Ragnar Nymoen 29/9 2009. Updated 9/10 2009
Note to IAM Chapter 20.1 (supplement to slide
sets to lecture 7)
The formal analysis of stabilization policy states that the government and/or
central bank aims to minimize the social loss function
SL =
2
y
2
+
,
> 0,
(1)
is a parameter that measures the degree to which the society (through the
mandated decisions makers) values stable in‡ation relative to output stability.
2
2
are the mathematical variances of output and in‡ation that are
y and
implied by our model of the economy.
Since 2y and 2 are derived from the model, we need to consider the solution
of the AD-AS model consisting of
(yt
y)
=
(
t
=
t 1
t
+ (yt
) + zt , and
(2)
y) + st ;
(3)
In the lectures we have already worked with the …nal equation for (yt
is:
(yt y) = (yt 1 y) + (zt zt 1 )
st ,
where
=
1
1+
; 0<
<1
y). It
(4)
(5)
given the assumptions of the model.
As stated in IAM p 612, 2y and 2 are interpreted as the unconditional
variances of the output and in‡ation, also called the marginal variances, or
asymptotical variances.
Finding
2
y
The most direct way of obtaining an expression for 2y is to use (4), and note
that asymptotically the variance should not depend on time, so:
V ar [(yt
y)] = V ar [(yt
1
y)] =
2
y.
We then take the variance on each side of “=” in (4). This gives:
2
y
=
2 2
y
+
2
V ar[zt
zt
1]
+(
)2 V ar[st ]
where we haved used that the variance of a sum of uncorrelated variables is the
sum of the variances, and also have used the rule that V ar[aU ] = V ar[ aU ] =
a2 V ar[U ], where U is a stochastic variable and a is a constant. (NB: How can
1
yt 1 be uncorrelated with zt 1 ? Probably an inconcistency here but we use that
assumption to obtain the formulae in the book). From this we obtain
(1
2
)
2
y
=
2
V ar[zt
)
=
zt
1]
)2 V ar[st ]:
+(
(6)
Note that
2
(1
2
=
(1 +
=
2
1
1+
+ ( )2
1
2
)
2
2
)2
+(
so that (6) can be written as
2
y
=
V ar[zt
+ 2 V ar[st ]
:
+ ( )2
zt
1]
2
(7)
Finally, introduce the notation that V ar[zt ] = 2z and V ar[st ] =
the same expression as in equation (11) on page 612 in IAM:
2
y
=
2
2
2
z
+
+(
2
s,
2 2
s
)2
and obtain
(8)
An alternative derivation uses repeated backward substitution in the …nal equation, and
lim j ((yt j 1 y) ! 0
j!1
to obtain:
(yt
y) =
1
X
j
(zt
zt
j
1 j)
1
X
+
j=0
y)]
=
2
=
2
2
2
z
2
2
z
1
X
2 j
+(
)
2
2
s
2
1
This is the same expression for
1
X
2 j
j=0
1
+(
2
)
2
s
1
2
1
2
2
=
j,
y):
j=0
=
st
j=0
which we can use to calculate the variance of (yt
V ar[(yt
j
2 2z
( ) 2s
+
+ ( )2 ] 2
[2 + ( )2 ]
2 2
2 2z
s
+
.
2
+( )
2 + ( )2
[2
2
2
y
as we found in equation (8).
2
2
NOTE added 9/10 2009:
Tord Krogh has pointed out that this derivation presumes that zt j is
uncorrelated with zt j . This is correct, so the above is only valid if zt follows
a random walk so that zt is without any autocorrelation. In that interpretation
2 2z can be replaced by 2 z for example. Random walk for zt does however
imply that the demand shocks are (fully) persistent.
To obtain the expression that is consistent with intended interpetation that
1
P
j
the demand shocks are temporaty, express
(zt j zt 1 j ), in terms of
j=0
the individual zt
1
X
j
(zt
j ’s:
zt
j
1 j)
=
zt
=
zt
(1
)zt
2
+
1
(1
)zt
2
+ :::
j=0
(1
)[ zt
1
2
+
zt
2
+ :::]:
The variance
V ar[
1
X
j
(zt
zt
j
1 j )]
=
2 2
z
=
2
2
z(
)2
+ (1
2
2
Z[
+(
2 2
) +(
2 3
) + :::]
j=0
=
replaces
2
2
2 1
z1
2
2
)2 (
+ (1
2 2 (1
z
y)] =
(1 +
2
1
1)
)
2
1
in the expression for V ar[(yt
V ar[(yt
1
2 2z
)(2 +
)
y)] above. This then gives
+
2 2
s
2
+(
)2
as the expression that is consistent with the assumption that zt and zt j (j 1)
2
1
are uncorrelated. (Again using = 1+1 and (1 2 ) = 2 +(
)2 .)
I also found the following link with corrections of typos to IAM
http://www.econ.ku.dk/okojacob/MAKRO-2-F07/typos-gbc-…rst-and-secondprint.pdf.
The above is noted on page 7.
Finding
To derive
of (2) in (3):
t
=
2
2
we use the …nal equation for
1
(1 +
)
t 1
+
(1 +
)
zt +
3
t,
1
(1 +
which we obtain by insertion
)
st +
(1 +
)
Taking the variance on both sides, and collecting terms gives:
2
+ ( )2
(1 + )2
2
2
=
(1 +
and then
2
2
z
)
+
2 2
z
=
+ 2s
+ ( )2
2
1
(1 +
2
)
2
s
(9)
which is (12) p 612 in IAM.
Both 2y and 2 are functions of the parameters b and h in the Taylor-rule
function. This is because, as we have seen
=
and
zt =
which means that the variance
1
2
z
is
2
z
=
2h
(1 +
2 b)
(gt g) + vt
(1 + 2 b)
2
v
(1 +
4
2
2 b)