Problem Set 3

ECN/ARE 240C - Time Series Econometrics

Professor Òscar Jordà

Winter 2002

1 Theoretical Exercises

1. Let

yt = βtα + εt εt ∼ N(0, σ2)

and α known. Given the normality of εt, the exact distribution of β̂LS

can be obtained.

(a) What is the distribution of β̂ for a generic α?

β̂ − β =

(T∑t=1

t2α

)−1T∑t=1

tαεt

so that

E(β̂ − β

)= 0 and E

[(β̂ − β

)2]= σ2ε

(T∑t=1

t2α

)−1and

β̂ ∼ N

β, σ2ε

(T∑t=1

t2α

)−1(b) What is the distribution of β̂ for α = 0?

√T(β̂ − β

)∼ N

(0, σ2

ε

)(c) What is the distribution of β̂ for α = 1/2? How else can you rescale the problem to obtain this

distribution?

Note:T∑t=1

t2α =T∑t=1

t =T (T + 1)

2≈ O(T 2)

Thusp limT→∞

(β̂ − β

)= p lim

T→∞

√T (β̂ − β) = 0

However,

T(β̂ − β

)∼ N

[0,

2σ2ε(1 + T−1)

]→ N

[0, 2σ2

ε

]1

(d) What is the distribution of β̂ for α = −1/2? Appropriately rescale to obtain the distribution.

This could arise from trending heteroscedasticity when y∗t= β + ε∗

twhere

ε∗t∼ N

[0, σ2

εt]and yt =

y∗t√t

then (β̂ − β

)=

(T∑t=1

t2α

)−1 [T∑t=1

(εt√t

)]Although

∑t−1 ≈ O(logT )→∞, it converges very slowly (see, e.g. Binmore, 1983). Thus:

(logT )1

2

(β̂ − β

)∼ N

0, σ2ε( T∑

t=1

t−1

)−1

log T

→ N[0, σ2εV

]where V is a finite constant.

(e) What is the distribution of β̂ for α = −1? Hint:

limT→∞

∑t−2 = π2/6

(β̂ − β

)d→ N

[0,

6σ2επ2

]Thus β̂ is unbiased but inconsistent for β, since the variance does not go to zero as T →∞.

2. Let the true D.G.P. beyt = yt−1 + ut

where ut is a MDS with variance σ2u and also suptE |ut|2+δ

<∞ for some δ > 0. Also, assume that y0 = 0.Consider the regression,

yt = Xtβ + vt

where Xt = (1 t)′.

(a) Let β̂ denote the OLS estimator, show that,[T−1/2(β̂1 − β1)

T−1/2(β̂2 − β2)

]=⇒

[1 1/21/2 1/3

]−1[ ∫

1

0W (λ)dλ∫ 1

0λW (λ)dλ

]σu

T−1DT (β̂ − β) =[D−1T

∑XtX

′tD−1T

]−1 [T−1D−1

T

∑Xtvt

]vt = yt − βXt

D−1T

∑XtX

′tD−1T

=

[1√T

0

0 1

T3/2

] [(T − 1)

∑t∑

t∑t2

] [1√T

0

0 1

T 3/2

]=[

T−1T

1

T2

∑t

1

T2

∑t 1

T3

∑t2

]→

[1 1/21/2 1/3

]T−1D−1

T

∑Xtvt =

[1

T3/2

∑vt

1

T5/2

∑tvt

]→

[σu∫W (λ)dλ

σu∫λW (λ)dλ

]

by FCLT and CMT where

1

T 5/2

∑tvt =

1

T

∑ t

T

1√T

[Tλ]∑1

εt

=

∫ 1

0

λVT (λ)

To show convergence, we have CMT∣∣∣∣∫ 1

0

λVT (λ)−∫ 1

0

λW (λ)

∣∣∣∣ =

∣∣∣∣∫ 1

0

λ (VT (λ)−W (λ))dλ

∣∣∣∣ ≤∫ 1

0

λ |VT (λ)−W (λ)| < ε∗t

∫ 1

0

λ <∞

which will hold since, by the statement of the problem, supλ |VT (λ)−W (λ)| < ε∗t . Now,

T−1DT

(β̂ − β

)=

[T−1/2β̂1

T 1/2β̂2

]Therefore,

β̂1 diverges to ±∞β̂

p→ 0

Solving out, we have that,

T−1/2β̂1 → σu

[4

∫W (λ)dλ− 6

∫λW (λ)dλ

]T 1/2β̂2 → σu

[12

∫W (λ)dλ− 6

∫λW (λ)dλ

]What is happening? As T →∞, yt drifts to ±∞, taking the mean with it.

(b) What are the probability limits of β̂1 and β̂2? Note: the problem is easier to attack by considering thefollowing

T−1DT (β̂ − β) =[D−1T

(∑XtX

′

t

)D−1

T

]−1 [

T−1D−1T

∑Xtvt

]where vt =

∑ts=1 ut and

DT =

[T 1/2 00 T 3/2

]v̂t = yt − β̂xt

1

σu√Tv̂[Tλ] =

1

σu√Tyt − 1

σu√Tβ̂1 −

1

σu√Tβ̂2t =

=1

σu

[VT (λ)− T−1/2β̂1 − T 1/2β̂2

t

T

]→

σu

σu

[W (λ)− {4 ∫ W (s)ds− 6

∫sW (s)ds

}−{12∫sW (s)ds− 6

∫W (s)ds

}λ

]≡ W (λ)− {4− 6λ}

∫W (s)ds− {6− 12λ}

∫sW (s)ds

therefore, a1(λ) = 4− 6λ, and a2(λ) = 6− 12λ

(c) Derive the limiting behavior of the t-statistic for the null β̂2 = 0. What implications does this have forthe usual procedure of comparing the t-statistic to ±1.96 (for a 5% test)?

t =β̂2√T[

σ̂2vM

2,2]1/2

where M22 is the (2,2) element of [1

T

∑XtX

′

t

]−1

M2,2 =

[[T−1T

1T

∑t

1T

∑t 1

T

∑t2

]−1]2,2

=

T−1T(

T−1T

1

T

∑t2)− ( 1

T

∑t)2 + op(1)

=1

1

T

∑t2 − ( 1T ∑ t

)2 + op(1)

Therefore,

T 2M2,2 =1

1

T

∑t2 − ( 1

T

∑t)2 → 1

1

3− 1

4

= 12

which meansM2,2 → 0

σ̂2

v =1

T

∑v̂2t =

∑(1√Tv̂t

)2

→∞

therefore

T−1σ̂2

v =1

T

∑(1√Tv̂t

)2

→ σu

∫W τ (λ)2dλ

and

T−1/2t =

√T β̂

2[T−1σ̂

2

vT2M2,2

]1/2 →σu

[12∫λW (λ)dλ− 6

∫W (λ)dλ

][12σ2u

∫W τ (λ)dλ

] ≡ 12∫λW (λ)dλ− 6

∫W (λ)dλ

12∫W τ (λ)dλ

and t→±∞. The t-statistic is not valid. Asymptotically, although β̂2→ 0, we will always reject that

β̂2 = 0. Notice that the trick in working out the result is to find the correct rates of convergence forall the parts, and then see what result falls out.

(d) Show that the R2 from the regression in (2) has the random limit

R2 =⇒ 1−∫W τ (λ)2dλ∫Wµ(λ)2dλ

where Wµ(λ) =W (λ)− ∫ W (s)ds.

R2 = 1−∑

v̂2t∑(yt − y)2

=

= 1−1

T 2

∑v̂2t

1

T2

∑(yt − y)2

→ 1−

∫W τ (λ)2dλ∫Wµ(λ)2dλ

by FCLT and CMT. Notice that the result for the numerator was shown above. The CMT holdsbecause it is a ratio and the denominator is positive, almost surely. This result shows that the R2 is arandom variable (between 0 and 1) and therefore, the actual R2 is random, asymptotically.

3. Show that ifyt = α+ ρyt−1 + ut

withut = ψ1ut−1 + ψ2ut−2 + ψ3ut−3 + εt

then yt can be expressed as

yt = µ+ βyt−1 + φ1∆yt−1 + φ

2∆yt−2 + φ

3∆yt−3 + εt

Find the values for φ1, φ2 and φ3, µ and β.

Solution

(1− ψ1L− ψ

2L2 − ψ

3L3)ut = εt

so that

(1− ψ1L− ψ2L2− ψ3L

3)yt = α(1− ψ1 − ψ2 − ψ3)

+ρ(1− ψ1L− ψ

2L2 − ψ

3L3)yt−1 + εt

or

yt = α(1− ψ1 − ψ2 − ψ3) + (ψ1 + ρ)yt−1 + (ψ2 − ρψ1)yt−2 +

(ψ3− ρψ

2)yt−3 − ρψ

4yt−4 + εt

Let

γ1 = (ψ1 + ρ)

γ2 = (ψ2 − ρψ1)

γ3 = (ψ3 − ρψ2)

γ4 = −ρψ4

Then, from the Beveridge-Nelson decomposition,

(1− βL)− (φ1L+ φ2L2 + φ3L

3)(1− L)

= γ1L+ γ

2L2 + γ

3L3 + γ

4L4

where

β = γ1 + ...+ γ4

φ1

= − [γ2+ ...+ γ

4]

...

φ3

= −γ4

Therefore,

µ = α(1− ψ1 − ψ2 − ψ3)

β = ρ+ (1− ρ)ψ1 + (1− ρ)ψ2 + ψ3 − ρψ4

φ1 = ρψ1 + ψ2(ρ− 1)− ψ3 + ρψ4

φ2 = ρψ2 − ψ3 + ρψ4

φ3 = ρψ4

4. First consider the numerator. Using the hint, all we need to do is find the distribution of

1

T 2

T∑t=1

y2t =1

T 2

T∑t=1

(t∑

s=1

es

)2

From the FCLT we know

1

σ√T

t∑s=1

esd→W (.)

Therefore, using FCLT and CMT, notice

1

T 2

T∑t=1

y2t =1

T 2

σ2

T

T∑t=1

(1

σ√T

t∑s=1

es

)2d→ σ2

∫ 1

0

W (r)2dr

Next, consider the denominator. Notice that ∆yt = et,hence it is trivial to show that

1

T

T∑t=1

∆y2tp→ σ2

Putting these two results together completes the proof. Note that if the residuals are not white noise but

they are covariance stationary, then we need to apply FCLT for dependent processes. Hence

1

T 2

T∑t=1

y2td→ σ2ψ(1)2

∫1

0

W (r)2dr

However, notice that the denominator will now converge in probability to

1

T

T∑t=1

∆y2tp→ γ0

so that the limiting distribution becomes

SBd→ ψ(1)2σ2

γ0

∫1

0

W (r)2dr