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1 Midterm Review. 2 Outline The Triangle of Stability - Power 7 The Triangle of Stability - Power 7...
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Transcript of 1 Midterm Review. 2 Outline The Triangle of Stability - Power 7 The Triangle of Stability - Power 7...
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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
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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)
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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
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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
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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
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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
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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
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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
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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)
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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)
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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
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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
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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
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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
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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)
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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
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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
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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.
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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
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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
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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
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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
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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)
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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
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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
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Fig. 2.1: Debt Service Payments as a Percent of Disposable Personal Income
2004.3
1993.1
0
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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
Quarter
Per
cen
t
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Figure 4.1 Monthly Change in Capacity Utilization, Total industry, 1967.02-2005.03
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-1
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Date