to approximate y as a linear function of x:
h
θ
(x) = θ
0
+ θ
1
x
1
+ θ
2
x
2
Here, the θ
i
’s are the parameters (also called weights) parameterizing the
space of linear functions mapping from X to Y. When there is no risk of
confusion, we will drop the θ subscript in h
θ
(x), and write it more simply as
h(x). To simplify our notation, we also introduce the convention of letting
x
0
= 1 (this is the intercept term), so that
h(x) =
n
X
i=0
θ
i
x
i
= θ
T
x,
where on the right-hand side above we are viewing θ and x both as vectors,
and here n is the number of input variables (not counting x
0
).
Now, given a training set, how do we pick, or learn, the parameters θ?
One reasonable method seems to be to make h(x) close to y, at least for
4
the training examples we have. To formalize this, we will define a function
that measures, for each value of the θ’s, how close the h(x
(i)
)’s are to the
corresponding y
(i)
’s. We define the cost function:
J(θ) =
1
2
m
X
i=1
(h
θ
(x
(i)
) − y
(i)
)
2
.
If you’ve seen linear regression before, you may recognize this as the familiar
least-squares cost function that gives rise to the ordinary least squares
regression model. Whether or not you have seen it previously, lets keep
going, and we’ll eventually show this to be a special case of a much broader
family of algorithms.
1 LMS algorithm
We want to choose θ so as to minimize J(θ). To do so, lets use a search
algorithm that starts with some “initial guess” for θ, and that repeatedly
changes θ to make J(θ) smaller, until hopefully we converge to a value of
θ that minimizes J(θ). Specifically, lets consider the gradient descent
algorithm, which starts with some initial θ, and repeatedly performs the
update:
θ
j
:= θ
j
− α
∂
∂θ
j
J(θ).
(This update is simultaneously performed for all values of j = 0, . . . , n.)
Here, α is called the learning rate. This is a very natural algorithm that
repeatedly takes a step in the direction of steepest decrease of J.
In order to implement this algorithm, we have to work out what is the
partial derivative term on the right hand side. Lets first work it out for the
case of if we have only one training example (x, y), so that we can neglect
the sum in the definition of J. We have:
∂
∂θ
j
J(θ) =
∂
∂θ
j
1
2
(h
θ
(x) − y)
2
= 2 ·
1
2
(h
θ
(x) − y) ·
∂
∂θ
j
(h
θ
(x) − y)
= (h
θ
(x) − y) ·
∂
∂θ
j
n
X
i=0
θ
i
x
i
− y
!
= (h
θ
(x) − y) x
j
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