# Quantitative Methods Questionnaire

Description

Section B
1. We consider an ARMA(1,1) model:
Xt
‘Xt 1 = c + “t
“t 1
where “t is a weak white noise and:
j’j 1
‘h Cov (ut 1 h ; ut ) = Cov (ut 1 ; ut ) = Cov (“t 1
“t 2 ; “t
“t 1 ) =
h=0
Therefore:
Cov (Xt 1 ; ut ) =
V ar (“t 1 )
The necessary and su¢ cient condition for consistency of OLS in (1), that is
Cov (Xt 1 ; ut ) = 0, is tantamount to = 0.
1
V ar (“t 1 )
(c) We want to get a consistent estimator of ‘ from (1) even when 6= 0.
Show that Xt 2 provides a valid instrumental variable to instrument Xt 1 . (25
marks)
Solution:
We know from the course on 2SLS (lecture notes 1 and 2) that the validity of
Xt 2 as an instrumental variable to instrument Xt 1 in (1) is the orthogonality
condition:
Cov (Xt 2 ; Xt
c
‘Xt 1 ) = 0
that is:
Cov (Xt 2 ; ut ) = 0
Since:
1
X
Xt 2 =
h
h
‘ L
h=0
!
(c + ut 2 )
depends only on ut j ; j 2; and Cov (ut ; ut j ) = 0; 8j > 1; we do have the
condition Cov (Xt 2 ; ut ) = 0:
(d) How would you measure the strength (relevance) of this instrument? Explain intuitively how this measure may depend on parameters ‘ and . However,
you do not need to prove an exact formula (25 marks)
Solution:
We know that the strength (relevance) of Xt 2 as an instrumental variable to
instrument Xt 1 is akin to have Cov (Xt 1 ; Xt 2 ) far from zero. Obviously, this
covariance is far from zero insofar as the persistence parameters of the process
Xt ;namely ‘ or are far from zero.
2. We consider the stochastic volatility model:
We assume that
rt+1
=
Et (“t+1 )
=
t “t+1
0; Et “2t+1 = 1
2
t is an AR(1) process:
2
t+1
=
!+
Et ( t+1 )
=
0
2
t +
t+1
(a) We de…ne:
s2t = V art (rt+1 + rt+2 )
Show that:
s2t = ! + (1 + ) 2t
(35 marks)
Solution:
We …rst note that by the law of iterated expectations:
Et (rt+1 + rt+2 )
=
=
Et (rt+1 ) + Et [Et+1 (rt+2 )]
t Et (“t+1 ) + Et [ t+1 Et+1 (“t+2 )] = 0
2
Hence, again by using the law of iterated expectations:
h
i
2
V art (rt+1 + rt+2 ) = Et (rt+1 + rt+2 )
h
i
h
i
2
2
= Et (rt+1 ) + Et (rt+2 ) + 2Et [(rt+1 rt+2 )]
h
i
h
i
2
2
2
2
=
E
(”
)
+
E
E
(”
)
+ 2Et [rt+1 t+1 Et+1 (“t+2 )]
t+1
t
t+2
t t
t+1 t+1
2
t + Et
=
2
t+1
2
t + [! +
+0=
2
t ] = ! + (1 +
) 2t
(b) Deduce that:
Et s2t+2 = 2! (1 + ) +
2 2
st
(30 marks)
Solution:
By application of the law of iterated expectations, we …rst compute:
Et
2
t+2
2
= Et
t+2
2 2
)+
t
= Et Et+1
= ! (1 +
!+
2
t+1
=!+
!+
2
t+2
that:
2
t
However, we know from the previous question that:
s2t
2
t =
!
1+
Therefore, we deduce from the above formula for Et
Et
s2t+2 !
= ! (1 + ) +
1+
2
2 st
!
1+
Therefore:
Et s2t+2
2
= ! + ! (1 + ) +
=
2
s2t
!
2 2
st
2! (1 + ) +
(c) What conclusion do you draw from this result for empirical work on
volatility? (35 marks)
Solution:
If r+1 is the log-return on period [t; t+1], then (rt+1 + rt+2 ) is the log-return
on period [t; t + 2]. Therefore, this exercise shows that if on short time intervals,
volatility persistence is well-described by an AR(1) process , then it is still the
case on longer time intervals. However, volatility persistence decrease with the
length of the interval since:
2]0; 1[=)
3
2