Chapter 6

Estimation of the mean and

covariance

Prerequisites

• Some idea of what a cumulant is.

Objectives

• To derive the sample autocovariance of a time series, and show that this is a positive definite

sequence.

• To show that the variance of the sample covariance involves fourth order cumulants, which

can be unwielding to estimate in practice. But under linearity the expression for the variance

greatly simplifies.

• To show that under linearity the correlation does not involve the fourth order cumulant. This

is the Bartlett formula.

• To use the above results to construct a test for uncorrelatedness of a time series (the Port-

manteau test). And understand how this test may be useful for testing for independence in

various di↵erent setting. Also understand situations where the test may fail.

Here we summarize the Central limit theorems we will use in this chapter. The simplest is the

case of iid random variables. The first is the classical central limit theorem. Suppose that {Xi} are

171

iid random variables with mean µ and variance �2 < 1. Then

1pn

nX

i=1

(Xi � µ)D! N (0,�2).

A small variant on the classical CLT is the case that {Xi} are independent random variables (but

not identically distributed). Suppose E[Xi] = µi, var[Xi] = �2i < 1 and for every " > 0

1

s2n

nX

i=1

E�

(Xi � µi)2I(s�1

n |Xi � µi| > ")�! 0

where s2n =Pn

i=1

�2i , which is the variance ofPn

i=1

Xi (the above condition is called the Lindeberg

condition). Then

1q

Pni=1

�2i

nX

i=1

(Xi � µi)D! N (0, 1).

The Lindeberg condition looks unwieldy, however by using Chebyshev’s and Holder inequality it

can be reduced to simple bounds on the moments.

Note that the Lindeberg condition generalizes to the conditional Lindeberg condition when

dealing with martingale di↵erences.

We now state a generalisation of this central limit to triangular arrays. Suppose that {Xt,n}are independent random variables with mean zero. Let Sn =

Pnt=1

Xt,n we assume that var[Sn] =Pn

t=1

var[Xt,n] = 1. For example, in the case that {Xt} are iid random variables and Sn =

1pn

Pnt=1

[Xt � µ] =Pn

t=1

Xt,n, where Xt,n = ��1/2n�1/2(Xt � µ). If for all " > 0

nX

t=1

E�

X2

t,nI(|Xt,n| > ")�! 0,

then SnD! N (0, 1).

6.1 An estimator of the mean

Suppose we observe {Yt}nt=1

, where

Yt = µ+Xt,

172

where µ is the finite mean, {Xt} is a zero mean stationary time series with absolutely summable

covariances (P

k |cov(X0

, Xk)| < 1). Our aim is to estimate the mean µ. The most obvious

estimator is the sample mean, that is Yn = n�1

Pnt=1

Yt as an estimator of µ.

6.1.1 The sampling properties of the sample mean

We recall from Example 1.5.1 that we obtained an expression for the sample mean. We showed

that

var(Yn) =1

nvar(X

0

) +2

n

nX

k=1

�n� k

n

�

c(k).

Furthermore, ifP

k |c(k)| < 1, then in Example 1.5.1 we showed that

var(Yn) =1

nvar(X

0

) +2

n

1X

k=1

c(k) + o(1

n).

Thus if the time series has su�cient decay in it’s correlation structure a mean squared consistent

estimator of the sample mean can be achieved. However, one drawback is that the dependency

means that one observation will influence the next, and if the influence is positive (seen by a positive

covariance), the resulting estimator may have a (much) larger variance than the iid case.

The above result does not require any more conditions on the process, besides second order

stationarity and summability of its covariance. However, to obtain confidence intervals we require

a stronger result, namely a central limit theorem for the sample mean. The above conditions are

not enough to give a central limit theorem. To obtain a CLT for sums of the formPn

t=1

Xt we

need the following main ingredients:

(i) The variance needs to be finite.

(ii) The dependence between Xt decreases the further apart in time the observations. However,

this is more than just the correlation, it really means the dependence.

The above conditions are satisfied by linear time series, if the co�cients �j decay su�cient fast.

However, these conditions can also be verified for nonlinear time series (for example the (G)ARCH

and Bilinear model described in Chapter 4).

We now state the asymptotic normality result for linear models.

173

Theorem 6.1.1 Suppose that Xt is a linear time series, of the form Xt =P1

j=�1 j"t�j, where "t

are iid random variables with mean zero and variance one,P1

j=�1 | j | < 1 andP1

j=�1 j 6= 0.

Let Yt = µ+Xt, then we have

pn�

Yn � µ�

= N (0,�2)

where �2 = var(X0

) + 2P1

k=1

c(k).

PROOF. Later in this course we will give precise details on how to prove asymptotic normality of

several di↵erent type of estimators in time series. However, we give a small flavour here by showing

asymptotic normality of Yn in the special case that {Xt}nt=1

satisfy an MA(q) model, then explain

how it can be extended to MA(1) processes.

The main idea of the proof is to transform/approximate the average into a quantity that we

know is asymptotic normal. We know if {✏t}nt=1

are iid random variables with mean µ and variance

one then

pn(✏n � µ)

D! N (0, 1). (6.1)

We aim to use this result to prove the theorem. Returning to Yn by a change of variables (s = t�j)

we can show that

1

n

nX

t=1

Yt = µ+1

n

nX

t=1

Xt = µ+1

n

nX

t=1

qX

j=0

j"t�j

= µ+1

n

n�qX

s=1

"s

0

@

qX

j=0

j

1

A+0

X

s=�q+1

"s

0

@

qX

j=q�s

j

1

A+nX

s=n�q+1

"s

0

@

n�sX

j=0

j

1

A

= µ+n� q

n

0

@

qX

j=0

j

1

A

1

n� q

n�qX

s=1

"s +1

n

0

X

s=�q+1

"s

0

@

qX

j=q+s

j

1

A+1

n

nX

s=n�q+1

"s

0

@

n�sX

j=0

j

1

A

:= µ+ (n� q)

n"n�q + E

1

+ E2

, (6.2)

where =Pq

j=0

j . It is straightforward to show that E|E1

| Cn�1 and E|E2

| Cn�1.

Finally we examine (n�q)n "n�q. We note that if the assumptions are not satisfied and

Pqj=0

j =

174

0 (for example the process Xt = "t � "t�1

), then

1

n

nX

t=1

Yt = µ+1

n

0

X

s=�q+1

"s

0

@

qX

j=q�s

j

1

A+1

n

nX

s=n�q+1

"s

0

@

n�sX

j=0

j

1

A .

This is a degenerate case, since E1

and E2

only consist of a finite number of terms and thus if "t are

non-Gaussian these terms will never be asymptotically normal. Therefore, in this case we simply

have that 1

n

Pnt=1

Yt = µ+O( 1n) (this is why in the assumptions it was stated that 6= 0).

On the other hand, if 6= 0, then the dominating term in Yn is "n�q. From (6.1) it is

clear thatpn� q"n�q

P! N (0, 1) as n ! 1. However, for finite q,p

(n� q)/nP! 1, therefore

pn"n�q

P! N (0, 1). Altogether, substituting E|E1

| Cn�1 and E|E2

| Cn�1into (6.2) gives

pn�

Yn � µ�

= pn"n�q +Op(

1

n)

P! N �

0, 2

�

.

With a little work, it can be shown that 2 = �2.

Observe that the proof simply approximated the sum by a sum of iid random variables. In the

case that the process is a MA(1) or linear time series, a similar method is used. More precisely,

we have

pn�

Yn � µ�

=1pn

nX

t=1

1X

j=0

j"t�j =1pn

1X

j=0

j

n�jX

s=1�j

"s

=1pn

1X

j=0

j

nX

t=1

"t +Rn

where

Rn =1pn

1X

j=0

j

0

@

n�jX

s=1�j

"s �nX

s=1

"s

1

A

=1pn

nX

j=0

j

0

@

0

X

s=1�j

"s �nX

s=n�j

"s

1

A+1pn

1X

j=n+1

j

0

@

n�jX

s=1�j

"s �nX

s=1

"s

1

A

:= Rn1 +Rn2 +Rn3 +Rn4.

175

We will show that E[R2

n,j ] = o(1) for 1 j 4. We start with Rn,1

E[R2

n,1] =1

n

nX

j1

,j2

=0

j1

j2

cov

0

@

0

X

s1

=1�j1

"s1

,0

X

s2

=1�j2

"s2

1

A

=1

n

nX

j1

,j2

=0

j1

j2

min[j1

� 1, j2

� 1]

=1

n

nX

j=0

2

j (j � 1) +2

n

nX

j1

=0

j1

,j1

�1

X

j2

=0

j2

min[j2

� 1]

1

n

nX

j=0

2

j (j � 1) +2

n

nX

j1

=0

|j1

j1

|.

SinceP1

j=0

| j | < 1 and, thus,P1

j=0

| j |2 < 1, then by dominated convegencePn

j=0

[1�j/n] j !P1

j=0

j andPn

j=0

[1 � j/n] 2

j ! P1j=0

2

j as n ! 1. This implies thatPn

j=0

(j/n) j ! 0 andPn

j=0

(j/n) 2

j ! 0. Substituting this into the above bounds for E[R2

n,1] we immediately obtain

E[R2

n,1] = o(1). Using the same argument we obtain the same bound for Rn,2, Rn,3 and Rn,4. Thus

pn�

Yn � µ�

= 1pn

nX

j=1

"t + op(1)

and the result then immediately follows. ⇤

Estimation of the so called long run variance (given in Theorem 6.1.1) can be di�cult. There

are various methods that can be used, such as estimating the spectral density function (which we

define in Chapter 8) at zero. Another approach proposed in Lobato (2001) and Shao (2010) is to

use the method of so called self-normalization which circumvents the need to estimate the long run

mean, by privotalising the statistic.

6.2 An estimator of the covariance

Suppose we observe {Yt}nt=1

, to estimate the covariance we can estimate the covariance c(k) =

cov(Y0

, Yk) from the the observations. A plausible estimator is

cn(k) =1

n

n�|k|X

t=1

(Yt � Yn)(Yt+|k| � Yn), (6.3)

176

since E[(Yt � Yn)(Yt+|k| � Yn)] ⇡ c(k). Of course if the mean of Yt is known to be zero (Yt = Xt),

then the covariance estimator is

cn(k) =1

n

n�|k|X

t=1

XtXt+|k|. (6.4)

The eagle-eyed amongst you may wonder why we don’t use 1

n�|k|Pn�|k|

t=1

XtXt+|k|, when cn(k) is a

biased estimator, whereas 1

n�|k|Pn�|k|

t=1

XtXt+|k| is not. However cn(k) has some very nice properties

which we discuss in the lemma below.

Lemma 6.2.1 Suppose we define the empirical covariances

cn(k) =

8

<

:

1

n

Pn�|k|t=1

XtXt+|k| |k| n� 1

0 otherwise

then {cn(k)} is a positive definite sequence. Therefore, using Lemma 1.6.1 there exists a stationary

time series {Zt} which has the covariance cn(k).

PROOF. There are various ways to show that {cn(k)} is a positive definite sequence. One method

uses that the spectral density corresponding to this sequence is non-negative, we give this proof in

Section 8.3.1.

Here we give an alternative proof. We recall a sequence is semi-positive definite if for any vector

a = (a1

, . . . , ar)0 we have

rX

k1

,k2

=1

ak1

ak2

cn(k1 � k2

) =nX

k1

,k2

=1

ak1

ak2

cn(k1 � k2

) = a0b⌃na � 0

where

b⌃n =

0

B

B

B

B

B

B

@

cn(0) cn(1) cn(2) . . . cn(n� 1)

cn(1) cn(0) cn(1) . . . cn(n� 2)...

.... . .

......

cn(n� 1) cn(n� 2)...

... cn(0)

1

C

C

C

C

C

C

A

,

noting that cn(k) =1

n

Pn�|k|t=1

XtXt+|k|. However, cn(k) =1

n

Pn�|k|t=1

XtXt+|k| has a very interesting

construction, it can be shown that the above convariance matrix is b⌃n = XnX0n, where Xn is a

177

n⇥ 2n matrix with

Xn =

0

B

B

B

B

B

B

@

0 0 . . . 0 X1

X2

. . . Xn�1

Xn

0 0 . . . X1

X2

. . . Xn�1

Xn 0...

......

......

......

......

X1

X2

. . . Xn�1

Xn 0 . . . . . . 0

1

C

C

C

C

C

C

A

Using the above we have

a0b⌃na = a0XnX0na = kX0ak2

2

� 0.

This this proves that {cn(k)} is a positive definite sequence.

Finally, by using Theorem 1.6.1, there exists a stochastic process with {cn(k)} as its autoco-

variance function. ⇤

6.2.1 Asymptotic properties of the covariance estimator

The main reason we construct an estimator is either for testing or constructing a confidence interval

for the parameter of interest. To do this we need the variance and distribution of the estimator. It

is impossible to derive the finite sample distribution, thus we look at their asymptotic distribution.

Besides showing asymptotic normality, it is important to derive an expression for the variance.

In an ideal world the variance will be simple and will not involve unknown parameters. Usually

in time series this will not be the case, and the variance will involve several (often an infinite)

number of parameters which are not straightforward to estimate. Later in this section we show

that the variance of the sample covariance can be extremely complicated. However, a substantial

simplification can arise if we consider only the sample correlation (not variance) and assume linearity

of the time series. This result is known as Bartlett’s formula (you may have come across Maurice

Bartlett before, besides his fundamental contributions in time series he is well known for proposing

the famous Bartlett correction). This example demonstrates, how the assumption of linearity can

really simplify problems in time series analysis and also how we can circumvent certain problems

in which arise by making slight modifications of the estimator (such as going from covariance to

correlation).

The following theorem gives the asymptotic sampling properties of the covariance estimator

(6.3). One proof of the result can be found in Brockwell and Davis (1998), Chapter 8, Fuller

178

(1995), but it goes back to Bartlett (indeed its called Bartlett’s formula). We prove the result in

Section 6.2.2.

Theorem 6.2.1 Suppose {Xt} is a mean zero linear stationary time series where

Xt = µ+1X

j=�1 j"t�j ,

whereP

j | j | < 1, {"t} are iid random variables with E("t) = 0 and E("4t ) < 1. Suppose we

observe {Xt : t = 1, . . . , n} and use (6.3) as an estimator of the covariance c(k) = cov(X0

, Xk).

Define ⇢n(r) = cn(r)/cn(0) as the sample correlation. Then for each h 2 {1, . . . , n}

pn(⇢n(h)� ⇢(h))

D! N (0,Wh) (6.5)

where ⇢n(h) = (⇢n(1), . . . , ⇢n(h)), ⇢(h) = (⇢(1), . . . , ⇢(h)) and

(Wh)ij =1X

k=�1{⇢(k + i)⇢(k + j) + ⇢(k � i)⇢(k + j) + 2⇢(i)⇢(j)⇢2(k)� 2⇢(i)⇢(k)⇢(k + j)� 2⇢(j)⇢(k)⇢(k + i)}.(6.6)

Equation (6.6) is known as Bartlett’s formula.

In Section 6.3 we apply the method for checking for correlation in a time series. We first show

how the expression for the asymptotic variance is obtained.

6.2.2 Proof of Bartlett’s formula

What are cumulants?

We first derive an expression for cn(r) under the assumption that {Xt} is a strictly stationary time

series with finite fourth order moment,P

k |c(k)| < 1 and for all r1

, r2

2 Z,P

k |4(r1, k, k+r2

)| <1 where

4

(k1

, k2

, k3

) = cum(X0

, Xk1

, Xk2

, Xk3

).

It is reasonable to ask what cumulants are. Cumulants often crops up in time series. To

understand what they are and why they are used, we focus the following discussion on just fourth

order cumulants.

The joint cumulant of Xt, Xt+k1

, Xt+k2

, Xt+k3

(denoted as cum(Xt, Xt+k1

, Xt+k2

, Xt+k3

)) is the

179

coe�cient of s1

s2

s3

s4

in the power series expansion of

log E[eis1Xt

+is2

Xt+k

1

+is3

Xt+k

2

+is4

Xt+k

4 ].

It looks very similar to the definition of moments. Indeed there is a one to one correpondence

between the moments and the cumulants, which is why they arise in time series. However, there

are important di↵erences

• If Xt is independent of Xt+k1

, Xt+k2

, Xt+k3

then

cum (Xt, Xt+k1

, Xt+k2

, Xt+k3

) = 0.

This is because the log of the corresponding characteristic function is

log E[eis1Xt

+is2

Xt+k

1

+is3

Xt+k

2

+is4

Xt+k

4 ] = log E[eis1Xt ] + log[E[eis2Xt+k

1

+is3

Xt+k

2

+is4

Xt+k

4 ].

Thus we see that the coe�cient of s1

s2

s3

s4

in the above expansion is zero.

• If Xt, Xt+k1

, Xt+k2

, Xt+k3

is multivariate Gaussian, then all cumulants higher than order 2

are zero.

Neither of the above two properties hold for moments.

We can see from the definition of the characteristic function, if the time series is strictly sta-

tionary then

log E[eis1Xt

+is2

Xt+k

1

+is3

Xt+k

2

+is4

Xt+k

4 ] = log E[eis1X0

+is2

Xk

1

+is3

Xk

2

+is4

Xk

4 ].

Thus the cumulants are invariant to shift

cum(Xt, Xt+k1

, Xt+k2

, Xt+k3

) = cum(X0

, Xk1

, Xk2

, Xk3

) = 4

(k1

, k2

, k3

).

Thus like the autocovariance functions for stationary processes it does not depend on t.

The cumulant is similar to the covariance in that

(a) The covariance measures the dependence between Xt and Xt+k. Note that cov[Xt, Xt+k] =

cov[Xt+k, Xt], hence the covariance is invariant to order (if the random variables are real).

The covariance is the second order cumulant.

180

Like the covariance, the joint cumulant cum[Xt, Xt+k1

, Xt+k2

, Xt+k3

] is also invariant to order.

(b) The cumulant is measuring the dependence between cum[Xt, Xt+k1

, Xt+k2

, Xt+k3

] in “all

directions”. For example, suppose {Xt} has zero mean then

cum[Xt, Xt+k1

, Xt+k2

, Xt+k3

] = E[XtXt+k1

Xt+k2

Xt+k3

]� E[XtXt+k1

]E[Xt+k2

Xt+k3

] (6.7)

�E[XtXt+k2

]E[Xt+k1

Xt+k3

]� E[XtXt+k3

]E[Xt+k1

Xt+k2

].(6.8)

(c) In time series we usually assume that the covariance decays over time i.e. if k > 0

|cov[Xt, Xt+k]| ↵(k)

where ↵(k) is a positive sequence such that ↵(k) ! 0 as k ! 1. We showed this result was

true for linear time series. The same is true of cumulants i.e. assume k1

k2

k3

then

|cum[Xt, Xt+k1

, Xt+k2

, Xt+k3

]| ↵(k1

)↵(k2

� k1

)↵(k3

� k2

).

(d) Often in proofs we can the assumptionP

r |c(r)| < 1. An analogous assumption isP

k1

,k2

,k3

|4

(k1

, k2

, k3

)| < 1.

Example 6.2.1 For the causal AR(1) model Xt = �Xt�1

+"t (where {"t} are iid random variables

with finite fourth order cumulant 4

) by using the MA(1) representation (assuming 0 k1

k2

k3

) we have

cum[Xt, Xt+k1

, Xt+k2

, Xt+k3

= 4

1X

j=0

�j�j+k1�j+k

2�j+k3 =

4

�k1+k2

+k3

1X

j=0

�4j .

Observe that the fourth order dependence decays as the lag increases.

The variance of the sample covariance in the case of strict stationarity

We will consider

var[cn(r)] =1

n2

n�|r|X

t,⌧=1

cov(XtXt+r, X⌧X⌧+r).

181

One approach for the analysis of cov(XtXt+r, X⌧X⌧+r) is to expand it in terms of expectations

cov(XtXt+r, X⌧X⌧+r) = E(XtXt+r, X⌧X⌧+r)�E(XtXt+r)E(X⌧X⌧+r), however it not clear how this

will give var[XtXt+r] = O(n�1). Instead we observe that cov(XtXt+r, X⌧X⌧+r) is the covariance

of the product of random variables. This belongs to the general class of cumulants of products

of random variables. We now use standard results on cumulants. The most important is that if

X,Y, U and V are mean zero random variables, then

cov[XY,UV ] = cov[X,U ]cov[Y, V ] + cov[X,V ]cov[Y, U ] + cum(X,Y, U, V )

(this result can be seen from (6.7)). This result can be generalized to higher order cumulants, see

Brillinger (2001). Using this result we have

var[cn(r)]

=1

n2

n�|r|X

t,⌧=1

�

cov(Xt, X⌧ )| {z }

=c(t�⌧) by stationarity

cov(Xt+r, X⌧+r) + cov(Xt, X⌧+r)cov(Xt+r, X⌧ ) + cum(Xt, Xt+r, X⌧ , X⌧+r)�

=1

n2

n�|r|X

t,⌧=1

⇥

c(t� ⌧)2 + c(t� ⌧ � r)c(t+ r � ⌧) + k4

(r, ⌧ � t, ⌧ + r � t)⇤

:= I + II + III,

where the above is due to strict stationarity of the time series. We analyse the above term by

term. Either (i) by changing variables and letting k = t � ⌧ and thus changing the limits of the

summand in an appropriate way or (ii) observing thatPn�|r|

t,⌧=1

c(t� ⌧)2 is the sum of the elements

in the Toeplitz matrix

0

B

B

B

B

B

B

@

c(0)2 c(1)2 . . . c(n� |r|� 1)2

c(�1)2 c(0)2 . . . c(n� |r|� 2)2

......

. . ....

c(�(n� |r|� 1))2 c(�(n� |r|� 2))2 . . . c(0)2

1

C

C

C

C

C

C

A

,

(noting that c(�k) = c(k)) the sum I can be written as

I =1

n2

n�|r|X

t,⌧=1

c(t� ⌧)2 =1

n2

(n�|r|�1)

X

k=�(n�|r|�1)

c(k)2n�|r|�|k|X

t=1

1 =1

n

n�|r|X

k=�(n�|r|)

✓

n� |r|� |k|n

◆

c(k)2.

182

For all k, (1�|k|/n)c(k)2 ! c(k)2 and |Pn�|r|k=�(n�|r|)(1�|k|/n)c(k)2| Pk c(k)

2, thus by dominated

convergence (see Appendix A)Pn

k=�(n�|r|)(1� |k|/n)c(k)2 !P1k=�1 c(k)2. This gives

I =1

n

1X

k=�1c(k)2 + o(

1

n).

Using a similar argument we can show that

II =1

n

1X

k=�1c(k + r)c(k � r) + o(

1

n).

To derive the limit of III, again we use a change of variables to give

III =1

n

n�|r|X

k=�(n�|r|)

✓

n� |r|� |k|n

◆

k4

(r, k, k + r).

To bound III we note that for all k, (1� |k|/n)k4

(r, k, k+r) ! k4

(r, k, k+r) and |Pn�|r|k=�(n�|r|)(1�

|k|/n)k4

(r, k, k+r)| Pk |k4(r, k, k+r)|, thus by dominated convergence we havePn

k=�(n�|r|)(1�|k|/n)k

4

(r, k, k + r) !P1k=�1 k

4

(r, k, k + r). This gives

III =1

n

1X

k=�14

(r, k, k + r) + o(1

n).

Therefore altogether we have

nvar[cn(r)] =1X

k=�1c(k)2 +

1X

k=�1c(k + r)c(k � r) +

1X

k=�14

(r, k, k + r) + o(1).

Using similar arguments we obtain

ncov[cn(r1), cn(r2)] =1X

k=�1c(k)c(k + r

1

� r2

) +1X

k=�1c(k � r

1

)c(k + r2

) +1X

k=�14

(r1

, k, k + r2

) + o(1).

We observe that the covariance of the covariance estimator contains both covariance and cumulants

terms. Thus if we need to estimate them, for example to construct confidence intervals, this can

be di�cult. However, there does exist methods for estimating the variance, the most popular is to

use a block type bootstrap, another method is to exploit properties of Fourier transforms and use

the method of orthogonal samples (see Subba Rao (2016)).

183

Below that under linearity the above fourth order cumulant term has a simpler form.

The covariance of the sample covariance under linearity

We recall that

1X

k=�1c(k + r

1

� r2

)c(k) +1X

k=�1c(k � r

1

)c(k + r2

) +1X

k=�14

(r1

, k, k + r2

) + o(1) = T1

+ T2

+ T3

+ o(1).

We now show that under linearity, T3

(the fourth order cumulant) has a much simpler form. Let

us suppose that the time series is linear

Xt =1X

j=�1 j"t�j

whereP

j | j | < 1, {"t} are iid, E("t) = 0, var("t) = 1 and 4

= cum4

("t). Then T3

is

T3

=1X

k=�1cum

0

@

X

j1

=�1 j

1

"�j1

,X

j2

=�1 j

2

"r1

�j2

,X

j3

=�1 j

3

"k�j3

,X

j4

=�1 j

4

"k+r2

�j1

1

A

=1X

k=�1

X

j1

,...,j4

=�1 j

1

j2

j3

j4

cum ("�j1

, "r1

�j2

, "k�j3

, "k+r2

�j1

) .

Standard results in cumulants, state that if any a variable is independent of all the others (see Sec-

tion 6.2.2), then cum[Y1

, Y2

, . . . , Yn] = 0. Applying this result to cum ("�j1

, "r1

�j2

, "k�j3

, "k+r2

�j1

)

reduces T3

to

T3

= 4

1X

k=�1

1X

j=�1 j j�r

1

j�k j�r2

�k.

Using a change of variables j1

= j and j2

= j � k we have

4

�

1X

j1

=�1 j j�r

1

��

1X

j2

=�1 j

2

j2

�r2

�

= 4

c(r1

)c(r2

),

recalling the covariance of a linear process in Lemma 3.1.1 (and assuming var["t] = 1).

Altogether this gives

ncov[cn(r1), cn(r2)] =1X

k=�1c(k)c(k + r

1

� r2

) +1X

k=�1c(k � r

1

)c(k + r2

) + 4

c(r1

)c(r2

) + o(1). (6.9)

184

Thus in the case of linearity our expression for the variance is simpler, and the only di�cult

parameter to estimate of 4

.

The variance of the sample correlation under linearity

A suprisingly twist in the story is that (6.9) can be reduced further, if we are interested in estimating

the correlation rather than the covariance. We recall the sample correlation is

⇢n(r) =cn(r)

cn(0),

which is an estimator of ⇢(r) = c(r)/c(0).

Lemma 6.2.2 (Bartlett’s formula) Suppose {Xt} is a linear time series, whereP

j | (j)| < 1.

Then the variance of the distribution of ⇢n(r) is

1X

k=�1{⇢(k + r)2 + ⇢(k � r)⇢(k + r) + 2⇢(r)2⇢2(k)� 4⇢(r)⇢(k)⇢(k + r)}.

PROOF. We use that

var[cn(r)] = O(n�1) and E[cn(r)] =

✓

n� |r|n

◆

c(r).

Thus for fixed r

|cn(0)� c(0)| = Op(n�1/2), |cn(r)� c(r)| = Op(n

�1/2) and |cn(0)� c(0)| |cn(r)� c(r)| = Op(n�1).

By making a Taylor expansion of cn(0)�1 about c(0)�1 we have

⇢n(r)� ⇢(r) =cn(r)

cn(0)� c(r)

c(0)

=[cn(r)� c(r)]

c(0)� [cn(0)� c(0)]

cn(r)

c(0)2+ [cn(0)� c(0)]2

cn(r)

cn(0)3| {z }

=O(n�1

)

(6.10)

=[cn(r)� c(r)]

c(0)� [cn(0)� c(0)]

c(r)

c(0)2+ 2 [cn(0)� c(0)]2

cn(r)

cn(0)3� [cn(0)� c(0)]

[cn(r)� c(r)]

c(0)2| {z }

O(n�1

)

:= An +Op(n�1), (6.11)

185

where cn(0) lies between cn(0) and c(0) and

An =[cn(r)� c(r)]

c(0)� [cn(0)� c(0)]

c(r)

c(0)2.

Thus the dominating term in ⇢n(r) � ⇢(r) is An, which is of order O(n�1/2) (by (6.9)). Thus the

limiting distribution of ⇢n(r)�⇢(r) is determined by An and the variance of the limiting distribution

is also determined by An. It is straightforward to show that

nvar[An] = nvar[cn(r)]

c(0)2� 2ncov[cn(r), cn(0)]

c(r)2

c(0)3+ nvar[cn(0)]

c(r)2

c(0)4. (6.12)

By using (6.9) we have

nvar

0

@

cn(r)

cn(0)

1

A

=

0

@

P1k=�1 c(k)2 +

P1k=�1 c(k)c(k � r) +

4

c(r)2 2P1

k=�1 c(k)c(k � r) + 4

c(r)c(0)

2P1

k=�1 c(k)c(k � r) + 4

c(r)c(0)P1

k=�1 c(k)2 +P1

k=�1 c(k)c(k � r) + 4

c(0)2

1

A

+o(1).

Substituting the above into (6.12) gives us

nvar[An] =

1X

k=�1c(k)2 +

1X

k=�1c(k)c(k � r) +

4

c(r)2!

1

c(0)2�

2

21X

k=�1c(k)c(k � r) +

4

c(r)c(0)

!

c(r)2

c(0)3+

1X

k=�1c(k)2 +

1X

k=�1c(k)c(k � r) +

4

c(0)2!

c(r)2

c(0)4+ o(1).

Focusing on the fourth order cumulant terms, we see that these cancel, which gives the result. ⇤

To prove Theorem 6.2.1, we simply use the Lemma 6.2.2 to obtain an asymptotic expression

for the variance, then we use An to show asymptotic normality of cn(r) (under linearity).

Exercise 6.1 Under the assumption that {Xt} are iid random variables show that cn(1) is asymp-

totically normal.

186

Hint: Let m = n/(B + 1) and partition the sumPn�1

k=1

XtXt+1

as follows

n�1

X

t=1

XtXt+1

=BX

t=1

XtXt+1

+XB+1

XB+2

+2B+1

X

t=B+2

XtXt+1

+X2B+2

X2B+3

+

3B+2

X

t=2B+3

XtXt+1

+X3B+3

X3B+4

+4B+3

X

t=3B+4

XtXt+1

+ . . .

=m�1

X

j=0

Um,j +m�1

X

j=0

X(j+1)(B+1)

X(j+1)(B+1)+1

where Um,j =Pj(B+1)+B

t=j(B+1)+1

XtXt+1

. Show that the second term in the above summand is asymp-

totically negligible and show that the classical CLT for triangular arrays can be applied to the first

term.

Exercise 6.2 Under the assumption that {Xt} is a MA(1) process, show that bcn(1) is asymptoti-

cally normal.

Exercise 6.3 The block bootstrap scheme is a commonly used method for estimating the finite

sample distribution of a statistic (which includes its variance). The aim in this exercise is to see

how well the bootstrap variance approximates the finite sample variance of a statistic.

(i) In R write a function to calculate the autocovariance bcn(1) =1

n

Pn�1

t=1

XtXt+1

.

Remember the function is defined as cov1 = function(x){...}

(ii) Load the library boot library("boot") into R. We will use the block bootstrap, which parti-

tions the data into blocks of lengths l and then samples from the blocks n/l times to construct

a new bootstrap time series of length n. For each bootstrap time series the covariance is

evaluated and this is done R times. The variance is calculated based on these R bootstrap

estimates.

You will need to use the function tsboot(tseries,statistic,R=100,l=20,sim="fixed").

tseries refers to the original data, statistic to the function you wrote in part (i) (which should

only be a function of the data), R=is the number of bootstrap replications and l is the length

of the block.

Note that tsboot(tseries,statistic,R=100,l=20,sim="fixed")$t will be vector of length

R = 100 which will contain the bootstrap statistics, you can calculate the variance of this

vector.

187

(iii) Simulate the AR(2) time series arima.sim(list(order = c(2, 0, 0), ar = c(1.5,�0.75)), n =

128) 500 times. For each realisation calculate the sample autocovariance at lag one and also

the bootstrap variance.

(iv) Calculate the mean of the bootstrap variances and also the mean squared error (compared

with the empirical variance), how does the bootstrap perform?

(iv) Play around with the bootstrap block length l. Observe how the block length can influence the

result.

Remark 6.2.1 The above would appear to be a nice trick, but there are two major factors that

lead to the cancellation of the fourth order cumulant term

• Linearity of the time series

• Ratio between cn(r) and cn(0).

Indeed this is not a chance result, in fact there is a logical reason why this result is true (and is

true for many statistics, which have a similar form - commonly called ratio statistics). It is easiest

explained in the Fourier domain. If the estimator can be written as

1

n

Pnk=1

�(!k)In(!k)1

n

Pnk=1

In(!k),

where In(!) is the periodogram, and {Xt} is a linear time series, then we will show later that the

asymptotic distribution of the above has a variance which is only in terms of the covariances not

higher order cumulants. We prove this result in Section 9.5.

6.3 Checking for correlation in a time series

Bartlett’s formula if commonly used to check by ‘eye; whether a time series is uncorrelated (there

are more sensitive tests, but this one is often used to construct CI in for the sample autocovariances

in several statistical packages). This is an important problem, for many reasons:

• Given a data set, we need to check whether there is dependence, if there is we need to analyse

it in a di↵erent way.

188

• Suppose we fit a linear regression to time series data. We may to check whether the residuals

are actually uncorrelated, else the standard errors based on the assumption of uncorrelated-

ness would be unreliable.

• We need to check whether a time series model is the appropriate model. To do this we fit

the model and estimate the residuals. If the residuals appear to be uncorrelated it would

seem likely that the model is correct. If they are correlated, then the model is inappropriate.

For example, we may fit an AR(1) to the data, estimate the residuals "t, if there is still

correlation in the residuals, then the AR(1) was not the correct model, since Xt � �Xt�1

is

still correlated (which it would not be, if it were the correct model).

We now apply Theorem 6.2.1 to the case that the time series are iid random variables. Suppose {Xt}are iid random variables, then it is clear that it is trivial example of a (not necessarily Gaussian)

linear process. We use (6.3) as an estimator of the autocovariances.

To derive the asymptotic variance of {cn(r)}, we recall that if {Xt} are iid then ⇢(k) = 0 for

k 6= 0. Thus by making a Taylor expansion (and noting that c(k)=0) (and/or using (6.6)) we see

that

pn⇢n =

pn

c(0)cn + op(1)

D! N (0,Wh),

where b⇢n = (b⇢n(1), . . . , b⇢n(m)), similar with bcn and

(Wh)ij =

8

<

:

1 i = j

0 i 6= j

In other words,pn⇢n

D! N (0, Ih). Hence the sample autocovariances at di↵erent lags are asymp-

totically uncorrelated and have variance one. This allows us to easily construct error bars for the

sample autocovariances under the assumption of independence. If the vast majority of the sample

autocovariance lie inside the error bars there is not enough evidence to suggest that the data is

a realisation of a iid random variables (often called a white noise process). An example of the

empirical ACF and error bars is given in Figure 6.1. We see that the empirical autocorrelations of

the realisation from iid random variables all lie within the error bars. In contrast in Figure 6.2

we give a plot of the sample ACF of an AR(2). We observe that a large number of the sample

autocorrelations lie outside the error bars.

189

0 5 10 15 20

0.00.2

0.40.6

0.81.0

Lag

ACF

Series iid

Figure 6.1: The sample ACF of an iid sample with error bars (sample size n = 200).

0 5 10 15 20

−0.4

0.00.4

0.8

Lag

ACF

Series ar2

5 10 15 20

−0.4

0.00.4

0.8

lag

acf

Figure 6.2: Top: The sample ACF of the AR(2) process Xt = 1.5Xt�1

+ 0.75Xt�2

+ "t witherror bars n = 200. Bottom: The true ACF.

190

Of course, simply checking by eye means that we risk misconstruing a sample coe�cient that

lies outside the error bars as meaning that the time series is correlated, whereas this could simply

be a false positive (due to multiple testing). To counter this problem, we construct a test statistic

for testing uncorrelatedness. Since under the nullpn(⇢n(h) � ⇢(h))

D! N (0, I), one method of

testing is to use the square correlations

Sh = nhX

r=1

|⇢n(r)|2, (6.13)

under the null it will asymptotically have a �2-distribution with h degrees of freedom, under the

alternative it will be a non-central (generalised) chi-squared. The non-centrality is what makes us

reject the null if the alternative of correlatedness is true. This is known as the Box-Pierce test.

Of course, a big question is how to select h. In general, we do not have to use large h since most

correlations will arise when r is small, However the choice of h will have an influence on power. If

h is too large the test will loose power (since the mean of the chi-squared grows as h ! 1), on

the other hand choosing h too small may mean that certain correlations at higher lags are missed.

How to selection h is discussed in several papers, see for example Escanciano and Lobato (2009).

6.4 Checking for partial correlation

We recall that the partial correlation of a stationary time series at lag t is given by the last coe�cient

of the best linear predictor of Xm+1

given {Xj}mj=1

i.e. �m where bXm+1|m =Pm

j=1

�jXm+1�j . Thus

�m can be estimated using the Yule-Walker estimator or least squares (more of this later) and the

sampling properties of the estimator are determined by the sampling properties of the estimator of

an AR(m) process. We state these now. We assume {Xt} is a AR(p) time series of the form

Xt =pX

j=1

�jXt�j + "t

where {"t} are iid random variables with mean zero and variance �2. Suppose an AR(m) model is

fitted to the data using the Yule-Walker estimator, we denote this estimator as b�m = b⌃�1

m rm. Let

b�m = (b�m1

, . . . , b�mm), the estimator of the partial correlation at lag m is b�mm. Assume m � p.

191

Then by using Theorem 7.2.1 (see also Theorem 8.1.2, Brockwell and Davis (1998)) we have

pn⇣

b�m � �m

⌘

P! N(0,�2⌃�1

m ).

where �m are the true parameters. Ifm > p, then �m = (�1

, . . . ,�p, 0, . . . , 0) and the last coe�cient

has the marginal distribution

pnb�mm

P! N(0,�2⌃mm).

Since m > p, we can obtain a closed for expression for ⌃mm. By using Remark 3.2.2 we have

⌃mm = ��2, thus

pnb�mm

P! N(0, 1).

Therefore, for lags m > p the partial correlations will be asymptotically pivotal. The errors bars in

the partial correlations are [�1.96n�1/2, 1.96n�1/2] and these can be used as a guide in determining

the order of the autoregressive process (note there will be dependence between the partial correlation

at di↵erent lags).

This is quite a surprising result and very di↵erent to the behaviour of the sample autocorrelation

function of an MA(p) process.

Exercise 6.4

(a) Simulate a mean zero invertible MA(1) process (use Gaussian errors). Use a reasonable sample

size (say n = 200). Evaluate the sample correlation at lag 2, drhon(2). Note the sample correlation

at lag two is estimating 0. Do this 500 times.

• Calculate of proportion of sample covariances |b⇢n(2)| > 1.96/pn

• Make a QQplot of b⇢n(2)/pn against a standard normal distribution. What do you observe?

(b) Simulate a causal, stationary AR(1) process (use Gaussian errors). Use a reasonable sample

size (say n = 200). Evaluate the sample partial correlation at lag 2, b�n(2). Note the sample partial

correlation at lag two is estimating 0. Do this 500 times.

• Calculate of proportion of sample partial correlations |b�n(2)| > 1.96/pn

• Make a QQplot of b�n(2)/pn against a standard normal distribution. What do you observe?

192

6.5 Checking for Goodness of fit

To check for adequency of a model, after fitting a model to the data the sample correlation of the

estimated residuals is evaluated. If there appears to be no correlation in the estimated residuals

(so the residuals are near uncorrelated) then the model is determined to adequately fit the data.

Consider the general model

Xt = g(Yt, ✓) + "t

where {"t} are iid random variables and "t is independent of Yt, Yt�1

, . . .. Note Yt can be a vector,

such as Yt�1

= (Xt�1

, Xt�2

, . . . , Xt�p) and examples of models which satisfy the above include the

AR(p) process. We will assume that {Xt, Yt} is a stationary ergodic process. Further to simplify

the discussion we will assume that ✓ is univariate, it is straightforward to generalize the discussion

below to the multivariate case.

Let b✓ denote the least squares estimator of ✓ i.e.

b✓ = argminnX

t=1

(Xt � g(Yt, ✓))2 . (6.14)

Using the “usual” Taylor expansion methods (and assuming all the usual conditions are satisfied,

such as |b✓ � ✓| = Op(n�1/2) etc) then it can be shown that

pn⇣

b✓ � ✓⌘

= I�1

1pn

nX

t=1

"t@g(Yt, ✓)

@✓+ op(1) where I = E

✓

@g(Yt, ✓)

@✓

◆

2

.

{"t @g(Yt

,✓)@✓ } are martingale di↵erences, which is why

pn⇣

b✓ � ✓⌘

is asymptotically normal, but more

of this in the next chapter. Let Ln(✓) denote the least squares criterion. Note that the above is

true because

@Ln(✓)

@✓= �2

nX

t=1

[Xt � g(Yt, ✓)]@g(Yt, ✓)

@✓

and

@2Ln(✓)

@✓2= �2

nX

t=1

[Xt � g(Yt, ✓)]@2g(Yt, ✓)

@✓2+ 2

nX

t=1

✓

@g(Yt, ✓)

@✓

◆

2

,

193

thus at the true parameter, ✓,

1

n

@2Ln(✓)

@✓2P! 2I.

Based on (6.14) we estimate the residuals using

b"t = Xt � g(Yt, b✓)

and the sample correlation with b⇢(r) = bc(r)/bc(0) where

bc(r) =1

n

n�|r|X

t=1

X

t

b"tb"t+r.

Often it is (wrongly) assumed that one can simply apply the results in Section 6.3 when checking

for adequacy of the model. That is make an ACF plot of b⇢(r) and use [�n�1/2, n1/2] as the error

bars. However, since the parameters have been estimated the size of the error bars will change. In

particular, under the null that the model is correct we will show that

pnb⇢(r) = N

0

B

B

B

@

0, 1|{z}

iid part

� �2

c(0)JrI�1Jr

| {z }

due to parameter estimation

1

C

C

C

A

where c(0) = var[Xt], �2 = var("t) and Jr = E[@g(Yt+r

,✓)@✓ "t] and I = E

⇣

@g(Yt

,✓)@✓

⌘

2

(see, for example,

Li (1992)). Thus the error bars under the null are

±✓

1pn

1� �2

c(0)JrI�1Jr

�◆�

.

Estimation of the parameters means the inclusion of the term �2

c(0)JrI�1Jr. If the lag r is not too

small then Jr will be close to zero and the [±1/pn] approximation is fine, but for small r, JrI�1Jr

can be large and positive, thus the error bars, ±n�1/2, are too wide. Thus one needs to be a little

cautious when interpreting the ±n�1/2 error bars. Note if there is no dependence between "t and

Yt+r then using the usual error bars is fine.

Remark 6.5.1 The fact that the error bars get narrower after fitting a model to the data seems

a little strange. However, it is far from unusual. One explanation is that the variance of the

194

estimated residuals tend to be less than the true residuals (since the estimated residuals contain

less information about the process than the true residuals). The most simplest example are iid

observations {Xi}ni=1

with mean µ and variance �2. The variance of the “estimated residual”

Xi � X is (n� 1)�2/n.

We now derive the above result (using lots of Taylor expansions). By making a Taylor expansion

similar to (6.10) we have

pn [b⇢n(r)� ⇢(r)]

pn[cn(r)� c(r)]

c(0)�p

n [cn(0)� c(0)]c(r)

c(0)2+Op(n

�1/2).

However, under the “null” that the correct model was fitted to the data we have c(r) = 0 for |r| > 0,

this gives

pnb⇢n(r) =

pnbcn(r)

c(0)+ op(1),

thus the sampling properties of b⇢n(r) are determined by bcn(r), and we focus on this term. It is

easy to see that

pnbcn(r) =

1pn

n�rX

t=1

⇣

"t + g(✓, Yt)� g(b✓, Yt)⌘⇣

"t+r + g(✓, Yt+r)� g(b✓, Yt+r)⌘

.

Heuristically, by expanding the above, we can see that

pnbcn(r) ⇡ 1p

n

n�rX

t=1

"t"t+r +1pn

nX

t=1

"t+r

⇣

g(✓, Yt)� g(b✓, Yt)⌘

+1pn

nX

t=1

"t⇣

g(✓, Yt+r)� g(b✓, Yt+r)⌘

,

then by making a Taylor expansion of g(b✓, ·) about g(✓, ·) (to take (b✓ � ✓) out of the sum)

pnbcn(r) ⇡ 1p

n

n�rX

t=1

"t"t+r +(b✓ � ✓)p

n

"

nX

t=1

"t+r@g(✓, Yt)

@✓+ "t

@g(✓, Yt+r)

@✓

#

+ op(1)

=1pn

n�rX

t=1

"t"t+r +pn(b✓ � ✓)

1

n

"

nX

t=1

"t+r@g(✓, Yt)

@✓+ "t

@g(✓, Yt+r)

@✓

#

+ op(1).

195

We make this argument precise below. Making a Taylor expansion we have

pnbcn(r) =

1pn

n�rX

t=1

"t � (b✓ � ✓)@g(✓, Yt)

@✓+

(b✓ � ✓)2

2

@2g(✓t, Yt)

@✓2

!

⇥

"t+r � (b✓ � ✓)@g(✓, Yt+r)

@✓+

(b✓ � ✓)2

2

@2g(✓t+r, Yt+r)

@✓2

!

=pnecn(r)�

pn(b✓ � ✓)

1

n

n�rX

t=1

✓

"t@g(✓, Yt+r)

@✓+ "t+r

@g(✓, Yt)

@✓

◆

+Op(n�1/2)(6.15)

where ✓t lies between b✓ and ✓ and

ecn(r) =1

n

n�rX

t=1

"t"t+r.

We recall that by using ergodicity we have

1

n

n�rX

t=1

✓

"t@g(✓, Yt+r)

@✓+ "t+r

@g(✓, Yt)

@✓

◆

a.s.! E

✓

"t@g(✓, Yt+r)

@✓

◆

= Jr,

where we use that "t+r and @g(✓,Yt

)

@✓ are independent. Subsituting this into (6.15) gives

pnbcn(r) =

pnecn(r)�

pn(b✓ � ✓)Jr + op(1)

=pnecn(r)� I�1Jr

1pn

n�rX

t=1

@g(Yt, ✓)

@✓"t

| {z }

=�p

n

2

@Ln

(✓)

@✓

+op(1).

Asymptotic normality ofpnbcn(r) can be shown by showing asymptotic normality of the bivariate

vectorpn(ecn(r),

@Ln

(✓)@✓ ). Therefore all that remains is to obtain the asymptotic variance of the

above (which will give the desired result);

var

pnecn(r) +

pn

2I�1Jr

@Ln(✓)

@✓

�

var�p

necn(r)�

| {z }

=1

+2I�1Jrcov

✓pnecn(r),

pn

2

@Ln(✓)

@✓

◆

+ I�2J 2

r var

✓pn

2

@Ln(✓)

@✓

◆

(6.16)

196

We evaluate the two covariance above;

cov

✓pnecn(r),�

pn

2

@Ln(✓)

@✓

◆

=1

n

n�rX

t1

,t2

=1

cov

⇢

"t1

"t1

+r, "t2

@g(Yt2

, ✓)

@✓

��

=1

n

n�rX

t1

,t2

=1

cov {"t1

, "t2

} cov⇢

"t1

+r,@g(Yt

2

, ✓)

@✓

�

+ cov {"t1

+r, "t2

} cov⇢

"t1

,@g(Yt

2

, ✓)

@✓

�

+cum

⇢

"t1

, "t1

+r, "t2

,@g(Yt

2

, ✓)

@✓

��

= �2E

"t@g(Yt+r, ✓)

@✓

�

= �2Jr.

Similarly we have

var

✓pn

2

@Ln(✓)

@✓

◆

=1

n

nX

t1

,t2

=1

cov

✓

"t1

@g(Yt1

, ✓)

@✓, "t

2

@g(Yt2

, ✓)

@✓

◆

= �2E

✓

@g(Yt1

, ✓)

@✓

◆

2

= �2I.

Substituting the above into (6.16) gives the asymptotic variance ofpnbc(r) to be

1� �2JrI�1Jr.

Thus we obtain the required result

pnb⇢(r) = N

✓

0, 1� �2

c(0)JrI�1Jr

◆

.

6.6 Long range dependence (long memory) versus changes

in the mean

A process is said to have long range dependence if the autocovariances are not absolutely summable,

i.e.P

k |c(k)| = 1. A nice historical background on long memory is given in this paper.

From a practical point of view data is said to exhibit long range dependence if the autocovari-

ances do not decay very fast to zero as the lag increases. Returning to the Yahoo data considered

in Section 4.1.1 we recall that the ACF plot of the absolute log di↵erences, given again in Figure

6.3 appears to exhibit this type of behaviour. However, it has been argued by several authors that

the ‘appearance of long memory’ is really because of a time-dependent mean has not been corrected

for. Could this be the reason we see the ‘memory’ in the log di↵erences?

We now demonstrate that one must be careful when diagnosing long range dependence, because

a slow/none decay of the autocovariance could also imply a time-dependent mean that has not been

197

0 5 10 15 20 25 30 35

0.00.2

0.40.6

0.81.0

Lag

ACF

Series abs(yahoo.log.diff)

Figure 6.3: ACF plot of the absolute of the log di↵erences.

corrected for. This was shown in Bhattacharya et al. (1983), and applied to econometric data in

Mikosch and Starica (2000) and Mikosch and Starica (2003). A test for distinguishing between long

range dependence and change points is proposed in Berkes et al. (2006).

Suppose that Yt satisfies

Yt = µt + "t,

where {"t} are iid random variables and the mean µt depends on t. We observe {Yt} but do not

know the mean is changing. We want to evaluate the autocovariance function, hence estimate the

autocovariance at lag k using

cn(k) =1

n

n�|k|X

t=1

(Yt � Yn)(Yt+|k| � Yn).

Observe that Yn is not really estimating the mean but the average mean! If we plotted the empirical

ACF {cn(k)} we would see that the covariances do not decay with time. However the true ACF

would be zero and at all lags but zero. The reason the empirical ACF does not decay to zero is

198

because we have not corrected for the time dependent mean. Indeed it can be shown that

cn(k) =1

n

n�|k|X

t=1

(Yt � µt + µt � Yn)(Yt+|k| � µt+k + µt+k � Yn)

⇡ 1

n

n�|k|X

t=1

(Yt � µt)(Yt+|k| � µt+k) +1

n

n�|k|X

t=1

(µt � Yn)(µt+k � Yn)

⇡ c(k)|{z}

true autocovariance=0

+1

n

n�|k|X

t=1

(µt � Yn)(µt+k � Yn)

| {z }

additional term due to time-dependent mean

Expanding the second term and assuming that k << n and µt ⇡ µ(t/n) (and is thus smooth) we

have

1

n

n�|k|X

t=1

(µt � Yn)(µt+k � Yn)

⇡ 1

n

nX

t=1

µ2

t �

1

n

nX

t=1

µt

!

2

+ op(1)

=1

n2

nX

s=1

nX

t=1

µ2

t �

1

n

nX

t=1

µt

!

2

+ op(1)

=1

n2

nX

s=1

nX

t=1

µt (µt � µs) =1

n2

nX

s=1

nX

t=1

(µt � µs)2 +

1

n2

nX

s=1

nX

t=1

µs (µt � µs)

| {z }

=� 1

n

2

Pn

s=1

Pn

t=1

µt

(µt

�µs

)

=1

n2

nX

s=1

nX

t=1

(µt � µs)2 +

1

2n2

nX

s=1

nX

t=1

µs (µt � µs)� 1

2n2

nX

s=1

nX

t=1

µt (µt � µs)

=1

n2

nX

s=1

nX

t=1

(µt � µs)2 +

1

2n2

nX

s=1

nX

t=1

(µs � µt) (µt � µs) =1

2n2

nX

s=1

nX

t=1

(µt � µs)2 .

Therefore

1

n

n�|k|X

t=1

(µt � Yn)(µt+k � Yn) ⇡ 1

2n2

nX

s=1

nX

t=1

(µt � µs)2 .

Thus we observe that the sample covariances are positive and don’t tend to zero for large lags.

This gives the false impression of long memory.

It should be noted if you study a realisation of a time series with a large amount of dependence,

it is unclear whether what you see is actually a stochastic time series or an underlying trend. This

199

makes disentangling a trend from data with a large amount of correlation extremely di�cult.

200