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BEE2006 Statistics and Econometrics
Tutorial 4
February 2013
BEE2006 Statistics and Econometrics
Question 11.4
I Let {yt : t = 1, 2, ...} follow a random walk with y0 = 0.I Show that Corr (yt, yt+h) =
√t/ (t+ h) for t ≥ 1, h > 0.
I Random walk process, for t = 1, 2, ...
yt = yt−1 + et
= yt−2 + et + et−1
= yt−3 + et + et−1 + et−2
...= y0 + et + et−1 + et−2 + · · ·+ e1.
I Assume that et ∼ i.i.d.(0, σ2
e)
for t ≥ 1.
BEE2006 Statistics and Econometrics 2012
I Properties of the random work process: Expectation
E (yt) = E (y0 + et + et−1 + et−2 + · · ·+ e1)
= E (y0) + E (et) + · · ·+ E (e1)
= E (y0)
I E (yt) is not depending on time, t.I With popular assumption of y0 = 0,I E (yt) = 0 for all t.
BEE2006 Statistics and Econometrics 2012
I Properties of the random work process: VarianceI Assume that y0 is nonrandom, (y0 = 0) ,I therefore Var (y0) = 0, then
Var (yt) = Var (y0 + et + et−1 + et−2 + · · ·+ e1)
= Var (0) +Var (et) + · · ·+Var (e1)
= 0+ σ2e + σ2
e + · · ·+ σ2e︸ ︷︷ ︸
t
= tσ2e .
I On the other hand, for Var (yt+h)
Var (yt+h) = Var (et+h + et+h−1 + · · ·+ et + et−1 + · · ·+ e1)
= Var (et+h) + · · ·+Var (et) + · · ·+Var (e1)
= σ2e + σ2
e + · · ·+ σ2e︸ ︷︷ ︸
t+h
= (t+ h) σ2e .
BEE2006 Statistics and Econometrics 2012
I Properties of the random work process: CovarianceI Under the assumption of et ∼ i.i.d.
(0, σ2
e)
for t ≥ 1.I Cov (et, es) = 0 for t 6= s, y0 = 0, E (yt+h) = 0.
Cov (yt, yt+h) = E (ytyt+h)
= E [(et + · · ·+ e1) (et+h + · · ·+ et + · · ·+ e1)]
= E(
e2t
)+ E
(e2
t−1
)+ · · ·+ E
(e2
1
)= Var (et) +Var (et−1) + · · ·+Var (e1)
= σ2e + σ2
e + · · ·+ σ2e︸ ︷︷ ︸
t
= tσ2e
BEE2006 Statistics and Econometrics 2012
I Properties of the random work process: Correlation
Corr (yt, yt+h) =Cov (yt, yt+h)√
Var (yt)Var (yt+h)
=tσ2
e√tσ2
e × (t+ h) σ2e
=tσ2
e√t (t+ h)σ2
e
=t√
t (t+ h)=
√t2
t (t+ h)
=
√t
t+ h
BEE2006 Statistics and Econometrics 2012
Question 11.6
I Let hy6t denote the 3-month holding yield (in percent)from buying a six-month T-bill at time (t− 1) and selling itat time t (three month hence) as a 3-month T-bill.
I Let hy3t−1 be the 3-month holding yield from buying a3-month T-bill at time (t− 1) .
I At time (t− 1), hy3t−1 is known, whereas hy6t is unknownbecause p3t (the price of 3-month T-bill) is unknown attime (t− 1) .
I The Expectation Hypothesis says that these two different3-month investments should be the same, on average.
BEE2006 Statistics and Econometrics 2012
Question 11.6
I We can write this as a conditional expectation:
E (hy6t|It−1) = hy3t−1,
I It−1: all observable information up through time t− 1.
I This suggests estimating the model
hy6t = β0 + β1hy3t−1 + ut
I Also, suggests testing H0 : β1 = 1.
BEE2006 Statistics and Econometrics 2012
Question 11.6 (a)
I Estimated model using given data is given by
hy6t = −0.058(0.070)
+ 1.104(0.039)
hy3t−1,
n = 123, R2 = 0.866.I Do you reject H0 : β1 = 1 against H1 : β1 6= 1 at the 1%
significance level?I Does the estimate seem practically different from one?
BEE2006 Statistics and Econometrics 2012
Question 11.6 (a): Solutions
I t− statistics for the null (H0 : β1 = 1)
t =1.104− 1
0.039≈ 2.67,
I df = 123− 2 = 121.I Critical Value at α = 1% : about 2.62.
I We can reject the null.I However, it would be difficult to strongly believe in that β1
is not the unity because 2.62 ≈ 2.67I Practically, we can observe that β1 is 10% higher than the
theoretical value (β1 = 1).
BEE2006 Statistics and Econometrics 2012
Question 11.6 (b)
I Another implication of the Expectation Hypothesis:I no other variables dated as t− 1 or earlier should help
explain hy6t, once hy3t−1 has been controlled for.
I Including one lag of the spread between 6-month and3-month T-bill rates gives
hy6t = −0.123(0.067)
+ 1.053(0.039)
hy3t−1 + 0.480(0.109)
(r6t−1 − r3t−1) ,
n = 123, R2 = 0.885.I (1) Testing the null H0 : β1 = 1.I (2) Is the lagged spread term significant?I (3) If, at time t− 1, r6 is above r3, should you invest in
6-month or-3-month T-bills?
BEE2006 Statistics and Econometrics 2012
Question 11.6 (b): Solution for (1)
I t− statistics for the null (H0 : β1 = 1)
t =1.053− 1
0.039≈ 1.36,
I df = 123− 2 = 121.I Critical Values for this two-tailed test:
α = 1% α = 5% α = 10%2.62 1.98 1.66
I We fail to reject the null at any significance level.I Therefore, it is able to believe in that β1 = 1.
BEE2006 Statistics and Econometrics 2012
Question 11.6 (b): Solution for (2)
I Significance of lagged variable: spread = r6t−1 − r3t−1
I The null hypothesis: H0 : γ2 = 0
t =0.480− 0
0.039≈ 4.40,
I df = 123− 2 = 121.I Critical Values for this two-tailed test:
α = 1% α = 5% α = 10%2.62 1.98 1.66
I We can strongly reject the null at any significance level.I Therefore, the lagged variable, spread is very significant.
BEE2006 Statistics and Econometrics 2012
Question 11.6 (b): Solution for (3)
I Recall,
hy6t = −0.123(0.067)
+ 1.053(0.039)
hy3t−1 + 0.480(0.109)
(r6t−1 − r3t−1) ,
I At time t− 1, ifr6t−1 > r3t−1,
I Then, 0.480× (r6t−1 − r3t−1) is positive. Therefore, we cananticipate that
E (hy6t|It−1) > hy3t−1.
I Thus, we have to invest in 6-month T-bills rather than toinvest in 3-month T-bills.
BEE2006 Statistics and Econometrics 2012
Question 11.6 (c)
I The sample correlation between hy3t and hy3t−1 is 0.914.I Why might this raise some concerns with the previous
analysis?
I Persistence in Time SeriesI Weakly dependent process: Integrated of order zero, I (0) .
(or almost independent process)I Unit root process: Integrated of order one, I (1) .I For I (1) , the first difference of the process is weakly
dependent process.
BEE2006 Statistics and Econometrics 2012
Question 11.6 (c)
I Detecting a time series I (1) : First-order autocorrelation
ρ1 = Corr (yt, yt−1) .
I If it is very closed to 1, we would think the stochasticprocess {yt} follows the unit root.
I Therefore, {hy3t} has high possibility to be I (1) because
Corr (hy3t, hy3t−1) = 0.914.
I So that, the conducted t− tests in earlier stages would notbe valid in this context.
BEE2006 Statistics and Econometrics 2012
Question 11.6 (d)
I How would you test for seasonality in the equationestimated in part (b)?
I Adding three quarterly dummy (seasonal dummy)variables
I The unrestricted model is given by
hy6t = α0 + β1hy3t−1 + γ2 (r6t−1 − r3t−1)
+δ2Q2t + δ3Q3t + δ4Q4t + ut
I The null hypothesis: H0 : δ2 = 0, δ3 = 0, δ4 = 0.I Construct F− test with q = 3 and dfur = 123− 5− 1 = 117.I Find the critical value from F3,117 distribution.
BEE2006 Statistics and Econometrics 2012
Question 11.7
I A partial adjusted model is
y∗t = γ0 + γ1xt + et
yt − yt−1 = λ (y∗t − yt−1) + at,
I y∗t is the desired or optimal level of y.I yt is the actual (observed) level of y.I γ1 measures the effect of xt on y∗t .
I The second equation describes how the actual y adjustsdepending on the relationship between y∗t and yt−1.
I λ measures the speed of adjustment and satisfies 0 < λ < 1.
BEE2006 Statistics and Econometrics 2012
Question 11.7 (a)
I Plug the first equation for y∗t into the second equation andshow that we can write
yt = β0 + β1yt−1 + β2xt + ut.
I Find βj in terms of γj and λ.I Find ut in terms of et and at.I Therefore, the partial adjustment model leads to a model
with lagged dependent variable and a contemporaneous x.
BEE2006 Statistics and Econometrics 2012
Question 11.7 (a): Solution
I Plugging the first equation for y∗t into the second equation,
yt − yt−1 = λ (γ0 + γ1xt + et − yt−1) + at
I Then, we can get
yt = λγ0 + (1− λ) yt−1 + λγ1xt + at + λet
I Let β0 = λγ0, β1 = (1− λ) , β2 = λγ1, and ut = at + λet.
I Finally, we can see the following equation consists of yt−1and xt.
yt = β0 + β1yt−1 + β2xt + ut.
BEE2006 Statistics and Econometrics 2012
Question 11.7 (b)
I Suppose that
E [et|xt, yt−1, xt−1, ...] = E [at|xt, yt−1, xt−1, ...] = 0.
I In addition, assume that all series are weakly dependent.(No unit root)
I How would you estimate βj?
BEE2006 Statistics and Econometrics 2012
Question 11.7 (b): Solution
I An OLS regression of yt on yt−1 and xt produces consistent,asymptotically normal estimator of βj.
yt = β0 + β1yt−1 + β2xt
I For ut, under the assumption of
E [et|xt, yt−1, xt−1, ...] = E [at|xt, yt−1, xt−1, ...] = 0.
BEE2006 Statistics and Econometrics 2012
Question 11.7 (b): Solution
I We can derive that
E [ut|xt, yt−1, xt−1, ...] = 0.
I Since ut is the function of yt, yt−1 and xt, for t 6= s,
Cov (ut, u|xt, yt−1, xt−1, ...) = 0.
I Therefore, it is obvious that the errors are seriallyuncorrelated and the model is dynamically completed.
BEE2006 Statistics and Econometrics 2012
Question 11.7 (b)
I If β1 = 0.7 and β2 = 0.2,I What are the estimates of γ1 and λ?
I Since we defined that
β1 = 1− λ and β2 = λγ1
I Therefore,
λ = 1− β1
= 0.3
γ1 =β2
λ= 0.67
BEE2006 Statistics and Econometrics 2012
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