BEE2006 Statistics and...

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.


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 .


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


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


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.


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.


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?


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).


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?


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.



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.



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.


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.


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.


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.


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.


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.


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.


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?


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.


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.


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


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