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Matrix ManipulationRegression

Jon Shen Week 13 Tutorials

1st June 2010

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Summary

Multiple Regression: Fit y = 0 + 1x1 + 2x2 + . . . to data For example, x1 = x, x2 = x

2

or x1 = ex, x2 = ln x, x3 = x

5

The equation is linear in the coefficients Know how to:

Express a given linear model in matrix-vector form; * Obtain least squares estimates if given relevant matrices; Determine E[] and Var[]; * Hypothesis tests: i = c for some constant c; * Determine confidence intervals for i; Calculate R2 and adjusted R2; Construct an ANOVA table for the regression; Interpret the coefficients * Determine a confidence intervals for the mean Y value or a

prediction interval for the individual Y value corresponding to

a given x value (see PASS questions);Jon Shen Week 13 Tutorials
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Consultation

Week 13: 4-5pm today in Quad2061

Week 13: 6pm+ today?

Week 13: Thursday 11-1 via email / www.scriblink.com Stuvac/Exam period: email - slow reply before June 17th

Exam period: 17/18th June: In person consultation at uniQuad2061 (email to arrange appointment)

Exam period: 17/18th June: Online consultation via email /www.scriblink.com

[email protected]

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Preparation

1. FORMULA BOOK! Know whats in here.

2. Starting point: lecture slides and tutorial problems

3. Revise class quizzes! (Quiz 1 Q5 and Quiz 2 Q4...)

4. Past Quizzes/Final exams

5. PASS questions - extra revision

6. IAA papers - even more revision http://www.actuaries.org.uk/students/exams/

preparing/exam_papers 2005-2009 - CT3 (most relevant) 2000-2004 - Subject 101 1997-1999 - Subject C


S
http://www.actuaries.org.uk/students/exams/preparing/exam_papershttp://www.actuaries.org.uk/students/exams/preparing/exam_papershttp://www.actuaries.org.uk/students/exams/preparing/exam_papershttp://www.actuaries.org.uk/students/exams/preparing/exam_papershttp://find/


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Technique

1. READ THE QUESTION

Understand what is being asked - highlight key words2. Demonstrate that you understand the material

If short on time, indicate how you would attempt to solve theproblem

3. Do questions you are most confident with first, and look at

the number of marks allocated


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Properties

Multiplication:

a b

c d

across

f g

h k

down=

af+ bh ag + bkcf+ dh cg + dk

Differentiation:

TXTY = XTY and

(XTY)T = XTY

TXTX = 2XTX

Expectation: If B is random and A is non-random, then E[AB] = AE[B] Var[AB] = AVar[B]AT


Summary
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Example 1Example 2Example 3

Linear Model

Consider the model y =

30 + 1x + 2x21. Is this a linear model? If not, explain how you can transform

it into a linear model.

2. Write the (transformed) model in matrix form.


Summary


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Linear Model - Solution

1. y3 = 0 + 1x + 2x2

2. Y = X + where

Y =

y31

y32

.

.

.y3n

, X =

1 x1 x21

1 x2 x22

. . .

. . .

1 xn x2n

, =

012

, =

12.

.

.n

,


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and ANOVA Table

You decide to fit the model from the previous slide to data. Youhave 103 data points and are given that

(X

T

X)

1

= 10

638580 2237 1040

2237 230 341040 34 223 andXTY =

7651.7291333.6053992.96

.

Also

123

i=1 y2

i = 501813.55,

123

i=1 y2

i = 500881.11, y = 62.21.

a) Determine the least squares estimates .b) Construct an ANOVA table for this model, and hence calculatethe R2 and adjusted R2.


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and ANOVA Table - Solution

a) = (XTX)1XTYUsing matrix multiplication:

= 10

638580 2237 10402237 230 341040 34 223

7651.72

91333.6053992.96= 106

38580 7651.72 2237 91333.60 1040 53992.962237 7651.72 + 230 91333.60 34 53992.96

1040

7651.72

34

91333.60 + 223

53992.96

= 106

347374162054069.72

977298.88

=

34.7374162.05406972

0.97729888

=

01

2


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b) N.B. The notation used here is slightly different from thenotation used in calculating ANOVA tables elsewhere.SST =

123

i=1 y2

i n(y)2 = 501813.5512362.212 = 25793.2057SSR = y

2

i

n(y)2 = 500881.11

123

62.212 = 24860.7657

There are 123 data points and 3 regression parameters (s)Hence, the total degrees of freedom is 123 - 1 = 122, and theregression degrees of freedom is 3 - 1 = 2.This gives us the following entries in the table:

Source df SS MS = SS/df F-StatRegression 2 24860.7657 25793.2057

2= 12896.60285 ?

Error ? ? ?

Total 122 25793.2057


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b) We can fill in the second row of the table:122 - 2 = 120

25793.2057 - 24860.7657 = 932.44932.44

120= 7.7703

Source df SS MS = SS/df F-Stat

Regression 2 24860.7657 12896.60285 ?

Error 120 932.44 7.7703Total 122 25793.2057


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Exam TipsMatrix Manipulation

Regression



b) Finally, you can calculate the F-statistic:12896.60285

7.7703= 1659.723244

Source df SS MS = SS/df F-StatRegression 2 24860.7657 12896.60285 1659.723244

Error 120 932.44 7.7703

Total 122 25793.2057

R2

=SSR

SST =24860.7657

25793.2057 = 0.963849395

R2a = 1MSESSTn1

= 1 122 7.770325793.2057

= 0.963247042


SummaryExample 1
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Regression


Hypothesis Testing and Confidence Interval

From the previous slides, you have calculated

=

34.742.050.98

, (XTX)1 = 106

38580 2237 10402237 230 34

1040

34 223

You have also constructed the ANOVA table

Source df SS MS = SS/df F-Stat

Regression 2 24860.77 12896.60 1659.72

Error 120 932.44 7.77Total 122 25793.21

a) Determine a 95% confidence interval for 2.b) Test the hypothesis H0 : 1 = 2 against H1 : 1 > 2 at 1%

significance.Jon Shen Week 13 Tutorials

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Example 1
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Regression


Hypothesis Testing and Confidence Interval - Solution

We need estimates of Var[1] and Var[2].From slide 8, Var[] = 2(XTX)1.

2

is unknown, so we estimate this with s2

= MSE = 7.77 fromthe ANOVA table.

Then our estimate is 7.77 10638580 2237 10402237 230 341040 34 223

In particular, we want Var[1] = 7.77 106 230 = 0.0017871and Var[2] = 7.77 106 223 = 0.00173271


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Example 1
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Regression

pExample 2Example 3


a) We have2 2Var[2] tnp = t1233 = t120

Then

Pr1.98 2 2Var[2] 1.98

= 0.95

Pr1.98

0.98

2

0.00173271 1.98 = 0.95

Pr(0.89758085 2 1.06241915) = 0.95

So a 95% confidence interval for 2 is (0.89758085, 1.06241915)


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Example 1
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Regression

pExample 2Example 3


b) H0 : 1 = 2 against Ha : 1 > 2Under the null, the test statistic is

1 2Var[1] = 2.05 20.0017871 = 0.002113705

This is a one-sided test. The 99th quantile for a t120 distribution is2.358. The observed value of 0.002113705 is less than thetheoretical value of 2.358, so fail to reject the null as there isinsufficient evidence to suggest that 1 is significantly differentfrom two.

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