Oct 25, 2001 - The latter can be evaluated from the following well known series representation of the logarithmic integral (e.g., Gradshteyn and Ryzhik, 1967,.
Regression with an Evaporating Logarithmic Trend∗ Peter C. B. Phillips Cowles Foundation, Yale University, University of Auckland & University of York and Yixiao Sun Department of Economics Yale University
October 25, 2001
Abstract Linear regression on an intercept and an evaporating logarithmic trend is shown to be asymptotically collinear to the second order. Consistency results for least squares are given, rates of convergence are obtained and asymptotic normality is established for short memory errors.
JEL Classification: C22 Key words and phrases: Asymptotic theory, asymptotic expansions, collinear regressors, linear process, linear regression, logarithmic factors, logarithmic integral. ∗
Phillips thanks the NSF for research support under Grant Nos. SBR 97-30295 and SES 0092509. Sun thanks the Cowles Foundation for support under a Cowles Prize Fellowship. The paper was typed by the authors in SW2.5. Computing was done in GAUSS.
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1. Problem Part A The time series Xt is generated by the model Xt = α +
β + ut , ln t
t = 2, ..., n
(1)
where α and β are unknown parameters whose least squares regression estimates are b respectively. The error u in (1) is assumed to be iid (0, σ 2 ) with b and β, denoted by α t finite fourth moment. b are strongly consistent for α and β as n → ∞. b and β 1. Show that α b b and β. 2. Find the asymptotic distribution of α
Part B Suppose that ut in (1) is the linear process ut =
∞ X
ci εt−i ,
with
i=0
∞ X
j |cj | < ∞,
(2)
j=0
and where εt is iid (0, σ 2 ) with finite fourth moment. Explain how you would modify your derivations in Part A to allow for such an error process in the regression model (1).
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2. Solution Part A Let zt =
1 ln t
1 n−1
−
Pn
1 t=2 ln t .
Then Pn
βb − β
t=2 zt ut , 2 t=2 zt Pn t=2 ut
=
Pn
b −α = α
ln−1 t . n−1
Pn
t=2
− (βb − β)
n−1
We start with βb and find an asymptotic representation of the components nt=2 1/ ln t P P and nt=2 1/ ln2 t that appear in nt=2 zt2 . Using Euler summation (which justifies (3) below) and partial integration (which justifies (4) below) we obtain an asymptotic series representation as follows: P
n X 1
n
1 dx + O (1) 2 ln x t=2 ln t Z n m m X X (k − 1)! (k − 1)! 1 −2 + m! = n m+1 dx + O(1) k k 2 ln x k=1 ln n k=1 ln 2 Z m n X (k − 1)! 1 + m! = n m+1 dx + O(1). k 2 ln x k=1 ln n Z
=
(3) (4) (5)
To show that (5) is a valid asymptotic series, we can ignore the O(1) term and by a further application of partial integration to the second term we see that the remainder is proportional to Rm+1 =
Z
n
1 m+1
ln
2
x
dx =
n m+1
ln
−
n
2 m+1
ln
2
+ (m + 1)
Z
n
1 m+2
ln
2
x
dx.
(6)
Let L = nα for some α ∈ (0, 1) and write Z 2
n
1
Z
L
1
n
1
dx = O (nα ) +
Z
n
1
(7)
1 Zn 1 1 Zn 1 1 Rm+1 , m+2 dx ≤ m+1 dx ≤ m+1 dx = ln L L ln ln L 2 ln ln L ln x x x
(8)
x
n
1
ln
2
m+2
ln
x
dx +
Z
dx.
m+2
dx =
L
m+2
ln
x
L
m+2
ln
x
But Z
L
It follows from (6)-(8) that Rm+1 ≤
n m+1
ln
n
+ (m + 1)
Z
n
1 m+2
dx ≤
2 ln x m+1 = + Rm+1 + O (nα ) ln L lnm+1 n
n
3
n m+1
ln
n
+
Z 2
L
m+1 m+1 Rm+1 m+2 dx + ln L ln x
so that
n
Rm+1 = O and thus
n X 1 t=2
ln t
=n
,
lnm+1 n
m X (k − 1)! k
k=1
ln n
+ O(
(9) n m+1
ln
n
),
(10)
showing that (5) is a valid asymptotic series. In a similar way we may establish that n X
Z n 1 1 = 2 2 dx + O (1) 2 ln x t=2 ln t m X n (k − 1)! +O = n k lnm+1 n k=2 ln n
(11) (12)
is a valid asymptotic series. Combining (10) and (12), we get n X
zt2 =
t=2
= =
=
!2
n X
n X 1 1 1 − 2 n − 1 t=2 ln t t=2 ln t Z n 2 Z n 1 1 1 dx − dx + O (1) 2 n − 1 2 ln x 2 ln x n 2n 6n n + + +O ln2 n ln3 n ln4 n ln5 n 2 1 n n 2n n − + 2 + 3 +O n ln n ln n ln n ln4 n n n +O →a.s. ∞, 4 ln n ln5 n
(13) (14)
(15)
observing that we have to go to the third order terms in expansion (14) to avoid degeneracy. We can obtain higher order asymptotic expansions by taking the above process to further terms, leading to the following explicit expression to order n/ ln10 n n X
8 56 408 3228 28032 267264 n n + 2 + 3 + 4 + 5 + = 4 1+ +O . 6 11 ln n ln n ln n ln n ln n ln n ln n ln n t=2 (16) It is immediately apparent from the size of the coefficients in (16) that very large values of n are required before these approximations can be expected to work well. P Of course, such approximations are hardly necessary since nt=2 zt2 is amenable to direct calculation or to direct approximation using the logarithmic integrals given in (3) and (11) above. The latter can be evaluated from the following well known series representation of the logarithmic integral (e.g., Gradshteyn and Ryzhik, 1967, 8.213.2, p. 926) zt2
li(y) =
Z 0
y
∞ X 1 (ln y)k dx = γ + ln ln y + , ln x k=1 kk!
4
for y > 1,
(17)
where γ is Euler’s constant. In the present case we have Z
n
2
1 dx = li (n) − li (2) , ln x
and Z 2
n
Z n 1 2 n 1 2 n dx = − + dx = − + [li (n) − li (2)] . 2 ln 2 ln n ln 2 ln n 2 ln x ln x
Observe that βb − β is the same as the error in the OLS estimator of β in the regression Xt = zt β + ut , where {ut } is a martingale difference sequence with respect to the natural filtration Ft . The persistent excitation condition (15) holds in this regression and so βb →a.s. β and βb is strongly consistent. Now observe that n ut ln−1 t b −α = − (βb − β) t=2 α n−1 n−1 = oa.s. (1) + oa.s. (1)Oa.s. (ln−1 n) = oa.s. (1),
Pn
P
t=2
b →a.s. α. and so α b We have b and β. Next, turn to the asymptotic distribution of α Pn
βb − β = =
t=2 zt ut 2 t=2 zt Pn t=2 zt ut
Pn
sn
√ !−1 n X n 2 (1 + o(1)), s = zt2 . n ln2 n t=2
Let ynt = zt ut /sn , then yn2, yn3 , ..., ynn are independent random variables with zero means and variances that sum to σ 2 . To apply the Liapounoff central limit theorem P for nt=2 ynt , we need to show that n X
E|ynt |3 → 0, as n → ∞.
t=2
But n X
3
E|ynt | =
Pn
t=2
t=2
and so
n X t=2
ln4 n |zt |3 E|ut |3 ≤ Const.n |sn |3 n Pn
ynt =
t=2 zt ut
sn 5
⇒ N (0, σ 2 ).
!3/2
= o(1),
Hence
√
n b 2 (β − β) = ln n
Pn
t=2 zt ut
sn
(1 + op (1)) ⇒ N (0, σ 2 ).
(18)
Note that −1 n ut t t=2 ln b = − (β − β) n−1 n−1 −1/2 −1/2 = Op (n ) − Op (n ln2 n ln−1 n) = Op (n−1/2 ln n).
Pn
b −α α
P
t=2
P b and we Thus, the term (βb − β)(n − 1)−1 nt=2 ln−1 t dominates the asymptotics of α deduce that √ √ n b −1 b − α) = − 2 (β n ln n(α − β) + op (1) =d N (0, σ 2 ). (19) ln n
In view of (18) and (19), we have the following joint asymptotics: b − α) n1/2 ln−1 n(α −2 1/2 n ln n(βb − β)
!
⇒ N (0, Σ) with Σ = σ
2
1 −1 −1 1
!
.
(20)
Remarks b − α) 1. The limit distribution (20) is singular and the components n1/2 ln−1 n(α −2 1/2 b and n ln n(β − β) are perfectly negatively correlated as n → ∞. Observe b by virtue of the fact that b exceeds that of β, that the rate of convergence of α the signal from the intercept is stronger than the signal from the evaporating logarithmic regressor ln−1 t.
2. Result (20) gives first order asymptotics. As is apparent from (16), asymptotic series expansions in the present model involve factors of ln1n in contrast to the more usual √1n and, correspondingly, they deliver only very slow improvements 2
4
n in the first order asymptotics. The asymptotic variance σ ln and higher order n approximations based on (16) are therefore poor approximations to the variance of βb even for quite large n. To illustrate, for values of the sample size n ∈ b [102 , 104 ] and for σ 2 = 1, Fig. 1 provides graphs of the exact variance of β, the asymptotic variance, the two-term and three-term series approximations to the variance based on the first few terms of (16), and direct calculation of the logarithmic integral representations of the variance obtained from (13) and (17). Apparently, only the latter are adequate for sample sizes n in this range.
3. The theory developed here is part of a general theory of regression on slowly varying regressors, a subject that has recently been studied in Phillips (2000).
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Figure 1: σ 2 /
Pn
2 t=1 zt
and Asymptotic and Integral Approximations
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Part B Using the Phillips-Solo (1992) device, we have ut = C(L)εt = C(1)εt + eεt−1 − eεt , Pn Pn e e e u e where eεt = C(L)ε t , C(L) = u=0 Cu L , Cu = s=u+1 cs . Then, we can write n X
zt ut =
t=2
n X
n X
zt C(1)εt +
t=2
= C(1)
n X
zt εt +
t=2
zt eεt−1 −
t=2 n−1 X
n X
zt eεt
t=2
(zt+1 − zt )eεt + z2 eε1 − zn eεn .
t=2
It follows that Pn−1 P (zt+1 − zt )eεt z2 eε1 C(1) nt=2 zt εt b + t=2 Pn 2 + Pn β−β = Pn 2 t=2 zt
2 t=2 zt
t=2 zt
zn eεn − Pn 2 . t=2 zt
Note that the first term satisfies the limit theory given in the earlier part. If we prove that the last three terms converge strongly to zero, then we are done with the b To obtain the asymptotic distribution, we need to consider strong consistency of β. Pn 2 1/2 b ( t=2 zt ) (β − β). We prove that the last three terms, so normalized, are op (1), and then the asymptotic distribution is determined by the first term. The following three results are given first. (a) z2 eε1 z2 eε1 = oa.s (1) and Pn 2 1/2 = op (1). 2 ( t=2 zt ) t=2 zt
Pn
Proof. These are immediate since
Pn
Pn
2 t=2 zt
→ ∞ and (
2 1/2 t=2 zt )
→ ∞.
(b) zn eεn zn eεn = oa.s (1) and Pn 2 1/2 = op (1). 2 ( t=2 zt ) t=2 zt
Pn
Proof. Note that zn is bounded and for any δ > 0 zn eεn zn2 E(eε2n ) P (| Pn 2 | > δ) ≤ Pn 2 2 2 ( t=2 zt ) δ t=2 zt n X
= O((
zt2 )−2 ),
t=2
so
P
n
P (| Pznneεnz2 | > δ) < ∞ and the first result follows. The second follows t=2 t
Pn
because P (| (Pnzneεzn2 )1/2 | > δ) = O(( t=2 t
2 −1 t=2 zt ) )
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→ 0.
(c) Pn−1 t=2
(zt+1 − zt )eεt = oa.s (1) and Pn 2 t=2 zt
Pn−1
(zt+1 − zt )eεt = op (1) ( t=2 zt2 )1/2
t=2
Pn
Proof. Note that {eεt } has finite fourth moment because t t X X 4 1/4 e e e ||εt ||4 := E(εt ) = || Ct−j εj ||4 ≤ |Cet−j | · ||εj ||4 < ∞. j=0
j=0
Pt
Here we have employed the fact that P assumption that ∞ j=0 j |cj | < ∞.
j=0
|Cet−j | < ∞, which follows from the
Next, n X
E
!2 e ε2t
n X
= E
t=2
n X
e ε4t + 2
t=2
s, t=2 sδ Pn 2 t=2 zt
i2 n−1 2 Pn−1 2 εt ) t=2 (zt+1 − zt ) ( t=2 e P ( nt=2 zt2 )4 δ 4 P 2 P 2 P n−1 n−1 n−1 e2 2 2 2 t=2 (zt+1 − zt ) t=2 εt t=2 (zt+1 − zt ) = P P ( nt=2 zt2 )4 δ 4 ( nt=2 zt2 )4 δ 4
E εt )4 t=2 (zt+1 − zt )e ≤ Pn ( t=2 zt2 )4 δ 4
Pn−1
≤ =
E( E
!
hP
Since n−1 X
(zt+1 − zt )2 =
t=2
=
n−1 X
(ln−1 (t + 1) − ln−1 t)2
t=2 n−1 X
1 ln2 (1 + )[ln−2 t][ln−2 (t + 1)] t t=2
< Const.
n X
t−2 < ∞,
t=2
we have
Pn−1
P (|
t=2
(zt+1 − zt )eεt | > δ) = O(n−2 ln8 n) 2 t=2 zt
Pn
9
O(n2 )
.
and so
Pn−1 t=2
(zt+1 − zt )eεt /
Pn
2 t=2 zt
= oa.s (1). Similarly,
P ! n−1 (z εt t=2 t+1 − zt )e P >δ P ( nt=2 zt2 )1/2 hP i2 e E n−1 (z − z ) ε ) t+1 t t t=2 ≤ Pn 2 2 t=2 zt δ
≤ ≤
E
Pn−1 t=2
(zt+1 − zt )2 eε2t + 2
P
s>t (|zt+1 2 2 t=2 zt δ
− zt ||zs+1 − zs |E|eεt eεs |)
Pn
E
2 2 εt + 2 t=2 (zt+1 − zt ) e
Pn−1
P
s>t
|zt+1 − zt ||zs+1 − zs |(E|eεt |2 )1/2 (E|eεs |2 )1/2
2 2 t=2 zt δ 2
Pn Pn−1
= Const.
(
t=2
|zt+1 − zt |)2 ln n = Const. Pn 2 2 → 0. 2 2 t=2 zt δ t=2 zt δ
Pn
The last equality follows as n−1 t=2 |zt+1 − zt | = O( Pn 2 −1/2 Pn−1 e ( t=2 zt ) t=2 (zt+1 − zt )εt = op (1).
Pn
P
t=2
1/t) = O(ln n). Hence
Combining results (a), (b), and (c), we have βb − β = oa.s. (1),
(21)
and then n X
(
zt2 )1/2 (βb
− β) =
t=2
n X
Pn t=2 zt εt 2 1/2 C(1) ( zt ) Pn 2 t=2 zt t=2
+ op (1) =d N (0, C (1)2 σ 2 ).
(22)
With (21), (22) and nt=2 ut /(n − 1) →a.s. 0 (e.g. Phillips and Solo, 1992), the previous arguments can now be repeated to obtain the strong consistency and the b . Specifically asymptotic distribution of α P
b − α) n1/2 ln−1 n(α −2 1/2 n ln n(βb − β)
! 2 2
⇒ N (0, Σ) with Σ = (C(1)) σ
1 −1 −1 1
!
.
3. References Gradshteyn I. S. and I. M. Ryzhik (1965). Tables of Integrals, Series and Products. New York: Academic Press. Phillips, P. C. B. (2000). “Regression with Slowly Varying Regressors,” Cowles Foundation Discussion Paper #1310, Yale University. Phillips, P. C. B. and V. Solo (1992). “Asymptotics for Linear Processes,” Annals of Statistics 20, 971–1001.
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