(7)Log_linearization

There are several ways to log-linearize, some more cumbersome than others, depending on the speci…c equation:
1. Linearize and substract and divide by steady state (example 1 below) 2. Linearize in logs: let x = log X, so that X = ex and linearize around
where y = Ak k^t+1 = y y c kct y c c cbt by trick 3 = y y c kbt y c c cbt by trick 4
Often, we like to simplify notation a bit by writing something like k^t+1 = (1 + ) kbt cbt
Same example, but linearizing in logs
– Directly log-linearize (using Taylor approximation):
h
i
u0 cec^t
=
Et AeA^t+1 F1 KeK^t+1 ; heh^t+1 u0 cec^t+1
h
i
c e c^t =
Et
c^t+1
– With CRRA utility and Cobb Douglas production function
u (c) = 1 c1 ; 1
F (K; h) = K h1
c^t ' EtA^t+1 (1 ) EtK^t+1 Eth^t+1
Etc^t+1
– U¤!
1.2 Example 2
– A technicality: if a variable has a zero steady state (like e.g. the nominal interest rate in a new Keynesian model), then we cannot log-linearize around the steady state for that variable. In that case, we normally log-linearize around 1+ the variable, so that the original variable reappears in the log-linearized model. 1d + i = log (1 + i) log (1 + {) ' i
– The variables in the log-linearized model are easy to interpret because they are relative deviations from steady state. x^ = log x log x ' x x x
– Parameters in the log-linearized model are easy to interpret because they are elasticities.
Log-linearizing
Thijs van Rens
January 23, 2008
1 What log-linearization is and how to do it
Log-linearization means nothing else than linearization (i.e. …rst-order Taylor approximation) in logs.
2.1 Example 3
This is the neoclassical growth model from question 1 in the …rst problem set. I log-linearize the equations using the tricks.
– Euler equation
+ u0 (c) A F11 K; h EtKt+1 K + F12 K; h Etht+1 h + AF1 K; h u00 (c) (Etct+1 c)
– By the original equation, the steady state terms drop out on both sides
+
AF1 AF1
K; h u00 (c) K; h u0 (c) (Etct+1
c)
2
– Making it look nice:
cu00 (c) u0 (c) c^t
'
EtA^t+1+
KF11 K; h F1 K; h
Et
K^ t+1
+
hF12 F1
K; h K; h
Et h^ t+1 +
cu00 (c) u0 (c)
2 Some useful tricks
1. z = xy ) z^ = x^ + y^
2. z = x=y ) z^ = x^ y^
3. z = ax ) z^ = x^
4. z = xa ) z^ = ax^
5.
z = x + y ) z^ =
x z
x^
+
y z
y^
3
These are just tricks that you may want to learn by heart in order to save time. There is nothing mystical about them: all are easily proven by linearizing and dividing by the steady state.
– Just like for linearization, the …rst question is “log-linearize around what?”. We typically (log-)linearize our models around the steady state or (for more complicated models with frictions) around the frictionless steady state.
Et AeA^t+1 K (1 )e (1 h )K^t+1 1 e(1 c )h^t+1 e c^t+1
h
i
c e c^t =
A K (1
)h1
c
Et eA^t+1 e (1
e )K^ t+1 (1
e )h^ t+1
c^t+1
h
i
e c^t = Et eA^t+1 e (1 e )K^ t+1 (1 e )h^t+1 c^t+1
Compared to linearization in levels log-linearization has some advantages:
– Many models in economics are approximately log-linear, so log-linearization might be a better approximation than linearization.
h
i
1 c^t ' Et 1 + A^t+1 1 (1 ) K^t+1 1 + (1 ) h^t+1 (1 c^t+1)
h
i
' Et 1 + A^t+1 (1 ) K^t+1 + (1 ) h^t+1 c^t+1
c^t ' EtA^t+1 (1 ) EtK^t+1 Eth^t+1
Etc^t+1
– Note: it looks like we could have just taken logs, but then how do we deal with the expectation operator? So, even though the original equation is multiplicative, this is still an approximation!
c^t+1 =
Akd t+(11 )ct
= kt+(d 11 )ct by trick 3
= kt+d (11 ) + c^t by trick 1
= (1 ) k^t+1 + c^t by trick 4
– Resource constraint
k^t+1 = Akd t ct = y y c Adkt + y cc dct by trick 5
– Divide by steady state (left and right steady state are the same)
u00 (c) u0 (c) (ct c) '
F1 K; h u0 (c) AF1 K; h u0 (c)
EtAt+1
A
u0 (c) A + AF1 K; h u0 (c) F11 K; h EtKt+1 K + F12 K; h Etht+1 h
x
1
– A variant of this is to substitute
x = elog x = elog xelog x log x = xex^
for all variables and linearize around x^ (see example 2 below) – It is sometimes useful to know some Taylor approximations by
– Linearizing left-hand side: u0 (ct) ' u0 (c) + u00 (c) (ct c)
– Linearizing right-hand side:
Et [At+1F1 (Kt+1; ht+1) u0 (ct+1)] ' AF1 K; h u0 (c) + F1 K; h u0 (c) EtAt+1 A
where = c= (y c).
4
heart, in particular (Uhlig, see Canova p.53):
ex^ ' 1 + x^ x^y^ ' 0
– Learn a few tricks by heart (see below)
For all methods, it is important to realize that we can switch derivatives and integrals and are therefore allo the expectations operator.
1.1 Example 1
Log-linearizing the Euler equation in the RBC model by linearizing and then substracting and dividing by the steady state.
– The Euler equation u0 (ct) = Et [At+1F1 (Kt+1; ht+1) u0 (ct+1)]