in this chapter we will cover

Slide 6.11Linear HypothesesLinear Hypotheses

MathematicalMathematicalMarketingMarketing

In This Chapter We Will Cover

Deductions we can make about even though it is not observed. These include

Confidence Intervals

Hypotheses of the form H0: i = c

Hypotheses of the form H0: i c

Hypotheses of the form H0: a′ = c

Hypotheses of the form A = c

We also cover deductions when V(e) 2I (Generalized Least Squares)



The Variance of the Estimator

.)(ˆ 1 yXXXβ

V(y) = V(X + e) = V(e) = 2I

12

112

121

)(

)()(

])[(])[()ˆ(V

XX

XXIXXXX

XXXIXXXβ

From these two raw ingredients and a theorem:

we conclude



What of the Distribution of the Estimator?

As

1nn1 ab

n

normal

Central Limit Property of Linear Combinations



So What Can We Conclude About the Estimator?

])(,[N~)(ˆ 121 XXβyXXXβ

From the Central Limit Theorem

From the V(linear combo) +assumptions about e

From Ch 5- E(linear combo)



Steps Towards Inference About

df~)q(V

)q(Eqt

kn

i

ii ~)ˆ(V

ˆ

t

In general

In particular

(X′X)-1X′y But note the hat on the V!



Lets Think About the Denominator

thatsokn

e

kn

SSsˆ

n

i

2

iError22

ii2

i ds)ˆ(V

where dii are diagonal elements of

D = (XX)-1 = {dij}

ii2

i d)ˆ(V



Putting It All Together

knii2

ii ~ds

ˆ

t

Now that we have a t, we can use it for two types of inference about :

Confidence Intervals

Hypothesis Testing



A Confidence Interval for i

A 1 - confidence interval for i is given by

ii2

kn,2/i dstˆ

1dstˆdstˆPr ii2

kn,2/ii

ii2

kn,2/i

which simply means that



Graphic of Confidence Interval

)ˆPr( i

ii2

kn,2/i dstˆ

i

1.0

0

ii2

kn,2/i dstˆ

1 -



Statistical Hypothesis Testing: Step One

H0: i = c

HA: i ≠ c

Generate two mutually exclusive hypotheses:



Statistical Hypothesis Testing Step Two

Summarize the evidence with respect to H0:

ii2

i

i

ii

ds

cˆ

)ˆ(V

ˆˆ

t



Statistical Hypothesis Testing Step Three

,t| ˆ| k-n /2,t

reject H0 if the probability of the evidence given H0 is small



One Tailed Hypotheses

Our theories should give us a sign for Step One in which case we might have

k-n ,tt

H0: i c

HA: i < c

In that case we reject H0 if



A More General Formulation

Consider a hypothesis of the form

H0: a´ = c

so if c = 0…

00110a

00110a

012

1

2

10a

tests H0: 1= 2

tests H0: 1 + 2 = 0

tests H0: 3

21

2



A t test for This More Complex Hypothesis

aXXa

aβaβa

12 )(

)ˆ(V)ˆ(V

.)(s

cˆˆ

12 aXXa

βa

t

We need to derive the denominator of the t using the variance of a linear combination

which leads to



Multiple Degree of Freedom Hypotheses

q

2

1

.q

.2

.1

0

1q0

c

c

c

:H

:H

β

a

a

a

cAβ



Examples of Multiple df Hypotheses

0

0

1000

0100:H

3

2

1

0

0

0

0

1010

0110:H

3

2

1

0

0

tests H0: 2 = 3 = 0

tests H0: 1 = 2 = 3



Testing Multiple df Hypotheses

)ˆ()()ˆ(SS11

HcβAAXXAcβA

kn,q

Error

H F~kn/SS

q/SS

yXX)XX(yyy 1

ErrorSS



Another Way to Think About SSH

0

0

1000

0100:H

3

2

1

0

0

We could calculate the SSH by running two versions of the model: the full modeland a model restricted to just 1

SSH = SSError (Restricted Model) – SSError (Full Model)

so F is

Assume we have an A matrix as below:

kn/)Full(SS

2/)Full(SS)stricted(ReSSF

Error

ErrorError



A Hypothesis That All ’s Are Zero

0:H *k210

kn/)Full(SS

*k/)Full(SS)tostricted(ReSSF

Error

Error0Error

)tostricted(ReSS

)Full(SS)tostricted(ReSSR

0Error

Error0Error2

If our hypothesis is

Then the F would be

Which suggests a summary for the model



Generalized Least Squares

f = eV-1e

yVXXVXβ 111 ][ˆ

When we cannot make the Gauss-Markov Assumption that V(e) = 2I

Suppose that V(e) = 2V. Our objective function becomes



SSError for GLS

kn

SSs Error2

)ˆ(V)ˆ(SS 1

Error βXyβXy

with



GLS Hypothesis Testing

ii

2

i

ds

cˆt

H0: i = 0 where dii is the ith diagonal element of (XV-1X)-1

H0: a = c aXVXa

a112 )(s

cˆˆ

β

t

H0: A - c = 0 )ˆ(])([)ˆ(SS 111

H cβAAXVXAcβA

kn,q

Error

H F~kn/SS

q/SS )ˆ()ˆ(SSError βXyβXy



Accounting for the Sum of Squares of the Dependent Variable

e′e = y′y - y′X(X′X)-1X′y

SSError = SSTotal - SSPredictable

y′y = y′X(X′X)-1X′y + e′e

SSTotal = SSPredictable + SSError



SSPredicted and SSTotal Are a Quadratic Forms

PyyyXXXXy 1)(

And SSTotal

yy = yIy

SSPredicted is

Here we have defined P = X(X′X)-1X′



The SSError is a Quadratic Form

Having defined P = X(XX)-1 X, now define M = I – P, i. e. I - X(XX)-1X.

The formula for SSError then becomes

.

][

)( 1

Myy

yPIy

PyyIyy

yXXXXyyyee



Putting These Three Quadratic Forms Together

SSTotal = SSPredictable + SSError

yIy = yPy + yMy

I = P + M

here we note that



M and P Are Linear Transforms of y

y = Py and

e = My

so looking at the linear model:

and again we see that

I = P + M

eyy ˆ

Iy = Py + My



The Amazing M and P Matrices

y = Py and yy ˆˆ = SSPredicted = y′Py

e = My and = SSError = y′My

What does this imply about M and P?



The Amazing M and P Matrices

y = Py and yy ˆˆ = SSPredicted = y′Py

e = My and = SSError = y′My

PP = P

MM = M



In Addition to Being Idempotent…

.nn

n1nnn1

n1nnn1

0PM

0P1

0M1

in this chapter we will cover

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