Linearization Solution
Lecture: Solving RBC model
Advanced MacroeconomicsUniversity of Warsaw
Jacek Suda
March 27, 2018
RBC
Linearization Solution
Plan of the Presentation
1 Linearization
2 SolutionBlanchard and Kahn’s (1980) MethodSims’ (2001) Method
RBC
Linearization Solution
What we did
We built the model
We have a non-linear expectational difference model
Want to solve it. We can either1 use non-linear solution method,2 use (linear) approximation method.
Today:approximate (linearly) using Taylor-series expansions andsolve the resulting linear system.
If we log-linearize the system, variables are expressed as deviationsfrom steady state values.
RBC
Linearization Solution
Taylor series approximation
Consider an arbitrary univariate function f : R→ R.Taylor’s theorem says that if f is k-times differentiable then f (x) can beapproximated by a power series around a particular point x∗:
f (x) = f (x∗) +f ′(x∗)
1!(x− x∗) +
f ′′(x∗)2!
(x− x∗)2 +f (3)(x∗)
3!(x− x∗)3 + . . .
+f (k)(x∗)
k!(x− x∗)k + hk(x)(x− x∗)k
f ′(x∗) denotes the first derivative of f with respect to x, evaluated at x∗;f ′′(x∗) is the second derivative at x∗;f (n)(x∗) is n-th derivative at x∗
limx→x∗ hk(x) = 0.The order of approximation tells how many powers are used toapproximate
1st order (linear) approximationf (x) ' f (x∗) + f ′(x∗)(x− x∗)
2nd order (quadratic) approximation
f (x) ' f (x∗) + f ′(x∗)(x− x∗) +f ′′(x∗)
2(x− x∗)2
RBC
Linearization Solution
Taylor series approximation
Taylor’s theorem holds for multivariate functions.
Consider a function f (x) = f (x1, x2), x = [x1 x2]′ k−times
differentiable at x∗.
Taylor series around x∗ = (x∗1 , x∗2) can be written as
f (x1, x2) = f (x∗) +∂f∂x1
(x∗)(x1 − x∗1 ) +∂f∂x2
(x∗)(x2 − x∗2 )+
+12∂2f∂x2
1(x∗)(x1 − x∗1 )
2 +12∂2f∂x2
2(x∗)(x2 − x∗2 )
2
+∂2f
∂x2∂x1(x∗)(x1 − x∗1 )(x2 − x∗2 ) + . . .
∂f∂xi
denotes the partial derivative of f with respect to xi, evaluated at x∗;
RBC
Linearization Solution
Log-linearization
Sometimes it is more convenient to first take log of these equations andlinearize them.Consider an equation f (xt) = g(xt). Taking logs of both sides yields
ln f (xt) = ln g(xt)
Now, if we take first-order Taylor expansion around xt = x∗ we get
ln f (xt) ≈ ln f (x∗) +f ′(x∗)f (x∗)
(xt − x∗)
ln g(xt) ≈ ln g(x∗) +g′(x∗)g(x∗)
(xt − x∗)
as [ln f (x)]′ =f ′(x)f (x)
Substituting in
ln f (x∗) +f ′(x∗)f (x∗)
(xt − x∗) = ln g(x∗) +g′(x∗)g(x∗)
(xt − x∗)
and using ln f (x∗) = ln g(x∗) and rearranging terms we getf ′(x∗)x∗
f (x∗)xt − x∗
x∗=
g′(x∗)x∗
g(x∗)xt − x∗
x∗
RBC
Linearization Solution
Log-linearization
Define xt =xt−x∗
x∗ , or the percentage deviation of xt from x∗ to get
f ′(x∗)x∗
f (x∗)xt =
g′(x∗)x∗
g(x∗)xt
Note that f ′(x∗)x∗
f (x∗) is the elasticity of f with respect to xt at x∗.
Since ln(1 + ε) ≈ ε for small values of ε, we can write xt as
xt ≈ ln (xt + 1) = ln(
xt − x∗
x∗+ 1
)= ln
(xt
x∗
)= ln xt − ln x∗
and, as x = eln x = ln ex,
xt = x∗xt
x∗= x∗eln( xt
x∗ ) ≈ x∗ext .
Lastly, note that the Taylor expansions of exponential function and theproduct of two variables around xt = yt = 0 are
exp(αxt) ≈ exp(αxt)|xt=0 + α exp(αxt)|xt=0xt = 1 + αxt
xt yt ≈ 0 + yt |yt=0(xt − 0) + xt |xt=0(yt − 0) = 0
RBC
Linearization Solution
Log-linearization
Using these observations we obtain several useful expressions
xt = x∗ext ≈ x∗ · (1 + xt)
xtyt = x∗ext y∗eyt ≈ x∗y∗ · (1 + xt + yt)
xat ≈ (x∗)a · (1 + axt)
xat yb
t ≈ (x∗)a(y∗)b · (1 + axt + byt)
RBC
Linearization Solution
Back to model - summary
Recall we have a system of 8 equations with 8 endogenous variables (c,r, l, w, k, i, y, z):
c−σt = βEt[c−σt+1(1 + rt+1 − δ)] (1)
wt =lϕt
c−σt(2)
kt = (1− δ)kt−1 + it (3)yt = ct + it (4)
RBC
Linearization Solution
Back to model - summary
yt = ztkαt−1l1−αt (5)
wt = (1− α)yt
lt(6)
rt = αyt
kt−1(7)
zt = exp(εt)zρt−1 (8)
We computed steady-states last time.Now we will linearize each equation.
RBC
Linearization Solution
Log-linearize labor - consumption choice
Equation (2)lϕt
c−σt= wt
In the steady state:
(lss)ϕ
(csst )−σ = wss
Let’s log-linearize:
(lss)ϕ
(css)−σ (1 + ϕlt + σct) = wss(1 + wt)
Divide by steady state:
ϕlt + σct = wt
RBC
Linearization Solution
Log-linearized Euler equation
Equation (1):c−σt = βEt[c−σt+1(1 + rt+1 − δ)]
In the steady state:
1 = β(1 + rss − δ)
When linearized:
(css)−σ
(1− σct) = β (css)−σ Et [(1− σct+1)(1− δ + rss(1 + rt+1))]
RBC
Linearization Solution
Log-linearized Euler equation
Substitute for rss:
1− σct = βEt
[(1− σct+1)(1− δ + (
1β− 1 + δ)(1 + rt+1))
]
1− σct = Et [(1− σct+1)(β − βδ + (1− β + βδ)(1 + rt+1))]
1− σct = Et [(1− σct+1)(1 + (1− β(1− δ))rt+1)]
Multiply and drop higher order terms:
1− σct = 1− σEtct+1 + (1− β(1− δ))Et rt+1
Rearrange terms:
σ(Etct+1 − ct) = (1− β(1− δ))Et rt+1
RBC
Linearization Solution
Log-linearized market clearing condition
Equation (4):ct + it = yt
In the steady state:css + iss = yss
Linearize:css(1 + ct) + iss(1 + it) = yss(1 + yt)
Subtract steady state equation:
(1− iss
yss )ct +iss
yss it = yt
RBC
Linearization Solution
Log-linearized firm’s equilibrium conditions
For labor (equation 6):(1− α)yt
lt= wt
Linearized:yt − lt = wt
For capital (equation 7):α
yt
kt−1= rt
Linearized:yt − kt−1 = rt
RBC
Linearization Solution
Log-linearized shock process
Equation (8):zt = exp(εt)z
ρt−1
After linearization:zt = ρzt−1 + εt
RBC
Linearization Solution
Steady state values
Our equations contain steady state ratio iss
yss .These are determined by our parameters
From the Euler equation:
rss = β−1 − (1− δ)
and from the equilibrium condition for capital:
rsskss = αyss
kss
yss =α
rss =α
β−1 − (1− δ)
RBC
Linearization Solution
Steady state values cont’d
From the capital accumulation equation:
δkss = iss
thusiss
yss = δkss
yss =αδ
β−1 − (1− δ)
RBC
Linearization Solution
Log-linearized system
We now have a system of 8 linear (difference) equations and 8 variablesHave to solve it
RBC
Linearization Solution Blanchard and Kahn’s (1980) Method Sims’ (2001) Method
Plan of the Presentation
1 Linearization
2 Solution
RBC
Linearization Solution Blanchard and Kahn’s (1980) Method Sims’ (2001) Method
Solving linear DSGE models
Solving a DSGE model means changing the system of forward-lookingdifference equations that we have ...... into a VAR , a system of backward-looking difference equationsThere are several techniques for solving such systems:e.g. Blanchard and Kahn (1980), Sims (2001), Christiano (2002), Uhlig(2002), Evans and Honkapohja (2001),...(Dynare will solve it for you)
RBC
Linearization Solution Blanchard and Kahn’s (1980) Method Sims’ (2001) Method
Blanchard & Kahn condition
One very important issue is the existence and the stability conditionWrite the system in state space form:
A1
[Xt+1
EtPt+1
]= A0
[Xt
Pt
]+ γZt+1
where:Xt: vector n× 1 of state variables (backward-looking)Pt: vector m× 1 of jumpers (forward-looking)Zt: vector k × 1 of shocks (with mean equal to 0 every period)A1, A0: (n + m)× (n + m) matricesγ: (n + m)× k matrix
RBC
Linearization Solution Blanchard and Kahn’s (1980) Method Sims’ (2001) Method
Blanchard & Kahn condition cont’d
Assume that A1 is invertible. Then:[Xt+1
EtPt+1
]= A
[Xt
Pt
]+ A−1
1 γZt+1
where A = A−11 A0
The Blanchard-Kahn condition says that for the model to have a uniquesolution, the number of eigenvalues of A lying outside the unit circle(unstable roots) must equal the number of forward looking variables(jumpers)
RBC