Applications of EconometricsTutorial Sheet 4
Unit Roots
1. How would you test whether the following model has a unit root? Would this change if the model
contained a time trend?
yt = α + ρyt−1 + et , t = 1, 2, …,
Solution: Because there are multiple ways for the above model to be a stable AR(1) process (ρ =
0.2, ρ = 0.4, ρ = 0.57, …), and only one way for it to be a I(1) process (ρ = 1), it makes sense to
think about the null hypothesis as H0 : ρ = 1 and the alternative hypothesis as H1 : ρ < 1. If the
equation is transformed by subtracting yt−1 from both sides,
∆yt = α + θyt−1 + et ,
then this test can be carried out on θ instead of ρ where θ = ρ − 1. In this model the null hypothesis
is now θ = 0 and the alternative hypothesis is θ < 0. Because under the null hypothesis yt is I(1) the
series is not weakly dependent and the test statistics will not be standard normally distributed. This
means the normal critical values will not apply. We need to use critical values from a distribution
known as a Dickey-Fuller distribution (this test is also known as a Dickey-Fuller (DF) test). These
values can be found in the textbook.
If the model contained a time trend we would have to amend the test. The critical values for
the model with a time trend are larger in magnitude (more negative) than the critical values for the
test without a time trend. This is because time series which are trend stationary (they have a linear
trend but are I(0) around the trend) can appear as if they have a unit root.
2. How would you test whether the following model has a unit root?
yt = α + ρ1 yt−1 + ρ2 yt−2 + et , t = 1, 2, ...,
Solution: First, add and subtract ρ2 yt−1 from the equation
yt = α + ρ1 yt−1 + ρ2 yt−2 + et + ρ2 yt−1 − ρ2 yt−1
= α + (ρ1 + ρ2 )yt−1 − ρ2 (yt−1 − yt−2 ) + et
Now you can subtract yt−1 from both sides of the equation
yt − yt−1 = α + (ρ1 + ρ2 )yt−1 − ρ2 (yt−1 − yt−2 ) + et − yt−1
∆yt = α + (ρ1 + ρ2 − 1)yt−1 − ρ2 ∆yt−1 + et
∆yt = α + θyt−1 + γ∆yt−1 + et
Here θ = ρ1 + ρ2 − 1 and γ = −ρ2 . Testing θ = 0 is the same as testing that ρ1 + ρ2 = 1 which is
the requirement for a unit root in an AR(2) model. Here the null hypothesis, H0 : θ = 0, is that
there is a unit root in the AR(2) model. The alternative hypothesis, H1 : θ < 0, is that it is a stable
AR(2) process.
3. What is the name of the problem that occurs when two independent I(1) processes are used in a regression
together?
Solution: This is known as a spurious regression. The reason for this is that even though the series
are independent, and we should expect to find estimates of β1 close to zero, we will end up rejecting
the null hypothesis of no effect (H0 : β1 = 0) far more than 5% of the time.
The reason for this is that if an I(1) series yt is regressed on another I(1) series xt the errors in
the following equation,
yt = β0 + β1 xt + et
must follow a random walk under H0 : β1 = 0. This violates the Gauss-Markov assumptions we
used from Chapter 11.
Cointegration
4. What does it mean for two variables to be cointegrated?
Solution: If yt and xt are two I(1) processes then in general yt − βxt will also be an I(1) process
for any number β. Nevertheless, it is possible that for some β 6= 0, yt − βxt is an I(0) process. If
such a β exists, we say that y and x are cointegrated, and we call β the cointegration parameter.
Why is it useful to know this? Well, if two variables are cointegrated, then they have a long
run relationship. A good example of this is the interest rate on U.S. treasury bills. Treasury bills
are securities which are sold at maturity lengths of 1-month, 3-months, 6-months, and 12-months
with a promise to pay a face value amount (typically $100) at maturity. They are sold at less than
face value to create a positive return to those who hold them. For example, if you pay $90 for a
3-month treasury bill, the interest rate is 100−90
= 0.11 = 11%. We saw in Tutorial 2 that the
90
interest rate on 3-month treasury bills was I(1). The same also goes for 6-month treasury bills.
Because of the longer maturity on 6-month treasury bills they pay a higher return. This difference
in returns between a 3-month treasury bill and 6-month treasury bill is known as the yield spread.
However, this difference would not be expected to grow or shrink over time due to arbitrage. While
the difference between them will change over time it is more likely that it fluctuates around a mean.
Econometrically, this can be represented as
r6t = r3t + µ + et
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where et is a zero mean, I(0) process. At any time period, there can be deviations from equilibrium,
but they will be temporary: there are economic forces that drive r6 and r3 back toward the equilibrium relationship. This is the application exercise at the end of the tutorial.
Also, if we know that two series are cointegrated we can test hypotheses about the cointegration
parameter (using a leads and lags estimator) and we can use the cointegrated variables as additional
explanatory variables when examining short run dynamics (in an error correction model).
5. How do you test for cointegration?
Solution: Before testing for cointegration you should have some idea that both series are I(1). This
can be done by examining the series’ autocorrelation coefficients and performing DF tests on them.
To test for cointegration we perform a DF test on the residuals of the following regression,
yt = α + βxt + ut ,
since ut = yt − α − βxt is supposed to be I(0) for yt and xt to be cointegrated. The null hypothesis,
that there is a unit root, means that yt and xt are not cointegrated. If we reject the null hypothesis
then we find that yt and xt are cointegrated. Because we have to take into account the estimation
of β in this method this test is called an Engle-Granger test and uses different critical values from
the traditional DF test.
6. When yt and xt are cointegrated in the following equation
yt = α + βxt + ut ,
OLS produces consistent estimates of α̂ and β̂. However, because xt is I(1), we cannot use the traditional
inference techniques to test hypotheses about β as the estimator won’t have a normal distribution, even
asymptotically. However, if we are willing to make the assumption that xt is strictly exogenous, then the
estimators will be normally distributed provided that the errors are homoskedastic, serially uncorrelated,
and normally distributed, conditional on the explanatory variables (see Time Series assumptions from
Tutorial 1 for a review). How would you go about improving the model to try and ensure that xt is
strictly exogenous?
Solution: The solution to this is the leads and lags estimator of β. We construct a new set of errors
which are, at least approximately, strictly exogenous. However, because xt is I(1), we need these
new error terms to be strictly exogenous with regards to ∆xt (remember that xt being I(1) means
it takes a form similar to xt = xt−1 + vt ). This can be achieved by writing ut as a function of the
∆xs for all s close to t. For example,
ut = η + φ0 ∆xt + φ1 ∆xt−1 + φ2 ∆xt−2 + γ1 ∆xt+1 + γ2 ∆xt+2 + et
where, by construction, et is uncorrelated with each ∆xs appearing in the equation. The hope is
that et is uncorrelated with further lags and leads of ∆xs . We know that, as |s − t| gets large, the
correlation between et and ∆xs approaches zero, because these are I(0) processes. Now, if we plug
this into the original equation we get
yt = (α + η) + βxt + φ0 ∆xt + φ1 ∆xt−1 + φ2 ∆xt−2 + γ1 ∆xt+1 + γ2 ∆xt+2 + et
This equation looks a bit strange because future ∆xs appear with both current and lagged ∆xt .
The key is that the coefficient on xt is still β, and, by construction, xt is now strictly exogenous in
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this equation.
As an example, and a way to think about the leads and lags model conceptually, imagine we are
interested in examining the following model:
M urderRatet = α + βP oliceP Ct + ut
where M urderRatet is the murder rate in a given city at a time t, and P oliceP Ct represents the
number of police per capita at time t. Imagine there is a shock in the murder rate at time t so
that the murder rate increases by more than expected, given the PolicePC at time t. This shock
will be captured by ut . It is quite likely that this shock may result in a change in the number of
police per capita at time t + 1. Therefore, an increase in the error term at time t is correlated with
∆P oliceP Ct+1 . However, it is unlikely that this shock will feed into later changes in police per
capita. In other words, as we move further away from t, we expect the correlation to tend to zero
if both ∆P oliceP Ct and et are I(0). Therefore, we do not expect an increase in et to be correlated
with, say, ∆P oliceP Ct+5 . This explains why we include ∆xs for s close to t. The inclusion of
these terms accounts for possible ways in which the assumption that xt is strictly exogenous may
be violated.
7. If yt and xt are two I(1) processes, and they are NOT cointegrated, then both of the variables can be
used in a regression together when they are used in their difference form. For example,
∆yt = α0 + α1 ∆yt−1 + γ0 ∆xt + γ1 ∆xt−1 + ut
However, if you knew that yt and xt were cointegrated, how could this equation be improved?
Solution: If yt and xt are cointegrated with parameter β, then we have additional I(0) variables
that we can include in the above equation. Let st = yt − βxt , so that st is I(0), and assume for the
sake of simplicity that st has zero mean. This is known as the error correction term. Now, we can
include lags of st in the equation. In the simplest case, this gives
∆yt = α0 + α1 ∆yt−1 + γ0 ∆xt + γ1 ∆xt−1 + δst−1 + ut
∆yt = α0 + α1 ∆yt−1 + γ0 ∆xt + γ1 ∆xt−1 + δ(yt−1 − βxt−1 ) + ut ,
where δ < 0. If yt−1 > βxt−1 , then y in the previous period has overshot the equilibrium; because
δ < 0, the error correction term works to push y back toward the equilibrium. Similarly, if yt−1 <
βxt−1 , the error correction term induces a positive change in y back toward the equilibrium. This
is an example of an error correction model.
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Forecasting
8. What value should we choose for ft , the forecast of yt+1 at time t, in order to minimize the squared
error loss function?
E(e2t+1 |It ) = E[(yt+1 − ft )2 |It ]
Solution: In this case we should choose ft = E(yt+1 |It ) to minimze the squared error loss function.
This result is a common result in statistics: means minimize mean squared errors.
Consider the following example where E(X) = µ. You would like to choose a value c which minimizes
the following expression, E(X − c)2 . Start by exanding it out, and then adding and subtracting µ2 ,
E(X − c)2 = E(X 2 − 2Xc + c2 )
= E(X 2 − 2Xc + c2 + µ2 − µ2 )
= E(X 2 − µ2 + µ2 − 2Xc + c2 )
= E(X 2 ) − µ2 + µ2 − 2µc + c2
= V ar(X) + µ2 − 2µc + c2
= V ar(X) + (µ − c)2
So if we choose c = µ the expression collapses down to V ar(X). If c 6= µ then we are adding some
positive term on to it meaning we are not minimizing it. This is why choosing ft = E(yt+1 |It )
minimizes the squared error loss function.
9. Consider the following simple forecast model with n time periods,
yn = β0 + β1 yn−1 + δ1 xn−1 + un .
The point forecast of yn+1 , at time n, is fˆn :
fˆn = β̂0 + β̂1 yn + δ̂1 xn ,
where we use a hat on fn to emphasize that we have estimated the parameters used in the forecast in a
regression model. How would you construct a 95% prediction interval for this point forecast?
Solution: If the point forecast is fˆn then we can construct a 95% forecast interval using the traditional methods: fˆn ± 1.96 ∗ SE. However, there is an important difference to how we have done this
previously. Traditionally we have added and subtracted 1.96 time the standard of the estimator. In
this case that would be ±1.96 ∗ SE(fˆn ).
Constructing a forecast interval is slightly different. We have to take into acccount two types
of uncertainty: uncertainty arising from estimation of the regression coefficients and uncertainty
associated with the future unknown value of un+1 , the regression error. For regressions with few coefficients and many observations, the uncertainty arising from future un+1 can be much larger than
the uncertainty associated with estimation of the parameters. In general, however, both sources of
uncertainty are important, so we now develop an expression for the root mean squared error that
incorporates these two sources of uncertainty.
The forecast error - which we will not know until time n + 1 - is
ên+1 = yn+1 − fˆn
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The standard error of ên+1 , which is what we would like to construct our prediction interval with,
is an estimator of what is called the root mean squared forecast error (RMSFE). This can be found
by taking the square root of the variance of ên+1 , the mean squared forecast error E(ê2n+1 ).
Var(ên+1 ) = Var(yn+1 − fˆn )
= Var(β0 + β1 yn + δ1 xn + un+1 − fˆn )
As we are examining the variance conditional on In , we can treat the β0 + β1 yn + δ1 xn as constant:
= Var(un+1 − fˆn )
We can re-write the variance of the sum as the sum of the variance as un+1 and fˆn are independent.
They are independent because the estimators in fˆn depend only on previous error terms and not on
un+1 . Therefore, the variance of the difference is the sum of the variances.
= Var(un+1 ) + Var(fˆn )
If we find values for the two terms on the right hand side then we can calculate our standard error.
The first term is relatively straight forward. We can estimate this from the residuals in the model
using the following formula
P 2
n ûn
σ̂ 2 =
.
n−k−1
where k is the number of explanatory variables in the model. The second term is more complicated.
We first need to calculate the SE(fˆn ) and then square it to get the variance of fˆn . Now we write fn
as β0 = fn − β1 yn − δ1 xn and plug it into the original regression equation to get
yt = fn + β1 (yt−1 − yn ) + δ1 (xt−1 − xn ) + ut .
Note how I switched notation from n to t. This is to illustrate that we need to subtract the value
of yn and xn (where n is the most recent time period we have) from each yt−1 and xt−1 that we
will be using as part of the regression. Estimating this regression will give a standard error for the
constant term which is the standard error of fˆn . Plug both of these values into the above formula
and then take the square root.
q
q
p
SE(ên+1 ) = Var(ên+1 ) = Var(un+1 ) + Var(fˆn ) = σ̂ 2 + SE(fˆn )2
The 95% prediction interval is then fˆn ± 1.96 ∗ SE(ên+1 )
10. What is an example of in-sample criteria when choosing between different forecast methods? What
about out-of-sample criteria? Which do you think is most important?
Solution: In a regression context, in-sample criteria include R-squared and especially adjusted Rsquared. Out-of-sample criteria revolve around splitting your sample into an estimation sample and
a forecasting sample and finding different ways of minimizing the forecast errors. Some examples
of this are the root mean squared error which is essentially the sample standard deviation of the
forecast errors (without any degrees of freedom adjustment) and the mean absolute error which is
the average of the absolute forecast errors. For forecasting, it is better to use out-of-sample criteria,
as forecasting is essentially an out-of- sample problem.
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Application
11. The following link brings you to the FRED website with monthly interest rate data for 3-month T-bills
and 6-month T-bills for 1959 onwards.
https://fred.stlouisfed.org/graph/?id=TB3MS,TB6MS
Both series appear to follow random walks. However what do you notice about the random walks?
Download the data in Excel or csv format format (1959+) and load it into Stata and confirm that each
series follows a random walk but that the difference between the series is I(0).
Solution: Despite the fact that both series appear to follow random walks they follow the same
path. This means that the difference between them is the same throughout the whole time series
and follows an I(0) process. The solution to question 4 should explain the economic theory behind
why we expect both of these series to get cointegrated but this offers you the chance to download
the data and prove it for yourself. Autoregressions and DF tests should confirm both of the series
display unit root behavior. To test for cointegration you can use the ‘egranger’ command for Stata,
though you have to download it. Once you do though, executing ‘egranger tb6ms tb3ms’ should
confirm that the series are cointegrated.
. ssc install egranger
checking egranger consistency and verifying not already installed...
all files already exist and are up to date.
. import delimited using "fredgraph.csv", clear
(3 vars, 745 obs)
. g monthly_date = mofd(date(date,"YMD"))
. format %tm monthly_date
. tsset monthly_date
time variable:
delta:
monthly_date, 1959m1 to 2021m1
1 month
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. egranger tb6ms tb3ms
Engle-Granger test for cointegration
N (1st step) =
745
N (test)
=
744
-----------------------------------------------------------------------------Test
1% Critical
5% Critical
10% Critical
Statistic
Value
Value
Value
-----------------------------------------------------------------------------Z(t)
-8.376
-3.911
-3.344
-3.050
Critical values from MacKinnon (1990, 2010)
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