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METRICS TERM 2 > HANDOUT 4 > Flashcards

HANDOUT 4 Flashcards

1
Q

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

A

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
Q

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

A
  1. genuine cointegrating relations

2. nonsense spurious regressions

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3
Q

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

A

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

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4
Q

Dog/owner example for spurious regression

A

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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5
Q

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

A

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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6
Q

Dog/owner example for cointegration

A

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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7
Q

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

A

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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8
Q

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

A

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

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9
Q

general definition of cointegration

A

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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10
Q

What are d and b usually?

A

d=b=1

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11
Q

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

A

I(0) i.e. stationary

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12
Q

Beta =

A

The cointegrating vector

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

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13
Q

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

A

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

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14
Q

Unbalanced regression =

A

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

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15
Q

A linear combination of I(1) series will…

A

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

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16
Q

Stochastic trend formula

A

Sum j=1,…,t €j

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17
Q

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

A

E(€1t, €2t) = 0

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18
Q

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

A

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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19
Q

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

A

Engel and Granger

20
Q

Step 1 of Engel and Granger =

A

Estimate a LR equilibrium equation

21
Q

2 types of LR equilibrium equations we can estimate

A
  1. static

2. dynamic (include lags)

22
Q

3 steps for static LR equation engel and granger

A
  1. Reg Yt = d0 + d1Xt + €t by OLS
  2. save et = Yt - (d0 + d1Xt)
  3. Test if et - I(0) using ADF model B
23
Q

ADF test on residuals for static LR equation

A

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

24
Q

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

A

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

25
Q

3 steps for dynamic LR equation engel and granger

A
  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
26
Q

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

A

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

27
Q

Solution to problem with our CVs for ADF test on residuals

A

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.

28
Q

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

A

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

29
Q

Limitation of ADF test =

A

LOW POWER

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

30
Q

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

A

SUPER-CONSISTENT

31
Q

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

A

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.

32
Q

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

A

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

33
Q

Do we have super-consistency with stationary series?

A

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

34
Q

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

A

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.

35
Q

Step 2 of Engel Granger =

We can only do step 2 if…

A

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

36
Q

ECM equation

A

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

37
Q

Is the ECM equation balanced?

A

YES - Yt and Xt are first differenced so I(0)

And since we find cointegration et-1 - I(0)

38
Q

et-1 in ECM is a measure of…

A

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

39
Q

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

A

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.

40
Q

gamma = -1 in ECM means

A

100% of disequilibrium corrected in 1 period

41
Q

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

A

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

42
Q

Can we carry out tests on ECM?

A

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

43
Q

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

A

YES - all terms I(0) = stationary.

44
Q

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

A

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.

45
Q

“ECM” equation if we find NO cointegration

A

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

46
Q

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

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

METRICS TERM 2 (6 decks)

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