11

Midterm ReviewMidterm Review

22

Outline Outline The Triangle of Stability - Power 7The Triangle of Stability - Power 7Augmented Dickey-Fuller Tests – Power 10Augmented Dickey-Fuller Tests – Power 10Lab FiveLab Five2005 Midterm2005 Midterm

33

I. The Triangle of StabilityI. The Triangle of Stability

44

Autoregressive ProcessesAutoregressive ProcessesOrder two: x(t) = bOrder two: x(t) = b1 1 * x(t-1) + b* x(t-1) + b2 2 * x(t-2) + wn(t)* x(t-2) + wn(t)

If bIf b2 2 = 0, then of order one: x(t) = b= 0, then of order one: x(t) = b1 1 * x(t-1) + * x(t-1) +

wn(t)wn(t) If bIf b1 1 = b= b2 2 = 0, then of order zero: x(t) = wn(t)= 0, then of order zero: x(t) = wn(t)

55

Quadratic Form and RootsQuadratic Form and Roots x(t) = bx(t) = b1 1 * x(t-1) + b* x(t-1) + b2 2 * x(t-2) + wn(t) is a second * x(t-2) + wn(t) is a second

order stochastic difference equation. If you drop order stochastic difference equation. If you drop the stochastic term, wn(t), then you have a the stochastic term, wn(t), then you have a second order deterministic difference equation:second order deterministic difference equation:

x(t) = bx(t) = b1 1 * x(t-1) + b* x(t-1) + b2 2 * x(t-2) ,* x(t-2) ,

Or, x(t) - bOr, x(t) - b1 1 * x(t-1) - b* x(t-1) - b2 2 * x(t-2) = 0* x(t-2) = 0 And substituting yAnd substituting y2-u 2-u for x(t-u),for x(t-u), yy2 2 – b– b11*y – b*y – b22 = 0, we have a quadratic equation = 0, we have a quadratic equation

66

Roots and stabilityRoots and stability Recall, for a first order autoregressive process, Recall, for a first order autoregressive process,

x(t) – bx(t) – b1 1 *x(t-1)= wn(t), i.e. where b*x(t-1)= wn(t), i.e. where b2 2 = 0, then if we = 0, then if we

drop the stochastic term, and substitute ydrop the stochastic term, and substitute y1-u 1-u for x(t-u), then y – b1 =0, and the root is b root is b1 1 . This root is . This root is

unstable, i.e. a random walk if bunstable, i.e. a random walk if b1 1 = 1.= 1. In an analogous fashion, for a second order In an analogous fashion, for a second order

process, if bprocess, if b1 1 = 0, then x(t) = b= 0, then x(t) = b2 2 *x(t-2) + wn(t), *x(t-2) + wn(t),

similar to a first order process with a time interval similar to a first order process with a time interval twice as long, and if the root btwice as long, and if the root b2 2 =1, we have a =1, we have a

random walk and instabilityrandom walk and instability

77

Unit Roots and InstabilityUnit Roots and Instability Suppose one root is unity, i.e. we are on the Suppose one root is unity, i.e. we are on the

boundary of instability, then factoring the quadratic boundary of instability, then factoring the quadratic from above, yfrom above, y2 2 – b– b11*y – b*y – b22 = (y -1)*(y –c), where (y-1) = (y -1)*(y –c), where (y-1) is the root, y=1, and c is a constant. Multiplying out is the root, y=1, and c is a constant. Multiplying out the right hand side and equating coefficients on the the right hand side and equating coefficients on the terms for yterms for y22, y, and y, y, and y00 on the left hand side: on the left hand side:

yy2 2 – b– b11*y – b*y – b22 = y = y2 2 –y –c*y + c = y–y –c*y + c = y2 2 –(1+c)*y + c–(1+c)*y + c So – bSo – b1 1 = -(1+c), and –b= -(1+c), and –b2 2 = c= c Or bOr b1 1 + b+ b2 2 =1, and b=1, and b2 2 = 1 – b= 1 – b1 1 , the equation for a , the equation for a

boundary of the triangleboundary of the triangle

88

Triangle of Stable Parameter Triangle of Stable Parameter SpaceSpace

If bIf b2 2 = 0, then we have a first order = 0, then we have a first order

process, ARTWO(t) = bprocess, ARTWO(t) = b1 1 ARTWO(t-1) + ARTWO(t-1) +

WN(t), and for stability -1<bWN(t), and for stability -1<b1 1 <1<1

01-1 b1

99

Triangle of Stable Parameter Triangle of Stable Parameter SpaceSpace

If bIf b1 1 = 0, then we have a process, = 0, then we have a process,

ARTWO(t) = bARTWO(t) = b2 2 ARTWO(t-2) + WN(t), that ARTWO(t-2) + WN(t), that

behaves like a first order process with the time behaves like a first order process with the time interval two periods instead of one period.interval two periods instead of one period.

For example, starting at ARTWO(0) at time For example, starting at ARTWO(0) at time zero, ARTWO(2) = bzero, ARTWO(2) = b2 2 ARTWO(0) , ignoring the ARTWO(0) , ignoring the

white noise shocks, and white noise shocks, and ARTWO(4) = bARTWO(4) = b2 2 ARTWO(2) = bARTWO(2) = b22

22 ARTWO(0) ARTWO(0)

and for stability -1<band for stability -1<b2 2 <1<1

1010

Triangle of Stable Parameter Triangle of Stable Parameter SpaceSpace

b1 = 0

b2

1

-1

1111

Triangle of Stable Parameter Triangle of Stable Parameter Space: Space:

b1 = 0

b2

1

-1

Draw a horizontal line through (0, -1) for (b1, b2)

b2 = 1 – b1

If b1 = 0, b2 = 1,If b1 = 1, b2

=0,If b2 = -1, b1

=2

Boundary points

+1-1

1212

Triangle of Stable Parameter Triangle of Stable Parameter Space: Space:

b1 = 0

b2

(0, 1)

(0, -1)

Draw a line from the vertex, for (b1=0, b2=1), though theend points for b1, i.e. through (b1=1, b2= 0) and (b1=-1, b2=0),

(1, 0)(-1, 0)

1313

Triangle of Stable Parameter Triangle of Stable Parameter Space: Space:

b1 = 0

b2

(0,1)

-1

(1, 0)(-1, 0)

b2 = 1 - b1

Note: along the boundary, when b1 = 0, b2 = 1, when b1 = 1, b2 = 0,and when b2 = -1, b2 = 2.

(2, -1)

1414

Triangle of Stable Parameter Triangle of Stable Parameter Space: Space:

b1 = 0

b2

(0, 1)

(0, -1)

(1, 0)(-1, 0) b1>0 b2>0

b1<0b2>0

b1<0b2<0 b1>0

b2<0

1515

Is the behavior different in each Is the behavior different in each Quadrant? Quadrant?

b1 = 0

b2

(0, 1)

(0, -1)

(1, 0)(-1, 0) b1>0 b2>0

b1<0b2>0

b1<0b2<0 b1>0

b2<0

I

II

III IV

1616

We could study with simulationWe could study with simulation

1717

Is the behavior different in each Is the behavior different in each Quadrant? Quadrant?

b1 = 0

b2

(0, 1)

(0, -1)

(1, 0)(-1, 0) b1= 0.3 b2= 0.3

b1= -0.3b2= 0.3

b1= -0.3b2= -0.3 b1= 0.3

b2= -0.3

I

II

III IV

1818

SimulationSimulationSample 1 1000Sample 1 1000Genr wn=nrndGenr wn=nrndSample 1 2Sample 1 2Genr artwo =wnGenr artwo =wnSample 3 1000Sample 3 1000Genr artwo = 0.3*artwo(-1)+0.3*artwo(-2) + Genr artwo = 0.3*artwo(-1)+0.3*artwo(-2) +

wnwnSample 1 1000Sample 1 1000

1919

II. Augmented Dickey-Fuller TestsII. Augmented Dickey-Fuller Tests

2020

Part II: Unit RootsPart II: Unit Roots

First Order Autoregressive or First Order Autoregressive or RandomWalk?RandomWalk?y(t) = b*y(t-1) + wn(t)y(t) = b*y(t-1) + wn(t)y(t) = y(t-1) + wn(t)y(t) = y(t-1) + wn(t)

2121

Unit RootsUnit Rootsy(t) = b*y(t-1) + wn(t)y(t) = b*y(t-1) + wn(t)we could test the null: b=1 against b<1we could test the null: b=1 against b<1 instead, subtract y(t-1) from both sides:instead, subtract y(t-1) from both sides:y(t) - y(t-1) = b*y(t-1) - y(t-1) + wn(t)y(t) - y(t-1) = b*y(t-1) - y(t-1) + wn(t)or or y(t) = (b -1)*y(t-1) + wn(t) y(t) = (b -1)*y(t-1) + wn(t)so we could regress so we could regress y(t) on y(t-1) and y(t) on y(t-1) and

test the coefficient for y(t-1), i.e.test the coefficient for y(t-1), i.e. y(t) = g*y(t-1) + wn(t), where g = (b-1)y(t) = g*y(t-1) + wn(t), where g = (b-1) test null: (b-1) = 0 against (b-1)< 0test null: (b-1) = 0 against (b-1)< 0

2222

Unit RootsUnit Roots

i.e. test b=1 against b<1i.e. test b=1 against b<1This would be a simple t-test except for a This would be a simple t-test except for a

problem. As b gets closer to one, the problem. As b gets closer to one, the distribution of (b-1) is no longer distributed distribution of (b-1) is no longer distributed as Student’s t distributionas Student’s t distribution

Dickey and Fuller simulated many time Dickey and Fuller simulated many time series with b=0.99, for example, and series with b=0.99, for example, and looked at the distribution of the estimated looked at the distribution of the estimated coefficient coefficient g

2323

III. Lab FiveIII. Lab Five

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2727

2828

2929

3030

3131

1.05-0.660.32=0.71

3232

3333

3434

3535

Unit RootsUnit Roots

Dickey and Fuller tabulated these Dickey and Fuller tabulated these simulated results into Tablessimulated results into Tables

In specifying Dickey-Fuller tests there are In specifying Dickey-Fuller tests there are three formats: no constant-no trend, three formats: no constant-no trend, constant-no trend, and constant-trend, and constant-no trend, and constant-trend, and three sets of tables.three sets of tables.

3636

Unit RootsUnit Roots

Example: the price of gold, weekly data, Example: the price of gold, weekly data, January 1992 through December 1999January 1992 through December 1999

3737

250

300

350

400

450

1/06/92 12/06/93 11/06/95 10/06/97 9/06/99

Trace of the Weekly Closing Price of Gold

3838

0

10

20

30

40

50

60

260 280 300 320 340 360 380 400

Series: GOLDSample 1/06/1992 12/27/1999Observations 417

Mean 345.7320Median 352.5000Maximum 414.5000Minimum 253.8000Std. Dev. 41.31026Skewness -0.514916Kurtosis 1.999475

Jarque-Bera 35.82037Probability 0.000000

3939

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4141

4242

4343

0.0

0.1

0.2

0.3

0.4

0.5

-4 -2 0 2 4

TVAR

TDE

NS

Simulated Student's t-Distribution, 415 Degrees of Freedom

5%

- 1.65

4444

Dickey-Fuller TestsDickey-Fuller Tests

The price of gold might vary around a The price of gold might vary around a “constant”, for example the marginal cost of “constant”, for example the marginal cost of productionproduction

PPGG(t) = MC(t) = MCG G + RW(t) = MC+ RW(t) = MCG G + WN(t)/[1-Z]+ WN(t)/[1-Z]

PPGG(t) - MC(t) - MCG G = RW(t) = WN(t)/[1-Z]= RW(t) = WN(t)/[1-Z]

[1-Z][P[1-Z][PGG(t) - MC(t) - MCG G ] = WN(t)] = WN(t)

[P[PGG(t) - MC(t) - MCG G ] - [P] - [PGG(t-1) - MC(t-1) - MCG G ] = WN(t)] = WN(t)

[P[PGG(t) - MC(t) - MCG G ] = [P] = [PGG(t-1) - MC(t-1) - MCG G ] + WN(t)] + WN(t)

4545

Dickey-Fuller TestsDickey-Fuller Tests

Or: [POr: [PGG(t) - MC(t) - MCG G ] = b* [P] = b* [PGG(t-1) - MC(t-1) - MCG G ] + WN(t)] + WN(t)

PPGG(t) = MC(t) = MCG G + b* P + b* PGG(t-1) - b*MC(t-1) - b*MCG G + WN(t) + WN(t)

PPGG(t) = (1-b)*MC(t) = (1-b)*MCG G + b* P + b* PGG(t-1) (t-1) + WN(t) + WN(t)

subtract Psubtract PGG(t-1) (t-1)

PPGG(t) - P(t) - PGG(t-1) = (1-b)*MC(t-1) = (1-b)*MCG G + b* P + b* PGG(t-1) (t-1) - P - PGG(t-1) + (t-1) +

WN(t)WN(t) or or PPGG(t) = (1-b)*MC(t) = (1-b)*MCG G + (1-b)* P + (1-b)* PGG(t-1) + WN(t)(t-1) + WN(t) Now there is an intercept as well as a slopeNow there is an intercept as well as a slope

4646

4747

4848

Augmented Dickey- Fuller TestsAugmented Dickey- Fuller Tests

4949

ARTWO’s and Unit RootsARTWO’s and Unit Roots Recall the edge of the triangle of stability: bRecall the edge of the triangle of stability: b2 2 = 1 – b= 1 – b1 1 , ,

so for stability bso for stability b1 1 + b+ b2 2 < 1< 1 x(t) = bx(t) = b1 1 x(t-1) + bx(t-1) + b2 2 x(t-2) + wn(t)x(t-2) + wn(t) Subtract x(t-1) from both sidesSubtract x(t-1) from both sides x(t) – x(-1) = (bx(t) – x(-1) = (b1 1 – 1)x(t-1) + b– 1)x(t-1) + b2 2 x(t-2) + wn(t)x(t-2) + wn(t) Add and subtract bAdd and subtract b2 2 x(t-1) from the right side: x(t-1) from the right side: x(t) – x(-1) = (bx(t) – x(-1) = (b1 1 + b+ b2 2 - 1) x(t-1) - b- 1) x(t-1) - b2 2 [x(t-1) - x(t-2)] + wn(t)[x(t-1) - x(t-2)] + wn(t) Null hypothesis: (bNull hypothesis: (b1 1 + b+ b2 2 - 1) = 0- 1) = 0 Alternative hypothesis: (bAlternative hypothesis: (b1 1 + b+ b2 2 -1)<0-1)<0

5050

Lab FiveLab Five

5151

2005 Midterm2005 Midterm

5252

5353

Fig. 2.1: Debt Service Payments as a Percent of Disposable Personal Income

2004.3

1993.1

0

2

4

6

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1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47

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Per

cen

t

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5555

5656

5757

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5959

6060

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6262

Figure 4.1 Monthly Change in Capacity Utilization, Total industry, 1967.02-2005.03

-4

-3

-2

-1

0

1

2

Date

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6666

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