In This Chapter We Will Cover

Slide 9.Slide 9.11

Confirmatory Confirmatory Factor AnalysisFactor Analysis

MathematicalMathematicalMarketingMarketing

In This Chapter We Will Cover

Models with multiple dependent variables, where the independent variablesare not observed. This is called Factor Analysis. We cover

The factor analysis model

A factor analysis example

Measurement properties of the unobserved variables

Maximum Likelihood estimation of the model

Some interesting special cases

When statistical reasoning is applied to factor analysis, as it will be inthis chapter, we often call this Confirmatory Factor Analysis.

Slide 9.Slide 9.22



Regression with Multiple Dependent Variables

Y = XB +

np2n1n

p22221

p11211

p*k2*k1*k

p11211

p00201

*nk1n

*k221

*k111

np2n1n

p22221

p11211

eee

eeeeee

xx1

xx1xx1

yyy

yyyyyy

These matrices have only one columnin univariate regression analysis

Slide 9.Slide 9.33



Comparing Regression with Factor Analysis

ip2i1i

p*k2*k1*k

p11211

p00201

*ik1iip2i1i eeexx1yyy

Looking at a typical row corresponding to the data from subject i:

iii eBxy

Slide 9.Slide 9.44



We Transpose It and Drop the Subscript i

iii eBxy

y = Bx + e

Then dropping the subscript i altogether gets us to

iii exBy

From the previous slide we have

Transpose both sides to get

Slide 9.Slide 9.55



The Factor Analysis Model

.εηληληλy

εηληληλyεηληληλy

pmpm22p11pp

2mm22221212

1mm12121111

.

y

yy

p

2

1

m

2

1

pm2p1p

m22221

m11211

p

2

1

ΛηyObserved variables

Factor Loadings Common Factors

Unique Factors

Slide 9.Slide 9.66



Assumptions of the Model

Ληy

Random inputs of the model:

~ N(0, )

~ N(0, )

Cov(, ) = 0

Slide 9.Slide 9.77



Now We Can Deduce the V(y)

)(E)(E)(E)(E

))((E

)(E)(V

εεΛηεεηΛΛηηΛ

εΛηεΛη

yyΣy

Named

Assumed 0

Named

We end up with only components 1 and 4 from the above equation

V(y) = +

Slide 9.Slide 9.88



A Simple Example to Get Us Going

Variables Description

y1 Measurement 1 of B



y4 Measurement 1 of C



Slide 9.Slide 9.99



The Pretend Example in Matrices

6

5

4

3

2

1

2

1

62

52

42

31

21

11

6

5

4

3

2

1

000

000

yyyyyy

εΛηy

.

00

0000

66

22

11

2221

11

Θ

Ψ

Slide 9.Slide 9.1010



Graphical Conventions of Factor Analysis

y2

y3

y4

y5

y6

y1

1 2

11

21

31

42

52

62

21

Note use of

boxes circles single-headed arrows double-headed arrows unlabeled arrows




Assume I have a model with just one y and one .

My model is then

y = +

Now assume you have a model y = ** +

where * = a∙ and

* = /a

Whose model is right?

Two Alternative Models




Ambiguity in the Model

alsoandaa

**y

2

2

22 a

a**)y(V

My Model

y = +

V(y) = 2 +

Your Model

y = ** +

where * = a∙ and* = /a

but




Resolving the Ambiguity by Setting the Metric

2221

11

62

52

31

21

and,

00

100001

ΨΛ

1

1and,

000

000

21

62

52

42

31

21

11

ΨΛ

Plan A

Plan B




Degrees of Freedom

2)1p(p

4 ’s3 ’s6 ’s

13 parameters

The General Alternative The Model

H0: = + HA: = S




ML Estimation of the Factor Analysis Model

i1

i2/12/pi 21exp

||)2(1)Pr( yΣy

Σy

n

ii

1

i2/n2/np

n

ii0 2

1exp||)2(

1)yPr( yΣyl

]n[TrTrTr 11n

iii

1

i

n

ii

1

i

SΣΣyyyΣyyΣy i

The likelihood of observation i is

The likelihood of the sample is

Because eaeb = ea+b




The Log of the Likelihood

)(tr||lnn21ttancons

)(trn21||lnn

21)2(lnpn

21Lln

1

1

00

SΣΣ

SΣΣl

n

ii

1

i2/n2/np

n

ii0 2

1exp||)2(

1)yPr( yΣyl




The Log of the Likelihood Under HA

p||lnn21

SLA = constant -

)(tr||lnn21ttanconsln 1

A

SSSl

)(tr||lnn21ttanconsln 1

0

SΣΣl




The Likelihood Ratio

)df(~LL2ln2 2A0

A

0

l

l

p)(tr||ln||lnnˆ 12 SΣSΣ

From L0 From LA

S, 0 2ˆ

n , 2ˆ




The Single Factor Model

p

2

1

p

2

1

p

2

1

y

yy

= +

if V() = 11 = 1

The latent variable is called a true score

The model is called congeneric tests




Even More Restrictive Models with More Degrees of Freedom

p21

pp2211

-equivalent tests

Parallel tests

00

0000

Σ




Multi-Trait Multi-Method Models

y21 y31y11 y22 y32y12 y32 y33y13

1 2 2

4 5 6




The MTMM Model in Equations

33

23

13

32

22

12

31

21

11

6

5

4

3

2

1

9693

8583

7473

6662

5552

4442

3631

2521

1411

33

23

13

32

22

12

31

21

11

000000000000

000000000000

000000000000

yyyyyyyyy

10001000

10001

11

)(V

3231

21

3231

21

Ψη

β0α




Goodness of Fit According to Bentler and Bonett (1980)

j

2

j dfˆ

Q

2

s

2

0

2

s0s ˆ

ˆˆ

1Q

QQ

S

0S0s

Define

HA: = S

H0: = +

HS: = (with diagonal)

Then we could have

Perfect Fit (1)

No Fit (0)(for off-diagonal)




Goodness of Fit

)(tr

tr1GFI1

21

SΣISΣ

)GFI1(df2

)1p(p1AGFI0

2/)1p(p

)s(RMSE

p

1i

i

1j

2

ijij




Modification Indices

22

22

)ˆ(

ˆ2n

MI

Documents

In This Chapter We Will Cover