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Econ 5410, Wednesday August 30, class 3

Robert de Jong1

1Department of Economics Ohio State University

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Two groups class split for computer class and test of Friday September 1 and Friday September 8 Exercise Review Calculating the regression line by hand Calculating the regression line by computer Another exercise Three more factoids

Today’s class 1 Hand out our first takehome 2 Two groups class split for computer class and test of Friday

September 1 and Friday September 8 3 Exercise 4 Short review 5 Calculation of the regression line by hand 6 Calculation of the regression line with the computer 7 Another exercise 8 Three more factoids about the regression line

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Outline

1 Econ 5410, Wednesday August 30, class 3 Two groups class split for computer class and test of Friday September 1 and Friday September 8 Exercise Review Calculating the regression line by hand Calculating the regression line by computer Another exercise Three more factoids

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Computer classes September 1 and September 8: If you specified a preference on the sheet that circulated last week, then please go to the time slot you specified (either 12:45pm-1:25pm, or 1:25pm-2:05pm) Otherwise, please come 1:25pm-2:05pm.

In other words: come 1:25pm-2:05pm, unless you put your name down for the 12:45pm-1:25pm time slot

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Outline

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Exercise

1 Find a counterexample that shows that in general, ( ∑N

i=1 Xi ) 2 6=

∑N i=1 X

2 i .

Example: N = 2, X1 = 1, X2 = 2. Then N∑

i=1

Xi = 2∑

i=1

Xi = X1 + X2 = 3,

so ( ∑N

i=1 Xi ) 2 = 32 = 9, while

N∑ i=1

X 2i = X 2 1 + X

2 2 = 1

2 + 22 = 1 + 4 = 5.

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Exercise

1 Find a counterexample that shows that in general, ( ∑N

i=1 Xi ) 2 6=

∑N i=1 X

2 i .

Example: N = 2, X1 = 1, X2 = 2. Then N∑

i=1

Xi = 2∑

i=1

Xi = X1 + X2 = 3,

so ( ∑N

i=1 Xi ) 2 = 32 = 9, while

N∑ i=1

X 2i = X 2 1 + X

2 2 = 1

2 + 22 = 1 + 4 = 5.

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

2 Calculate ∑4

i=1 i .

4∑ i=1

i = 1 + 2 + 3 + 4 = 10.

3 X̄ denotes the average of the Xi . Why is∑N i=1 Xi − NX̄ = 0?

X̄ = N−1 N∑

i=1

Xi,

so N∑

i=1

Xi −NX̄ = N∑

i=1

Xi −N ·N−1 N∑

i=1

Xi = N∑

i=1

Xi − N∑

i=1

Xi = 0.

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

2 Calculate ∑4

i=1 i . 4∑

i=1

i = 1 + 2 + 3 + 4 = 10.

3 X̄ denotes the average of the Xi . Why is∑N i=1 Xi − NX̄ = 0?

X̄ = N−1 N∑

i=1

Xi,

so N∑

i=1

Xi −NX̄ = N∑

i=1

Xi −N ·N−1 N∑

i=1

Xi = N∑

i=1

Xi − N∑

i=1

Xi = 0.

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

2 Calculate ∑4

i=1 i . 4∑

i=1

i = 1 + 2 + 3 + 4 = 10.

3 X̄ denotes the average of the Xi . Why is∑N i=1 Xi − NX̄ = 0?

X̄ = N−1 N∑

i=1

Xi,

so N∑

i=1

Xi −NX̄ = N∑

i=1

Xi −N ·N−1 N∑

i=1

Xi = N∑

i=1

Xi − N∑

i=1

Xi = 0.

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

4 Consider the following data points Xi and Yi :

Xi Yi 7 3 6 1 5 4

For those data points, calculate∑N i=1 Xi ;∑N i=1 Yi ;∑N i=1 Y

2 i ;∑N

i=1 Xi Yi .

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

4 We have

Xi Yi Y 2i Xi Yi 7 3 9 21 6 1 1 6 5 4 16 20

18 8 26 47

Therefore,∑N i=1 Xi = 18;∑N i=1 Yi = 8;∑N i=1 Y

2 i = 26;∑N

i=1 Xi Yi = 47.

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Outline

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Suppose that we try the line y = β0 + β1x point # vertical distance squared vertical

total: |Y1 −β0 −β1X1|2 +|Y2 −β0 −β1X2|2 +|Y3 −β0 −β1X3|2

In general:

|Y1 −β0 −β1X1|2 + |Y2 −β0 −β1X2|2

+ . . . + |YN −β0 −β1XN|2

where N denotes sample size Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Using mathematics, we can find the values of β0 and β1 that minimize

|Y1 −β0 −β1X1|2 + |Y2 −β0 −β1X2|2

+ . . . + |YN −β0 −β1XN|2.

Question: why not minimize the sum of absolute values, or fourth powers?

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Minimizing N∑

i=1

(Yi − (β0 + β1Xi ))2

over all possible values of β0 and β1 gives

β̂1 = N ·

∑N i=1 Xi Yi −

∑N i=1 Xi ·

∑N i=1 Yi

N · ∑N

i=1 X 2 i − (

∑N i=1 Xi )

and β̂0 = Ȳ − β̂1X̄

Note Ȳ = N−1 ∑N

i=1 Yi , the average of the Yi

The mathematical calculation requires being able to find the minimum of a function of two variables using differentiation

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Outline

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Last class’ in-class example

Xi Yi Xi Yi X 2i 1 2 2 1 4 2 8 16 1 3 3 1 6 7 13 18

β̂1 = N ·

∑N i=1 Xi Yi −

∑N i=1 Xi ·

∑N i=1 Yi

N · ∑N

i=1 X 2 i − (

∑N i=1 Xi )

= 3 · 13 − 7 · 6 3 · 18 − 62

= 39 − 42 54 − 36

= −3 18

= − 1 6 ≈−0.167

Also

β̂0 = Ȳ − β̂1X̄ = 7/3 − (− 1 6

)(6/3) = 8/3 ≈ 2.67

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Q: does the regression line y = (8/3) − (1/6)x pass through the middle of the points (1,2) and (1,3)?

A: Yes, if x = 1, the corresponding y value for the regression line is

(8/3) − (1/6) · 1 = (16/6) − (1/6)

= 15/6 = 2.5

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Outline

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Computer software: Gretl will calculate the regression line for us

Other packages that can do this: Excel, or other spreadsheet program SAS, SPSS Stata, Eviews, etc.

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Goal of this class: be able to interpret output for regressions; examples:

1 A “wage equation": explaining individuals’ wages from characteristics; http://www.econ.ohio-state.edu/ dejong/files/wage2.gdt

2 Teenage pregnancy rates in Ohio counties; http://www. econ.ohio-state.edu/dejong/teenpreg.gdt

3 George W. Bush approval rating vs. oil price http://www.econ.ohio-state.edu/dejong/ bushapproval2.gdt

Robert de Jong Econ 5410, Wednesday August 30, class 3

http://www.econ.ohio-state.edu/dejong/files/wage2.gdt

http://www.econ.ohio-state.edu/dejong/teenpreg.gdt

http://www.econ.ohio-state.edu/dejong/bushapproval2.gdt

Econ 5410, Wednesday August 30, class 3

Outline

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

In-class exercise

Xi Yi Xi Yi X 2i 1 3 3 1 3 4 12 9 2 5 10 4 6 12 25 14

Question 1

β̂1 = N ·

∑N i=1 Xi Yi −

∑N i=1 Xi ·

∑N i=1 Yi

N · ∑N

i=1 X 2 i − (

∑N i=1 Xi )

= 3 · 25 − 6 · 12

3 · 14 − 62 =

75 − 72 42 − 36

= 3 6

= 0.5

β̂0 = Ȳ − 1.4X̄ = 4 − 0.5 · 2 = 3 Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Question 2

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Questions 3 and 4: calculate

residuals ei = Yi − β̂0 − β̂1Xi = Yi − 3 − 0.5Xi

and

predictions Ŷi = β̂0 + β̂1Xi = 3 + 0.5Xi

Xi Yi Yi − 3 − 0.5Xi 3 + 0.5Xi 1 3 3 − 3 − 0.5 · 1 = −0.5 3 + 0.5 · 1 = 3.5 3 4 4 − 3 − 0.5 · 3 = −0.5 3 + 0.5 · 3 = 4.5 2 5 5 − 3 − 0.5 · 2 = 1 3 + 0.5 · 2 = 4 6 12 0 12

Question 5 and 6: residuals add up to 0; predictions sum to the same value as Yi ; why ?Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Residual: ei = Yi − β̂0 − β̂1Xi

Average of residuals:

N−1 N∑

i=1

ei = N −1

N∑ i=1

(Yi − β̂0 − β̂1Xi )

= (N−1 N∑

i=1

Yi ) − β̂0 − β̂1(N−1 N∑

i=1

Xi ) = Ȳ − β̂0 − β̂1X̄

but β̂0 = Ȳ − β̂1X̄,

N−1 N∑

i=1

ei = Ȳ − β̂0 − β̂1X̄ = 0 Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Average of predictions = average of Yi

average of predictions Ŷi

= N−1 N∑

i=1

Ŷi = N −1

N∑ i=1

(β̂0 + β̂1Xi )

= β̂0 + β̂1X̄

= (Ȳ − β̂1X̄ ) + β̂1X̄

= Ȳ .

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Outline

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Why does the regression line pass through the point (X̄, Ȳ ) ?

Earlier: regression line was y = 3 + 0.5x , X̄ = 2, Ȳ = 4

This regression line passed through (2,4) because

3 + 0.5X̄ = 3 + 0.5 · 2

= 3 + 1 = 4 = Ȳ

Reason: we draw the line

y = β̂0 + β̂1x = (Ȳ − β̂1X̄ ) + β̂1x,

and for x = X̄ , we get

y = (Ȳ − β̂1X̄ ) + β̂1X̄ = Ȳ .

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Last week’s in-class example; Xi and Yi reversed: Yi Xi Xi Yi X 2i 1 2 2 4 4 2 8 4 1 3 3 9 6 7 13 17

β̂1 = N ·

∑N i=1 Xi Yi −

∑N i=1 Xi ·

∑N i=1 Yi

N · ∑N

i=1 X 2 i − (

∑N i=1 Xi )

= 3 · 13 − 7 · 6 3 · 17 − 72

= 39 − 42 51 − 49

= −3 2

= −1.5

Also

β̂0 = Ȳ − β̂1X̄ = 2 − (−3/2)(7/3) = 2 + (7/2) = 5.5 Regression line: y = 5.5 − 1.5x

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Other example

Yi Xi Xi Yi X 2i 3 2 6 4 4 2 8 4 5 2 10 4 12 6 24 12

So,

β̂1 = N ·

∑N i=1 Xi Yi −

∑N i=1 Xi ·

∑N i=1 Yi

N · ∑N

i=1 X 2 i − (

∑N i=1 Xi )

= 3 · 24 − 6 · 12

3 · 12 − 62 =

72 − 72 36 − 36

= 0 0

=???

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Robert de Jong Econ 5410, Wednesday August 30, class 3

Econ 5410, Wednesday August 30, class 3

Two groups class split for computer class and test of Friday September 1 and Friday September 8
Exercise
Review
Calculating the regression line by hand
Calculating the regression line by computer
Another exercise
Three more factoids