Things to do in Lecture 1

• Outline basic concepts of causality• Go through the ideas & principles

underlying ordinary least squares (OLS) estimation

• Derive formally how to estimate the value of the coefficients we are interested using OLS

• Run an OLS regression

• Regression analysis is primarily concerned with quantifying the relationship between variables

• Does more than measures of association like the correlation coefficient since it implies that one variable depends on another

• hence there is a causal relationship between the variable whose behaviour we would like to explain - the dependent variable (which label y)and the explanatory (or independent) variable

that we think can explain this behaviour (which label X)

DERIVING OLS REGRESSION COEFFICIENTS

- Given data on both the dependent and explanatory variable we can establish that causal relationship using regression analysis

- How?- fit a straight line through the data that best

summarises the relationship- Since the equation of a straight line is given by

Y= b0 + b1X

we can obtain an estimate of the values for the intercept b0 (the constant)

and the slope b1

that together give the “line of best fit”How ?

#

#

#

#

#

#

Y

X

Which line (and hence which slope & intercept) to choose?

- The one that minimises the sum of squared residuals

do

b0

b1

d1

Suppose we plot a set of observations on the Y variable and the X variable of interest - a scatter diagram

Using the principle of Ordinary Least Squares (OLS)

- Try to fit a (straight) line through the data based on “minimising the sum of squared residuals”

What is a residual?

If we fit a line through some data then this will give a predicted value for the dependent variable based on the value of the X variable and the values for the constant and the slope

#

#

#

#

#

#

y

X

b0

Xi

yi

iy^

b1

Given any straight line can always read off actual and predicted values of variable of interest for every individual i in the data set

iXbbiy 1^

0^^

↓

For example given the equation of a straight line

Y = b0 + b1X

we know

when X = 1, then the predicted value of Y

)1(1^

0^^

bby

and when X = 2, then the predicted value of Y

)2(1^

0^^

bby

We can then compare this predicted value with the actual value of the dependent variable and the difference between the actual and predicted value gives the residual

iXbbiy 1^

0^^

Which since

gives

iyiyiu^^

iXiyiyiyiu 1^

0^^^

#

#

#

#

#

#

y

X

b0

Xi

yi

The difference between the actual and predicted value is the residual

iu^

b1

iy^

predicted

actual

We can then compare this predicted value with the actual value of the dependent variable and the difference between the actual and predicted value gives the residual

iXbbyiyiyiu 1^

0^^^

(where the i subscript refers to the ith individual or firm or time period in the data set)

Things to know about residuals

So intuitively then the line of best fit should be the one that delivers the smallest residual values for the each observation in the data set

1. The larger the residual the worse the prediction

iXbbyiyiyiu 1^

0^^^

2. Since the difference between the actual and predicted value gives the residual

0^^

iyiyiu

a positive residual means

The model underpredicts

and similarly The model over-predicts

(larger than actual)

iXbbyiyiyiu 1^

0^^^

0^^

iyiyiu

Given this..Suppose we tried to minimise the sum of all the

residuals in the data in an attempt to get the line of best fit

Whilst this might seem intuitive it will not work because it is possible that any positive residual will be offset by a negative residual in the summation and so the sum could be close to zero even if the overall fit of the regression were poor

N

iiu

1

^

We can avoid this problem is use instead the principle of Ordinary Least Squares (OLS)

Rather than minimise the sum of residuals, minimise the sum of squared residuals

- Squaring ensures that values are always positive and so can never cancel each other out

-Also gives more “weight” to larger residuals and so harder to get away with a poor fit, (the larger any one residual in absolute value, the larger is the sum of squared residuals (RSS)

N

iiu

1

2^

Things to do in Lecture 2

• Derive formally how to estimate the value of the coefficients we are interested using OLS

• Run an OLS regression• Interpret the regression output• Measure how well the model fits

the data

#

#

#

#

#

#

y

X

Which line (and hence which slope & intercept) to choose?

- The one that minimises the sum of squared residuals

do

b0

b1

d1

The Idea Behind Ordinary Least Squares (OLS)

#

#

#

#

#

#

y

X

b0

Xi

yi

The difference between the actual and predicted value is the residual

iu^

b1

iy^

predicted

actual

Consider the following simple example

N=2 and want to fit a straight line y=b0 +b1Xthru’ the following data points using principle of OLS (min sum of squared residuals)(Y1=3 X1 =1) (Y2=5 X2=2)

1

^11

^yyu

It follows that can write estimated residual for the 1st observation

and similarly for the 2nd observation )2(1

^0

^52

^22

^bbyyu

)1(1^

0^

31^

bbu and y1=3

)1(1^

0^

1^

bby )1(1^

0^

1^

Xbby using

as

OLS: minimise the sum of squared residuals

2)1^

20^

5(2)1^

0^

3( bbbbS

)42010425(

)21669(

1^

0^

1^

0^

21

^2

0^

1^

0^^

0^

21

^2

0^

bbbbbb

bbbbbbS

)2561261652( 1^

0^^

0^

21

^2

0^

bbbbbbS (A)

Expanding the terms in brackets

Adding together like terms

2

2^2

1^

uuS

Now need to find values of and 1^b

which minimise this sum,

Using the rules of calculus we know the first order condition for minimisation are:

0

0^

bd

dS0

1^

bd

dS

0^b

→

01664 1^

0

^ bb 026106 1

^

0

^ bb

)2561261652( 1^

0^^

0^

21

^2

0^

bbbbbbS

This gives 2 simultaneous equations

832 1^

0

^ bb

1353 1^

0

^ bb

which can solve for unknown values of 0^b and 1

^b

using rules for simultaneous equations

10^

b 21^

b

So the estimated regression line becomes XY 21^

ie the intercept (constant) with the y axis is at 1 and the slope of the straight line is 2

Basic idea underlying OLS is to choose a “line of best fit”

-Choose a straight line that passes through the data and minimises the sum of squared residuals

Now need to do this more generally so can apply the technique to any possible combination of (x, y) data pairs and any number of observations

If we wish to fit a (straight) line through N (rather than 2) observations, then the OLS principle is still the same ie choose

and to minimise

N

iii

N

iiN yyuuuuS

1

2^

1

2^2^2

2^2

1^

)(.....

0^b

1^b

(where now the summation runs from 1 to N rather than 1 to 2 )

sub. in ii Xbby^

1

^

0

^

iiiiii

nnnnnn

Nn

XbbYXbYbXbbNY

XbbYXbYbXbbY

XbbYXbYbXbbY

XbbYXbbYS

^

1

^

0

^

1

^

02

2^

1

2^

02

^

1

^

0

^

1

^

02

2^

1

2^

02

1

^

1

^

011

^

11

^

021

2^

1

2^

02

1

2^

1

^

02

1

^

1

^

01

222

222

...

222

)(...)(

This is just a generalised version of (A) above

Again, find values of 0^b and 1

^b

which minimise this sum, using the same simple calculus rules

0

0^

bd

dSand 0

1^

bd

dS

01^

220^

20

0^

iXbiYbN

b

S

00^

2221

^20

1^

iXbiYiXiXb

b

S

Now these two (1st order) minimisation conditions give

and again we have 2 simultaneous equations (called the normal equations) which can again solve for

0^b and 1

^b

(1)

(2)

Using the fact that the sample means of Y and X

N

iiyYNN

yY

N

ii

1

__1

N

iixXNN

xX

N

ii

1

__1

can re-write (1)

0_

1^

2_

20^

2 XNbYNbN

and so obtain the formula to calculate the OLS estimate of the intercept

_1

^_0

^XbYb

(3)(*** learn this **)

01^

220^

2 iXbiYbN

Sub.

0)1^

(21

^ iXXbYiYiXiXb

00^

2221

^2 iXbiYiXiXb

_1

^_0

^XbYb into (2)

gives

0)1^

(21

^ XNXbYiYiXiXb

and simplifying

YXNiYiXXNiXb

22

^

1

collecting terms

Dividing both sides by 1/N

YXiYiXNXiXN

b

1221^

1

which gives the formula to calculate the OLS estimate of the slope

),(Cov)(Var1^

YXXb

)(Var),(Cov

1^

XYX

b

))((12)(

1^

1YiYXiXN

XiXNb

(**** learn this ****)

)(Var),(Cov

1^

XYX

b

The OLS equations give a nice, clear intuitive meaning about the influence of the variable X on the size of the slope, since it shows that:

So_

1^_

0^

XbYb

are how the computer determines the size of the intercept and the slope respectively in an OLS regression

i) the greater the covariance between X and Y, the larger the (absolute value of) the slope

ii) the smaller the variance of X, the larger the (absolute value of) the slope

It is equally important to be able to interpret the effect of an estimated regression coefficient

Given OLS essentially passes a straight line through the data, then given

1b

Xd

dY

So the OLS estimate of the slope will give an estimate of the unit change in the dependent variable y following a unit change in the level of the explanatory variable

(so you need to be aware of the units of measurement of your variables in order to be able to interpret what the OLS coefficient is telling you)

Xbby 1^

0^^