A Note on Lemma 10
1
Introduction
As it stands, Lemma 10 on p 46 is only an approximation when 2 enters
into the score. The exact result, as it applies to Corollary 9 on p 47 is set
out below. This result necessitates modi…cations to some later results.
2
Information matrix for Corollary 9
The information matrix for ; the parameters that appear only in the dynamic equation for a single time-varying parameter tpt 1 ; is given by I D( );
as in Theorem 1 on p37. Lemma 10 extends Theorem 1 to include a second
set of parameters, originally denoted 2 but now called ; to avoid subscripts
and because the result is di¤erent from what it was before. The score for
breaks down into two parts, the …rst is conditional on tpt 1 and the second
is conditional on past observations, Yt 1 : The chain rule1 gives
@ ln f (yt j
d ln f (yt j Yt 1 ; ; )
=
d
@
tpt 1 ;
)
+
@ ln f (yt j Yt 1 ;
@ tpt 1
tpt 1 ) d tpt 1
d
:
(1)
The …rst derivative of (1) is conditioned on tpt 1 ; as in the static model.
Conditioning on tpt 1 automatically implies conditioning on Yt 1 but the
converse is not true because even with Yt 1 …xed, tpt 1 may depend on
through ut 1: Hence the second term. (In the derivative of ln f (yt j Yt 1 ; ; )
with respect to
only the second term appears.) The expectation of each
part of (1) is zero.
In what follows it is assumed that, for the static model, the score and its
…rst derivative with respect to and have …nite time invariant …rst and
second moments that do not depend on and that the same is true for the
cross-product of the score and these …rst derivatives. This condition is an
extension of Condition 2 on p 35.
(x;g)
(x;g) dg(x)
In our notation, the chain rule reads d f (x;g(x))
= @f@x
+ @f @g
dx
dx ; i.e. the total
derivative d takes into account all dependencies, while the partial derivative @ treats all
input parameters that are not di¤erentiated as constants.
1
1
The information matrix, (2.55), for Corollary 9 will be written
=
;
(2)
where
=E
d ln ft (yt j Yt 1 ; ; ) d ln ft (yt j Yt 1 ; ; )
;
d
d 0
d ln ft (yt j Yt 1 ; ; ) d ln ft (yt j Yt 1 ; ; )
d
d 0
d tpt 1 d tpt 1
= I +I E
d
d 0
d tpt 1
d tpt 1
+I E
+E
I ;
0
d
d
= E
where I
(3)
and I ; like I ; are as in the static information matrix, and
d ln ft (yt j Yt 1 ; ; ) d ln ft (yt j Yt 1 ; ; )
@
@ 0
d tpt 1 @ tpt 1
d tpt 1
= I E
+I E
0
d
@
d 0
= E
(4)
In (2.55) - which holds when ut does not depend on
the entries are as
in (2) but with
= I and
= I E [d tpt 1 =d 0 ] : Note that in the
general case additional terms, involving I ; remain even when I = 0:
3
First-order model
In the …rst-order model
= I D( ) as in (2.56). The other terms require
evaluation of the derivatives of t+1pt : If xt is de…ned as on p 35,
d
t+1pt
d
=
d
tpt 1
d
d tpt
= xt
d
1
@ut
@ut d tpt 1
+
@
@ tpt 1 d
@ut
+
;
t = 1; :::; T:
@
+
2
(5)
The chain rule is used to take the total derivative of ut ; thus tpt 1 is …xed
when @ut =@ is evaluated. Taking expectations conditional on Yt 1 gives
Et
d
1
t+1pt
= Et
d
d
xt
1
tpt 1
@ut
@
+
d
=a
@
tpt 1
@
+ Et
@ut ( tpt 1 )
:
@
1
We can take the unconditional expectation of @ut =@ when, as assumed, it
does not depend on tpt 1 : Thus
E
d
t+1pt
=
d
1
a
E
@ut
=
@
1
a
(6)
I :
(When ut is k times the score, as on p32, rather than the score, the last term
has to be multiplied by k:) Furthermore
Et
d
1
t+1pt
d
d
t+1pt
0
= Et
x2t
1
d
tpt 1
d
tpt 1
0
2
+
Et
1
@ut @ut
@ @ 0
d tpt 1 @ut
xt
d
@ 0
d
d
@ut d tpt 1
+ Et 1 xt
+ Et 1
@
d 0
@ut @ut
d tpt 1 d tpt 1
+ 2 Et 1
= b
0
d
@ @ 0
d
d tpt 1 0
@ tpt 1
+ c
+
c;
0
d
@
d
(7)
where
c = Et
1
xt
@ut
@
= Et
@ut @ut
@ut
+
@
@ tpt 1 @
1
=
I + Et
1
@ut @ut
@ @
The conditional expectations can be replaced by the corresponding unconditional expectations so taking unconditional expectations in (7) gives
E
d
t+1pt
d
d
t+1pt
0
d
=
1
1
2
b
2
=
1
b
E
E
@ut @ut
@ @ 0
@ut @ut
@ @ 0
3
+ cE
1
1
a
@
tpt 1
0
@
+ E
c I + I c0
:
@
tpt 1
@
c0
(8)
Therefore
d ln ft (yt j Yt 1 ; ; ) d ln ft (yt j Yt 1 ; ; )
d
d 0
2
@ut @ut
1
= I +I
E
+
2 I I
0
1 b
@ @
1 a
2
I I
1 a
E
The individual terms for ;
and ! in
@ut @ut
I E
@ @ 0
@ut @ut
E
@ @ 0
I
(9)
are shown in Appendix A to
be
=
I
g
1 b
1
a
1
1
g
a
g! + 0 0
1
1
a
I
;
with
g = E ut
@ut
@
c
1
Thus evaluation of
E
a
I
and
and
@ut @ut
@ @ 0
g! =
E
@ut @ut
@ @
(1 + )I
requires that we determine
; E
@ut @ut
@ @ 0
and E ut
@ut
@
;
all of which are independent of by assumption.
When ut does not depend on , the information matrix in (2) has
I and
1
= 0 0
I ;
1 a
=
(10)
which is as in (2.56).
4
Beta-t-EGARCH
In Beta-t-EGARCH tpt 1 becomes tpt 1 : The matrix in Proposition 20 on
p 116 needs to be modi…ed by adding a term to I = h( )=2 so that
" "
#
#
2
2
1
@ut
2 I2
2 I
@ut @ut
2
= h( )+I
E
E
I2 ;
2
1 b
@
1 a
1 a
@ @
1 a
4
0
where
I
E
"
@ut
@
=
2
#
2
( + 3) ( + 1)
= Eb4t
2Eb3t (1
I
=
2
+3
bt )= + Eb2t (1
bt ) 2 =
2
and
E
@ut @ut
@ @
=
2( + 1)Eb3t (1
bt ) + 2( + 1)Eb2t (1
bt ) 2 =
The expectations of the terms involving the beta variables, bt ; depend only
on and can be evaluated using (2.5) on p 23. The elements in the last
column (and bottom row) of the information matrix, that is the elements of
; are
g
g
I
;
I
1 b
1 b1 a
and
1
1
@ut @ut
+ I
I
E
(1 + )I
;
1 a
1 b1 a
@ @
where
cI
@ut
g = E ut
@
1 a
with
E ut
@ut
@
= ( + 1)b3t
b2t
( + 1)b2t (1
bt )= + bt (1
bt )= :
Table 4.1 on p 116 needs to be amended. However, because the additional
terms depend on ; which is typically rather small, the di¤erences are small.
The original table also had I set to the correct expression divided by minus
2. This correction was made when the approximate ASEs shown in brackets
in the table below were calculated. As can be seen the e¤ect of the omitted
terms on the ASEs of and is negligible. However, the exact ASE(!)
is very close to the RMSE. On the other hand ASE( ) is smaller than the
RMSE but this underestimation is not unusual with shape parameters and
it also happens with location in the next section, where the exact results
are much closer to the RMSEs. The extra terms in the information for
must be positive and this can be expected to translate into a smaller ASE.
Calculations for the other combinations of parameters show a similar pattern.
5
Parameter
ML estimates for T=10,000
RMSE( ) ASE( ) RMSE( ) ASE( )
0.95 0.10
0.0053
0.0051
0.0048
0.0050
(0.0053)
(0.0048)
0.99 0.05
0.0018
0.0018
0.0030
0.0031
(0.0018)
(0.0031)
Parameter
ML estimates for T=10,000
RMSE(!) ASE(!) RMSE( ) ASE( )
0.032
0.032
0.325
0.280
(0.043)
(0.345)
0.065
0.066
0.343
0.332
(0.091)
(0.345)
0.95 0.10
0.99 0.05
Corollary 21 is a¤ected in that the limiting distribution of e is not
exact.
Proposition 27 on p137-8 needs to be modi…ed as follows. The matrix
I( ; ) is as in the amended Proposition 20 and the term ( + 1)=( + 3) is
multiplied by
2
2
1+
1
b
8( + 1)E[bt (1
bt )3 ] = 1 +
1
8( + 4)( + 2)
b ( + 7)( + 5)( + 3)
The block diagonality remains.
Proof There is some simpli…cation because I
@ut
=
@
2"t e
2
(1
bt )2 ( + 1)
= 0 and
1
is an odd function. Because
2
I
@ut
1
= 4( + 1)2 e 2
bt (1 bt )3
@
#
"
2
2
2
@ut
+1
8( + 1)2
+I
E
=
+
E[bt (1
1 b
@
+3
1 b ( + 3)
bt )3 ]
The in‡ation in the information quantity coming from tpt 1 is typically
rather small, for example with = 0:05; = 6; it is 4:03%. Since the information matrix is still block diagonal, this translates into a slightly smaller
asymptotic variance for the ML estimator of :
6
Proposition 32 and 34 in Chapter 5 need to be modi…ed in a similar way
to Proposition 20.
Remark 1 In classic EGARCH ut is replaced by j"t j and this does not depend on : Thus
= h( )=2 and
= [0 0 I (1
)=1 a] :
5
Dynamic Location
Proposition 9 is modi…ed by setting the elements in the information matrix
for and to
2
I
+I
1
2
I +I
1
b
b
E
"
@ut
@
E
@ut
@
2
#
@ut
@
2
+
+3
=
=
+1
+31
2
b
4 Eb3t (1
+1
2
+
( + 1)( + 3)
+31
bt )
2
b
2Eb3t (1 bt )
and
2
I
+I
1
b
E
"
@ut
@
2
#
h( )
+
2
=
+1
+31
2
1
b
Eb3t (1
bt )
As regards the last two columns, the entries are
3
2
+1
+1
2Eb2t (1 bt )
Eb2t (1 bt )
+3 1 b
+3 1 b
4 +1
2Eb2t (1 bt ) +1
Eb2t (1 bt ) 5
+3 1 b 1 a
+3 1 b 1 a
0
0
2
3
2
1
+1
2
2
=
Eb (1 bt ) 4 1 a 1 a 5
+31 b t
0
0
The last two rows of the last two columns could be written as
I
I
Proof
We have
I
I
+
+1
+31
@ut
= (y
@
2
b
Eb3t (1
)bt (1
7
bt )
bt )=
4
2
2
1=
so
2
@ut
@
E
= e2 Eb3t (1
and
bt )=
E ut
@ut
@
= e2 E(1
bt )b2t
Also
@ut
@ut
= 2(y
)bt (1 bt ) = 2
@
@
Functions of exp(2 ) cancel.
Note that the ML estimator of ! is distributed independently of the other
parameters in large samples so Corollary 12 remains valid.
The Table shows some of the results from Harvey and Luati (2014) together with the correct ASEs. On the whole the new ASEs are closer to the
RMSEs
Parameter
ML estimates for T = 1000
!
0.8 1.0
RMSE
0.025 0.067 0.031 0.144 0.920
ASE jasa 0.025 0.045 0.038 0.147 1.092
ASE new 0.022 0.058 0.032 0.147 0.855
0.95 1.3
RMSE
0.011 0.071 0.029 0.445 0.740
ASE jasa 0.010 0.043 0.038 0.596 1.092
ASE new 0.009 0.069 0.028 0.596 0.672
A
Derivation of the …rst-order information
matrix
The terms involving
Et
d
1
t+1pt
d
d
t+1pt
0
d
and : For
= Et
1
+Et
= b
d
x2t
tpt 1
+Et
d
d
d
8
tpt 1
0
d
tpt 1
0
[xt ut ]
tpt 1
0
+ Et
d
d
1
d
tpt 1
xt ut
1
d
d
d
+ Et
+ Et
tpt 1
0
d
1
ut
+ Et
@ut
@ 0
1
ut
d
tpt 1
1
d
xt
@ut
@ 0
@ut
@ 0
1
xt
@ut d tpt
d
@ 0
1
Taking unconditional expectations and noting that E(d
d
E
t+1pt
d
d
t+1pt
0
=
d
1
E ut
b
@ut
@ 0
t+1pt =d
c
1
Hence
d ln ft (yt j Yt 1 ; ; ) d ln ft (yt j Yt 1 ; ; )
E
=I
d
1
d 0
a
b
) = 0;
I
@ut
@ 0
E ut
c
1 a
(11)
I
For
Et
d
1
t+1pt
d
d
t+1pt
0
= Et
d
x2t
1
+Et
1
d
tpt 1
d
d
xt (
tpt 1
0
+ Et
d
!)
tpt 1
@
tpt 1
0
1
(
!)
tpt 1
+ Et
d
tpt 1
1
d
@
@ut
d tpt 1 d tpt 1
= b
+ E
( tpt 1 !)
0
d
d
@ 0
d tpt 1
@ut d tpt
+ Et 1 xt 0
+a( tpt 1 !)
0
d
d
@
Now Ed
follows:
t+1pt =d
Et 2 (
!)
tpt 1
d
= 0 and E(
tpt 1
0
d
= Et
= a
2
d
d
1
t 1pt 2
d
+
Et 2 (
xt
@ut
@ 0
1
!) = 0: The third term is found as
tpt 1
(xt
@ut
@ 0
t 1pt 2
d
(
t 1pt 2
@ut 1
)(
@ 0
@ut 1
)( ( t
@ 0
d t 1pt
!) + c
d
+
!) + ut 1 )
1pt 2
2
+
2
Et 2 ut
!)
t 1pt 2
On taking unconditional expectations, the last term disappears. As regards
the …rst term
d tpt 1
d t 1pt 2
@ut 1
E( tpt 1 !)
= Et 2 (xt 1
+
)( ( t 1pt 2 !) + ut 1 )
0
d
d
@ 0
Taking unconditional expectations
E(
tpt 1
!)
d
tpt 1
0
d
=
=
1
1
@ut 1
d
@ 0
@ut 1
c
I + 2 Eut 1
1 a
@ 0
cE
a
1
1
a
9
d
t 1pt 2
+
2
Eut
1
1
@ut 1
@ 0
Hence
2
1
d ln ft (yt j Yt 1 ; ; ) d ln ft (yt j Yt 1 ; ; )
=I
E
0
d
1 b1
d
Eut
a
@ut
@ 0
c
1
a
I
For !
Et
d
1
t+1pt
d
d!
t+1pt
= Et
d
1
+Et
= b
d
x2t
1
tpt 1
d
+a(1
d
tpt 1
d
d!
d
d
)
xt (1
d
tpt 1
0
d
d
)
tpt 1
0
+ Et
tpt 1
0
tpt 1
0
d
+ Et
d
+ (1
1
(1
)
d
1
@ut
@ 0
tpt 1
d!
@ut
@ 0
@ut d tpt
xt
@
d!
)Et
+ Et
1
xt
@ut
@
1
1
so
E
d
t+1pt
d!
d
t+1pt
0
d
=
1
b
(1
@ut
@ 0
)E
+
(1
1
)a
E
a
@ut
@ 0
+ E xt
Thus when ut is the score
d ln ft (yt j Yt 1 ; ; ) d ln ft (yt j Yt 1 ; ; )
d!
d 0
1
@ut @ut
1
+I
(1 + )I + E
= I
1 a
1 b1 a
@ @ 0
E
10
@ut 1
@ 0 1
a