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3 March 2014 1
CDS M Phil Econometrics
Vijayamohanan Pillai N
CDS M Phil Econometrics Vijayamohan
Vijayamohan: CDS MPhil: Econometrics
3 March 2014 Vijayamohan: CDS MPhil: Econometrics
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Old Least Squares Old Least Squares (OLS)(OLS)
CDS M Phil Econometrics Vijayamohan
3 March 2014 Vijayamohan: CDS MPhil: Econometric s 3
Multiple Regression Analysis
• y = β0 + β1x1 + β2x2 + . . . βkxk + u
3 March 2014 Vijayamohan: CDS MPhil: Econometrics
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Multiple Regression Analysis
Define column vectors of n observations on yand K – 1 variables.
iikiii uxxxfy += ),...,,( 32
iikkiii uxxxy +++++= ββββ ...33221
This can be expressed as
i
k
jijji uxy +=∑
=1
β
General form of the multiple linear regression mode l:
3 March 2014 Vijayamohan: CDS MPhil: Econometrics
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iikk3i32i21i ux...xxy +β++β+β+β=i = 1, …, n
1k1k1331221111 ux...xxxy +β++β+β+β=
2k2k2332222112 ux...xxxy +β++β+β+β=
3k3k3333223113 ux...xxxy +β++β+β+β=
……………………….
nnkk3n32n21n1n ux...xxxy +β++β+β+β=
Multiple Regression Analysis
3 March 2014 Vijayamohan: CDS MPhil: Econometrics
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+
β
ββ
=
=
n
2
1
k
2
1
nk2n1n
k22221
k11211
n
2
1
u
...
u
u
...
x...xx
............
x...xx
x...xx
y
...
y
y
y
= Xβ + u
x1 is a column of ones:
[ ] [ ]TT1n11 11xx LL =
K – 1 variables and 1 constant
Multiple Regression Analysis
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The Classical Assumptions
A1. Linearity
Linearity of the parameters (and disturbance)Linearity of the parameters (and disturbance)
Sometimes models which appear to be nonSometimes models which appear to be non--linear linear can be estimated using the leastcan be estimated using the least--squares procedure . squares procedure.
∏=
ββ=n
2k
uk expxexpy k1
u)xln()yln(n
2kkk1 ∑
=+β+β=
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The Classical Assumptions
A2. E(u) = 0
The disturbance term has a zero expectation
=
=
0
0
0
u
u
u
E)u(E
n
2
1
MM
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The Classical Assumptions
A3. Nonstochastic Regressors
X is a non-stochastic n x k matrix.
That is, it consists of a set of fixed numbers.
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The Classical Assumptions
What does ‘Nonstochastic’ Mean?
Assumption A3 comes in two forms:
a weak version and a strong version.
• Strong Version: The explanatory variables
should be non - stochastic.
• Weak Version: The explanatory variables are random but distributed independently of the error term.
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The Classical Assumptions
What does ‘Nonstochastic’ Mean?
The term ‘nonstochastic’ essentially means
that the variables are determined “outside”
the context of the regression
(this is another reason why we use the term
independent when referring to explanatory
variables)
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The Classical Assumptions: Homoscedasticity
Variance
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The Classical Assumptions
A4. Spherical Disturbances
n2T I)uu(E)u(Var σ==
=
)u(E)uu(E)uu(E
)uu(E)u(E)uu(E
)uu(E)uu(E)u(E
)uu(E
2n2n1n
n22212
n12121
T
L
MOMM
L
L
σ
σσ
=
2
2
2
00
00
00
L
MOMM
L
L
n2Iσ=
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The Classical Assumptions: Spherical disturbances
n2T I)uu(E)u(Var σ==
=
)u(E)uu(E)uu(E
)uu(E)u(E)uu(E
)uu(E)uu(E)u(E
)uu(E
2n2n1n
n22212
n12121
T
L
MOMM
L
L
σ
σσ
=
2
2
2
00
00
00
L
MOMM
L
L
n2Iσ=
Spherical disturbances ⇒⇒
(i)
and
(ii)
22ii )u(E)u(Var σ=≡ n,...,1i =∀ homoskedasticity
0)u,u(E)u,u(Cov jiji =≡ ji ≠∀ No autocorrelation
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..
x1 x2
Homoscedasticity
E(y|x) = β0 + β1x
y
The Classical Assumptions: Homoscedasticity
x
f(y|x)
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.
xx1 x2
f(y|x)
x3
.. E(y|x) = β0 + β1x
The Classical Assumptions: Homoscedasticity
Heteroscedasticity
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The Classical Assumptions
A5. Identifiability
No exact linear relationships among the variables (no perfect multicollinearity).
Specifically,
X is n x K with rank K. (X has full (column) rank)
⇒ the columns of X are linearly independent.
NonNon--singular Xsingular X
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The Classical Assumptions
A5. Identifiability
No exact linear relationships among the variables (no perfect multicollinearity).
Implicit within this assumption are the requirements of more observations than variables (micronumerosity) and sufficient variability in the values of the regressors.
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The Classical Assumptions
A6. Normality
Final assumption: the disturbances are normally distributed.
useful for the purposes of statistical inference
but not necessary for analysing the properties of the estimators.
),0(~ 2nN Iu σ
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The Classical / Old / Ordinary Least squares (OLS)
Sample counterpart of the k-variable regression mod el:
uˆXy +β=
Where and are the sample counterparts of β and u.
OLS aims to minimise the difference between
an observed value of yi and its predicted value.
⇒ the error be the least.
β u
yyu −=β= ˆXy
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The Classical / Old / Ordinary Least squares (OLS)
Specifically, the problem is to find an estimator that minimises the error sum of squares:
)yy()yy(uu −′−=′
yXˆ2ˆXXˆyy ′β′−β′β′+′=
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A necessary condition for a minimum is that the fir st-order conditions equal zero.
The Classical / Old / Ordinary Least squares (OLS)
0yX2ˆXX2ˆ
uu=′−β′=
β∂′∂
Therefore, rearrange to give the normal equation
yXˆX)X( ′=β′
yXX)X(ˆ 1 ′′=β −
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The Classical / Old / Ordinary Least squares (OLS)
yXX)X(ˆ 1 ′′=β −
For a 2-variable model,
0ar(X)V ; )X(Var
)y,X(Covˆ >=β
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The Classical / Old / Ordinary Least squares (OLS)
XTX is invertible provided X has full rank.
(Why?)
XX2ˆˆ
uu2
′=β′∂β∂′∂
XTX is positive definite for a minimum.
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Statistical Properties of Least Squares Estimator
Mean of β
uXX)X(β
yXX)X(ˆ
1
1
′′+=
′′=β−
−
β)(E =β Property of unbiasedness
[key assumptions: E(u) = 0 and non-stochastic regressors]
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Statistical Properties of Least Squares Estimator
Variance of β [ ])βˆ)(β(E)(Var ′−β−β=β
Noting that uXX)X(ˆ 1 ′′=β−β −
then])XX(XuuX)XX[(E)(Var 11 −− ′′′′=β
[key assumptions: 22ii )u(E)u(Var σ=≡
0)u,u(E)u,u(Cov jiji =≡and
for all i, i≠j.
12 )XX()(Var −′σ=β
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Statistical Properties of Least Squares Estimator
Variance of β
12 )XX()(Var −′σ=β
In the scalar case∑
σ=β2i
2
2x
)(Var
MinimumVariance (best)
property
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Statistical Properties of Least Squares Estimator
Unbiased Estimator of σ2
For statistical inference we require an estimate of
Variance of
kn
u
sˆ
n
1i
2i
22
−==σ∑=
and therefore, σ2.
An unbiased estimator of σ2 is:
β
12 )XX()(Var −′σ=β
Its square root = the standard error of the regression
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( )
( )
(ESS) squares of sum error the is u
(RSS) squares of sum regression )(explained the is yy
(TSS) squares of sum total the is yy
2i
2
i
2
i
∑
∑
∑
−
−
OLS: Goodness of Fit
( ) ( ) ∑∑∑ +−=− 2i
2i
2i uyyyy
What proportion of the total variation of y is accounted for by the variation in X?
In terms of sums of squares
TSS = RSS + ESS
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OLS: Measure of Goodness of Fit
Coefficient of determination: TSS
RSSR2 =
or TSS
ESS1R2 −=
• R2 never decreases when a new X variable is added to the model
– This can be a disadvantage when comparing models
• What is the net effect of adding a new variable?
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• What is the net effect of adding a new
variable?
– We lose a degree of freedom when a new X
variable is added
– Did the new X variable add enough
explanatory power to offset the loss of one
degree of freedom?
OLS: Measure of Goodness of Fit
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OLS: Measure of Goodness of Fit
Adjusted RAdjusted R22
• Shows the proportion of variation in Y explained by all X variables adjusted for the number of X variables used
(where n = sample size, k = number of parameters)
)1n/(TSS
)kn/(ESS1R2
adj −−
−=
−−−−=
kn
1n)R1(1 2
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OLS: Measure of Goodness of Fit
Adjusted RAdjusted R22
(where n = sample size, k = number of parameters)
– Penalize excessive use of unimportant independent variables
– Smaller than r2
– Useful in comparing among models
)1n/(TSS
)kn/(ESS1R2
adj −−−=
−−−−=
kn
1n)R1(1 2
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OLS: Measure of Goodness of Fit
The relationship between R 2 and 2R
222 Rkn
1n
kn
k1)R1(
kn
1n1R
−−
+−−
=−−−
−=
Two other measures of fit are the Schwartz criterion (SC)
nlnn
k
n
ESSlnSC +=
and the Akaike information criterion (AIC)
n
k2
n
ESSlnAIC +=
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OLS: Statistical Inference
we know that the expected value is
β)(E =β
And we know that a suitable estimate of the variance of
the slope parameter is
12 )XX()(Var −′σ=β
12 )XX(s)(Var −′=β
kn
u
sˆ
n
1i
2i
22
−==σ∑=
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OLS: Statistical Inference
In order to make meaningful inference the variable must be normally distributed.
One of the assumptions introduced above was:
)I,0(N~u n2σ
This means that u has a multivariate normal distribution.
(If it is not, then n must be large – i.e. becomes normally distributed by virtue of the central limit theorem.
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OLS: Statistical Inference
This implies that the sampling error ( - β) is related to u as follows:
β
uXX)X(
βu)(XβXX)X(
βyXX)X(βˆ
1
1
1
′′⇒
−+′′⇒
−′′=−β
−
−
−
which implies that the sampling error is also multivariate normally distributed:
))XX(,0(N~)β( 12 −′σ−β
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OLS: Statistical Inference
In hypothesis testing we are concerned about testing whether a particular finding is compatible with some stated hypothesis or not.
A common hypothesis test is formulated as
kk0 β:H β=
kk1 :H β≠β
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OLS: Statistical Inference
Using our earlier result, if we were looking at the kth diagonal element of (X TX)-1 we obtain
)))XX((,0(N~)( kk12
kk−′σβ−β
where kk1))XX(( −′ is the (k,k) element of (XTX)-1
If we define the ratio z k by dividing
by its standard deviation
kk12
kkk
))XX((
ˆz
−′σ
β−β=
where )1,0(N~zk
kˆ β−β
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OLS: Statistical Inference
If we do not know σ2, it is natural to replace it with the estimate s2.
This generates the t-ratio
)(SE
ˆ
))XX((s
ˆt
k
kk
kk1T2
kkk β
β−β=β−β=−
where SE( k) denotes the standard error of k. This is distributed as t c
α/2;(n-k) .β β
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OLS: Statistical Inference
An Example: Investment Function
Theory of the behaviour of investors:They care only about real interest rates.
Real I = f(real gdp, an interest rate (the 90-day T-bill rate), inflation (change in the log of CPI), real disposable personal income, trend)
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OLS: Statistical Inference
An Example: Investment Function
Hypothesis:
Equal increases in interest rates and rate of
inflation would have no independent effect on
investment.
So the Ho : β2 + β3 = 0.The test statistic is:
)ˆ(SE
0ˆˆt
32
32
β+β−β+β
=
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OLS: Statistical Inference
An Example: Investment Function
SE(.) = [0.003192 + 0.002342 +2(–3.718 x 10–6 )]1/2
= 0.002866
845.1002866.0
00331.00086.0t −=+−=
95% critical value: t(203 – 5) = 1.96
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OLS: Statistical Inference
Testing Linear Restrictions
As well as testing restrictions on individual regression coefficients we may wish to test linear combinations of them. For example
1:H 320 =β+β
430 :H β=β
(c)
(b)
(a)
0...:H k4320 =β==β=β=β
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OLS: Statistical Inference
Testing Linear Restrictions: The F Test
The t-distribution can be used to test a single null hypothesis
If we want to conduct a “joint” test then we can no longer use the t distribution
For example, suppose we want to test whether all the explanatory variables in the model are significantly different from zero
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OLS: Statistical Inference
Testing Linear Restrictions: The F Test
F-Test of Entire Equation (“Testing the Joint Significance of the Explanatory Variables”)
H0: β2 = β3=…= βk= 0 Equivalently: H 0: R2 = 0
H1: H0 not true (at least one of the β is nonzero)
Cannot say: “if the coefficients are individuallyinsignificant this means they must be jointly
insignificant”
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OLS: Statistical Inference
Testing Linear Restrictions: The F Test
F-Test of Entire Equation (“Testing the Joint Significance of the Explanatory Variables”)
H0: β2 = β3=…= βk= 0 Equivalently: H 0: R2 = 0
H1: H0 not true (at least one of the β is nonzero)
kn
ESS1k
RSS
F
−
−=
kn
R11k
R
F 2
2
−−
−=or
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OLS: Statistical Inference
Testing Linear Restrictions: Multiple Restriction F Test
This time suppose we wish to test whether a subset of regression coefficients are zero. In this case
[ ]
uˆXˆX
uˆ
ˆXXy
3322
3
232
+β+β⇒
+
ββ
= H0: β3= 0
H1: H0 not true
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OLS: Statistical Inference
Testing Linear Restrictions: Multiple Restriction F Test
[ ]
uˆXˆX
uˆ
ˆXXy
3322
3
232
+β+β⇒
+
ββ
= H0: β3= 0
H1: H0 not true
To find the change in the fit of a multiple regress ion when an additional variable x 3 is added to a model that already contains K – 1 variables:
J = 1 linear restriction
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OLS: Statistical Inference
Testing Linear Restrictions: Multiple Restriction F Test
Thus the unrestricted regression is:
[ ]
uˆXˆXy
uˆ
ˆXXy
3322
3
232
+β+β=
+
ββ=
H0: β3= 0
H1: H0 not true
[ ]
*22
*2
32
uˆXy
u0
ˆXXy
+β=
+
β=And the restricted regression:
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OLS: Statistical Inference
The residuals from the restricted regression:
22*ˆXyu β−=
whereas the residuals from the unrestricted regress ion:
3322ˆXˆXyu β−β−=
Essentially we wish to see whether the reduction in the ESS is 'large enough' to suggest that X 3 is significant.
Testing Linear Restrictions: Multiple Restriction F Test
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OLS: Statistical Inference
Testing Linear Restrictions: Multiple Restriction F Test
The test statistic is
)kn/(uu
J/)uuuu(F
T
T*
T*
−−= J = 1; K= 2
)kn/()R1(
J/)RR(F
2
2*
2
−−−=or
Residuals from the restricted regression: 22*ˆXyu β−=
Residuals from the unrestricted regression:
3322ˆXˆXyu β−β−=
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OLS: Statistical Inference
An Example: Production Function
Testing Linear Restrictions: Multiple Restriction F Test
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An Example: Production Function
• Cobb-Douglas:
• Translog:
• Translog function relaxes the CD assumption of a unitary elasticity of substitution
• CD obtained by the restriction:
uKlnLlnyln 21 +β+β+α=
Unrestricted model: K = 6
uKlnLln)K(ln2
1)L(ln
2
1KlnLlnyln 5
24
2321 +β+β+β+β+β+α=
0543 =β=β=β
J = 3
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Estimated production Functions
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An Example: Production Function
CD or Translog?
The F-statistic for the hypothesis of CD model:
Ho: CD model is appropriate
Critical value: F(3,21) = 3.07
)kn/(uu
J/)uuuu(F
T
T*
T*
−−=
768.121/67993.0
3/)67993.085163.0(F =
−=
Conclusion?
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An Example: Production Function
CD function: Constant Returns to Scale?
1:Ho 32 =β+β
22
t1157.0)00961.0(200728.001586.0
)13757.0603.0(F ==
−+−+=
• Hypothesis of constant returns to scale:
• Equivalent to a restriction that the two coefficients of CD function sum to unity
•F-test with J = 1 and K = 3
• Critical value: F(1,24) = 4.26 Conclusion?3 March 2014 Vijayamohan: CDS MPhil:
Econometrics58
An Example: Production Function
CD function: Constant Returns to Scale?
• Hypothesis of constant returns to scale:
• Can have t-test:
• Critical value: ‘rule of thumb’
1:Ho 32 =β+β
3402.0)]00961.0(200728.001586.0[
)13757.0603.0(t
2/1−=
−+−+=
)1ˆ(SE
1ˆˆt
32
32
−β+β−β+β
=
Conclusion?
= – (F1/2)
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Test of Structural Break: The Chow test
• To test structural change in time series data
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5010
015
020
025
030
0S
avin
gs
1000 2000 3000 4000 5000Income
Test of Structural Break: The Chow test
Savings and personal disposable income (billions of dollars) US, 1970 – 1995.
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5010
015
020
0S
avin
gs
500 1000 1500 2000 2500Income
5010
015
020
0S
avin
gs
500 1000 1500 2000 2500Income
1970 – 1981
Test of Structural Break: The Chow test
150
200
250
300
Sav
ing
s
2000 3000 4000 5000 6000Income
150
200
250
300
Sav
ing
s
2000 3000 4000 5000 6000Income
1982 – 1995
Point of structural break: 1982 (assumed)
5010
015
020
025
030
0S
avin
gs
1000 2000 3000 4000 5000Income
5010
015
020
025
030
0S
avin
gs
1000 2000 3000 4000 5000Income
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Test of Structural Break: The Chow test
Steps
1. Run the two sub-period regressions and the full-period regressionFull-period: St = a0 + b0Yt n = 26 (n1 + n2)Sub-period 1: St = a1 + b1Yt n1 = 12 Sub-period 2: St = a2 + b2Yt n2 = 14
2. The pooled regression = restricted regression, obtained under the restrictions that a1 = a2 (= a0) and b1 = b2 (= b0).
3. Get the restricted ESS from it (ESSR): )uu( **′
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Test of Structural Break: The Chow test
1. Run the two sub-period regressions and the full-period regressionFull-period: St = a0 + b0Yt n = 26 (n1 + n2)Sub-period 1: St = a1 + b1Yt n1 = 12 Sub-period 2: St = a2 + b2Yt n2 = 14
4. The sub-period regressions = unrestricted regressions.
5. Get the unrestricted ESS from them: : ESSUR = ESS1 + ESS2.
)uu( ′
Steps
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Test of Structural Break: The Chow test
Sub-period 1: St = a1 + b1Yt n1 = 12 Sub-period 2: St = a2 + b2Yt n2 = 14
4. The sub-period regressions = unrestricted regressions.
5. Get the unrestricted ESS from them: : ESSUR = ESS1 + ESS2.
)uu( ′
)k2n/(uu
k/)uuuu(F
T
T*
T*
−−
= k = j; n – 2k = (n1 – k) + (n2 – k)
Steps
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Test of Structural Break: The Chow test
Pooled Regression Result: 1970 – 1995
)uu( **′Restricted ESS: = 23248.3
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Sub-period 1: 1970 – 1981
Test of Structural Break: The Chow test
Unrestricted ESS1 = 1785.033
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Sub-period 2: 1982 – 1995
Test of Structural Break: The Chow test
)uu( ′
Unrestricted ESS2 = 10005.221
Unrestricted ESS = ESS1 + ESS2 = 1785.033 + 10005.221 = 11790.254
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Test of Structural Break: The Chow test
)k2n/(uu
k/)uuuu(F
T
T*
T*
−−
=
69.1022/254.11790
2/)254.117903.23248(F =
−=
k = j; n – 2k = (n1 – k) + (n2 – k)
Critical value: F(2,22) at 1% α = 5.72
H0 = a1 = a2 = a0 and b1 = b2 = b0: Parameter stability
Conclusion?
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