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

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