Ch3 slides

‘Introductory Econometrics for Finance’ © Chris Brooks 2013 1
Chapter 3
A brief overview of the
classical linear regression model

Regression
• Regression is probably the single most important tool at the
econometrician’s disposal.
But what is regression analysis?
• It is concerned with describing and evaluating the relationship between
a given variable (usually called the dependent variable) and one or
more other variables (usually known as the independent variable(s)).

Some Notation
• Denote the dependent variable by y and the independent variable(s) by x1, x2,
... , xk where there are k independent variables.
• Some alternative names for the y and x variables:
y x
dependent variable independent variables
regressand regressors
effect variable causal variables
explained variable explanatory variable
• Note that there can be many x variables but we will limit ourselves to the
case where there is only one x variable to start with. In our set-up, there is
only one y variable.

Regression is different from Correlation
• If we say y and x are correlated, it means that we are treating y and x in
a completely symmetrical way.
• In regression, we treat the dependent variable (y) and the independent
variable(s) (x’s) very differently. The y variable is assumed to be
random or “stochastic” in some way, i.e. to have a probability
distribution. The x variables are, however, assumed to have fixed
(“non-stochastic”) values in repeated samples.

Simple Regression
• For simplicity, say k=1. This is the situation where y depends on only one x
variable.
• Examples of the kind of relationship that may be of interest include:
– How asset returns vary with their level of market risk
– Measuring the long-term relationship between stock prices and
dividends.
– Constructing an optimal hedge ratio

Simple Regression: An Example
• Suppose that we have the following data on the excess returns on a fund
manager’s portfolio (“fund XXX”) together with the excess returns on a
market index:
• We have some intuition that the beta on this fund is positive, and we
therefore want to find whether there appears to be a relationship between
x and y given the data that we have. The first stage would be to form a
scatter plot of the two variables.
Year, t Excess return
= rXXX,t – rft
Excess return on market index
= rmt - rft
1 17.8 13.7
2 39.0 23.2
3 12.8 6.9
4 24.2 16.8
5 17.2 12.3

Graph (Scatter Diagram)
0
5
10
15
20
25
30
35
40
45
0 5 10 15 20 25
Excess return on market portfolio
Excess
return
on
fund
XXX

Finding a Line of Best Fit
• We can use the general equation for a straight line,
y=a+bx
to get the line that best “fits” the data.
• However, this equation (y=a+bx) is completely deterministic.
• Is this realistic? No. So what we do is to add a random disturbance
term, u into the equation.
yt =  + xt + ut
where t = 1,2,3,4,5

Why do we include a Disturbance term?
• The disturbance term can capture a number of features:
- We always leave out some determinants of yt
- There may be errors in the measurement of yt that cannot be
modelled.
- Random outside influences on yt which we cannot model

Determining the Regression Coefficients
• So how do we determine what  and  are?
• Choose  and  so that the (vertical) distances from the data points to the
fitted lines are minimised (so that the line fits the data as closely as
possible): y
x

Ordinary Least Squares
• The most common method used to fit a line to the data is known as
OLS (ordinary least squares).
• What we actually do is take each distance and square it (i.e. take the
area of each of the squares in the diagram) and minimise the total sum
of the squares (hence least squares).
• Tightening up the notation, let
yt denote the actual data point t
denote the fitted value from the regression line
denote the residual, yt - t
ŷ
t
ŷ
t
û

Actual and Fitted Value
y
i
x x
i
y
i
ŷ
i
û

How OLS Works
• So min. , or minimise . This is known
as the residual sum of squares.
• But what was ? It was the difference between the actual point and
the line, yt - .
• So minimising is equivalent to minimising
with respect to and .
$
 $

2
5
2
4
2
3
2
2
2
1
ˆ
ˆ
ˆ
ˆ
ˆ u
u
u
u
u 



t
ŷ
t
û


5
1
2
ˆ
t
t
u
 2
ˆ
  t
t y
y  2
ˆt
u

Deriving the OLS Estimator
• But , so let
• Want to minimise L with respect to (w.r.t.) and , so differentiate L
w.r.t. and
(1)
(2)
• From (1),
• But and .
$
 $

$
 $

t
t x
y 
 ˆ
ˆ
ˆ 

 




t
t
t x
y
L
0
)
ˆ
ˆ
(
2
ˆ





 




t
t
t
t x
y
x
L
0
)
ˆ
ˆ
(
2
ˆ





0
ˆ
ˆ
0
)
ˆ
ˆ
( 





  
 t
t
t
t
t x
T
y
x
y 



  y
T
yt   x
T
xt
  




t i
t
t
t
t x
y
y
y
L 2
2
)
ˆ
ˆ
(
)
ˆ
( 


Deriving the OLS Estimator (cont’d)
• So we can write or (3)
• From (2), (4)
• From (3), (5)
• Substitute into (4) for from (5),
$

0
ˆ
ˆ 

 x
y 

 


t
t
t
t x
y
x 0
)
ˆ
ˆ
( 

x
y 
 ˆ
ˆ 

 
 















t
t
t
t
t
t
t
t
t
t
t
t
t
t
x
x
T
x
y
T
y
x
x
x
x
x
y
y
x
x
x
y
y
x
0
ˆ
ˆ
0
ˆ
ˆ
0
)
ˆ
ˆ
(
2
2
2






0
ˆ
ˆ 

 x
T
T
y
T 


Deriving the OLS Estimator (cont’d)
• Rearranging for ,
• So overall we have
• This method of finding the optimum is known as ordinary least squares.
$


 

 t
t
t y
x
x
y
T
x
x
T )
(
ˆ 2
2

x
y
x
T
x
y
x
T
y
x
t
t
t


 ˆ
ˆ
and
ˆ
2
2








What do We Use and For?
• In the CAPM example used above, plugging the 5 observations in to make up
the formulae given above would lead to the estimates
= -1.74 and = 1.64. We would write the fitted line as:
• Question: If an analyst tells you that she expects the market to yield a return
20% higher than the risk-free rate next year, what would you expect the return
on fund XXX to be?
• Solution: We can say that the expected value of y = “-1.74 + 1.64 * value of x”,
so plug x = 20 into the equation to get the expected value for y:
$
 $

$
 $

06
.
31
20
64
.
1
74
.
1
ˆ 




i
y
t
t x
y 64
.
1
74
.
1
ˆ 



Accuracy of Intercept Estimate
• Care needs to be exercised when considering the intercept estimate,
particularly if there are no or few observations close to the y-axis:
y
0 x

The Population and the Sample
• The population is the total collection of all objects or people to be studied,
for example,
• Interested in Population of interest
predicting outcome the entire electorate
of an election
• A sample is a selection of just some items from the population.
• A random sample is a sample in which each individual item in the
population is equally likely to be drawn.

The DGP and the PRF
• The population regression function (PRF) is a description of the model that
is thought to be generating the actual data and the true relationship
between the variables (i.e. the true values of  and ).
• The PRF is
• The SRF is
and we also know that .
• We use the SRF to infer likely values of the PRF.
• We also want to know how “good” our estimates of  and  are.
t
t x
y 
 ˆ
ˆ
ˆ 

t
t
t u
x
y 

 

t
t
t y
y
u ˆ
ˆ 


Linearity
• In order to use OLS, we need a model which is linear in the parameters (
and  ). It does not necessarily have to be linear in the variables (y and x).
• Linear in the parameters means that the parameters are not multiplied
together, divided, squared or cubed etc.
• Some models can be transformed to linear ones by a suitable substitution
or manipulation, e.g. the exponential regression model
• Then let yt=ln Yt and xt=ln Xt
t
t
t u
x
y 

 

t
t
t
u
t
t u
X
Y
e
X
e
Y t




 ln
ln 




Linear and Non-linear Models
• This is known as the exponential regression model. Here, the coefficients
can be interpreted as elasticities.
• Similarly, if theory suggests that y and x should be inversely related:
then the regression can be estimated using OLS by substituting
• But some models are intrinsically non-linear, e.g.
t
t
t u
x
y 




t
t
x
z
1

t
t
t u
x
y 





Estimator or Estimate?
• Estimators are the formulae used to calculate the coefficients
• Estimates are the actual numerical values for the coefficients.

The Assumptions Underlying the
Classical Linear Regression Model (CLRM)
• The model which we have used is known as the classical linear regression model.
• We observe data for xt, but since yt also depends on ut, we must be specific about
how the ut are generated.
• We usually make the following set of assumptions about the ut’s (the
unobservable error terms):
• Technical Notation Interpretation
1. E(ut) = 0 The errors have zero mean
2. Var (ut) = 2 The variance of the errors is constant and finite
over all values of xt
3. Cov (ui,uj)=0 The errors are statistically independent of
one another
4. Cov (ut,xt)=0 No relationship between the error and
corresponding x variate

The Assumptions Underlying the
CLRM Again
• An alternative assumption to 4., which is slightly stronger, is that the
xt’s are non-stochastic or fixed in repeated samples.
• A fifth assumption is required if we want to make inferences about the
population parameters (the actual  and ) from the sample parameters
( and )
• Additional Assumption
5. ut is normally distributed
$
 $


Properties of the OLS Estimator
• If assumptions 1. through 4. hold, then the estimators and determined by
OLS are known as Best Linear Unbiased Estimators (BLUE).
What does the acronym stand for?
• “Estimator” - is an estimator of the true value of .
• “Linear” - is a linear estimator
• “Unbiased” - On average, the actual value of the and ’s will be equal to
the true values.
• “Best” - means that the OLS estimator has minimum variance among
the class of linear unbiased estimators. The Gauss-Markov
theorem proves that the OLS estimator is best.
$
 $

$

$

$

$

$


Consistency/Unbiasedness/Efficiency
• Consistent
The least squares estimators and are consistent. That is, the estimates will
converge to their true values as the sample size increases to infinity. Need the
assumptions E(xtut)=0 and Var(ut)=2 <  to prove this. Consistency implies that
• Unbiased
The least squares estimates of and are unbiased. That is E( )= and E( )=
Thus on average the estimated value will be equal to the true values. To prove
this also requires the assumption that E(ut)=0. Unbiasedness is a stronger
condition than consistency.
• Efficiency
An estimator of parameter  is said to be efficient if it is unbiased and no other
unbiased estimator has a smaller variance. If the estimator is efficient, we are
minimising the probability that it is a long way off from the true value of .
$
 $

$
 $

$
 $

$

  0
0
ˆ
Pr
lim 










T

Precision and Standard Errors
• Any set of regression estimates of and are specific to the sample used in
their estimation.
• Recall that the estimators of  and  from the sample parameters ( and ) are
given by
• What we need is some measure of the reliability or precision of the estimators
( and ). The precision of the estimate is given by its standard error. Given
assumptions 1 - 4 above, then the standard errors can be shown to be given by
where s is the estimated standard deviation of the residuals.
$
 $

$

$

$

$

x
y
x
T
x
y
x
T
y
x
t
t
t


 ˆ
ˆ
and
ˆ
2
2





















2
2
2
2
2
2
2
2
2
1
)
(
1
)
ˆ
(
,
)
(
)
ˆ
(
x
T
x
s
x
x
s
SE
x
T
x
T
x
s
x
x
T
x
s
SE
t
t
t
t
t
t



Estimating the Variance of the Disturbance Term
• The variance of the random variable ut is given by
Var(ut) = E[(ut)-E(ut)]2
which reduces to
Var(ut) = E(ut
2)
• We could estimate this using the average of :
• Unfortunately this is not workable since ut is not observable. We can use
the sample counterpart to ut, which is :
But this estimator is a biased estimator of 2.
2
t
u

 2
2 1
t
u
T
s

 2
2
ˆ
1
t
u
T
s
t
û

Estimating the Variance of the Disturbance Term
(cont’d)
• An unbiased estimator of  is given by
where is the residual sum of squares and T is the sample size.
Some Comments on the Standard Error Estimators
1. Both SE( ) and SE( ) depend on s2 (or s). The greater the variance s2, then
the more dispersed the errors are about their mean value and therefore the
more dispersed y will be about its mean value.
2. The sum of the squares of x about their mean appears in both formulae.
The larger the sum of squares, the smaller the coefficient variances.
$
 $

2
ˆ2



T
u
s t
 2
ˆt
u

Consider what happens if is small or large:
y
y
0
x
x
y
y
0
x
x
 2
  x
xt

(cont’d)
3. The larger the sample size, T, the smaller will be the coefficient
variances. T appears explicitly in SE( ) and implicitly in SE( ).
T appears implicitly since the sum is from t = 1 to T.
4. The term appears in the SE( ).
The reason is that measures how far the points are away from the
y-axis.
$
 $

$

 2
  x
xt
 2
t
x
 2
t
x

Example: How to Calculate the Parameters and
Standard Errors
• Assume we have the following data calculated from a regression of y on a
single variable x and a constant over 22 observations.
• Data:
• Calculations:
• We write
$ ( * . * . )
*( . )
.
 



830102 22 4165 86 65
3919654 22 4165
035
2
$ . . * . .
   
8665 035 4165 5912
6
.
130
,
3919654
,
65
.
86
,
5
.
416
,
22
,
830102
2








RSS
x
y
x
T
y
x
t
t
t
t
t x
y 
 ˆ
ˆ
ˆ 

t
t x
y 35
.
0
12
.
59
ˆ 


Example (cont’d)
• SE(regression),
• We now write the results as
   
  0079
.
0
5
.
416
22
3919654
1
*
55
.
2
)
(
35
.
3
5
.
416
22
3919654
22
3919654
*
55
.
2
)
(
2
2











SE
SE
)
0079
.
0
(
35
.
0
)
35
.
3
(
12
.
59
ˆ t
t x
y 


55
.
2
20
6
.
130
2
ˆ2





T
u
s t

An Introduction to Statistical Inference
• We want to make inferences about the likely population values from
the regression parameters.
Example: Suppose we have the following regression results:
• is a single (point) estimate of the unknown population
parameter, . How “reliable” is this estimate?
• The reliability of the point estimate is measured by the coefficient’s
standard error.
$ .
  05091
)
2561
.
0
(
5091
.
0
)
38
.
14
(
3
.
20
ˆ t
t x
y 


Hypothesis Testing: Some Concepts
• We can use the information in the sample to make inferences about the
population.
• We will always have two hypotheses that go together, the null hypothesis
(denoted H0) and the alternative hypothesis (denoted H1).
• The null hypothesis is the statement or the statistical hypothesis that is actually
being tested. The alternative hypothesis represents the remaining outcomes of
interest.
• For example, suppose given the regression results above, we are interested in
the hypothesis that the true value of  is in fact 0.5. We would use the notation
H0 :  = 0.5
H1 :   0.5
This would be known as a two sided test.

One-Sided Hypothesis Tests
• Sometimes we may have some prior information that, for example, we
would expect  > 0.5 rather than  < 0.5. In this case, we would do a
one-sided test:
H0 :  = 0.5
H1 :  > 0.5
or we could have had
H0 :  = 0.5
H1 :  < 0.5
• There are two ways to conduct a hypothesis test: via the test of
significance approach or via the confidence interval approach.

The Probability Distribution of the
Least Squares Estimators
• We assume that ut  N(0,2)
• Since the least squares estimators are linear combinations of the random
variables
i.e.
• The weighted sum of normal random variables is also normally distributed, so
 N(, Var())
 N(, Var())
• What if the errors are not normally distributed? Will the parameter estimates
still be normally distributed?
• Yes, if the other assumptions of the CLRM hold, and the sample size is
sufficiently large.
$
  w y
t t
$

$


The Probability Distribution of the
Least Squares Estimators (cont’d)
• Standard normal variates can be constructed from and :
and
• But var() and var() are unknown, so
and
$
 $

 
 
1
,
0
~
var
ˆ
N


 
 
 
1
,
0
~
var
ˆ
N


 
2
~
)
ˆ
(
ˆ


T
t
SE 


2
~
)
ˆ
(
ˆ


T
t
SE 



Testing Hypotheses:
The Test of Significance Approach
• Assume the regression equation is given by ,
for t=1,2,...,T
• The steps involved in doing a test of significance are:
1. Estimate , and , in the usual way
2. Calculate the test statistic. This is given by the formula
where is the value of  under the null hypothesis.
test statistic
SE


$ *
( $)
 

 *
SE( $)
 SE( $)

$
 $

t
t
t u
x
y 

 


The Test of Significance Approach (cont’d)
3. We need some tabulated distribution with which to compare the estimated
test statistics. Test statistics derived in this way can be shown to follow a t-
distribution with T-2 degrees of freedom.
As the number of degrees of freedom increases, we need to be less cautious in
our approach since we can be more sure that our results are robust.
4. We need to choose a “significance level”, often denoted . This is also
sometimes called the size of the test and it determines the region where we
will reject or not reject the null hypothesis that we are testing. It is
conventional to use a significance level of 5%.
Intuitive explanation is that we would only expect a result as extreme as this
or more extreme 5% of the time as a consequence of chance alone.
Conventional to use a 5% size of test, but 10% and 1% are also commonly
used.

Determining the Rejection Region for a Test of
Significance
5. Given a significance level, we can determine a rejection region and non-
rejection region. For a 2-sided test:
f(x)
95% non-rejection
region
2.5%
rejection region
2.5%
rejection region

The Rejection Region for a 1-Sided Test (Upper Tail)
f(x)
95% non-rejection
region 5% rejection region

The Rejection Region for a 1-Sided Test (Lower Tail)
f(x)
95% non-rejection region
5% rejection region

The Test of Significance Approach: Drawing
Conclusions
6. Use the t-tables to obtain a critical value or values with which to
compare the test statistic.
7. Finally perform the test. If the test statistic lies in the rejection
region then reject the null hypothesis (H0), else do not reject H0.

A Note on the t and the Normal Distribution
• You should all be familiar with the normal distribution and its
characteristic “bell” shape.
• We can scale a normal variate to have zero mean and unit variance by
subtracting its mean and dividing by its standard deviation.
• There is, however, a specific relationship between the t- and the
standard normal distribution. Both are symmetrical and centred on
zero. The t-distribution has another parameter, its degrees of freedom.
We will always know this (for the time being from the number of
observations -2).

What Does the t-Distribution Look Like?
normal distribution
t-distribution

Comparing the t and the Normal Distribution
• In the limit, a t-distribution with an infinite number of degrees of freedom is
a standard normal, i.e.
• Examples from statistical tables:
Significance level N(0,1) t(40) t(4)
50% 0 0 0
5% 1.64 1.68 2.13
2.5% 1.96 2.02 2.78
0.5% 2.57 2.70 4.60
• The reason for using the t-distribution rather than the standard normal is that
we had to estimate , the variance of the disturbances.
t N
( ) ( , )
  01
2

The Confidence Interval Approach
to Hypothesis Testing
• An example of its usage: We estimate a parameter, say to be 0.93, and
a “95% confidence interval” to be (0.77,1.09). This means that we are
95% confident that the interval containing the true (but unknown)
value of .
• Confidence intervals are almost invariably two-sided, although in
theory a one-sided interval can be constructed.

How to Carry out a Hypothesis Test
Using Confidence Intervals
1. Calculate , and , as before.
2. Choose a significance level, , (again the convention is 5%). This is equivalent to
choosing a (1-)100% confidence interval, i.e. 5% significance level = 95%
confidence interval
3. Use the t-tables to find the appropriate critical value, which will again have T-2
degrees of freedom.
4. The confidence interval is given by
5. Perform the test: If the hypothesised value of  (*) lies outside the confidence
interval, then reject the null hypothesis that  = *, otherwise do not reject the null.
$
 $
 SE( $)
 SE( $)

))
ˆ
(
ˆ
),
ˆ
(
ˆ
( 


 SE
t
SE
t crit
crit 




Confidence Intervals Versus Tests of Significance
• Note that the Test of Significance and Confidence Interval approaches
always give the same answer.
• Under the test of significance approach, we would not reject H0 that  = *
if the test statistic lies within the non-rejection region, i.e. if
• Rearranging, we would not reject if
• But this is just the rule under the confidence interval approach.
 £

£ 
t
SE
t
crit crit
$ *
( $)
 

)
ˆ
(
*
ˆ
)
ˆ
( 


 SE
t
SE
t crit
crit 

£

£


)
ˆ
(
ˆ
*
)
ˆ
(
ˆ 



 SE
t
SE
t crit
crit 

£
£



Constructing Tests of Significance and
Confidence Intervals: An Example
• Using the regression results above,
, T=22
• Using both the test of significance and confidence interval approaches,
test the hypothesis that  =1 against a two-sided alternative.
• The first step is to obtain the critical value. We want tcrit = t20;5%
)
2561
.
0
(
5091
.
0
)
38
.
14
(
3
.
20
ˆ t
t x
y 


Determining the Rejection Region
-2.086 +2.086
2.5% rejection region
2.5% rejection region
f(x)

Performing the Test
• The hypotheses are:
H0 :  = 1
H1 :   1
Test of significance Confidence interval
approach approach
Do not reject H0 since Since 1 lies within the
test stat lies within confidence interval,
non-rejection region do not reject H0
test stat
SE




 
$ *
( $)
.
.
.
 

05091 1
02561
1917
)
0433
.
1
,
0251
.
0
(
2561
.
0
086
.
2
5091
.
0
)
ˆ
(
ˆ






 
 SE
tcrit

Testing other Hypotheses
• What if we wanted to test H0 :  = 0 or H0 :  = 2?
• Note that we can test these with the confidence interval approach.
For interest (!), test
H0 :  = 0
vs. H1 :   0
H0 :  = 2
vs. H1 :   2

Changing the Size of the Test
• But note that we looked at only a 5% size of test. In marginal cases
(e.g. H0 :  = 1), we may get a completely different answer if we use a
different size of test. This is where the test of significance approach is
better than a confidence interval.
• For example, say we wanted to use a 10% size of test. Using the test of
significance approach,
as above. The only thing that changes is the critical t-value.
test stat
SE




 
$ *
( $)
.
.
.
 

05091 1
02561
1917

Changing the Size of the Test:
The New Rejection Regions
-1.725 +1.725
5% rejection region
5% rejection region
f(x)

Changing the Size of the Test:
The Conclusion
• t20;10% = 1.725. So now, as the test statistic lies in the rejection region,
we would reject H0.
• Caution should therefore be used when placing emphasis on or making
decisions in marginal cases (i.e. in cases where we only just reject or
not reject).

Some More Terminology
• If we reject the null hypothesis at the 5% level, we say that the result
of the test is statistically significant.
• Note that a statistically significant result may be of no practical
significance. E.g. if a shipment of cans of beans is expected to weigh
450g per tin, but the actual mean weight of some tins is 449g, the
result may be highly statistically significant but presumably nobody
would care about 1g of beans.

The Errors That We Can Make
Using Hypothesis Tests
• We usually reject H0 if the test statistic is statistically significant at a
chosen significance level.
• There are two possible errors we could make:
1. Rejecting H0 when it was really true. This is called a type I error.
2. Not rejecting H0 when it was in fact false. This is called a type II error.
Reality
H0 is true H0 is false
Result of
Significant
(reject H0)
Type I error
= 

Test Insignificant
( do not
reject H0)

Type II error
= 

The Trade-off Between Type I and Type II Errors
• The probability of a type I error is just , the significance level or size of test we
chose. To see this, recall what we said significance at the 5% level meant: it is only
5% likely that a result as or more extreme as this could have occurred purely by
chance.
• Note that there is no chance for a free lunch here! What happens if we reduce the size
of the test (e.g. from a 5% test to a 1% test)? We reduce the chances of making a type
I error ... but we also reduce the probability that we will reject the null hypothesis at
all, so we increase the probability of a type II error:
• So there is always a trade off between type I and type II errors when choosing a
significance level. The only way we can reduce the chances of both is to increase the
sample size.
less likely
to falsely reject
Reduce size  more strict  reject null
of test criterion for hypothesis more likely to
rejection less often incorrectly not
reject

A Special Type of Hypothesis Test: The t-ratio
• Recall that the formula for a test of significance approach to hypothesis
testing using a t-test was
• If the test is H0 : i = 0
H1 : i  0
i.e. a test that the population coefficient is zero against a two-sided
alternative, this is known as a t-ratio test:
Since  i* = 0,
• The ratio of the coefficient to its SE is known as the t-ratio or t-statistic.
 
test statistic
SE
i i
i


$
$
*
 

test stat
SE
i
i

$
( $ )



The t-ratio: An Example
• Suppose that we have the following parameter estimates, standard errors
and t-ratios for an intercept and slope respectively.
Coefficient 1.10 -4.40
SE 1.35 0.96
t-ratio 0.81 -4.63
Compare this with a tcrit with 15-3 = 12 d.f.
(2½% in each tail for a 5% test) = 2.179 5%
= 3.055 1%
• Do we reject H0: 1 = 0? (No)
H0: 2 = 0? (Yes)

What Does the t-ratio tell us?
• If we reject H0, we say that the result is significant. If the coefficient is not
“significant” (e.g. the intercept coefficient in the last regression above), then
it means that the variable is not helping to explain variations in y. Variables
that are not significant are usually removed from the regression model.
• In practice there are good statistical reasons for always having a constant
even if it is not significant. Look at what happens if no intercept is included:
t
y
t
x

An Example of the Use of a Simple t-test to Test a
Theory in Finance
• Testing for the presence and significance of abnormal returns (“Jensen’s
alpha” - Jensen, 1968).
• The Data: Annual Returns on the portfolios of 115 mutual funds from
1945-1964.
• The model: for j = 1, …, 115
• We are interested in the significance of j.
• The null hypothesis is H0: j = 0 .
jt
ft
mt
j
j
ft
jt u
R
R
R
R 



 )
(



Frequency Distribution of t-ratios of Mutual Fund
Alphas (gross of transactions costs)
Source Jensen (1968). Reprinted with the permission of Blackwell publishers.

Frequency Distribution of t-ratios of Mutual Fund
Alphas (net of transactions costs)
Source Jensen (1968). Reprinted with the permission of Blackwell publishers.

Can UK Unit Trust Managers “Beat the Market”?
• We now perform a variant on Jensen’s test in the context of the UK market,
considering monthly returns on 76 equity unit trusts. The data cover the
period January 1979 – May 2000 (257 observations for each fund). Some
summary statistics for the funds are:
Mean Minimum Maximum Median
Average monthly return, 1979-2000 1.0% 0.6% 1.4% 1.0%
Standard deviation of returns over time 5.1% 4.3% 6.9% 5.0%
• Jensen Regression Results for UK Unit Trust Returns, January 1979-May
2000
R R R R
jt ft j j mt ft jt
    
  
( )

Can UK Unit Trust Managers “Beat the Market”?
: Results
Estimates of Mean Minimum Maximum Median
 -0.02% -0.54% 0.33% -0.03%
 0.91 0.56 1.09 0.91
t-ratio on  -0.07 -2.44 3.11 -0.25
• In fact, gross of transactions costs, 9 funds of the sample of 76 were
able to significantly out-perform the market by providing a significant
positive alpha, while 7 funds yielded significant negative alphas.

The Overreaction Hypothesis and
the UK Stock Market
• Motivation
Two studies by DeBondt and Thaler (1985, 1987) showed that stocks which
experience a poor performance over a 3 to 5 year period tend to outperform
stocks which had previously performed relatively well.
• How Can This be Explained?
2 suggestions
1. A manifestation of the size effect
DeBondt & Thaler did not believe this a sufficient explanation, but Zarowin
(1990) found that allowing for firm size did reduce the subsequent return on
the losers.
2. Reversals reflect changes in equilibrium required returns
Ball & Kothari (1989) find the CAPM beta of losers to be considerably
higher than that of winners.

The Overreaction Hypothesis and
the UK Stock Market (cont’d)
• Another interesting anomaly: the January effect.
– Another possible reason for the superior subsequent performance
of losers.
– Zarowin (1990) finds that 80% of the extra return available from
holding the losers accrues to investors in January.
• Example study: Clare and Thomas (1995)
Data:
Monthly UK stock returns from January 1955 to 1990 on all firms
traded on the London Stock exchange.

Methodology
• Calculate the monthly excess return of the stock over the market over a 12,
24 or 36 month period for each stock i:
Uit = Rit - Rmt n = 12, 24 or 36 months
• Calculate the average monthly return for the stock i over the first 12, 24, or
36 month period:
R
n
U
i it
t
n



1
1

Portfolio Formation
• Then rank the stocks from highest average return to lowest and from 5
portfolios:
Portfolio 1: Best performing 20% of firms
Portfolio 2: Next 20%
Portfolio 5: Worst performing 20% of firms.
• Use the same sample length n to monitor the performance of each
portfolio.

Portfolio Formation and
Portfolio Tracking Periods
• How many samples of length n have we got?
n = 1, 2, or 3 years.
• If n = 1year:
Estimate for year 1
Monitor portfolios for year 2
Estimate for year 3
...
Monitor portfolios for year 36
• So if n = 1, we have 18 INDEPENDENT (non-overlapping) observation /
tracking periods.

Constructing Winner and Loser Returns
• Similarly, n = 2 gives 9 independent periods and n = 3 gives 6 independent
periods.
• Calculate monthly portfolio returns assuming an equal weighting of stocks in
each portfolio.
• Denote the mean return for each month over the 18, 9 or 6 periods for the
winner and loser portfolios respectively as and respectively.
• Define the difference between these as = - .
• Then perform the regression
= 1 + t (Test 1)
• Look at the significance of 1.
Rp
W
Rp
L
RDt
Rp
L
Rp
W
RDt

Allowing for Differences in the Riskiness
of the Winner and Loser Portfolios
• Problem: Significant and positive 1 could be due to higher return being
required on loser stocks due to loser stocks being more risky.
• Solution: Allow for risk differences by regressing against the market risk
premium:
= 2 + (Rmt-Rft) + t (Test 2)
where
Rmt is the return on the FTA All-share
Rft is the return on a UK government 3 month t-bill.
RDt

Is there an Overreaction Effect in the
UK Stock Market? Results
Panel A: All Months
n = 12 n = 24 n =36
Return on Loser 0.0033 0.0011 0.0129
Return on Winner 0.0036 -0.0003 0.0115
Implied annualised return difference -0.37% 1.68% 1.56%
Coefficient for (3.47): 1
̂ -0.00031
(0.29)
0.0014**
(2.01)
0.0013
(1.55)
Coefficients for (3.48): 2
̂ -0.00034
(-0.30)
0.00147**
(2.01)
0.0013*
(1.41)
̂ -0.022
(-0.25)
0.010
(0.21)
-0.0025
(-0.06)
Panel B: All Months Except January
Coefficient for (3.47): 1
̂ -0.0007
(-0.72)
0.0012*
(1.63)
0.0009
(1.05)
Notes: t-ratios in parentheses; * and ** denote significance at the 10% and 5% levels
respectively. Source: Clare and Thomas (1995). Reprinted with the permission of Blackwell
Publishers.

Testing for Seasonal Effects in Overreactions
• Is there evidence that losers out-perform winners more at one time of the
year than another?
• To test this, calculate the difference between the winner & loser portfolios
as previously, , and regress this on 12 month-of-the-year dummies:
• Significant out-performance of losers over winners in,
– June (for the 24-month horizon), and
– January, April and October (for the 36-month horizon)
– winners appear to stay significantly as winners in
• March (for the 12-month horizon).
R M
Dt i i t
i
 

 
1
12
RDt

Conclusions
• Evidence of overreactions in stock returns.
• Losers tend to be small so we can attribute most of the overreaction in the
UK to the size effect.
Comments
• Small samples
• No diagnostic checks of model adequacy

The Exact Significance Level or p-value
• This is equivalent to choosing an infinite number of critical t-values from
tables. It gives us the marginal significance level where we would be
indifferent between rejecting and not rejecting the null hypothesis.
• If the test statistic is large in absolute value, the p-value will be small, and
vice versa. The p-value gives the plausibility of the null hypothesis.
e.g. a test statistic is distributed as a t62 = 1.47.
The p-value = 0.12.
• Do we reject at the 5% level?...........................No
• Do we reject at the 10% level?.........................No
• Do we reject at the 20% level?.........................Yes

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