HANDOUT 4 Flashcards by Ella Grant

What is the problem when trying to find LR equilibrium relationships between two non-stationary I(1) series?

If we regress any non-stationary series on any other non-stationary series –> always an apparent significant LR relation even if they’re not related.

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2 types of LR equilibrium relations between I(1) series.

genuine cointegrating relations

2. nonsense spurious regressions

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What do we find a significant LR relation between two I(1) series even if they’re not related?

because both are trending across time with a stochastic/random trend

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Dog/owner example for spurious regression

Both drunk = non-stationary
Yt and Xt start off at 0 i.e. on path
Each step the direction is randomly determined by each flipping a coin.

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Dog/owner example spurious reg - why do we end up with a significant relationship?

Non-stationary so as time increases, variance–> infinity. P(end up on path)=0 so always going to suggest a trend.

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Dog/owner example for cointegration

Both drunk = non-stationary
Yt and Xt start off at 0 i.e. on path
Dog connected to owner by elasticated lead and owner heavier = fixed max distance between Yt and Xt.

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Dog/owner example cointegration what is the residual from LR equation?

residual = actual Yt - predicted Yt

Difference between where dog actually is and where predicted to end up based on the owner’s location.

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The residual from dog/owner has to be integrated of order…

I(0) i.e. the residual MUST be stationary since the dog can only drift so far from the owner.

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general definition of cointegration

Yt and Xt are cointegrated of order d,b if:
a) Yt and Xt - I(d)
b) There exists a vector B such that
€t = Yt - BXt - I(d - b)

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What are d and b usually?

d=b=1

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So if Yt and Xt are I(1) and are cointegrated, the residuals are…

I(0) i.e. stationary

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Beta =

The cointegrating vector

B = the LR / equilibrium relationship between Yt and Xt.

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An arbitrary combination of I(1) series will be…

I(1), unless they’re cointegrated in which case the combo is I(0).

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Unbalanced regression =

When variables are NOT cointegrated of the same order - we cannot find cointegration between I(1) and I(0) series.

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A linear combination of I(1) series will…

Eliminate the stochastic trend = the linear combo - I(0) i.e. stationary

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Stochastic trend formula

Sum j=1,…,t €j

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What shows Yt and Xt have no link between them for spurious reg?

E(€1t, €2t) = 0

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In what % of cases do we reject H0 where H0 = no LR relation?

H0: B = 0
Reject H0 in 80% of cases
P(reject H0 I H0 false) should only = 5% for 5% significance level.

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Whose two step procedure allows us to distinguish between cointegrating and spurious relations?

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Engel and Granger

Step 1 of Engel and Granger =

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Estimate a LR equilibrium equation

2 types of LR equilibrium equations we can estimate

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static

2. dynamic (include lags)

3 steps for static LR equation engel and granger

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Reg Yt = d0 + d1Xt + €t by OLS
save et = Yt - (d0 + d1Xt)
Test if et - I(0) using ADF model B

ADF test on residuals for static LR equation

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change et = M + gamma et-1 + sum delta j
change et-j + Rt
H0: gamma = 0
H1: gamma ≠ 0

H0 and H1 for ADF test on residuals for static LR equation

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H0: gamma = 0 means et non-stationary = spurious reg.
H1: gamma < 0 means et - I(0) = cointegration

3 steps for dynamic LR equation engel and granger

1. reg Yt = d0 + dlXt + d2Xt-1 + d3Xt-2 + d4Yt-1 + d5Yt-2 + Ut 2. Solve for LR * 3. et = Yt - (predicted from LR) - do ADF test

What's the problem with our CVs for ADF test on residuals?

OLS minimises RSS = finds min variance for residuals = et look as stationary as possible = bias towards rejecting H0.

Solution to problem with our CVs for ADF test on residuals

Mackinnon's CVs for n=2 n = no unique non-stationary series in the reg that the residuals are from. Higher n = CVs more negative = harder to reject H0.

How do we determine p* for ADF test of et?

Max P* = T^1/3 - check no observations on residuals. Or look at PACF for d.resid - difference!! see no spikes.

Limitation of ADF test =

LOW POWER | often do not reject H0 = often find et are non-stationary = often find NO cointegration when there actually is.

If we find cointegration, our OLS estimators in our LR equation are...

SUPER-CONSISTENT

Why are OLS estimators super-consistent when we have cointegration?

In the LR equation, bias = COV(Xt, €t)/Var(Xt) Covariance between Xt - I(1) and €t - I(0) is small and constant. Var(X)-->infinity as time rises due to non-stationarity so bias-->0 very quickly.

What does super-consistency of estimators in our long run equation mean?

We do NOT need to worry about correct dynamics, omitted relevant variables, heteroscedasticity, endogeneity etc. our OLS coefficients are good regardless.

Do we have super-consistency with stationary series?

NO - we DO have to worry about term 1 issues such as endogeneity for stationary series.

What about our t-ratios from OLS estimation of LR equation?

t-ratios = NOT interpretable As we have a LR equation, there will be some serial correlation due to misspecified dynamics/omitted relevant variable. So we only use LR equation for testing et which only needs correct coefficients.

Step 2 of Engel Granger = | We can only do step 2 if...

Estimate a SR equation = Error-correction model - ONLY do if find cointegration i.e. reject H0 in step one.

ECM equation

change Yt = alpha + B1 change Xt-1 + B2 change Xt-2 + B3 change Yt-1 + gamma et-1 + Rt

Is the ECM equation balanced?

YES - Yt and Xt are first differenced so I(0) | And since we find cointegration et-1 - I(0)

et-1 in ECM is a measure of...

disequilibria of Y from its equilibrium path et-1 = yt-1 - y^t-1 Residuals from LR equation

What is the sign on the error correct term? Why?

Gamma < 0 Because if et-1 > 0, this means Y was above equilibrium last period so we want change Yt < 0 this period to correct the disequilibrium. And visa versa.

gamma = -1 in ECM means

100% of disequilibrium corrected in 1 period

How can we use ECM as another test for cointegration??

Test H0: gamma = 0 - no correction since no equilibria = non-stationary residuals H1: gamma ≠ 0 - correction of disequilibria = there must be an equilibrium relationship = cointegration

Can we carry out tests on ECM?

Yes -all CLRM assumptions hold and distributions of test stats well-behaved as equation is BALANCED.

Do we have to worry about term 1 issues with ECM?

YES - all terms I(0) = stationary.

If we express change Xt as a function of disequilibria in Yt, what is the sign on the error correction term?

et-1 = --bVt-1 If b<0, et-1>0 so we need gamma < 0 If b>0, et-1 < 0 so we need gamma > 0 So in the second case gamma can be positive.

"ECM" equation if we find NO cointegration

NO cointegration = NO equilibrium = get rid of error-correction term. change Yt = alpha + B1change Xt-1 + B2 change Xt-2 + B3 change Yt-1 + Rt

How do we deal with I(0) and I(1) variables in the ECM to make sure the equation is always balanced?

``` First difference I(1) variables Keep I(0) variables as they are. ```

HANDOUT 4 Flashcards

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