AD model

Min Kim (min.kim@rutgers.edu)

$\begin{aligned} y_{t} & = ϕ E_{t} y_{t + 1} + ϕ u_{t} \\ u_{t} & = ρ u_{t - 1} + e_{t} \end{aligned}$

which can be rewritten as

$[\begin{matrix} u_{t + 1} \\ E y_{t + 1} \end{matrix}] = F [\begin{matrix} u_{t} \\ y_{t} \end{matrix}] + G [\begin{matrix} e_{t + 1} \\ 0 \end{matrix}]$

where $F = [\begin{matrix} ρ & 0 \\ - 1 & 1 / ϕ \end{matrix}]$ and $G = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]$ .

Recall that we have one forward-looking variable, that is, $y$ . Hence, in order to satisfy BK condition, we need one eigen value outside of unit circle in matrix $F$ .

12.8 μs

139 s

 
begin
    ρ = 0.5
    θ = 1.0
    b = 0.5
    ϕ = θ/(θ+b)
end;

200 ns

x
 
F = [ρ 0;-1 1/ϕ];

11.3 μs

BK is satisfied?

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md""" BK is satisfied?"""

8.8 μs

Yes!

x
 
begin
    BK = 1 == sum(eigvals(F) .> 1)
    if BK
        md""" Yes!"""
    else
        md""" No!"""
    end
end

45.1 μs

Let's check the decomposition as well

x
 
md"""
Let's check the decomposition as well
"""

8.1 μs

2×2 Matrix{Float64}:
 0.5  0.0
 0.0  1.5

x
 
Λ = diagm(eigvals(F))

12.3 μs

2×2 Matrix{Float64}:
 0.707107  0.0
 0.707107  1.0

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D = eigvecs(F)

49.6 μs

true

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F ≈ (D)*Λ*inv(D)

8.8 μs