Notes on Solving the Linearized Model

Let us consider the following New-Keynesian model, in log-linearized form bt = w bt − cbt νh

ybt = cbt bt b t + αh ybt = a bt bt = a bt + sbt + (α − 1)h w b t = ybt + π bt m 

b t − Et π bt+1 cbt = Et cbt+1 − R



bt = κsbt + βEt π bt+1 π bt = ρa a bt−1 + εat a b t = ρm m b t−1 + εmt m

First of all, one has to understand where the difficulty of the problem is: Solving for the dynamics of expectations! The rest is just a matter of algebra. The strategy will therefore be to 1. Simplify the model to eliminate static equations 2. Solve the forward looking stochastic difference equation system to express variables as a function of the state variables. Before starting, it is important to understand the structure of the problem. First of all, note that the system is, by construction, linear. Second, when entering period t, agents have two bt , and the monetary shock, m b t . This piece of information informations: the technology shock, a

is then sufficient for the agent to take all other decisions. We therefore know that the solution of the problem will be a linear function of the two shocks x bt = γax a b t + γm b t for x ∈ {y, c, h, π, s, R, w} x m

(1)

Let us first start by reducing the size of the system, which, as stated in 1., amounts to use the

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set of static equations bt = w bt − cbt νh

(2)

ybt = cbt

(3)

bt b t + αh ybt = a

(4)

bt bt = a bt + sbt + (α − 1)h w

(5)

b t = ybt + π bt m

(6)

Our aim will be to express the marginal cost as a function of inflation and the shocks only, such that the NKPC reduces to an equation relating inflation to the shocks only. Plugging (3) and (4) in (2), we get bt = w

ν+α ν bt ybt − a α α

Plugging (3) and (4) in (5), we get bt + sbt = w

bt 1−α a ybt − α α

Combining these two equations we get sbt =

1+ν bt ) (ybt − a α

bt −π bt ) in the last equation to get Finally use (6) (ybt = m

sbt =

1+ν bt −a bt − π bt ) (m α

Note that at this stage we have bt −π bt ybt = m

(7)

cbt = ybt b t = 1 (ybt − a bt ) h α ν+α ν bt = bt w ybt − a α α 1+ν bt −a bt − π bt ) sbt = (m α

(8)

Let us now make use of this information plug the expression for sbt in the NKPC bt = κ π

1+ν bt −a bt − π bt ) + βEt π bt+1 (m α

which implies bt = π

Let us denote ϕ ≡

1 α+κ(1+ν)

κ(1 + ν) β bt −a bt ) + bt+1 (m Et π α + κ(1 + ν) α + κ(1 + ν) < 1, the NKPC rewrites bt = (1 − ϕ)(m bt −a bt ) + βϕEt π bt+1 π

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(9) (10) (11)

Note that, at this stage, this equation involves only inflation and the shocks, and takes the form of the prototypical linear rational expectation model we encountered earlier in class. Let us use the guess (1) and plug it in the last expression π π b t + γm b t = (1 − ϕ)(m bt −a bt ) + βϕEt [γaπ a bt+1 + γm b t+1 ] γaπ a m m

Remember that finding a solution to the model, just amounts to find the two parameters γaπ and π that characterize the solution for inflation. Using the AR(1) structure of the shock, we have γm

bt+1 = ρa a bt and Et m b t+1 = ρm m b t , such that the preceding equation rewrites that Et a π π b t + γm b t = (1 − ϕ)(m bt −a bt ) + βϕγaπ ρa a bt + βϕγm bt γaπ a m ρm m

or π bt + (γm bt = 0 (γaπ (1 − βϕρa ) + (1 − ϕ))a (1 − βϕρm ) − (1 − ϕ))m π bt and m b t , such that it has to be the case that γaπ and γm This equation has to hold for any a

satisfy γaπ (1 − βϕρa ) + (1 − ϕ) = 0 π γm (1 − βϕρm ) − (1 − ϕ) = 0

which implies

1−ϕ 1−ϕ π and γm = 1 − βϕρa 1 − βϕρm The solution for inflation is then given by γaπ = −

bt = − π

1−ϕ 1−ϕ bt + bt a m 1 − βϕρa 1 − βϕρm

Now that the solution for inflation is available, all the real quantities of the model can simply be obtained from the system (7)–(11). Plugging this in (7) and (8), we get ybt = cbt =

1−ϕ ϕ(1 − βρm ) bt + bt a m 1 − βϕρa 1 − βϕρm

Using (9), hours are given by bt = h

ϕ(1 − βρa ) ϕ(1 − βρm ) 1 bt ) = − bt + bt (ybt − a a m α α(1 − βϕρa ) α(1 − βϕρm )

Using (10), the real wage takes the form bt = w

α(1 − ϕ) − ϕν(1 − βρa ) ϕ(ν + α)(1 − βρm ) bt + bt a m α(1 − βϕρa ) α(1 − βϕρm )

Finally, using (11), the real marginal cost is sbt =

1+ν α



ϕ(1 − βρm ) ϕ(1 − βρa ) bt − bt m a 1 − βϕρm 1 − βϕρa



One, nominal, variable remains to be computed–the nominal interest rate, which is given by the second forward looking equation: b t = Et ybt+1 + Et π bt+1 − ybt R

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bt evaluated in t + 1 and compute It would be tempting to just plug the solution for ybt and π

expectations. But there is a much simpler approach in this case: Use Equation (6), such that this rewrites b t = Et m b t+1 − ybt R

Making use of the AR(1) process, this simplifies to b t = ρm m b t − ybt R

Now, making use of the solution for ybt , we get  b t = ρm − R

ϕ(1 − βρm ) 1−ϕ bt − bt m a 1 − βϕρm 1 − βϕρa 

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