Problem Set 1: Answer Key
1) To obtain the process for yt is very simple:
"t
= yt
=
(yt
=
(1
yt
yt
1
) + !t
) + yt + ! t
+1
So, fyt gt= 1 is also a stationary AR(1) process with the same autore+1
gressive coe¢ cient as f"t gt= 1 . However, as the "mutatis mutandis" clause
warns us, while the theoretical intercept in the original process is zero, it need
+1
not be for fyt gt= 1 :
This is how e-views understands the command "Y c ar(1)". It treats the
given series Y as a constant plus a series that exhibits autoregressive behaviour.
The autoregressive coe¢ cient is the same as we have seen, but special attention
must be paid to the interpretation of the intercept provided by e-views.
2) The best way to proceed is to rewrite the given process in deviation from its
unconditional mean (remember we are, for the time being, dealing with
stationary processes). Therefore, we shall …rst calculate the mean. Only
the case in ii. shall be solved, as the case in i. follows in the exact same
manner and is somewhat less cumbersome.
ii. Consider the following ARMA(2,1) process:
(1
yt
2 L )yt
2
1L
= c + 1 yt 1 + 2 yt
= c + (1 + L)"t
2
+ "t
1
+ "t
Let
(1
2
1
2L )
1L
be the lag operator polynomial that allows us to "solve" for yt in terms of "t
and its past values. Recalling the results reviewed in Problem Set 0:
yt
yt
=
=
(1
2
1
c
2L )
1L
c
1
1
+ (1
+ (1
2
1
(1
2L )
1L
2
1L
2L
)
1
+ L)"t
(1 + L)"t
2
Now, as yt is bounded, the terms of the lag polynomial on "t must not
diverge. Hence, we can take expectations on both sides of the equality and by
the assumption of White Noise on "t we get:
E [yt ] =
c
1
1
1
2
It is straightforward that the process in deviation from its mean is:
yet =
et 1
1y
et 2
2y
+
+ "t
1
+ "t
To calculate the autocorrelation function we …rst compute the autocovariance function and then divide by the variance.
(k)
= E [e
yt yet k ]
= E [( 1 yet 1 + 2 yet
=
yt 1 yet k ) +
1 E(e
+ "t 1 + "t )e
yt k ]
yt 2 yet k ) + E("t
2 E(e
2
et k )
1y
+ E("t yet
k)
In order to proceed, we need to give values to.k = 0; 1; 2; ::::until we …nd
a recursion which is valid for any natural number k.
= E(e
yt2 )
f or k = 0
=
yt 1 yet ) + 2 E(e
yt 2 yet ) + E("t 1 yet ) + E("t yet )
1 E(e
=
et ) + E("t yet )
1 (1) + 2 (2) + E("t 1 y
(0)
Take the last term:
E("t yet )
= E [( 1 yet 1 + 2 yet 2 + "t 1 + "t )"t ]
=
yt 1 "t ) + 2 E(e
yt 2 "t ) + E("t 1 "t ) + E("2t )
1 E(e
By stationarity, yt j can be written in terms of "t
past. By properties of White Noise, we then have:
E(e
yt
E("t
j "t )
1 "t )
j
= 0 8j > 0
= 0
) E("t yet ) =
and values further in the
2
"
Now take the previous to last term:
E("t
et )
1y
= E [( 1 yet 1 + 2 yet 2 + "t 1 + "t )"t 1 ]
=
yt 1 "t 1 ) + 2 E(e
yt 2 "t 1 ) + E("2t 1 ) + E("t "t
1 E(e
2
= ( 1+ ) "
So,
(0) =
1
(1) +
2
(2) + [1 + (
1
+ )]
2
"
Analogously,
(1)
= E(e
yt yet 1 )
f or k = 1
= E [( 1 yet 1 + 2 yet 2 + "t 1 + "t )e
yt 1 ]
=
et 1 ) + E("t yet
1 (0) + 2 (1) + E ("t 1 y
2
=
1 (0) + 2 (1) +
"
2
1)
1)
Note that for any k
2 both terms containing the expectation of the
products of " and y are zero. Therefore,
(2) =
(k) =
1
1
(1) + 2 (0)
f or k = 2:
(k 1) + 2 (k 2)
f or k > 2:
To solve for (0) , (1) and (2); we have to solve the following system of 3
linear equations:
2
32
3 2
3
1
(0)
[1 + ( 1 + )] 2"
1
2
2
4
5
1
0 5 4 (1) 5 = 4
1
2
"
1
(2)
0
2
1
It can be shown that the determinant of the above matrix is ( 2 + 1) ( 1
which is non-zero under stationarity. So, the system has only one solution given
by:
2
3 2
3 12
3
(0)
1
[1 + ( 1 + )] 2"
1
2
2
4 (1) 5 = 4
5
1
0 5 4
1
2
"
(2)
1
0
2
1
Once you have solved for (0), the autocorrelation function is simply
the autocovariance function divided by (0):
3) We shall derive this result in two steps, following Hamilton (1994).
step 1: the sum of a zero-mean MA(q1) process and a zero-mean MA(q2)
process that are independent is a zero-mean MA(maxfq1; q2g) process.
Let our zero-mean MA processes be:
1
u1t =
q1 (L)"t
2
u2t =
q2 (L)"t
cov(u1t ; u2t j ) = EE u1t u2t
k
= 0 8j
De…ne their sum and obtain its relevant moments:
wt = u1t + u2t
E (wt ) = 0
(k) = E [wt wt k ]
= E u1t + u2t
= E u1t u1t
=
1
(k) +
k
2
u1t
+E
2
k + ut k
u2t u2t k +
E u1t u2t
k
+ E u2t u1t
k
(k)
We know that for a MA(q) process (and only for such a process), (k) =
0 8k > q: In this case, (k) = 1 (k) + 2 (k) = 0 8k > max fq1; q2g : Hence,
the sum has to be a zero-mean MA(maxfq1; q2g) process.
3
2
+ 1) (1
2
1)
step 2: Consider the processes xt and yt in terms of deviations from their mean
and in the lag operator polynomial notation and consider their sum (also
in deviations from its mean):
zet
L)
x
e
t
x
L
y
e
t
y
(1
1
= x
et + yet
= ! xt
= ! yt
Cross apply the lag operator polynomial to each White Noise process
as follows:
1
yL
(1
x L)
et
x L) x
(1
1
yL
yet
=
1
=
(1
yL
! xt
y
x L) ! t
The order in which we apply the polynomial is irrelevant. So we can
write the sum of the above two processes as:
1
(1
yL
(1
x L)
1
et
x L) x
yL
+ (1
x L)
1
yL
x
et + (1
x L)
1
yL
(1
x L)
(1
1
yL
x L)
1
yet
yet
(e
xt + yet )
yL
zet
=
1
yL
! xt + (1
=
1
yL
! xt + (1
=
1
yL
! xt + (1
=
1
yL
! xt + (1
y
x L) ! t
y
x L) ! t
y
x L) ! t
y
x L) ! t
The right hand side consists of the sum of two zero-mean MA(1) processes,
so by step 1 it is a MA(1) process. On the left hand side we have that a second
order lag operator polynomial is applied to zet , giving it a second order autoregressive behaviour. A process that has both a second order autoregressive
behaviour and a MA(1) behaviour is an ARMA(2,1) process.
4) Consider a sample of size T. It should be clear that this sample CANNOT
BE a RANDOM SAMPLE.
i. The conditional probability distribution of every sample unit t
yt =yt
1; yt 2 ; :::
= yt =yt
1
N c + yt
1;
2 is:
2
"
This is because yt is a linear function of a gaussian random variable,
given yt 1 : The likelihood of the sample can only be written conditional of the
…rst observation of the sample, as the unconditional density is unknown. Let
Lc be the conditional likelihood and lc be the conditional log-likelihood:
Lc (c; ;
" ; y2 ; :::; yT =y1 )
lc (c; ;
" ; y2 ; :::; yT =y1 )
= fy2 =y1 (y2 ) :fy3 =y1 ;y2 (y3 ) :::fyT =yT 1 ;:::;y1 (yT )
= fy2 =y1 (y2 ) :fy3 =y2 (y3 ) :::fyT =yT 1 (yT )
(
)
T 1
T
2
1 X
1
2
exp
(yt c
yt 1 )
=
2 2"
2 2" t=2
=
=
ln fLc (c; ;
T
1
2
4
" ; y2 ; :::; yT =y1 )g
ln(2 )
(T
1) ln(
")
2
T
1 X
2
" t=2
(yt
c
yt
2
1)
It is clear that, for estimators e
c; e; e" :
arg max lc e
c; e; e" ; y2 ; :::; yT =y1 = arg min
e
c;e
T
X
e
c;e t=2
yt
e
c
e yt
2
1
This shows that the CMLE of the parameters of the autoregressive
structure are none others than the (by now very well known) OLS estimators:
# 1" P
#
"
PT
T
b
c
y
y
T 1
t=2 t 1
PT t=2 t
b = PT yt 1 PT y 2
t=2
t=2 t 1
t=2 yt yt 1
ii. Recall that the OLS estimator is consistent if the following condition holds:
E (xt ut ) = 0k
In this case:
k
=
2
xt
=
ut
= "t
1
yt 1
and so:
E ("t )
E [yt 1 "t ]
E (xt ut ) =
As "t is an instance of a White Noise, both expectations are zero, ensuring consistency. Of course, we are also assuming the (almost surely) full rank
condition. In a simple model, like this one, the full rank condition is equivalent
to asking that the variable yt 1 be non-constant. But this has to be so; otherwise the White Noise term would also have to be constant (violating the White
Noise assumption).
That the orthogonality condition ensures consistency is obtained in a
basic econometrics course resorting to the weak law of large numbers under the
random sampling hypothesis. In a time series course we cannot do that, but we
can resort to the ergodicity property (we might also resort to other versions of
the law of large numbers, but ergocity seems to be the most popular choice).
A stationary process is said to be ergodic for the moment m (time-independent
under stationarity) if that moment can be consistently estimates by its sample
analogue. A stationary process is ergodic for if the sequence of autocovariances
(indexed by k) is absolutely summable:
+1
X
k=0
j
kj
< +1
It turns out that for the processes we will handle in this course, stationarity is
not only a necessary but also a su¢ cient condition for ergodicity for . It also
turns out that if the White Noise term is gaussian, ergodicity for is su¢ cient
for ergodicity for every moment.
5
iii. Recall that, for a linear model y = X + u, under the Lindberg-Feller
theorem conditions:
p
n b
! D N 0k ; 2u Q 1
Q
=
XT X
n
p lim
For our AR(1) model and size T sample, it is easy to verify that the
Lindberg-Feller conditions hold (ask the TA if you can‘t see this), so:
p
T
b
c
b
1
Q = p lim
= p lim
c
1
T
"
1
PT
"
1
t=2
T
!D N 02 ;
T
PT
1
y
t=2 t
yt
1
1
1
PT
t=2 yt
PTT 12
t=2 yt
T 1
2
1
"Q
PT
yt
Pt=2
T
2
t=2 yt
#!
1
1
#!
1
1
The plim of a matrix the the matrix whose elements are the plims of
the elements of the original matrix. Under ergodicity:
1
Q =
Q
1
=
2
"
+
0
#
2
1+
0
1
0
0
0
And so,
p
T
1
b
c
b
c
0
!D N @
0
0
2
;4
2
"
2
1+
2
"
0
0
2
"
2
"
0
0
31
5A
We can further reduce the expression presented above. For a stationary
6
AR(1):
=
2
"
0
2
"
c
1
(1 + )
=
1
2
1+
=
5)
T
1
b
c
b
c
D
!
2
"
=
2
"
=
2
"
N
2
c
= (1 + ) c
2
"
1
2
"
1
1+
+ c2
0
0
;
s=t
2
c
6
41 +
"
i. Under the market e¢ ciency hypothesis,
(1
X
Pt = Et
=
2
"
2
1
2
0
p
2
"
=
0
=
0
2
"
;
2
7
5
c2
1+
1
1+
1
2
"
3
2
"
1+
1
+ c2
(1 + ) c
Ds
(1 + r)s
t
(1 + ) c
1
2
#!
)
1
X
Et [Ds ]
(1 + r)s t
s=t
The above equation is the result of forward looking behaviour on prices:
Pt =
1
1+r
Et [Pt+1 ] + Et [Dt ]
and a transversality condition that rules out speculative bubbles. From this last
equation,
(1 + r) Pt
1+r
= Et [Pt+1 ] + (1 + r) Et [Dt ]
Pt+1
Dt
= Et
+ (1 + r) Et
Pt
Pt
From the de…nition of returns we have:
Pt+1
Dt
Rt =
1 + (1 + r)
Pt
Pt
Pt+1
Dt
= 1 + Rt (1 + r)
Pt
Pt
Pt+1
Dt
Et
= 1 + Et [Rt ] (1 + r)Et
Pt
Pt
7
Combining our results so far,
1+r
=
1 + Et [Rt ]
(1 + r)Et
1+r
= 1 + Et [Rt ]
) Et [Rt r] = 0
Dt
Dt
+ (1 + r) Et
Pt
Pt
The condition derived states that under market e¢ ciency, the expected
excess return of the risky asset is zero.
ii. Let dividends follow a stationary AR(2) process:
Dt =
1 Dt 1
+
2 Dt 2
+ "t
The companion form of this process is:
zt
zt
Dt
;
Dt 1
= Azt 1 + ut
=
1
A=
1
2
0
;
ut =
"t
0
What we are looking for is Et [Dt+j ] for a non-negative j, the …rst
element of Et [zt+j ]:
Et [Dt+j ] = 1 0 Et [zt+j ]
It is straightforward that:
Et [zt+j ] = Aj zt
and so prices can be written as:
Pt
=
=
=
1
X
j=0
1
X
1
(1 + r)j
1
0
Et [zt+j ]
1
1 0 Aj zt
j
(1
+
r)
j=0
9
8
1
=
<X
1
j
1 0
zt
A
:
(1 + r)j ;
j=0
Take the sum. By excercise 6) in Problem Set 0 we know that stationarity of an AR(2) process implies that the eigenvalues of the matrix of the
companion form are inside the unit circle (and viceversa, they are equivalent
A
conditions). In addition, the eigenvalues of 1+r
are simply the eigenvalues of A
divided by 1 + r; which is greater than 1. So, under stationarity the eigenvalues
A
of 1+r
are inside the unit circle and the sum is convergent. This means that it
8
can be written as:
1
X
j=0
1
Aj
(1 + r)j
=
1
X
j=0
=
=
=
"
1
1
2
1+r
1
1+r
1+r
1
1+r
1
1+r
1
1+r
c11
c21
=
1
A
1+r
I
"
j
A
1+r
2
1+r
1
#
#
1
1
c12
c22
Going back to prices,
Pt
1
=
c11
c21
0
c12
c22
zt
c11 c12
c21 c22
= Dt c11 + c12 Dt 1
= (c11 1 + c12 L) Dt
1
=
Dt
Dt 1
0
Since we are assuming stationarity, we can write Dt as:
Dt = 1
2
2L
1L
1
"t
and so,
Pt
=
=
1
1L
2
2L
Pt
Pt
=
=
(c11 1 + c12 L) 1
1
1L
2
2L
1L
1
2
2L
1
"t
(c11 1 + c12 L) "t
(c11 1 + c12 L) "t
1 Pt 1 + 2 Pt 2 + c11 "t + c12 "t
1
The last line shows that stock prices follow an ARMA(2,1) process. It
also shows that its autoregressive structure is the same as it is for dividends.
9