APPENDIX: A Critique of Structural VARs Using Real Business Cycle Theory Ellen McGrattan Revised May 2005

Table of Contents

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2. The Economic Model . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1. Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . 3 2.2. Maximization problems . . . . . . . . . . . . . . . . . . . . 3 2.3. First-order conditions . . . . . . . . . . . . . . . . . . . . . 4 2.4. Log-linear computation . . . . . . . . . . . . . . . . . . . . 6

3. VARs and the 2-Shock Version of the Model . . . . . . . . . . . . . . 8 3.1. The Decision Functions . . . . . . . . . . . . . . . . . . . . 8 3.2. The Model’s Moving Average . . . . . . . . . . . . . . . . . . 11 3.3. Special Property of the D’s . . . . . . . . . . . . . . . . . . 12 3.4. VAR Coefficients . . . . . . . . . . . . . . . . . . . . . . . 13 3.5. Proposition 1: Model has infinite-order VAR . . . . . . . . . . . 13 3.6. Blanchard-Quah Identification . . . . . . . . . . . . . . . . . 15

3.6.1. Sign convention on A_{0}(1, 1) . . . . . . . . . . . . . . . . 17 3.6.2. Sign convention on Ā(1, 1) . . . . . . . . . . . . . . . . 17 3.6.3. Full solution . . . . . . . . . . . . . . . . . . . . . . 18 3.6.4. Cholesky decomposition . . . . . . . . . . . . . . . . . 18

3.7. Proposition 2: OLS Results . . . . . . . . . . . . . . . . . . 18 3.8. The Propositions for Two Special Cases . . . . . . . . . . . . . 21 3.8.1. Proposition 3a: No capital in the model . . . . . . . . . . 21 3.8.2. Proposition 3b: Only one shock . . . . . . . . . . . . . . 22

4. VARs and 3-Shock Versions of the Model . . . . . . . . . . . . . . . 25

4.1. Investment Tax Shock and Investment-Output in VAR . . . . . . 26 4.1.1. The Model’s Moving Average . . . . . . . . . . . . . . . 30 4.1.2. Special Property of the D’s . . . . . . . . . . . . . . . . 31 4.1.3. Proposition 4: Model has infinite-order VAR . . . . . . . . 32 4.1.4. A Way to Make M Singular . . . . . . . . . . . . . . . 34

4.2. Government Spending Shock and Investment-Output in VAR …. 35

4.2.1. The Model’s Moving Average . . . . . . . . . . . . . . . 35 4.3. Investment Tax Shock and Consumption-Output in VAR . . . . . 36

4.3.1. The Model’s Moving Average . . . . . . . . . . . . . . . 37 4.4. Adding Variables with Little Effect on Fluctuations . . . . . . . . 38

5. VARs and 4-Shock Versions of the Model . . . . . . . . . . . . . . . 39

1. Introduction

This appendix contains the proofs of Propositions 1-4 in the main text. I start by introducing the economic model and some preliminary algebra used in the proofs.

2. The Economic Model

2.1. Nomenclature Below I will use the following notation for the model variables:

N: population (N_{t} = (1 + g_{n})^{t}) c: per-capita consumption x: per-capita investment k: per-capita net capital stock (beginning of period t stock has subscript t) l: per-capita labor input tr: per-capita government transfers G: total government spending K: total stock of capital (K_{t} = N_{t}k_{t}) L: total labor input in production (L_{t} = N_{t}l_{t}) Z: labor-augmenting technical change (Z_{t} = Z_{t−1}z_{t}) z: the innovation to technology r: rental rate on capital w: wage rate τ_{v}: tax rate on v v̂: detrended, per-capita variable V (v̂_{t} = V_{t}/[N_{t}Z_{t}]) with the exception of k k̂: detrended, per-capita capital, k̂_{t} = K_{t}/[N_{t}Z_{t−1}]

2.2. Maximization problems

Consider an economy with households, firms, and the government. The representative household chooses consumption, investment, and labor to solve the following maximization problem:

max_{{c_{t},x_{t},l_{t}}}E ∑_{t=0}^{∞} β^{t} U(c_{t}, 1 − l_{t})N_{t}
subject to (1 + τ_{ct})c_{t} + (1 + τ_{xt})x_{t} = (1 − τ_{kt})r_{t}k_{t} + (1 − τ_{lt})w_{t}l_{t} + τ_{kt}δk_{t} + tr_{t}
N_{t+1}k_{t+1} = [(1 − δ)k_{t} + x_{t}]N_{t}
c_{t}, x_{t} ≥ 0 in all states

taking processes for the rental rate, wage rate, the tax rates, and transfers as given. The representative firm solves a simple static problem at t:

max_{{K_{t},L_{t}}} F (K_{t}, Z_{t}L_{t}) − r_{t}K_{t} − w_{t}L_{t}.

The government sets rates of taxes and transfers in such a way that their budget constraint at t, namely,

G_{t} + N_{t}tr_{t} = τ_{kt}(r_{t} − δ)N_{t}k_{t} + τ_{lt}w_{t}l_{t}N_{t} + τ_{ct}N_{t}c_{t} + τ_{xt}N_{t}x_{t}

is satisfied. In equilibrium, the following conditions must hold:

N_{t}(c_{t} + x_{t}) + G_{t} = F (K_{t}, Z_{t}L_{t})  (2.1)
N_{t}k_{t} = K_{t}
N_{t}l_{t} = L_{t}.

2.3. First-order conditions

I now derive first-order conditions in this economy. The Lagrangian for the household optimization problem is given by

L = E ∑_{t} β^{t}N_{t} {U(c_{t}, 1 − l_{t})
+ μ_{t}{(1 − τ_{kt})r_{t}k_{t} + (1 − τ_{lt})w_{t}l_{t} + τ_{kt}δk_{t} + tr_{t} − (1 + τ_{ct})c_{t} − (1 + τ_{xt})x_{t}}
+ λ_{t}{(1 − δ)k_{t} + x_{t} − (1 + g_{n})k_{t+1}}}

Here, I am assuming that the investment decision will be interior.

The relevant first-order conditions are found by taking derivatives of L with respect to c_{t}, l_{t}, x_{t}, and k_{t+1}:

0 = U_{1}(c_{t}, 1 − l_{t}) − μ_{t}(1 + τ_{ct})
0 = −U_{2}(c_{t}, 1 − l_{t}) + μ_{t}(1 − τ_{lt})w_{t}
0 = μ_{t}(1 + τ_{xt}) + λ_{t} = 0
0 = −(1 + g_{n})λ_{t} + E_{t}{μ_{t+1}[(1 − τ_{kt+1})r_{t+1} + δτ_{kt+1}] + λ_{t+1}(1 − δ)}

Eliminating multipliers yields:

\frac{U_{2}(c_{t}, 1 − l_{t})}{U_{1}(c_{t}, 1 − l_{t})} = \frac{1 − τ_{lt}}{1 + τ_{ct}} w_{t}  (2.2)

\frac{1 + τ_{xt}}{1 + τ_{ct}} U_{1}(c_{t}, 1 − l_{t}) = βE_{t}[\frac{U_{1}(c_{t+1}, 1 − l_{t+1})}{1 + τ_{ct+1}}\{(1 − τ_{kt+1})r_{t+1} + δτ_{kt+1}

+ (1 − δ)(1 + τ_{xt+1})}]. (2.3)

In addition, there are first-order conditions for the firm’s static problem. These are

r_{t} = F_{1}(K_{t}, Z_{t}L_{t})  (2.4)
w_{t} = F_{2}(K_{t}, Z_{t}L_{t})Z_{t}.  (2.5)

Finally, I have a resource constraint given by (2.1).

From here on, I make the following functional form assumptions and auxiliary choices:

F (k, l) = k^{θ}l^{1−θ}  (2.6)
U(c, 1 − l) = (c(1 − l)^{ψ})^{1−σ}/(1 − σ)  (2.7)
τ_{kt} = τ_{ct} = 0
s_{t} = [log z_{t}, τ_{lt}, τ_{xt}, log ĝ_{t}]′
s_{t+1} = P_{0} + P s_{t} + Qη_{s,t+1}, η_{s} ∼ N(0_{4×1}, I_{4×4}).  (2.8)

I have turned off τ_{c} since it plays a similar role to τ_{n} in distorting the labor-leisure choice. Similarly, I have turned off τ_{k} since it plays a similar role to τ_{x} in distorting the intertemporal margin.

If I substitute the choices (2.6)-(2.7) into (2.1) and (2.2)-(2.5), then substitute the equilibrium rates r_{t} and w_{t} into (2.2) and (2.3), I have:

N_{t}(c_{t} + g_{t}) + N_{t+1}k_{t+1} − (1 − δ)N_{t}k_{t} = (N_{t}k_{t})^{θ}(Z_{t}N_{t}l_{t})^{1−θ}  (2.9)

\frac{ψc_{t}}{1 − l_{t}} = (1 − τ_{lt})(1 − θ)(N_{t}k_{t})^{θ}Z_{t}^{1−θ}(N_{t}l_{t})^{−θ}  (2.10)
(1 + τ_{xt})c_{t}^{−σ}(1 − l_{t})^{ψ(1−σ)}
= βE_{t}[c_{t+1}^{−σ}(1 − l_{t+1})^{ψ(1−σ)}{θ(N_{t+1}k_{t+1})^{θ−1}(Z_{t+1}N_{t+1}l_{t+1})^{1−θ}
+ (1 − δ)(1 + τ_{xt+1})}].  (2.11)

2.4. Log-linear computation

I first normalize the variables as follows:

ĉ_{t} = c_{t}/Z_{t}, x̂_{t} = x_{t}/Z_{t}, ĝ_{t} = g_{t}/Z_{t}, ŷ_{t} = y_{t}/Z_{t}, k̂_{t} = k_{t}/Z_{t−1}.

Using the functional forms for F and U in (2.6) and (2.7), respectively, the equilibrium rental and wage rates are:

\begin{aligned}r_{t} & = θK_{t}^{θ−1}(Z_{t}L_{t})^{1−θ} = θk̂_{t}^{θ−1}(z_{t}l_{t})^{1−θ} \\ w_{t} & = (1 − θ)K_{t}^{θ}(Z_{t}L_{t})^{−θ}Z_{t} = (1 − θ)k̂_{t}^{θ}(z_{t}l_{t})^{−θ}Z_{t}. \end{aligned}

This implies the following first-order conditions

ĉ_{t} + ĝ_{t} + (1 + g_{n})k̂_{t+1} − (1 − δ)z_{t}^{−1}k̂_{t} = ŷ_{t} = k̂_{t}^{θ}l_{t}^{1−θ}z_{t}^{−θ}  (2.12)
\frac{ψĉ_{t}}{1 − l_{t}} = (1 − τ_{lt})(1 − θ)k̂_{t}^{θ}(z_{t}l_{t})^{−θ}  (2.13)

(1 + τ_{xt})ĉ_{t}^{−σ}(1 − l_{t})^{ψ(1−σ)}

= βz_{t+1}^{−σ}E_{t}ĉ_{t+1}^{−σ}(1 − l_{t+1})^{ψ(1−σ)}[θk̂_{t+1}^{θ−1}(z_{t+1}l_{t+1})^{1−θ} + (1 − δ)(1 + τ_{xt+1})]. (2.14)

Next, I compute the steady state of the system for constant values for z, the taxes, and government spending:

k̂/l = (\frac{(1 + τ_{x})(1 − βz^{−σ}(1 − δ))}{βz^{−σ}θz^{1−θ}})^{1/(θ−1)}

ĉ = [(k̂/l)^{θ−1}z^{−θ} − (1 + g_{n}) + (1 − δ)z^{−1}] k̂ − ĝ = ξ_{1}k̂ − ĝ

ĉ = [(1 − τ_{l})(1 − θ)(k̂/l)^{θ}z^{−θ}/ψ] (1 − 1/(k̂/l) k̂) = ξ_{2} − ξ_{3}k̂

where the last 2 equations imply k̂ = (ξ_{2} + ĝ)/(ξ_{1} + ξ_{3}), ĉ = ξ_{1}k̂ − ĝ, l = (1/(k̂/l))k̂.

Assume that the solution for the capital decision takes the form:

log k̂_{t+1} = γ_{k} log k̂_{t} + γ [ log z_{t} τ_{lt} τ_{xt} log ĝ_{t}]′ + constant,  (2.15)

where γ_{k} is a scalar and γ is 1 × 4 and equal to [γ_{z}, γ_{l}, γ_{x}, γ_{g}]. Assume the residual from the dynamic first-order condition (2.14) can be written (after substitutions from (2.12) and (2.13)):

f(E_{t} log k̂_{t+2}, log k̂_{t+1}, log k̂_{t}, log z_{t+1}, log z_{t}, τ_{lt+1}, τ_{lt}, τ_{xt+1}, τ_{xt}, log ĝ_{t+1}, log ĝ_{t})
≈ a_{0}E_{t} log k̂_{t+2} + a_{1} log k̂_{t+1} + a_{2} log k̂_{t} + b_{0}E_{t}s_{t+1} + b_{1}s_{t}.

Then the general solution algorithm is to find γ_{k} that solves the quadratic equation

a_{0}γ_{k}^{2} + a_{1}γ_{k} + a_{2} = 0,

and γ that solves the linear equations:

a_{0}γ_{k}γ + a_{0}γP + a_{1}γ + b_{0}P + b_{1} = 0_{1×4}.

Note that this implies:

γ = −[(a_{0}a + a_{1})I_{4×4} + a_{0}P′]^{−1}(b_{0}P + b_{1}I_{4×4})′.  (2.16)

Once I have values for the the coefficients γ_{k} and γ, I can use (2.12) and (2.13) to back out c_{t} and l_{t} (either nonlinearly or by way of a log-linear approximation).

One property of the solution that I use later is the fact that γ_{k} = −γ_{z}. A second look at This is true because k̂_{t} is everywhere divided by z_{t} in the first-order conditions (2.12)-(2.13). Thus, when the first-order conditions are log-linearized, the same coefficients hit log(k̂_{t}) and − log(z_{t}).

Given values for the coefficients in (2.15), I can derive expressions for labor, consumption, and investment using the static first-order conditions. In particular, I log-linearize (2.13) after substituting in for consumption from (2.12):

0 ≈ ψ{k̂^{θ}l^{1−θ}z^{−θ}[θ(log k̂_{t} − log z_{t}) + (1 − θ) log l_{t}]
− (1 + g_{n})k̂ log k̂_{t+1} + (1 − δ)z^{−1}k̂(log k̂_{t} − log z_{t}) − ĝ log ĝ_{t}}
+ (1 − θ)(1 − τ_{l})k̂^{θ}(zl)^{−θ}(1 − l){1/(1 − τ_{l}) τ_{lt}
− θ log k̂_{t} + θ(log l_{t} + log z_{t}) + l/(1 − l) log l_{t}}.

which I write succinctly as

log l_{t} = ϕ_{lk} log k̂_{t} + ϕ_{lz} log z_{t} + ϕ_{ll}τ_{lt} + ϕ_{lg} log ĝ_{t} + ϕ_{lk′} log k̂_{t+1}.

With this equation for log l, I use the production relation and the capital accumulation equation to write log ŷ and log x̂ as follows:

log ŷ_{t} = (θ + (1 − θ)ϕ_{lk}) log k̂_{t} + ((1 − θ)ϕ_{lz} − θ) log z_{t}
+ (1 − θ)[ϕ_{ll}τ_{lt} + ϕ_{lg} log ĝ_{t} + ϕ_{lk′} log k̂_{t+1}]
≡ ϕ_{yk} log k̂_{t} + ϕ_{yz} log z_{t} + ϕ_{yl}τ_{lt} + ϕ_{yg} log ĝ_{t} + ϕ_{yk′} log k̂_{t+1}  (2.17)
log x̂_{t} ≈ (1 + g_{n})k̂/x̂ log k̂_{t+1} − (1 − δ)z^{−1}k̂/x̂(log k̂_{t} − log z_{t})
≡ ϕ_{xk} log k̂_{t} + ϕ_{xz} log z_{t} + ϕ_{xk′} log k̂_{t+1}.  (2.18)

Finally, I can log-linearize (2.12) to get

log ĉ_{t} ≈ {ŷ[θ(log k̂_{t} − log z_{t}) + (1 − θ) log l_{t}] − ĝ log ĝ_{t}
− (1 + g_{n})k̂ log k̂_{t+1} + (1 − δ)z^{−1}k̂[log k̂_{t} − log z_{t}]}/ĉ
= [θŷ/ĉ + (1 − θ)ϕ_{lk}ŷ/ĉ + (1 − δ)k̂/(ĉz)] log k̂_{t}
− [θŷ/ĉ − (1 − θ)ϕ_{lz}ŷ/ĉ + (1 − δ)k̂/(ĉz)] log z_{t}
+ [(1 − θ)ϕ_{ll}ŷ/ĉ]τ_{lt}
+ [(1 − θ)ϕ_{lg}ŷ/ĉ − ĝ/ĉ] log ĝ_{t}
+ [(1 − θ)ϕ_{lk′} ŷ/ĉ − (1 + g_{n})k̂/ĉ] log k̂_{t+1}
≡ ϕ_{ck} log k̂_{t} + ϕ_{cz} log z_{t} + ϕ_{cl}τ_{lt} + ϕ_{cg} log ĝ_{t} + ϕ_{ck′} log k̂_{t+1}.  (2.19)

3. VARs and the 2-Shock Version of the Model

3.1. The Decision Functions

Assume the economy has only two shocks and they are orthogonal: a unit root in technology log z and an AR(1) in the tax rate on labor τ. (For convenience I drop l on τ_{lt} throughout this section.) The capital decision function has the form:

log k̂_{t+1} = γ_{0} + γ_{k} log k̂_{t} + γ_{z} log z_{t} + γ_{l}τ_{t}

and the labor decision function can be written:

log l_{t} = ϕ_{lz} log z_{t} + ϕ_{ll}τ_{t} + ϕ_{lk} log k̂_{t} + ϕ_{lk′} log k̂_{t+1}
= ϕ_{lz} log z_{t} + ϕ_{ll}τ_{t} + ϕ_{lk} log k̂_{t} + ϕ_{lk′}[γ_{0}γ_{k} log k̂_{t} + γ_{z} log z_{t} + γ_{l}τ_{t}]
= (ϕ_{lk} + ϕ_{lk′}γ_{k}) log k̂_{t} + (ϕ_{lz} + ϕ_{lk′}γ_{z}) log z_{t} + (ϕ_{ll} + ϕ_{lk′}γ_{l})τ_{t}.

These imply that output from a Cobb-Douglas production technology with capital share θ is:

log ŷ_{t} = θ(log k̂_{t} − log z_{t}) + (1 − θ) log l_{t}
= (θ + (1 − θ)ϕ_{lk}) log k̂_{t} − (θ − (1 − θ)ϕ_{lz}) log z_{t} + (1 − θ)ϕ_{ll}τ_{t}
+ (1 − θ)ϕ_{lk′} log k̂_{t+1}
= (θ + (1 − θ)(ϕ_{lk} + ϕ_{lk′}γ_{k})) log k̂_{t} − (θ − (1 − θ)(ϕ_{lz} + ϕ_{lk′}γ_{z})) log z_{t}
+ (1 − θ)(ϕ_{ll} + ϕ_{lk′}γ_{l})τ_{t}

I can write the capital stock in terms of all lagged shocks as follows:

log k̂_{t} = γ_{0} + γ_{k}(γ_{0} + γ_{k} log k̂_{t−2} + γ_{z} log z_{t−2} + γ_{l}τ_{t−2}) + γ_{z} log z_{t−1} + γ_{l}τ_{t−1}
= γ_{0}[1 + γ_{k} + γ_{k}^{2} + …]
+ γ_{z}[log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …]
+ γ_{l}[τ_{t−1} + γ_{k}τ_{t−2} + γ_{k}^{2}τ_{t−3} + …]

or in differences as follows:

log k̂_{t} − log k̂_{t−1} = γ_{z}[log z_{t−1} + (γ_{k} − 1){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}]
+ γ_{l}[τ_{t−1} + (γ_{k} − 1){τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2}τ_{t−4} + …}]

or in quasi-differences as follows:

log k̂_{t} − α log k̂_{t−1} = γ_{z}[log z_{t−1} + (γ_{k} − α){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}]
+ γ_{l}[τ_{t−1} + (γ_{k} − α){τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2}τ_{t−4} + …}]

I can also write hours in terms of past shocks as follows:

log l_{t} = ϕ_{lz} log z_{t} + ϕ_{ll}τ_{t} + ϕ_{lk} log k̂_{t} + ϕ_{lk′} log k̂_{t+1}

= ϕ_{lz} log z_{t} + ϕ_{ll}τ_{t}

+ ϕ_{lk}γ_{z}[log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …]
+ ϕ_{lk}γ_{l}[τ_{t−1} + γ_{k}τ_{t−2} + γ_{k}^{2}τ_{t−3} + …]
+ ϕ_{lk′}γ_{z}[log z_{t} + γ_{k} log z_{t−1} + γ_{k}^{2} log z_{t−2} + …]
+ ϕ_{lk′}γ_{l}[τ_{t} + γ_{k}τ_{t−1} + γ_{k}^{2}τ_{t−2} + …]

= [(ϕ_{lz} + ϕ_{lk′}γ_{z}) log z_{t} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{z} log z_{t−1} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{k}γ_{z} log z_{t−2} + …]

+ [(ϕ_{ll} + ϕ_{lk′}γ_{l})τ_{t} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{l}τ_{t−1} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{k}γ_{l}τ_{t−2} + …]

where I have ignored constant terms.

I can write logged hours in differences as follows:

log l_{t} − log l_{t−1} = ϕ_{lz}(log z_{t} − log z_{t−1}) + ϕ_{ll}(τ_{t} − τ_{t−1})

+ ϕ_{lk′}(log k̂_{t+1} − log k̂_{t}) + ϕ_{lk}(log k̂_{t} − log k̂_{t−1}) = ϕ_{lz}(log z_{t} − log z_{t−1}) + ϕ_{ll}(τ_{t} − τ_{t−1})

+ ϕ_{lk′}γ_{z}[log z_{t} + (γ_{k} − 1){log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …}] + ϕ_{lk′}γ_{l}[τ_{t} + (γ_{k} − 1){τ_{t−1} + γ_{k}τ_{t−2} + γ_{k}^{2}τ_{t−3} + …}] + ϕ_{lk}γ_{z}[log z_{t−1} + (γ_{k} − 1){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}] + ϕ_{lk}γ_{l}[τ_{t−1} + (γ_{k} − 1){τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2}τ_{t−4} + …}]

= [ϕ_{lz} + ϕ_{lk′}γ_{z}] log z_{t} − [ϕ_{lz} − ϕ_{lk}γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − 1)] log z_{t−1} + γ_{z}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …} + [ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{t} − [ϕ_{ll} − ϕ_{lk}γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − 1)]τ_{t−1} + γ_{l}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2}τ_{t−4} + …}

or in quasi-difference form as follows:

log l_{t} − α log l_{t−1} = ϕ_{lz}(log z_{t} − α log z_{t−1}) + ϕ_{ll}(τ_{t} − ατ_{t−1})

+ ϕ_{lk′}(log k̂_{t+1} − α log k̂_{t}) + ϕ_{lk}(log k̂_{t} − α log k̂_{t−1}) = ϕ_{lz}(log z_{t} − α log z_{t−1}) + ϕ_{ll}(τ_{t} − ατ_{t−1})

+ ϕ_{lk′}γ_{z}[log z_{t} + (γ_{k} − α){log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …}] + ϕ_{lk′}γ_{l}[τ_{t} + (γ_{k} − α){τ_{t−1} + γ_{k}τ_{t−2} + γ_{k}^{2}τ_{t−3} + …}] + ϕ_{lk}γ_{z}[log z_{t−1} + (γ_{k} − α){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}] + ϕ_{lk}γ_{l}[τ_{t−1} + (γ_{k} − α){τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2}τ_{t−4} + …}]

= [ϕ_{lz} + ϕ_{lk′}γ_{z}] log z_{t} − [αϕ_{lz} − ϕ_{lk}γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − α)] log z_{t−1} + γ_{z}(γ_{k} − α)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …} + [ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{t} − [αϕ_{ll} − ϕ_{lk}γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − α)]τ_{t−1} + γ_{l}(γ_{k} − α)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2}τ_{t−4} + …}

I can use the expressions for output and hours to write out the change in productivity as follows:

log(y_{t}/l_{t}) − log(y_{t−1}/l_{t−1})

= log ŷ_{t} − log ŷ_{t−1} + log z_{t} − log l_{t} − log l_{t−1} = log z_{t} + θ(log k̂_{t} − log k̂_{t−1} − log l_{t} + log l_{t−1} − log z_{t} + log z_{t−1}) = (1 − θ) log z_{t} + θ log z_{t−1}

− θ(log l_{t} − log l_{t−1} − log k̂_{t} + log k̂_{t−1}) = (1 − θ) log z_{t} + θ log z_{t−1} − θ{[ϕ_{lz} + ϕ_{lk′}γ_{z}] log z_{t}

− [ϕ_{lz} − (ϕ_{lk} − 1)γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − 1)] log z_{t−1} + γ_{z}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …] + [ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{t} − [ϕ_{ll} − (ϕ_{lk} − 1)γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − 1)]τ_{t−1} + γ_{l}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2} log z_{t−4} + …]}

= {1 − θ − θ[ϕ_{lz} + ϕ_{lk′}γ_{z}]} log z_{t} + θ[1 + ϕ_{lz} − (ϕ_{lk} − 1)γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − 1)] log z_{t−1} − θγ_{z}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …] − θ[ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{t} + θ[ϕ_{ll} − (ϕ_{lk} − 1)γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − 1)]τ_{t−1} − θγ_{l}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][τ_{t−2} + γ_{k}τ_{t−3} + γ_{k}^{2}τ_{t−4} + …]

3.2. The Model’s Moving Average The moving average for the model is given by:

[ \begin{matrix}(1 − L) log y_{t}/l_{t} \\ (1 − αL) log l_{t} \end{matrix} ] ≡ X_{t} = D_{0}ω_{t} + D_{1}ω_{t−1} + D_{2}ω_{t−2} + …

where ω_{t} = [log z_{t}, τ_{t}]′ and

\begin{aligned}D_{0} & = [ \begin{matrix}1 − θ − θ(ϕ_{lz} + ϕ_{lk′}γ_{z}) & −θ(ϕ_{ll} + ϕ_{lk′}γ_{l}) \\ ϕ_{lz} + ϕ_{lk′}γ_{z} & ϕ_{ll} + ϕ_{lk′}γ_{l} \end{matrix} ] \\ D_{1} & = [ \begin{matrix}θ(1 + ϕ_{lz} − (ϕ_{lk} − 1)γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − 1)) & θ(ϕ_{ll} − (ϕ_{lk} − 1)γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − 1)) \\ −αϕ_{lz} + (ϕ_{lk} + ϕ_{lk′}(γ_{k} − α))γ_{z} & −αϕ_{ll} + (ϕ_{lk} + ϕ_{lk′}(γ_{k} − α))γ_{l} \end{matrix} ] \\ D_{2} & = [ \begin{matrix}−θγ_{z}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1] & −θγ_{l}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1] \\ (ϕ_{lk} + ϕ_{lk′}γ_{k})(γ_{k} − α)γ_{z} & (ϕ_{lk} + ϕ_{lk′}γ_{k})(γ_{k} − α)γ_{l} \end{matrix} ] \end{aligned}

and D_{j} = γ_{k}D_{j−1} for j ≥ 3.

Let a = ϕ_{lk} + ϕ_{lk′}γ_{k} and b = ϕ_{ll} + ϕ_{lk′}γ_{l}. Also, note that ϕ_{lz} = −ϕ_{lk} and γ_{z} = −γ_{k} hold in the model economy with a unit root in technology.

\begin{aligned}D_{0} & = [ \begin{matrix}1 − θ + θa & −θb \\ −a & b \end{matrix} ] \\ D_{1} & = [ \begin{matrix}θ(1 − γ_{k})(1 − a) & θ(b + (1 − a)γ_{l}) \\ (α − γ_{k})a & −αb + γ_{l}a \end{matrix} ] \\ D_{2} & = [ \begin{matrix}γ_{k}(1 − a)θ(1 − γ_{k}) & −γ_{l}(1 − a)θ(1 − γ_{k}) \\ γ_{k}a(α − γ_{k}) & −γ_{l}a(α − γ_{k}) \end{matrix} ] \end{aligned}

and, again, D_{j} = γ_{k}^{j−2}D_{2} for j ≥ 3. Note that D_{2} is singular.

If τ_{t} is an AR(1), it is more convenient to write the MA process in terms of η_{t} = [log z_{t}, η_{lt}] rather than in terms of ω_{t}. In this case,

X_{t} = D_{0}η_{t}+(D_{0}P +D_{1})η_{t−1}+(D_{0}P^{2}+D_{1}P +D_{2})η_{t−2}+(D_{0}P^{3}+D_{1}P^{2}+D_{2}P +D_{3})η_{t−3}….

I normalize the MA so it has an identity for the first coefficient. That is, I set C_{0} = I, C_{1} = (D_{0}P + D_{1})D_{0}^{−1}, and C_{j} = C_{j−1}D_{0}P D_{0}^{−1} + D_{j}D_{0}^{−1}.

3.3. Special Property of the D’s

Next, I will show that the D matrices have a special property that will be exploited when I characterize coefficients of the VAR found by regressing X_{t} on lags of itself. The D’s for the RBC model satisfy the relation:

(γ_{k}I − (D_{0}P^{2} + D_{1}P + D_{2})(D_{0}P + D_{1})^{−1}) D_{2} = 0.  (3.1)

One method of proof is to multiply all terms of the matrices in (3.1) and show that all elements are zero. I have done this but the algebra is messy.

A simpler proof is as follows. Note that

D_{2} = [ \begin{matrix}(1 − a)θ(1 − γ_{k}) \\ (α − γ_{k})a \end{matrix} ] [ γ_{k} −γ_{l}] ≡ gh′.  (3.2)

Thus, I can rewrite the left hand side of (3.1) as follows

(γ_{k}I − (D_{0}P^{2} + D_{1}P + D_{2})(D_{0}P + D_{1})^{−1})D_{2}
= [γ_{k}(gh′) − (gh′)(D_{0}P + D_{1})^{−1}(gh′)] − [(D_{0}P + D_{1})P (D_{0}P + D_{1})^{−1}gh′]. (3.3)

I will prove that both terms in (3.3) in square brackets is equal to 2×2 zero matrices. The first step of the proof is to show that

(D_{0}P + D_{1})^{−1}g = [ \begin{matrix}1 \\ 0 \end{matrix} ] .  (3.4)

The proof of this step is trivial since the first column of D_{0}P +D_{1} is equal to g. Substituting (3.4) into (3.3), the result (3.1) follows immediately from the fact that h′[1, 0]′ = γ_{k} and P [1, 0]′ = 0.

3.4. VAR Coefficients

Given expressions for the D coefficients in the model MA, and thus the normalized C coefficients, I can directly write out expressions for the coefficients in the VAR of X_{t} regressed on lags of itself. I will denote the VAR coefficients by B_{j}, j = 1, 2, . . .. They are related to the MA coefficients as follows:

B_{j} = C_{j} − B_{1}C_{j−1} − B_{2}C_{j−2} − … B_{j−1}C_{1}.  (3.5)

3.5. Proposition 1: Model has infinite-order VAR

Proposition 1. The model described above has a VAR representation with coefficients B_{j} that satisfy

B_{j} = MB_{j−1}  (3.6)

for j ≥ 2, with B_{1} = C_{1} = (D_{0}P + D_{1})D_{0}^{−1}. The matrix M is a 2×2 matrix with eigenvalues equal to α and

\frac{γ_{k} − γ_{l}a/b − θ}{1 − θ} ,

where a = ϕ_{lk} + ϕ_{lk′}γ_{k} and b = ϕ_{ll} + ϕ_{lk′}γ_{l} are the coefficients on k and τ_{l} in the labor decision function.

Proof of Proposition 1. Choose M = C_{2}C_{1}^{−1} − C_{1}. Using the formula (3.5) for the VAR coefficient, it is easy to show that M = B_{2}B_{1}^{−1}. Therefore, B_{2} = MB_{1} holds. Consider the next coefficient. Using the formula (3.5), I have

B_{3} − MB_{2} = C_{3} − B_{1}C_{2} − B_{2}C_{1} − M(C_{2} − B_{1}C_{1})
= C_{3} − B_{1}C_{2} − MC_{2}
= C_{3} − C_{1}C_{2} − (C_{2}C_{1}^{−1} − C_{1})C_{2}

= C_{3} − C_{2}C_{1}^{−1}C_{2}

= C_{2}D_{0}P D_{0}^{−1} + γ_{k}D_{2}D_{0}^{−1} − C_{2}C_{1}^{−1}(C_{1}D_{0}P D_{0}^{−1} + D_{2}D_{0}^{−1}) = γ_{k}D_{2}D_{0}^{−1} − C_{2}C_{1}^{−1}D_{2}D_{0}^{−1}) = (γ_{k}I − C_{2}C_{1}^{−1})D_{2}D_{0}^{−1} = (γ_{k}I − (D_{0}P^{2} + D_{1}P + D_{2})(D_{0}P + D_{1})^{−1})D_{2}D_{0}^{−1} = 0

where the last relation follows from intermediate calculations done in Section 3. The same calculation can be done for any j^{∗} using the fact that (3.6) holds for all j < j^{∗}, namely

B_{j} − MB_{j−1} = C_{j} − B_{1}C_{j−1} … − B_{j−1}C_{1} − M(C_{j−1} − … B_{j−2}C_{1})

= C_{j} − B_{1}C_{j−1} − MC_{j−1} = C_{j} − C_{1}C_{j−1} − (C_{2}C_{1}^{−1} − C_{1})C_{j−1} = C_{j} − C_{2}C_{1}^{−1}C_{j−1} = C_{j−1}D_{0}P D_{0}^{−1} + D_{j}D_{0}^{−1} − C_{2}C_{1}^{−1}(C_{j−2}D_{0}P D_{0}^{−1} + D_{j−1}D_{0}^{−1}) = C_{j−1}D_{0}P D_{0}^{−1} + γ_{k}^{j−2}D_{2}D_{0}^{−1} − C_{2}C_{1}^{−1}(C_{j−2}D_{0}P D_{0}^{−1} + γ_{k}^{j−3}D_{2}D_{0}^{−1}) = (C_{j−1} − C_{2}C_{1}^{−1}C_{j−2})D_{0}P D_{0}^{−1} + γ_{k}^{j−3}(γ_{k}I − C_{2}C_{1}^{−1})D_{2}D_{0}^{−1} = (B_{j−1} − MB_{j−2})D_{0}P D_{0}^{−1} + γ_{k}^{j−3}(γ_{k}I − C_{2}C_{1}^{−1})D_{2}D_{0}^{−1} = γ_{k}^{j−3}(γ_{k}I − C_{2}C_{1}^{−1})D_{2}D_{0}^{−1} = γ_{k}^{j−3}(γ_{k}I − (D_{0}P^{2} + D_{1}P + D_{2})(D_{0}P + D_{1})^{−1})D_{2}D_{0}^{−1} = 0.

Next, I prove that the two eigenvalues of M are λ_{1} = α and λ_{2} = (γ_{k} − γ_{l}a/b − θ)/(1 − θ). One way to do this is to write out all of the terms for matrix M and derive expressions for the trace and the determinant. The trace is equal to the sum of the eigenvalues and the determinant is equal to the product of the eigenvalues. This is 2 equations and 2 unknowns. I have done this but the algebra is messy.

A simpler proof that the eigenvalues are λ_{1} = α and λ_{2} = (γ_{k} − γ_{l}a/b − θ)/(1 − θ) is as follows. Using (3.2) and the definitions of the C’s in terms of the D’s, I can derive the following expression for M in terms of the D’s, P, and h:

M = C_{2}C_{1}^{−1} − C_{1}

= (D_{0}P^{2} + D_{1}P + D_{2})(D_{0}P + D_{1})^{−1} − (D_{0}P + D_{1})D_{0}^{−1} = (D_{0}P + D_{1})P (D_{0}P + D_{1})^{−1} + D_{2}(D_{0}P + D_{1})^{−1}

− (D_{0}P + D_{1})D_{0}^{−1}(D_{0}P + D_{1})(D_{0}P + D_{1})^{−1}

= D_{2}(D_{0}P + D_{1})^{−1} − (D_{0}P + D_{1})D_{0}^{−1}D_{1}(D_{0}P + D_{1})^{−1}
= (D_{0}P + D_{1})[1, 0]′h′(D_{0}P + D_{1})^{−1} − (D_{0}P + D_{1})D_{0}^{−1}D_{1}(D_{0}P + D_{1})^{−1}
= (D_{0}P + D_{1})([1, 0]′h′ − D_{0}^{−1}D_{1})(D_{0}P + D_{1})^{−1}.

Appealing to standard results in linear algebra, the eigenvalues of M are equal to the eigenvalues of the simpler matrix [1, 0]′h′ − D_{0}^{−1}D_{1}, which is equal to

[1, 0]′h′ − D_{0}^{−1}D_{1} = \frac{1}{1 − θ} [ \begin{matrix}γ_{k} − θ(1 − a + aα) & −γ_{l} + θb(1 − α) \\ (γ_{k} − α − θ(1 − a)(1 − α))a/b & α − θ(a + α(1 − a)) − γ_{l}a/b \end{matrix} ]

Taking the trace, I get

trace([1, 0]′h′ − D_{0}^{−1}D_{1}) = α + \frac{γ_{k} − γ_{l}a/b − θ}{1 − θ} .  (3.7)

Taking the determinant, I get

det([1, 0]′h′ − D_{0}^{−1}D_{1}) = α × \frac{γ_{k} − γ_{l}a/b − θ}{1 − θ} .  (3.8)

The two equations (3.7) and (3.8) uniquely determine the two eigenvalues which are those proposed.

3.6. Blanchard-Quah Identification

I now consider the procedure of Blanchard and Quah (1989) when applied to data from the 2-shock version of the model described above.

Blanchard and Quah start, as I did, with a VAR

X_{t} = B_{1}X_{t−1} + B_{2}X_{t−2} + … B_{p}X_{t−p} + v_{t}, Ev_{t}v_{t} = Ω

which is estimated using time series {X_{t}}. As described above, this implies the MA

X_{t} = v_{t} + C_{1}v_{t−1} + C_{2}v_{t−2} + ….  (3.9)

Some structure is needed to derive a “structural MA” with shocks that have economic interpretation. In this case, I will use the following notation for the structural MA:

X_{t} = A_{0}ϵ_{t} + A_{1}ϵ_{t−1} + A_{2}ϵ_{t−2} + ….  (3.10)

where A_{0}ϵ_{t} = v_{t} and A_{j} = C_{j}A_{0}.

Because I will impose restrictions on the sums of the A’s and C’s, I define

C̄ = I + C_{1} + C_{2} + C_{3} + …
Ā = A_{0} + A_{1} + A_{2} + A_{3} + …
= C̄A_{0}.

Since A_{0}ϵ_{t} = v_{t}, it must be the case that A_{0}Eϵ_{t}ϵ_{t}A′_{0} = Ω. Blanchard and Quah assume that the elements of ϵ_{t} are orthogonal and demand shocks do not have a long-run effect on productivity. Without loss of generality, we can normalize the magnitude of the variances of the elements of ϵ_{t} and assume, therefore, that

A_{0}A′_{0} = Ω
C̄(1, 1)A_{0}(1, 2) + C̄(1, 2)A_{0}(2, 2) = 0  (3.11)

which is four equations in the four unknown elements of A_{0}. Condition (3.11) ensures that the demand shock does not have a long-run effect on productivity. Writing out the system of 4 equations and 4 unknowns yields:

ω_{11} = A_{0}(1, 1)^{2} + A_{0}(1, 2)^{2}
ω_{12} = A_{0}(1, 1)A_{0}(2, 1) + A_{0}(1, 2)A_{0}(2, 2)  (3.12)
ω_{22} = A_{0}(2, 1)^{2} + A_{0}(2, 2)^{2}
0 = C̄(1, 1)A_{0}(1, 2) + C̄(1, 2)A_{0}(2, 2)
Eliminate A_{0}(1, 2) using the fact that A_{0}(1, 2) = −C̄(1, 2)A_{0}(2, 2)/C̄(1, 1):

\begin{aligned}ω_{11} & = A_{0}(1, 1)^{2} + f^{2}A_{0}(2, 2)^{2} \\ ω_{12} & = A_{0}(1, 1)A_{0}(2, 1) + fA_{0}(2, 2)^{2} \\ ω_{22} & = A_{0}(2, 1)^{2} + A_{0}(2, 2)^{2} \end{aligned}

where f = −C̄(1, 2)/C̄(1, 1). Solve for A_{0}(1, 1) and A_{0}(2, 1):

A_{0}(1, 1) = [ω_{11} − f^{2}A_{0}(2, 2)^{2}]^{1/2}  (3.13)
A_{0}(2, 1) = [ω_{22} − A_{0}(2, 2)^{2}]^{1/2}  (3.14)

and substitute to get:

ω_{12} = fA_{0}(2, 2)^{2} + [ω_{11} − f^{2}A_{0}(2, 2)^{2}]^{1/2}[ω_{22} − A_{0}(2, 2)^{2}]^{1/2}.

Let λ = A_{0}(2, 2)^{2} and the result is a quadratic in λ:

(ω_{12} − fλ)^{2} = (ω_{11} − f^{2}λ)(ω_{22} − λ)

which can be written out:

ω_{12}^{2} − 2fλω_{12} + f^{2}λ^{2} = ω_{11}ω_{22} − f^{2}λω_{22} − ω_{11}λ + f^{2}λ^{2}

and simplified as follows:

λ = \frac{ω_{11}ω_{22} − ω_{12}^{2}}{ω_{11} + f^{2}ω_{22} − 2fω_{12}} .  (3.15)

In addition, I need to impose sign conventions since impulse responses can be either positive or negative. I will consider one sign convention for the demand shock and two different sign conventions for the technology shock.

The demand shock in our example is a shock to the tax rate on labor. For this choice of shock, I want to impose A_{0}(2, 2) < 0 so that hours fall with a positive shock to the tax rate on labor. For A_{0}(2, 2), it must be the case that A_{0}(2, 2) = −\sqrt{λ} since λ is positive. Given A_{0}(2, 2), it immediately follows that A_{0}(1, 2) = fA_{0}(2, 2).

3.6.1. Sign convention on A_{0}(1, 1)

For the technology shock, I first consider the sign convention that productivity rises on impact in response to a positive technology shock, namely A_{0}(1, 1) > 0. In this case, I need to use the positive root of A_{0}(1, 1)^{2} = ω_{11} − f^{2}λ:

A_{0}(1, 1) = \sqrt{ω_{11} − f^{2}λ}.

Given a value for A_{0}(1, 1), I have A_{0}(2, 1) from:

A_{0}(2, 1) = (ω_{12} − fλ)/A_{0}(1, 1).

3.6.2. Sign convention on Ā(1, 1)

The second sign convention assumes that productivity is positive in the long run so that Ā(1, 1) > 0 and therefore

C̄(1, 1)A_{0}(1, 1) + C̄(1, 2)A_{0}(1, 2) > 0.

This condition can also be written in terms of A_{0}(1, 1) and known parameters:

C̄(1, 1)A_{0}(1, 1) + C̄(1, 2)(ω_{1,2} − fλ)/A_{0}(1, 1) > 0.  (3.16)

In this case, I choose the sign on the square root of ω_{11} − f^{2}λ so that (3.16) is satisfied.

3.6.3. Full solution

The full solution is

A_{0}(2, 2) = −\sqrt{λ}
A_{0}(1, 2) = fA_{0}(2, 2)
A_{0}(1, 1) = root of ω_{11} − f^{2}λ satisfying sign convention
A_{0}(2, 1) = (ω_{12} − fλ)/A_{0}(1, 1)

where λ is defined in (3.15) and f = −C̄(1, 2)/C̄(1, 1).

3.6.4. Cholesky decomposition

In the literature, many report using the following formula for A_{0}:

A_{0} = C̄^{−1}L

where L is a lower triangular matrix such that with positive elements on the diagonal that satisfies LL′ = C̄ΩC̄′. This choice imposes the long-run restriction in (3.11) and the long-run sign convention Ā(1, 1) automatically.

It does not impose A_{0}(2, 2) < 0. However, in most cases, responses to demand shocks are not discussed.

3.7. Proposition 2: OLS Results

Proposition 1 says that the model has an infinite-lag vector autoregressive structure. The next proposition considers the outcome when OLS regressions are run with one lag. Let V_{0} = EX_{t}X′_{t} be the theoretical variance matrix for X_{t}. Let V_{1} = EX_{t}X′_{t−1} be the covariance matrix for X_{t} and its lag. If ED_{0}η_{t}η′_{t}D′_{0} = Ω is the theoretical variancecovariance of the model’s shock vector, then

V_{0} = Ω + C_{1}ΩC′_{1} + C_{2}ΩC′_{2} + …  (3.17)
V_{1} = C_{1}Ω + C_{2}ΩC′_{1} + C_{3}ΩC′_{2} + ….  (3.18)

Proposition 2. Assume that a regression is run of the form

X_{t} = B_{ols}X_{t−1} + v_{t}, Ev_{t}v′_{t} = Ω_{ols}

with X_{t} from the RBC model. Then, the variance-covariance matrix is

Ω_{ols} = V_{0} − V_{1}V_{0}^{−1}V′_{1}  (3.19)
= Ω + MΩM′ − MΩV_{0}^{−1}ΩM′  (3.20)

where M = C_{2}C_{1}^{−1} − C_{1} and the inverse of the sum of MA coefficients is

C̄_{ols}^{−1} = I − B_{ols}
= C̄^{−1} + M(I − M)^{−1}C_{1} + M(Ω − V_{0})V_{0}^{−1}.

In other words, the OLS matrices Ω_{ols} and C̄_{ols} are not equal to their theoretical counterparts, Ω and C̄.

Proof of Proposition 2. The relation (3.19) follows from the standard projection formulas,

B_{ols} = (EX_{t}X_{t−1})(EX_{t−1}X′_{t−1})^{−1} = V_{1}V_{0}^{−1}
Ev_{t}v′_{t} = E(X_{t} − B_{ols}X_{t−1})(X_{t} − B_{ols}X_{t−1})′ = V_{0} − V_{1}V_{0}^{−1}V′_{1}.  (3.21)

Before substituting in (3.17) and (3.18), I can exploit the nature of the model’s MA. In particular, I can use the fact that

C_{j} = (C_{1} + M)C_{j−1},  (3.22)

which follows from the formula (3.5) and Proposition 1. That is,

C_{j} − (C_{1} + M)C_{j−1} = (B_{j} + B_{j−1}C_{1} + B_{j−2}C_{2} + … + B_{1}C_{j−1})
− C_{1}C_{j−1} − M(B_{j−1} + B_{j−2}C_{1} + … + B_{1}C_{j−2})
= B_{1}C_{j−1} − C_{1}C_{j−1}
= 0.

Thus, I can write V_{0} as follows:

V_{0} = Ω + C_{1}ΩC′_{1} + (C_{1} + M)C_{1}ΩC′_{1}(C_{1} + M)′ + (C_{1} + M)^{2}C_{1}ΩC′_{1}(C_{1} + M)^{2}′ + …

which implies that

V_{0} = (C_{1} + M)V_{0}(C_{1} + M)′ + Ω + C_{1}ΩC′_{1} − (C_{1} + M)Ω(C_{1} + M)′.  (3.23)

For V_{1},

V_{1} = C_{1}Ω + (C_{1} + M)C_{1}ΩC′_{1} + (C_{1} + M)^{2}Ω(C_{1} + M) + …
= (C_{1} + M)V_{0} − MΩ.  (3.24)

Substituting (3.23) and (3.24) into (3.21) yields

V_{0} − V_{1}V_{0}^{−1}V′_{1} = V_{0} − [(C_{1} + M)V_{0} − MΩ]V_{0}^{−1}[(C_{1} + M)V_{0} − MΩ]′
= V_{0} − (C_{1} + M)V_{0}(C_{1} + M)′ + MΩ(C_{1} + M)′
+ (C_{1} + M)ΩM′ − MΩV_{0}^{−1}ΩM′
= Ω + C_{1}ΩC′_{1} − (C_{1} + M)Ω(C_{1} + M)′ + MΩ(C_{1} + M)′
+ (C_{1} + M)ΩM′ − MΩV_{0}^{−1}ΩM′
= Ω + MΩM′ − MΩV_{0}^{−1}ΩM′

which is the same as (3.20). This proves the first part of the proposition.

For the second part, I need to construct the matrix C̄_{ols} using the relation between the AR coefficients and the MA coefficients in (3.5). In this case,

C̄_{ols} = (I − B_{ols})^{−1}.

Thus, I have

C̄_{ols} = (I − V_{1}V_{0}^{−1})^{−1}
= (I − [(C_{1} + M)V_{0} − MΩ]V_{0}^{−1})^{−1}
= (I − C_{1} − M + MΩV_{0}^{−1})^{−1}
= (I − (I − M)^{−1}C_{1} + (I − M)^{−1}C_{1} − C_{1} − M + MΩV_{0}^{−1})^{−1}
= (C̄^{−1} + (I + M + M^{2} + …)C_{1} − C_{1} − M + MΩV_{0}^{−1})^{−1}
= (C̄^{−1} + M(I − M)^{−1}C_{1} + M(Ω − V_{0})V_{0}^{−1})^{−1}.  (3.25)

The term M(I − M)^{−1}C_{1} + M(Ω − V_{0})V_{0}^{−1} is not generically zero. This proves the second part of the proposition.

The term MΩM′ − MΩV_{0}^{−1}ΩM′ is zero if the RBC model’s VAR representation has only one lag (e.g., M = 0). It is close to zero if one of the shocks is close to 0. In this latter case, ΩV_{0}^{−1}Ω ≈ Ω and the SVAR user detects correctly the variance of the one shock driving the system. This is true even if the VAR coefficients are wrong (e.g., M is very different than 0).

What happens if we have a VAR with n lags? In this case, the formula is messy but Ev_{t}v′_{t} can be written

Ev_{t}v′_{t} = V_{0} − [ V_{1} V_{2} ⋯ V_{n}] [ \begin{matrix}V_{0} & V_{1} & ⋯ & V_{n−1} \\ V′_{1} & V_{0} & ⋯ & V_{n−2} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ V′_{n−1} & V′_{n−2} & ⋯ & V_{0} \end{matrix} ]^{−1} [ V′_{1} ; V′_{2} ; ⋮ ; V′_{n} ]

with V_{j} = (C_{1} + M)^{j−1}V_{1}, where V_{0} is the matrix in (3.23) and V_{1} is the matrix in (3.24).

3.8. The Propositions for Two Special Cases

In this section, I consider two special cases. The first has θ = 0. The second has σ_{τ} = 0. I show in these very special cases that the SVAR can uncover the true impulse response for hours in response to a technology shock even if only one lag is used in the VAR regression.

3.8.1. Proposition 3a: No capital in the model

Proposition 3a. Assume that θ is set to 0 in the RBC model. If a regression is run of the form

X_{t} = B_{ols}X_{t−1} + v_{t}

with X_{t} from the RBC model, then the Blanchard-Quah procedure recovers the true impulse response function for hours in response to technology, namely

A_{j}(2, 1) = 0  (3.26)

for all j.

Proof of Proposition 3a. It is important to note that C_{1} is singular in this case. Thus, I can’t write M as C_{2}C_{1}^{−1} − C_{1}, but rather, I simply work with:

M = [ \begin{matrix}0 & 0 \\ M(2, 1) & α \end{matrix} ]

for arbitrary M(2, 1) (which is what I would have in the limit as θ goes to 0). It is easy to show that B_{2} = C_{2} − C_{1}^{2} = MB_{1}, B_{3} = C_{3} − B_{2}C_{2} − B_{1}C_{1} = MB_{2} and so on. To prove the result in (3.26), I need to show that the errors that a SVAR user encounters in estimating Ev_{t}v′_{t} and C̄ do not affect the (2,1) elements of the A_{j}’s. Starting with Ev_{t}v′_{t}:

Ev_{t}v′_{t} = Ω + MΩM′ − MΩV_{0}^{−1}M′

= [ \begin{matrix}σ_{z}^{2} & 0 \\ 0 & σ_{τ}^{2} \end{matrix} ] + [ \begin{matrix}0 & 0 \\ 0 & x \end{matrix} ]

where the second matrix is MΩM′ − MΩV_{0}^{−1}M′ with a nonzero (2,2) element x. The value of x does not affect the result, so I don’t need to specify it precisely. Next, I consider the C̄ matrix:

C̄_{ols}^{−1} = (C̄^{−1} + M(I − M)^{−1}C_{1} + M(Ω − V_{0})V_{0}^{−1})^{−1}

= ([ \begin{matrix}1 & 0 \\ 0 & (1 − α)/(1 − ρ) \end{matrix} ]^{−1} + [ \begin{matrix}0 & 0 \\ 0 & y \end{matrix} ])^{−1}

where y is a nonzero term in the SVAR error. The magnitude of y does not affect the result. Notice that the (1,1) and (1,2) elements of C̄ are correctly computed. Notice also that this implies f = −C̄(1, 2)/C̄(1, 1) = 0 and therefore A_{0}(2, 1) = 0. For all higher terms, A_{j}(2, 1) = 0 because V_{1}V_{0}^{−1} has zeros in the first column.

3.8.2. Proposition 3b: Only one shock

Proposition 3b. Assume that σ_{τ} = 0 in the RBC model. If a regression is run of the form

X_{t} = B_{ols}X_{t−1} + v_{t}

with X_{t} from the RBC model, then the Blanchard-Quah procedure recovers the true impulse responses to technology, namely

A_{j} = D_{j}Q

for all j.

Proof of Proposition 3b. The first part of the proof is concerned with the impact coefficient A_{0}. I show that Ev_{t}v′_{t} = Ω if σ_{τ} = 0, where Ω is the true variance-covariance matrix for the model. This is the main step in showing that the impact coefficient is correct. Then I show that the other coefficients can also be recovered by the SVAR. From Proposition 2, the following holds for the one-lag regression regardless of the size of the shocks:

Ev_{t}v′_{t} = Ω + MΩM′ − MΩV_{0}^{−1}ΩM′.

I now show that Ω = ΩV_{0}^{−1}Ω if σ_{τ} = 0, and therefore Ev_{t}v′_{t} = Ω. I do this in three steps. First, I show that

Ω = Ω(1, 1) [ \begin{matrix}1 & ζ \\ ζ & ζ^{2} \end{matrix} ]  (3.27)

where ζ = −a/(1 − θ + θa). Second, I show that

\frac{1}{1 + ζν} [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] V_{0} = Ω  (3.28)

where ν = −θ(1 − γ_{k})(1 − a)/[(α − γ_{k})a]. Third, I show that (3.27) and (3.28) imply:

Ω − ΩV_{0}^{−1}Ω = 0.  (3.29)

Writing out Ω yields

Ω = D_{0}QQ′D′_{0}

= [ \begin{matrix}1 − θ + θa & −θb \\ −a & b \end{matrix} ] [ \begin{matrix}σ_{z}^{2} & 0 \\ 0 & 0 \end{matrix} ] [ \begin{matrix}1 − θ + θa & −a \\ −θb & b \end{matrix} ]

= [ \begin{matrix}(1 − θ + θa)^{2} & −(1 − θ + θa)a \\ −(1 − θ + θa)a & a^{2} \end{matrix} ] σ_{z}^{2}

= Ω(1, 1) [ \begin{matrix}1 & −a/(1 − θ + θa) \\ −a/(1 − θ + θa) & a^{2}/(1 − θ + θa)^{2} \end{matrix} ]

and (3.27) holds. Writing out V_{0} in (3.28) yields

\frac{1}{1 + ζν} [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] (Ω + C_{1}ΩC′_{1} + C_{2}ΩC′_{2} + C_{3}ΩC′_{3} …) .  (3.30)

The second term on the right hand side of (3.30) is equal to a 2×2 matrix of zeros:

[ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ]C_{1}ΩC′_{1}

= [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] [ \begin{matrix}θ^{2}(1 − γ_{k})^{2}(1 − a)^{2} & θ(1 − γ_{k})(1 − a)(α − γ_{k})a \\ θ(1 − γ_{k})(1 − a)(α − γ_{k})a & (α − γ_{k})^{2}a^{2} \end{matrix} ] σ_{z}^{2}

= [ \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ] .

All higher terms are also equal to 2×2 matrices of zeros:

[ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ]C_{j}ΩC′_{j}

= [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] (C_{j−1}D_{0}P D_{0}^{−1} + D_{j}D_{0}^{−1})(D_{0}QQ′D′_{0})(C_{j−1}D_{0}P D_{0}^{−1} + D_{j}D_{0}^{−1})′

= [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] (γ_{k}^{j−2}D_{2})(QQ′)(γ_{k}^{j−2}D_{2})′

= [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] γ_{k}^{2j−4} [ \begin{matrix}[γ_{k}(1 − a)θ(1 − γ_{k})]^{2} & γ_{k}^{2}(1 − a)θ(1 − γ_{k})a(α − γ_{k}) \\ γ_{k}^{2}(1 − a)θ(1 − γ_{k})a(α − γ_{k}) & [γ_{k}a(α − γ_{k})]^{2} \end{matrix} ] σ_{z}^{2}

= [ \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ] .

Thus,

\frac{1}{1 + ζν} [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] (Ω + C_{1}ΩC′_{1} + C_{2}ΩC′_{2} + C_{3}ΩC′_{3} …)

= \frac{1}{1 + ζν} [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] Ω

= \frac{Ω(1, 1)}{1 + ζν} [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] [ \begin{matrix}1 & ζ \\ ζ & ζ^{2} \end{matrix} ]

= Ω

which proves (3.28). Now, I am ready for

Ω − ΩV_{0}^{−1}Ω = Ω(1, 1) [ \begin{matrix}1 & ζ \\ ζ & ζ^{2} \end{matrix} ] − \frac{1}{1 + ζν} [ \begin{matrix}1 & ν \\ ζ & ζν \end{matrix} ] [ \begin{matrix}1 & ζ \\ ζ & ζ^{2} \end{matrix} ]

= Ω(1, 1) [ \begin{matrix}1 & ζ \\ ζ & ζ^{2} \end{matrix} ] − \frac{1}{1 + ζν} [ \begin{matrix}1 + ζν & ζ(1 + ζν) \\ ζ(1 + ζν) & ζ^{2}(1 + ζν) \end{matrix} ]

= [ \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ]

which proves that there is no error in computing Ev_{t}v′_{t}, that is Ev_{t}v′_{t} = Ω. The next step is to show that this is all that is needed for the correct inference. Recall the formulas for the elements of A_{0} and λ in Section 3.6. Because Ω_{ols} = Ω and det(Ω)=0, it must be the case that λ = 0. Thus, A_{0}(2, 1) found by the SVAR is

A_{0} = [ \begin{matrix}\sqrt{ω_{11}} & 0 \\ \sqrt{ω_{22}} & 0 \end{matrix} ]

where \sqrt{ω_{jj}} = \sqrt{Ω(j, j)}. Using the formulas above, we have A_{0}(1, 1) = (1 − θ + θa)σ_{z} and A_{0}(2, 1) = −aσ_{z}. Thus, A_{0} = D_{0}Q. This proves that there is no mistaken inference for the impact coefficient. Next, I check A_{j} for j > 1, which is equal to B_{ols}^{j}A_{0}. For j = 1,

B_{ols}A_{0} = V_{1}V_{0}^{−1}A_{0}
= V_{1}V_{0}^{−1}D_{0}Q
= (C_{1} + M − MΩV_{0}^{−1})D_{0}Q
= C_{1}D_{0}Q
= (D_{0}P + D_{1})D_{0}^{−1}D_{0}Q
= D_{1}Q

where I have used the fact that (I − ΩV_{0}^{−1})D_{0}Q = 0. For j = 2,

B_{ols}^{2}A_{0} = (V_{1}V_{0}^{−1})^{2}A_{0}

= (V_{1}V_{0}^{−1})C_{1}D_{0}Q
= (C_{1} + M − MΩV_{0}^{−1})C_{1}D_{0}Q
= C_{2}D_{0}Q − MΩV_{0}^{−1}C_{1}D_{0}Q
= C_{2}D_{0}Q
= (D_{0}P^{2} + D_{1}P + D_{2})D_{0}^{−1}D_{0}Q
= D_{2}Q

where I have used the fact that ΩV_{0}^{−1}C_{1}D_{0}Q = 0. Similarly, I can prove it for higher terms by noting that if B_{ols}^{j−1}A_{0} = C_{j−1}D_{0}Q holds, then B_{ols}^{j}A_{0} = C_{j}D_{0}Q holds and so does the following:

B_{ols}^{j}A_{0} = (V_{1}V_{0}^{−1})C_{j−1}D_{0}Q
= (C_{1} + M − MΩV_{0}^{−1})C_{j−1}D_{0}Q
= C_{j}D_{0}Q − MΩV_{0}^{−1}C_{j−1}D_{0}Q
= C_{j}D_{0}Q
= (D_{0}P^{j} + D_{1}P + … + D_{j})D_{0}^{−1}D_{0}Q
= D_{j}Q.

This establishes that in the case with σ_{τ} = 0, the SVAR uncovers the true impulse responses to technology.

What is interesting about the last two propositions is that the special cases are not relevant for modern business cycle theorists. Modern business cycle theorists assume that both capital accumulation and shocks in addition to technology (e.g., distortions to labor) are quantitatively important. Furthermore, adding these factors is not a recent phenomena. They are central to the work following Kydland and Prescott (1982) (which includes my thesis).

4. VARs and 3-Shock Versions of the Model

I consider several versions of an RBC model with three shocks and three variables in the VAR. The first has a investment tax shock and the log of the investment-output ratio in the VAR. The second has a government spending shock and the log of the investment-output ratio in the VAR. The third has a investment tax shock and the log of the consumptionoutput ratio in the VAR.

4.1. Investment Tax Shock and Investment-Output in VAR

Assume the economy is an RBC model with three orthogonal shocks: a unit root in technology log z, an AR(1) in the tax rate on labor τ_{l}, and an AR(1) in the tax rate on investment τ_{x}. The capital decision function has the form:

log k̂_{t+1} = γ_{0} + γ_{k} log k̂_{t} + γ_{z} log z_{t} + γ_{l}τ_{lt} + γ_{x}τ_{xt}  (4.1)

and the labor decision function can be written:

log l_{t} = ϕ_{lz} log z_{t} + ϕ_{ll}τ_{lt} + ϕ_{lx}τ_{xt} + ϕ_{lk} log k̂_{t} + ϕ_{lk′} log k̂_{t+1}

= ϕ_{lz} log z_{t} + ϕ_{ll}τ_{lt} + ϕ_{lx}τ_{xt} + ϕ_{lk} log k̂_{t} + ϕ_{lk′}[γ_{0}γ_{k} log k̂_{t} + γ_{z} log z_{t} + γ_{l}τ_{lt} + γ_{x}τ_{xt}] = (ϕ_{lk} + ϕ_{lk′}γ_{k}) log k̂_{t} + (ϕ_{lz} + ϕ_{lk′}γ_{z}) log z_{t} + (ϕ_{ll} + ϕ_{lk′}γ_{l})τ_{lt} + (ϕ_{lx} + ϕ_{lk′}γ_{x})τ_{xt}.

Note that I include the term ϕ_{lx}τ_{xt} here even though it is equal to 0 in equilibrium. I do so because the same mathematics will be used later in the case of the government spending shock.

Next, I write out output from a Cobb-Douglas production technology with capital share θ is:

log ŷ_{t} = θ(log k̂_{t} − log z_{t}) + (1 − θ) log l_{t}
= (θ + (1 − θ)ϕ_{lk}) log k̂_{t} − (θ − (1 − θ)ϕ_{lz}) log z_{t} + (1 − θ)ϕ_{ll}τ_{lt}
+ (1 − θ)ϕ_{lx}τ_{xt} + (1 − θ)ϕ_{lk′} log k̂_{t+1}
= (θ + (1 − θ)(ϕ_{lk} + ϕ_{lk′}γ_{k})) log k̂_{t} − (θ − (1 − θ)(ϕ_{lz} + ϕ_{lk′}γ_{z})) log z_{t}
+ (1 − θ)(ϕ_{ll} + ϕ_{lk′}γ_{l})τ_{lt}
+ (1 − θ)(ϕ_{lx} + ϕ_{lk′}γ_{x})τ_{xt}

I can write the capital stock in terms of all lagged shocks as follows:

log k̂_{t} = γ_{0} + γ_{k}(γ_{0} + γ_{k} log k̂_{t−2} + γ_{z} log z_{t−2} + γ_{l}τ_{lt−2}) + γ_{x}τ_{xt−2}) + γ_{z} log z_{t−1} + γ_{l}τ_{lt−1} + γ_{x}τ_{xt−1}

= γ_{0}[1 + γ_{k} + γ_{k}^{2} + …]

+ γ_{z}[log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …]
+ γ_{l}[τ_{lt−1} + γ_{k}τ_{lt−2} + γ_{k}^{2}τ_{lt−3} + …]
+ γ_{x}[τ_{xt−1} + γ_{k}τ_{xt−2} + γ_{k}^{2}τ_{xt−3} + …]

or in differences as follows:

log k̂_{t} − log k̂_{t−1} = γ_{z}[log z_{t−1} + (γ_{k} − 1){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}]
+ γ_{l}[τ_{lt−1} + (γ_{k} − 1){τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2}τ_{lt−4} + …}]
+ γ_{x}[τ_{xt−1} + (γ_{k} − 1){τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2}τ_{xt−4} + …}]

or in quasi-differences as follows:

log k̂_{t} − α log k̂_{t−1} = γ_{z}[log z_{t−1} + (γ_{k} − α){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}]
+ γ_{l}[τ_{lt−1} + (γ_{k} − α){τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2}τ_{lt−4} + …}]
+ γ_{x}[τ_{xt−1} + (γ_{k} − α){τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2}τ_{xt−4} + …}]

I can also write hours in terms of past shocks as follows:

log l_{t} = ϕ_{lz} log z_{t} + ϕ_{ll}τ_{lt} + ϕ_{lx}τ_{xt} + ϕ_{lk} log k̂_{t} + ϕ_{lk′} log k̂_{t+1}

= ϕ_{lz} log z_{t} + ϕ_{ll}τ_{lt} + ϕ_{lx}τ_{xt}

+ ϕ_{lk}γ_{z}[log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …] + ϕ_{lk}γ_{l}[τ_{lt−1} + γ_{k}τ_{lt−2} + γ_{k}^{2}τ_{lt−3} + …]

+ ϕ_{lk}γ_{x}[τ_{xt−1} + γ_{k}τ_{xt−2} + γ_{k}^{2}τ_{xt−3} + …]

+ ϕ_{lk′}γ_{z}[log z_{t} + γ_{k} log z_{t−1} + γ_{k}^{2} log z_{t−2} + …] + ϕ_{lk′}γ_{l}[τ_{lt} + γ_{k}τ_{lt−1} + γ_{k}^{2}τ_{lt−2} + …]

+ ϕ_{lk′}γ_{x}[τ_{xt} + γ_{k}τ_{xt−1} + γ_{k}^{2}τ_{xt−2} + …]

= [(ϕ_{lz} + ϕ_{lk′}γ_{z}) log z_{t} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{z} log z_{t−1} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{k}γ_{z} log z_{t−2} + …] + [(ϕ_{ll} + ϕ_{lk′}γ_{l})τ_{lt} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{l}τ_{lt−1} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{k}γ_{l}τ_{lt−2} + …] + [(ϕ_{ll} + ϕ_{lk′}γ_{x})τ_{xt} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{x}τ_{xt−1} + (ϕ_{lk} + ϕ_{lk′}γ_{k})γ_{k}γ_{x}τ_{xt−2} + …]

where I have ignored constant terms.

I can write logged hours in differences as follows:

log l_{t} − log l_{t−1} = ϕ_{lz}(log z_{t} − log z_{t−1}) + ϕ_{ll}(τ_{lt} − τ_{lt−1}) + ϕ_{lx}(τ_{xt} − τ_{xt−1})

+ ϕ_{lk′}(log k̂_{t+1} − log k̂_{t}) + ϕ_{lk}(log k̂_{t} − log k̂_{t−1})

= ϕ_{lz}(log z_{t} − log z_{t−1}) + ϕ_{ll}(τ_{lt} − τ_{lt−1}) + ϕ_{lx}(τ_{xt} − τ_{xt−1})

+ ϕ_{lk′}γ_{z}[log z_{t} + (γ_{k} − 1){log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …}]

+ ϕ_{lk′}γ_{l}[τ_{lt} + (γ_{k} − 1){τ_{lt−1} + γ_{k}τ_{lt−2} + γ_{k}^{2}τ_{lt−3} + …}]

+ ϕ_{lk′}γ_{x}[τ_{xt} + (γ_{k} − 1){τ_{xt−1} + γ_{k}τ_{xt−2} + γ_{k}^{2}τ_{xt−3} + …}]

+ ϕ_{lk}γ_{z}[log z_{t−1} + (γ_{k} − 1){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}]

+ ϕ_{lk}γ_{l}[τ_{lt−1} + (γ_{k} − 1){τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2}τ_{lt−4} + …}]

+ ϕ_{lk}γ_{x}[τ_{xt−1} + (γ_{k} − 1){τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2}τ_{xt−4} + …}]

= [ϕ_{lz} + ϕ_{lk′}γ_{z}] log z_{t} − [ϕ_{lz} − ϕ_{lk}γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − 1)] log z_{t−1}

+ γ_{z}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}

+ [ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{lt} − [ϕ_{ll} − ϕ_{lk}γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − 1)]τ_{lt−1}

+ γ_{l}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2}τ_{lt−4} + …}

+ [ϕ_{lx} + ϕ_{lk′}γ_{x}]τ_{xt} − [ϕ_{lx} − ϕ_{lk}γ_{x} − ϕ_{lk′}γ_{x}(γ_{k} − 1)]τ_{xt−1}

+ γ_{x}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2}τ_{xt−4} + …}

or in quasi-difference form as follows:

log l_{t} − α log l_{t−1} = ϕ_{lz}(log z_{t} − α log z_{t−1}) + ϕ_{ll}(τ_{lt} − ατ_{lt−1}) + ϕ_{lx}(τ_{xt} − ατ_{xt−1})

+ ϕ_{lk′}(log k̂_{t+1} − α log k̂_{t}) + ϕ_{lk}(log k̂_{t} − α log k̂_{t−1})

= ϕ_{lz}(log z_{t} − α log z_{t−1}) + ϕ_{ll}(τ_{lt} − ατ_{lt−1}) + ϕ_{lx}(τ_{xt} − ατ_{xt−1})

+ ϕ_{lk′}γ_{z}[log z_{t} + (γ_{k} − α){log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …}]

+ ϕ_{lk′}γ_{l}[τ_{lt} + (γ_{k} − α){τ_{lt−1} + γ_{k}τ_{lt−2} + γ_{k}^{2}τ_{lt−3} + …}]

+ ϕ_{lk′}γ_{x}[τ_{xt} + (γ_{k} − α){τ_{xt−1} + γ_{k}τ_{xt−2} + γ_{k}^{2}τ_{xt−3} + …}]

+ ϕ_{lk}γ_{z}[log z_{t−1} + (γ_{k} − α){log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}]

+ ϕ_{lk}γ_{l}[τ_{lt−1} + (γ_{k} − α){τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2}τ_{lt−4} + …}]

+ ϕ_{lk}γ_{x}[τ_{xt−1} + (γ_{k} − α){τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2}τ_{xt−4} + …}]

= [ϕ_{lz} + ϕ_{lk′}γ_{z}] log z_{t} − [αϕ_{lz} − ϕ_{lk}γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − α)] log z_{t−1}

+ γ_{z}(γ_{k} − α)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …}

+ [ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{lt} − [αϕ_{ll} − ϕ_{lk}γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − α)]τ_{lt−1}

+ [ϕ_{lx} + ϕ_{lk′}γ_{l}]τ_{xt} − [αϕ_{lx} − ϕ_{lk}γ_{l} − ϕ_{lk′}γ_{x}(γ_{k} − α)]τ_{xt−1}

+ γ_{l}(γ_{k} − α)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2}τ_{lt−4} + …}

+ γ_{x}(γ_{k} − α)[ϕ_{lk′}γ_{k} + ϕ_{lk}]{τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2}τ_{xt−4} + …} (4.2)

I can use the expressions for output and hours to write out the change in productivity as follows:

log(y_{t}/l_{t}) − log(y_{t−1}/l_{t−1})

= log ŷ_{t} − log ŷ_{t−1} + log z_{t} − log l_{t} − log l_{t−1} = log z_{t} + θ(log k̂_{t} − log k̂_{t−1} − log l_{t} + log l_{t−1} − log z_{t} + log z_{t−1}) = (1 − θ) log z_{t} + θ log z_{t−1}

− θ(log l_{t} − log l_{t−1} − log k̂_{t} + log k̂_{t−1}) = (1 − θ) log z_{t} + θ log z_{t−1}

− θ(log l_{t} − log l_{t−1} − log k̂_{t} + log k̂_{t−1}) = (1 − θ) log z_{t} + θ log z_{t−1} − θ{[ϕ_{lz} + ϕ_{lk′}γ_{z}] log z_{t}

− [ϕ_{lz} − (ϕ_{lk} − 1)γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − 1)] log z_{t−1} + γ_{z}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …] + [ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{lt} − [ϕ_{ll} − (ϕ_{lk} − 1)γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − 1)]τ_{lt−1} + [ϕ_{lx} + ϕ_{lk′}γ_{x}]τ_{xt} − [ϕ_{lx} − (ϕ_{lk} − 1)γ_{x} − ϕ_{lk′}γ_{x}(γ_{k} − 1)]τ_{xt−1} + γ_{l}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2} log τ_{lt−4} + …]} + γ_{x}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2} log τ_{xt−4} + …]}

= {1 − θ − θ[ϕ_{lz} + ϕ_{lk′}γ_{z}]} log z_{t} + θ[1 + ϕ_{lz} − (ϕ_{lk} − 1)γ_{z} − ϕ_{lk′}γ_{z}(γ_{k} − 1)] log z_{t−1} − θγ_{z}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][log z_{t−2} + γ_{k} log z_{t−3} + γ_{k}^{2} log z_{t−4} + …] − θ[ϕ_{ll} + ϕ_{lk′}γ_{l}]τ_{lt} + θ[ϕ_{ll} − (ϕ_{lk} − 1)γ_{l} − ϕ_{lk′}γ_{l}(γ_{k} − 1)]τ_{lt−1} − θγ_{l}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][τ_{lt−2} + γ_{k}τ_{lt−3} + γ_{k}^{2}τ_{lt−4} + …] − θ[ϕ_{lx} + ϕ_{lk′}γ_{x}]τ_{xt} + θ[ϕ_{lx} − (ϕ_{lk} − 1)γ_{x} − ϕ_{lk′}γ_{x}(γ_{k} − 1)]τ_{xt−1}

− θγ_{x}(γ_{k} − 1)[ϕ_{lk′}γ_{k} + ϕ_{lk} − 1][τ_{xt−2} + γ_{k}τ_{xt−3} + γ_{k}^{2}τ_{xt−4} + …]  (4.3)

Now I write out the log of the investment share:

log(x_{t}/y_{t}) = log x̂_{t} − log ŷ_{t}
= ϕ_{xk}(log k̂_{t} − log z_{t}) + ϕ_{xk′} log k̂_{t+1} − θ(log k̂_{t} − log z_{t}) − (1 − θ) log l_{t}
= (ϕ_{xk} − θ)(log k̂_{t} − log z_{t}) + ϕ_{xk′}[γ_{k} log k̂_{t} + γ_{z} log z_{t} + γ_{l}τ_{lt} + γ_{x}τ_{xt}]
− (1 − θ)[(ϕ_{lk} + ϕ_{lk′}γ_{k}) log k̂_{t} + (ϕ_{lz} + ϕ_{lk′}γ_{z}) log z_{t}
+ (ϕ_{ll} + ϕ_{lk′}γ_{l})τ_{lt} + (ϕ_{lx} + ϕ_{lk′}γ_{x})τ_{xt}]
= [−ϕ_{xk} + θ + ϕ_{xk′}γ_{z} − (1 − θ)(ϕ_{lz} + ϕ_{lk′}γ_{z})] log z_{t}
+ [ϕ_{xk′}γ_{l} − (1 − θ)(ϕ_{ll} + ϕ_{lk′}γ_{l})]τ_{lt}
+ [ϕ_{xk′}γ_{x} − (1 − θ)(ϕ_{lx} + ϕ_{lk′}γ_{x})]τ_{xt}
+ [ϕ_{xk} − θ + ϕ_{xk′}γ_{k} − (1 − θ)(ϕ_{lk} + ϕ_{lk′}γ_{k})]
{γ_{z}[log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …]
+ γ_{l}[τ_{lt−1} + γ_{k}τ_{lt−2} + γ_{k}^{2}τ_{lt−3} + …]
+ γ_{x}[τ_{xt−1} + γ_{k}τ_{xt−2} + γ_{k}^{2}τ_{xt−3} + …]}

4.1.1. The Model’s Moving Average The moving average for the model is given by:

[ \begin{matrix}(1 − L) log y_{t}/l_{t} \\ (1 − αL) log l_{t} \\ log x_{t}/y_{t} \end{matrix} ] ≡ X_{t} = D_{0}ω_{t} + D_{1}ω_{t−1} + D_{2}ω_{t−2} + …

where ω_{t} = [log z_{t}, τ_{lt}, τ_{xt}]′ and

D_{0} = [ \begin{matrix}1 − θ + θa & −θb & −θc \\ −a & b & c \\ −d & e & f \end{matrix} ]  (4.4)
D_{1} = [ \begin{matrix}θ(1 − a)(1 − γ_{k}) & θ(b + (1 − a)γ_{l}) & θ(c + (1 − a)γ_{x}) \\ (α − γ_{k})a & −αb + γ_{l}a & −αc + γ_{x}a \\ −dγ_{k} & dγ_{l} & dγ_{x} \end{matrix} ]  (4.5)
D_{2} = [ \begin{matrix}γ_{k}(1 − a)θ(1 − γ_{k}) & −γ_{l}(1 − a)θ(1 − γ_{k}) & −γ_{x}(1 − a)θ(1 − γ_{k}) \\ γ_{k}a(α − γ_{k}) & −γ_{l}a(α − γ_{k}) & −γ_{x}a(α − γ_{k}) \\ −dγ_{k}^{2} & dγ_{l}γ_{k} & dγ_{x}γ_{k} \end{matrix} ] , (4.6)

and D_{j} = γ_{k}D_{j−1} for j ≥ 3 where a = ϕ_{lk} + ϕ_{lk′}γ_{k}, b = ϕ_{ll} + ϕ_{lk′}γ_{l}, c = ϕ_{lx} + ϕ_{lk′}γ_{x}, d = ϕ_{xk} + ϕ_{xk′}γ_{k} − θ − (1 − θ)a, e = ϕ_{xk′}γ_{l} − (1 − θ)b, and f = ϕ_{xk′}γ_{x} − (1 − θ)c. Note that ϕ_{lz} = −ϕ_{lk}, ϕ_{xz} = −ϕ_{xk}, and γ_{z} = −γ_{k} hold in the model economy with a unit root in technology. Note also that D_{2} is singular for all parameterizations, and D_{1} is singular if α = 0.

If τ_{lt} and τ_{xt} are AR(1) processes, then it is more convenient to write the MA process in terms of η_{t} = [log z_{t}, η_{lt}, η_{xt}] rather than in terms of ω_{t}. In this case, X_{t} = D_{0}η_{t}+(D_{0}P +D_{1})η_{t−1}+(D_{0}P^{2}+D_{1}P +D_{2})η_{t−2}+(D_{0}P^{3}+D_{1}P^{2}+D_{2}P +D_{3})η_{t−3}…. I normalize the MA so it has an identity for the first coefficient. That is, I set C_{0} = I, C_{1} = (D_{0}P + D_{1})D_{0}^{−1}, and C_{j} = C_{j−1}D_{0}P D_{0}^{−1} + D_{j}D_{0}^{−1}.

4.1.2. Special Property of the D’s

As in the 2-shock case, it is the case that the D matrices have a special property that can be exploited when I characterize coefficients of the VAR of X_{t}. In other words, the D’s for the 3-shock RBC model also satisfy the relation:

(γ_{k}I − (D_{0}P^{2} + D_{1}P + D_{2})(D_{0}P + D_{1})^{−1}) D_{2} = 0.  (4.7)

which is the same as (3.1). Because D_{1} is singular when α = 0, I will assume that the choice of P and α is such that D_{0}P + D_{1} is invertible. This rules out the case with P and α identically equal to 0. If that is the case of interest, assume that α is positive but very close to zero.

The steps of the proof of (4.7) in the 3-variable case is the same as in the 2-variable case. First note from (4.6) that

D_{2} = [ \begin{matrix}(1 − a)θ(1 − γ_{k}) \\ (α − γ_{k})a \\ −dγ_{k} \end{matrix} ] [ γ_{k} −γ_{l} −γ_{x}] ≡ gh′

Thus, I can rewrite the left hand side as follows

(γ_{k}I − (D_{0}P^{2} + D_{1}P + D_{2})(D_{0}P + D_{1})^{−1})D_{2}
= [γ_{k}(gh′) − (gh′)(D_{0}P + D_{1})^{−1}(gh′)] − [(D_{0}P + D_{1})P (D_{0}P + D_{1})^{−1}gh′]. (4.8)

Both terms in (4.8) in square brackets are equal to 3×3 zero matrices. The first step is to show that

(D_{0}P + D_{1})^{−1}g = [ \begin{matrix}1 \\ 0 \\ 0 \end{matrix} ] .  (4.9)

The proof of this step is trivial since the first column of D_{0}P +D_{1} is equal to g. Substituting (4.9) into (4.8), the result (4.7) follows immediately from the fact that h′[1, 0, 0]′ = γ_{k} and P [1, 0, 0]′ = 0.

4.1.3. Proposition 4: Model has infinite-order VAR

The map between the theoretical MA and the VAR is the same as before. What is new is the VAR representation.

Proposition 4. The model described above has a VAR representation with coefficients B_{j} that satisfy

B_{j} = MB_{j−1}  (4.10)

for j ≥ 2, with B_{1} = C_{1} = (D_{0}P + D_{1})D_{0}^{−1}. The matrix M is 3×3 with eigenvalues equal to 0, α, and (1 − δ)/[z(1 + g_{n})].

Proof of Proposition 4. The first part of the proof is the same as for Proposition 1. The second part, involving the expressions of the eigenvalues, is different. In the three shock case, one can use the same derivations as those in Proposition 1 to show that [1, 0, 0]′h′ − D_{0}^{−1}D_{1} has the same eigenvalues as M. In this case, D_{0}^{−1} is given by:

D_{0}^{−1} = \frac{1}{|D_{0}|} [ \begin{matrix}bf − ce & θ(bf − ce) & 0 \\ af − cd & (1 − θ)f + θ(af − cd) & −(1 − θ)c \\ bd − ae & −(1 − θ)e + θ(bd − ae) & (1 − θ)b \end{matrix} ]

and the elements of [1, 0, 0]′h′ − D_{0}^{−1}D_{1} are given by

(1, 1) = γ_{k} − θ(1 − γ_{k} − a + aα)/(1 − θ)
(1, 2) = −γ_{l} − θ(γ_{l} + b − bα)/(1 − θ)
(1, 3) = −γ_{x} − θ(γ_{x} + c − cα)/(1 − θ)
(2, 1) = {(af − cd)(γ_{k} − θ + θa − θaα) − af(1 − θ)α]}/|D_{0}|
(2, 2) = {(af − cd)(−γ_{l} − θb + θbα) + bf(1 − θ)α}/|D_{0}|
(2, 3) = {(af − cd)(−γ_{x} − θc + θcα) + cf(1 − θ)α}/|D_{0}|
(3, 1) = {(bd − ae)(γ_{k} − θ + θa − θaα) + ae(1 − θ)α}/|D_{0}|
(3, 2) = {(bd − ae)(−γ_{l} − θb + θbα) − be(1 − θ)α}/|D_{0}|
(3, 3) = {(bd − ae)(−γ_{x} − θc + θcα) − ce(1 − θ)α}/|D_{0}|

where |D_{0}| = (1−θ)(bf −ce). To prove the proposition, I will show that trace([1, 0, 0]′h′ − D_{0}^{−1}D_{1}) equals the sum of the proposed eigenvalues, |[1, 0, 0]′h′ − D_{0}^{−1}D_{1}| = 0, and |[1, 0, 0]′h′ − D_{0}^{−1}D_{1} − αI| = 0. These three conditions uniquely determine the three eigenvalues.

To compute the trace, sum (1,1), (2,2), and (3,3):

trace([1, 0, 0]′h′ − D_{0}^{−1}D_{1}) = γ_{k} − θ(1 − γ_{k} − a + aα)/(1 − θ)

+ {(af − cd)(−γ_{l} − θb + θbα) + bf(1 − θ)α

+ (bd − ae)(−γ_{x} − θc + θcα) − ce(1 − θ)α}/|D_{0}| = α + {(γ_{k} − θ)(bf − ce) − γ_{l}(af − cd) − γ_{x}(bd − ae)}/|D_{0}| = α + {(γ_{k} − θ)ϕ_{xk′}(bγ_{x} − cγ_{l}) − γ_{l}(aϕ_{xk′}γ_{x} − c(ϕ_{xk} + ϕ_{xk′}γ_{k} − θ))

− γ_{x}(b(ϕ_{xk} + ϕ_{xk′}γ_{k} − θ) − aϕ_{xk′}γ_{l})}
/[(1 − θ)ϕ_{xk′}(bγ_{x} − cγ_{l})]
= α + \frac{θ(1 − ϕ_{xk′}) − ϕ_{xk}}{ϕ_{xk′}(1 − θ)}  (4.11)
= α + \frac{(1 − δ)}{z(1 + g_{n})}  (4.12)

where z without a subscript is the steady state value.

Next I compute the determinant of [1, 0, 0]′h′ − D_{0}^{−1}D_{1} and show it is 0. Denoting the matrix by M, I get

det(M) = M_{1,1}|M([2, 3], [2, 3])| − M_{1,2}|M([2, 3], [1, 3])| + M_{1,3}|M([2, 3], [1, 2])|

= (γ_{k} − θ(1 − γ_{k} − a + aα)/(1 − θ))(1 − θ)αd(fb − ec)(γ_{l}c − γ_{x}b)/|D_{0}|^{2} + (γ_{l} + θ(γ_{l} + b − bα)/(1 − θ))(1 − θ)αd(fb − ec)(−γ_{k}c + θc + γ_{x}a)/|D_{0}|^{2} − (γ_{x} + θ(γ_{x} + c − cα)/(1 − θ))(1 − θ)αd(fb − ec)(−γ_{k}b + θb + γ_{l}a)/|D_{0}|^{2}

= {(γ_{k}(1 − θ) − θ(1 − γ_{k} − a + aα))(γ_{l}c − γ_{x}b)

+ (γ_{l}(1 − θ) + θ(γ_{l} + b − bα))(−γ_{k}c + θc + γ_{x}a) − (γ_{x}(1 − θ) + θ(γ_{x} + c − cα))(−γ_{k}b + θb + γ_{l}a)}αd(fb − ec)/|D_{0}|^{2}

= {(γ_{k} − θ + θa(1 − α))(γ_{l}c − γ_{x}b)

+ (γ_{l} + θb(1 − α))(−(γ_{k} − θ)c + γ_{x}a) − (γ_{x} + θc(1 − α))(−(γ_{k} − θ)b + γ_{l}a)}αd(fb − ec)/|D_{0}|^{2}

= 0  (4.13)

Finally, I take the determinant of M − αI and show it is 0 as follows:

det(M − αI)

= (M_{1,1} − α)|M([2, 3], [2, 3]) − αI| − M_{1,2}(|M([2, 3], [1, 3])| − αM_{2,1})

+ M_{1,3}(|M([2, 3], [1, 2])| + αM_{3,1}) = α{α[M_{1,1} + M_{2,2} + M_{3,3} − α]

− M_{1,1}M_{2,2} − M_{1,1}M_{3,3} − M_{2,2}M_{3,3}
+ M_{1,2}M_{2,1} + M_{1,3}M_{3,1} + M_{2,3}M_{3,2}}
= α{α[trace(M) − α]
− (M_{1,1}M_{2,2} − M_{1,2}M_{2,1})
− (M_{1,1}M_{3,3} − M_{1,3}M_{3,1})
− (M_{2,2}M_{3,3} − M_{2,3}M_{3,2})}
= α{α[(bf − ce)(γ_{k} − θ + θa − θaα)
+ (af − cd)(−γ_{l} − θb + θbα) + bf(1 − θ)α
+ (bd − ae)(−γ_{x} − θc + θcα) − ce(1 − θ)α − α(1 − θ)(bf − ce)]
− fα[b(γ_{k} − θ) − aγ_{l}]
+ eα[c(γ_{k} − θ) − aγ_{x}]
− dα[γ_{l}c − γ_{x}b]}/|D_{0}|
= 0  (4.14)

The result in (4.14) implies that α is an eigenvalue. The result in (4.13) implies that 0 is an eigenvalue. Given these results, the fact that the trace is (4.12) implies that (1 − δ)/[z(1 + g_{n})] is the third eigenvalue. This completes the proof.

4.1.4. A Way to Make M Singular

Above I included the investment share in the VAR. The investment share is typically added to capture the capital dynamics if capital is unobserved. What if I assume that capital is observed and use the capital share instead?

I can see the answer directly from the proof in Proposition 4. At the step (equation (4.11)) that I fill in expressions for ϕ_{xk} and ϕ_{xk′} using (2.18), I could instead use ϕ_{xk} = 1 and ϕ_{xk′} = 0. This yields a third eigenvalue equal to -1/0 or −∞. This clearly doesn’t work since the MA is not invertible.

However, it shows me what would work: adding the capital next period relative to output, log(k_{t+1}/y_{t}), and, therefore, setting ϕ_{xk} = 0 and ϕ_{xk′} = 1. If α = 0, then the matrix M has 3 zero eigenvalues. A researcher running a VAR would find that B_{2} is singular and the rest of the B_{j}, j ≥ 3, are zero matrices. In fact the structure would be such that the second and third column of B_{2} would be equal and equal to the negative of the first column of B_{2}. That is how it works: certain lags are cancelling so it effectively mimics the model’s finite-lag VAR.

What is interesting is that it won’t work if I divide k_{t+1} by k_{t} and include the log of the growth rate of capital. If I add log(k_{t+1}/k_{t}) to the VAR, then I proceed the same way through the proof of Proposition 4 using d = γ_{k} − 1, e = γ_{l}, and f = γ_{x}. The result is that the eigenvalues of M are 0, α, and 1. The fact that one is 1 means that the MA is not invertible.

What these results tell me is that one has to proceed carefully, using lots of the details of the model, to determine if the SVAR has a short-lag representation. Since most business cycle models have a short-lag state-space representation, we advise using it directly. The state-space representation also allows us to treat the capital stocks as unobserved. This is certainly necessary in business cycle models with sticky price models and staggered constracts; the state vector includes the distribution of capital stocks which is unobserved.

4.2. Government Spending Shock and Investment-Output in VAR

Assume the economy is an RBC model with three orthogonal shocks: a unit root in technology log z, an AR(1) in the tax rate on labor τ_{l}, and an AR(1) in log ĝ_{t}. The capital decision function has the form:

log k̂_{t+1} = γ_{0} + γ_{k} log k̂_{t} + γ_{z} log z_{t} + γ_{l}τ_{lt} + γ_{g} log ĝ_{t}.

4.2.1. The Model’s Moving Average

I can borrow the expressions from above for this case, replacing γ_{x} by γ_{g} and τ_{xt} by log ĝ_{t}. In other words, the D matrices are given by:

D_{0} = [ \begin{matrix}1 − θ + θa & −θb & −θc \\ −a & b & c \\ −d & e & f \end{matrix} ]  (4.15)

D_{1} = [ \begin{matrix}θ(1 − a)(1 − γ_{k}) & θ(b + (1 − a)γ_{l}) & θ(c + (1 − a)γ_{g}) \\ (α − γ_{k})a & −αb + γ_{l}a & −αc + γ_{g}a \\ −dγ_{k} & dγ_{l} & dγ_{g} \end{matrix} ]  (4.16)

D_{2} = [ \begin{matrix}γ_{k}(1 − a)θ(1 − γ_{k}) & −γ_{l}(1 − a)θ(1 − γ_{k}) & −γ_{g}(1 − a)θ(1 − γ_{k}) \\ γ_{k}a(α − γ_{k}) & −γ_{l}a(α − γ_{k}) & −γ_{g}a(α − γ_{k}) \\ −dγ_{k}^{2} & dγ_{l}γ_{k} & dγ_{g}γ_{k} \end{matrix} ] , (4.17)

and D_{j} = γ_{k}D_{j−1} for j ≥ 3 where a = ϕ_{lk} + ϕ_{lk′}γ_{k}, b = ϕ_{ll} + ϕ_{lk′}γ_{l}, c = ϕ_{lg} + ϕ_{lk′}γ_{g}, d = ϕ_{xk} + ϕ_{xk′}γ_{k} − θ − (1 − θ)a, e = ϕ_{xk′}γ_{l} − (1 − θ)b, and f = ϕ_{xk′}γ_{g} − (1 − θ)c. Again, it is the case that ϕ_{lz} = −ϕ_{lk}, ϕ_{xz} = −ϕ_{xk}, and γ_{z} = −γ_{k}. Note also that D_{2} is singular for all parameterizations, and D_{1} is singular if α = 0.

As before, we can rewrite the theoretical MA in terms of η_{t} = [log z_{t}, η_{lt}, η_{gt}] where this time, the third shock is government spending shock η_{gt}.

Because the D matrices have the same form as in the case with the investment tax shock, it is trivial to prove that Proposition 4 also holds for the economy with government spending shocks. The matrix M in the two cases are different (i.e., have different elements) but the eigenvalues are the same. The C matrices (where C_{j} = C_{j−1}D_{0}P D_{0}^{−1} + D_{j}D_{0}^{−1}) also have different elements. So, the garbled terms in C̄ = I+C_{1}+C_{2}+… and the variancecovariance matrix Evv′ will differ depending on which shock I use. What Proposition 4 tells me is that the garbling will likely imply that the mistaken inference is quantitatively important if the third shock has a nonnegligible variance.

4.3. Investment Tax Shock and Consumption-Output in VAR

The decision function for next period capital is again (4.1), the formula for the change in productivity is again (4.3), and the formula for quasi-differenced hours is (4.2).

Suppose the third variable in the VAR is the log of the consumption share:

log(c_{t}/y_{t}) = log ĉ_{t} − log ŷ_{t}
= (ϕ_{ck} − (1 − θ)ϕ_{lk} − θ) log k̂_{t} + (ϕ_{cz} − (1 − θ)ϕ_{lz} + θ) log z_{t}
+ (ϕ_{cl} − (1 − θ)ϕ_{ll})τ_{lt} + (ϕ_{cx} − (1 − θ)ϕ_{lx})τ_{xt}
+ (ϕ_{ck′} − (1 − θ)ϕ_{lk′}) log k̂_{t+1}
= [ϕ_{ck} + ϕ_{ck′}γ_{k} − (1 − θ)(ϕ_{lk} + ϕ_{lk′}γ_{k}) − θ] log k̂_{t}
+ [ϕ_{cz} + ϕ_{ck′}γ_{z} − (1 − θ)(ϕ_{lz} + ϕ_{lk′}γ_{z}) + θ] log z_{t}
+ [ϕ_{cl} + ϕ_{ck′}γ_{l} − (1 − θ)(ϕ_{ll} + ϕ_{lk′}γ_{l})]τ_{lt}
+ [ϕ_{cx} + ϕ_{ck′}γ_{x} − (1 − θ)(ϕ_{lx} + ϕ_{lk′}γ_{x})]τ_{xt}
= [ϕ_{cz} + ϕ_{ck′}γ_{z} − (1 − θ)(ϕ_{lz} + ϕ_{lk′}γ_{z}) + θ] log z_{t}
+ [ϕ_{cl} + ϕ_{ck′}γ_{l} − (1 − θ)(ϕ_{ll} + ϕ_{lk′}γ_{l})]τ_{lt}
+ [ϕ_{cx} + ϕ_{ck′}γ_{l} − (1 − θ)(ϕ_{lx} + ϕ_{lk′}γ_{x})]τ_{xt}
+ [ϕ_{ck} + ϕ_{ck′}γ_{k} − (1 − θ)(ϕ_{lk} + ϕ_{lk′}γ_{k}) − θ]
{γ_{z}[log z_{t−1} + γ_{k} log z_{t−2} + γ_{k}^{2} log z_{t−3} + …]
+ γ_{l}[τ_{lt−1} + γ_{k}τ_{lt−2} + γ_{k}^{2}τ_{lt−3} + …]
+ γ_{x}[τ_{xt−1} + γ_{k}τ_{xt−2} + γ_{k}^{2}τ_{xt−3} + …]}

4.3.1. The Model’s Moving Average

The moving average can again be represented in terms of the D matrices in (4.4)-(4.6) with D_{j} = γ_{k}D_{j−1} for j ≥ 3 and a = ϕ_{lk} + ϕ_{lk′}γ_{k}, b = ϕ_{ll} + ϕ_{lk′}γ_{l}, c = ϕ_{lx} + ϕ_{lk′}γ_{x}, d = ϕ_{ck} + ϕ_{ck′}γ_{k} − θ − (1 − θ)a, e = ϕ_{cl} + ϕ_{ck′}γ_{l} − (1 − θ)b, and f = ϕ_{cx} + ϕ_{ck′}γ_{x} − (1 − θ)c. Note that ϕ_{lz} = −ϕ_{lk}, ϕ_{cz} = −ϕ_{ck}, and γ_{z} = −γ_{k} hold in the model economy with a unit root in technology. Note also that D_{2} is singular for all parameterizations, and D_{1} is singular if α = 0.

Because the D matrices have the same form, I can again find the eigenvalues of M by working with the simpler matrix [1, 0, 0]′h′ − D_{0}^{−1}D_{1}. I will show that the eigenvalues in this case are 0, α, and

\frac{1 − δ}{z(1 + g_{n})} + (\frac{θ}{1 − θ} ) \frac{ĝ}{k̂(1 + g_{n})}

I can follow the proof of proposition 8 to show this. There is only one deviation needed, which is the derivation of the third eigenvalue. Writing out the trace of [1, 0, 0]′h′ −D_{0}^{−1}D_{1} yields

trace([1, 0, 0]′h′ − D_{0}^{−1}D_{1}) = α + {(γ_{k} − θ)(bf − ce) − γ_{l}(af − cd) − γ_{x}(bd − ae)}/|D_{0}|

as before. Using the new arguments of the D’s, this trace can be simplified as follows:

trace([1, 0, 0]′h′ − D_{0}^{−1}D_{1})

= α − \frac{θ}{1 − θ} + \{γ_{k}(b(ϕ_{cx} + ϕ_{ck′}γ_{x}) − c(ϕ_{cl} + ϕ_{ck′}γ_{l}))

− γ_{l}(a(ϕ_{cx} + ϕ_{ck′}γ_{x}) − c(ϕ_{ck} + ϕ_{ck′}γ_{k} − θ))

− γ_{x}(b(ϕ_{ck} + ϕ_{ck′}γ_{k} − θ) − a(ϕ_{cl} + ϕ_{ck′}γ_{l}))}/|D_{0}|

= α − \frac{θ}{1 − θ} + \{−γ_{k}cϕ_{cl} + γ_{l}c(ϕ_{ck} − θ)

− γ_{x}(b(ϕ_{ck} − θ) − aϕ_{cl})}/|D_{0}|

= α − \frac{θ}{1 − θ} + \{−γ_{k}ϕ_{lk′}γ_{x}ϕ_{cl} + γ_{l}ϕ_{lk′}γ_{x}(ϕ_{ck} − θ)

− γ_{x}(ϕ_{ll} + ϕ_{lk′}γ_{l})(ϕ_{ck} − θ) + γ_{x}(ϕ_{lk} + ϕ_{lk′}γ_{k})ϕ_{cl})}/|D_{0}|

= α − \frac{θ}{1 − θ} + γ_{x}\{ϕ_{ll}θ − ϕ_{ll}ϕ_{ck} + ϕ_{lk}ϕ_{cl}\}/|D_{0}|

= α − \frac{θ}{1 − θ} + \frac{ϕ_{ll}θ − ϕ_{ll}ϕ_{ck} + ϕ_{lk}ϕ_{cl}}{(1 − θ)(ϕ_{ll}ϕ_{ck′} − ϕ_{cl}ϕ_{lk′}}

= α + \frac{1 − δ}{z(1 + g_{n})} + (\frac{θ}{1 − θ} ) \frac{ĝ}{k̂(1 + g_{n})}

where I am using ϕ_{cl} = (1 − θ)ϕ_{ll}ŷ/ĉ, ϕ_{ck′} = (1 − θ)ϕ_{lk′} ŷ/ĉ −(1 + g_{n})k̂/ĉ, and ϕ_{ck} = (1 − θ)ϕ_{lk}ŷ/ĉ +θŷ/ĉ +(1 − δ)k̂/(zĉ) to get the final result. The mathematics for the other steps of the proof are the same as in the case where investment-output is the third variable in the VAR.

I also considered a government spending shock with the third variable being the consumption-output ratio. The maximal eigenvalue of M is not the same as in the case of the investment tax shock because ϕ_{lg} and ϕ_{cg} is not zero. In fact the expression for the eigenvalue is messy and uninterpretable. I evaluated it numerically and found it large for reasonable parameters. This means that the garbling will be significant no matter what shock is included.

4.4. Adding Variables with Little Effect on Fluctuations

Conventional wisdom says that researchers interested in the effects of a particular shock, say technology, can ignore specifying anything about the other shocks. In this section, I use an example from above—the 3-variable example with investment-to-output in the VAR and a shock to the investment tax—to illustrate why one cannot ignore the other shocks even if they have a small effect on the variances of the observables.

Consider two economies. The first has a capital decision function with γ_{x} as the coefficient on τ_{xt} and a variance for the innovation equal to η_{x}. The second has γ̃_{x} and η̃_{x} with η̃_{x} chosen so that the following holds: γ_{x}η_{x} = γ̃_{x}η̃_{x}. Compute variance decompositions for the system

S_{t+1} = A_{S}S_{t} + B_{S}η_{t}
X_{t} = C_{S}S_{t}

with state vector S_{t} = [log k̂_{t}, log z_{t}, τ_{lt}, τ_{xt}, log k̂_{t−1}, log z_{t−1}, τ_{lt−1}, τ_{xt−1}, 1]′ and innovations η_{t} = [0, η_{zt}, η_{lt}, η_{xt}, 0, 0, 0, 0, 0]′ for the two economies. Also compute impulse responses applying the Blanchard-Quah methodology. Call these impulse responses the “empirical” impulse responses.

Proposition 5. The variance decompositions and empirical impulse responses are the same for the economy with γ_{x} and η_{x} as those for the economy with γ̃_{x} and η̃_{x}.

Proof of Proposition 5. The two economies have the same variances and covariances for the observed variables. That implies that the variance decomposition calculations are the same in the two economies. It also implies that the variance covariance of errors in a VAR of X on lags is the same. Thus, to prove the proposition, I need only show that the C matrices (and hence M) are not affected by the size of γ_{x}. To do this, I can write out the elements of D_{0}P D_{0}^{−1}, D_{1}D_{0}^{−1}, and D_{2}D_{0}^{−1} and show that they do not depend on γ_{x}. This is simple algebra and left for the reader.

Proposition 5 contradicts the view that “other shocks don’t matter.” In the example above, it could be that τ_{xt} contributes little to the total variation in X_{t}. But, if the eigenvalues in M are near 1, the garbled term in the expression for C̄ (see (3.25)) will still be large. This is what happens with shocks like τ_{xt} and log ĝ_{t}. They are not very important in accounting for business cycle fluctuations, but their variances are nonnegligible. Thus, they garble the signal of the other shocks.

There is also another point that should be made here. The other shocks may not be the “deep structural” shocks that researchers desire. In other words, likelihood ratio tests would reject the hypothesis that our shocks are uncorrelated. This is (maybe) a more important reason for not ignoring the other shocks.

5. VARs and 4-Shock Versions of the Model

In this section I consider a VAR with 4 variables and 4 shocks. The four variables are the change in the log of labor productivity, quasi-differenced logged hours, the log of the investment-output ratio, and the log of the consumption-output ratio.

Proposition 6. The model just described has a VAR representation with coefficients B_{j} that satisfy

B_{j} = MB_{j−1}

for j ≥ 2, with B_{1} = C_{1} = (D_{0}P + D_{1})D_{0}^{−1}. The matrix M is 4×4 with eigenvalues equal to 0, 0, α, and (1 − δ)/[z(1 + g_{n})].

Proof of Proposition 6. The first part of the proof, showing that the VAR has an infinite number of lags, follows exactly as in Proposition 1. The second part, which involves derivation of eigenvalues of M, is almost exactly as in Proposition 4. In this case, however, I need to work with different D matrices. The matrices are given by

D_{0} = [ \begin{matrix}1 − θ + θa & −θb & −θc & −θd \\ −a & b & c & d \\ −e & f & g & h \\ −i & j & k & l \end{matrix} ]  (5.1)

D_{1} = [ \begin{matrix}θ(1 − a)(1 − γ_{k}) & θ(b + (1 − a)γ_{l}) & θ(c + (1 − a)γ_{x}) & θ(d + (1 − a)γ_{g}) \\ (α − γ_{k})a & −αb + γ_{l}a & −αc + γ_{x}a & −αd + γ_{g}a \\ −eγ_{k} & eγ_{l} & eγ_{x} & eγ_{g} \\ −iγ_{k} & iγ_{l} & iγ_{x} & iγ_{g} \end{matrix} ](5.2)

D_{2} = [ \begin{matrix}θ(1 − a)(1 − γ_{k}) \\ (α − γ_{k})a \\ −eγ_{k} \\ −iγ_{k} \end{matrix} ] [ γ_{k} −γ_{l} −γ_{x} −γ_{g}] ≡ mn′  (5.3)

and D_{j} = γ_{k}D_{j−1} for j ≥ 3 where

a = ϕ_{lk} + ϕ_{lk′}γ_{k}
b = ϕ_{ll} + ϕ_{lk′}γ_{l}
c = ϕ_{lk′}γ_{x}
d = ϕ_{lg} + ϕ_{lk′}γ_{g}
e = ϕ_{xk} + ϕ_{xk′}γ_{k} − θ − (1 − θ)a
f = ϕ_{xk′}γ_{l} − (1 − θ)b
g = ϕ_{xk′}γ_{x} − (1 − θ)c
h = ϕ_{xk′}γ_{g} − (1 − θ)d
i = ϕ_{ck} + ϕ_{ck′}γ_{k} − θ − (1 − θ)a
j = ϕ_{cl} + ϕ_{ck′}γ_{l} − (1 − θ)b
k = ϕ_{ck′}γ_{x} − (1 − θ)c
l = ϕ_{cg} + ϕ_{ck′}γ_{g} − (1 − θ)d

Note that ϕ_{lz} = −ϕ_{lk}, ϕ_{xz} = −ϕ_{xk}, ϕ_{cz} = −ϕ_{ck}, and γ_{z} = −γ_{k} hold in the model economy with a unit root in technology. Note also that D_{2} is singular for all parameterizations, and D_{1} is singular if α = 0.

Using the same proof as in Proposition 1 or Proposition 4, it is easy to show that the eigenvalues of M are equal to the eigenvalues of

M = [1, 0, 0, 0]′n′ − D_{0}^{−1}D_{1}.

Therefore, to prove Proposition 6, I need to prove that |M − αI| = 0, |M| = 0, |M − (1 − δ)/(z(1 + g_{n}))I| = 0, and trace(M) = α + (1 − δ)/(z(1 + g_{n})).

Let me start with the inverse D_{0}^{−1}:

\frac{1}{|D_{0}|} [ \begin{matrix}|D_{0}|/(1 − θ) & θ|D_{0}|/(1 − θ) & 0 & 0 \\ 0 & (1 − θ)(gl − hk) & 0 & (1 − θ)(ch − dg) \\ {a(hj−fl)+b(el−hi) \atop +d(fi−ej)} & {(1−θ)(hj−fl) \atop +θ\{a(hj−fl)+b(el−hi)+d(fi−ej)\}} & (1 − θ)(bl − dj) & (1 − θ)(df − bh) \\ {a(fk−gj)+b(gi−ek) \atop +c(ej−fi)} & {(1−θ)(fk−gj) \atop +θ\{a(fk−gj)+b(gi−ek)+c(ej−fi)\}} & (1 − θ)(cj − bk) & (1 − θ)(bg − cf) \end{matrix} ]

where |D_{0}| = (1 − θ)(bgl − bhk + chj − dgj). In calculating this inverse, I use the following facts:

0 = a(gl − hk) + c(hi − el) + d(ek − gi)
0 = a(ch − dg) + c(ed − ah) + d(ag − ce)
0 = a(bl − dj) + b(di − al) + d(aj − bi)
0 = a(bh − df) + b(de − ah) + d(af − be)
0 = cl − dk.

Using D_{0}^{−1}, I have then that the elements of M = [1, 0, 0, 0]′n′ − D_{0}^{−1}D_{1} are

M_{1,1} = γ_{k} − θ(1 − γ_{k} − a + αa)/(1 − θ)
M_{1,2} = −γ_{l} − θ(γ_{l} + b − bα)/(1 − θ)
M_{1,3} = −γ_{x} − θ(γ_{x} + c − cα)/(1 − θ)
M_{1,4} = −γ_{g} − θ(γ_{g} + d − dα)/(1 − θ)
M_{2,1} = (1 − θ){(hk − gl)(aα − γ_{k}a) + (ch − dg)iγ_{k}}/|D_{0}|
M_{2,2} = (1 − θ){(hk − gl)(−αb + γ_{l}a) − (ch − dg)iγ_{l}}/|D_{0}|
M_{2,3} = (1 − θ){(hk − gl)(−αc + γ_{x}a) − (ch − dg)iγ_{x}}/|D_{0}|
M_{2,4} = (1 − θ){(hk − gl)(−αd + γ_{g}a) − (ch − dg)iγ_{g}}/|D_{0}|
M_{3,1} = {[a(hj − fl) + b(el − hi) + d(fi − ej)][γ_{k} − θ(1 − a + aα)] − a(hj − fl)(1 − θ)α}/|D_{0}|
M_{3,2} = {[a(hj − fl) + b(el − hi) + d(fi − ej)][−γ_{l} − θb + θbα)] + b(hj − fl)(1 − θ)α}/|D_{0}|
M_{3,3} = {[a(hj − fl) + b(el − hi) + d(fi − ej)][−γ_{x} − θc + θcα)] + c(hj − fl)(1 − θ)α}/|D_{0}|
M_{3,4} = {[a(hj − fl) + b(el − hi) + d(fi − ej)][−γ_{g} − θd + θdα)] + d(hj − fl)(1 − θ)α}/|D_{0}|
M_{4,1} = {[a(fk − gj) + b(gi − ek) + c(ej − fi)][γ_{k} − θ(1 − a + aα)] − a(fk − gj)(1 − θ)α}/|D_{0}|
M_{4,2} = {[a(fk − gj) + b(gi − ek) + c(ej − fi)][−γ_{l} − θb + θbα)] + b(fk − gj)(1 − θ)α}/|D_{0}|
M_{4,3} = {[a(fk − gj) + b(gi − ek) + c(ej − fi)][−γ_{x} − θc + θcα)] + c(fk − gj)(1 − θ)α}/|D_{0}|
M_{4,4} = {[a(fk − gj) + b(gi − ek) + c(ej − fi)][−γ_{g} − θd + θdα)] + d(fk − gj)(1 − θ)α}/|D_{0}|

Starting with the trace of M multiplied by |D_{0}| (to simplify the algebra), I find

trace(M)|D_{0}|

= [γ_{k} − θ(1 − a + αa)]|D_{0}|/(1 − θ)

+ (1 − θ){(hk − gl)(−αb + γ_{l}a) − (ch − dg)iγ_{l}}

+ {[a(hj − fl) + b(el − hi) + d(fi − ej)][−γ_{x} − θc + θcα)] + c(hj − fl)(1 − θ)α}

+ {[a(fk − gj) + b(gi − ek) + c(ej − fi)][−γ_{g} − θd + θdα)] + d(fk − gj)(1 − θ)α} = [γ_{k} − θ(1 − a + αa)][bgl − bhk + chj − dgj]

+ (1 − θ){(hk − gl)(−αb + γ_{l}a) − (ch − dg)iγ_{l}} + {[a(hj − fl) + b(el − hi) + d(fi − ej)][−γ_{x} − θc + θcα)] + c(hj − fl)(1 − θ)α} + {[a(fk − gj) + b(gi − ek) + c(ej − fi)][−γ_{g} − θd + θdα)] + d(fk − gj)(1 − θ)α}

= [bgl − bhk + chj − dgj]γ_{k} + [bgl − bhk + chj − dgj]α(1 − θ) − [bgl − bhk + chj − dgj]θ + [a(hk − gl) − i(ch − dg)](1 − θ)γ_{l} − [a(hj − fl) + b(el − hi) + d(fi − ej)]γ_{x} − [a(fk − gj) + b(gi − ek) + c(ej − fi)]γ_{g}

= [bgl − bhk + chj − dgj]γ_{k} [a(hk − gl) − i(ch − dg)](1 − θ)γ_{l} − [a(hj − fl) + b(el − hi) + d(fi − ej)]γ_{x} − [a(fk − gj) + b(gi − ek) + c(ej − fi)]γ_{g} + (α − θ/(1 − θ))|D_{0}|

= \{(\frac{θ − ϕ_{xk}}{ϕ_{xk′}(1 − θ)}) + (α − \frac{θ}{1 − θ} )\} |D_{0}|

where the last step involves writing out all terms and cancelling.

Next, I need to show that the determinant of M is 0. The proof of this is simple because it follows immediately from the fact that the last two rows of D_{1} are linearly dependent and proportional to n′. No matter what is the magnitude of the elements of D_{0}^{−1} or the first two rows of D_{1}, the determinant must be zero. (Formally, the determinant of the matrix AB + Cn′ is 0 where A, B, and C are arbitrary matrices of dimension 4 × 2, 2 × 4, and 4 × 1, respectively.)

Finally, I need to show that |M − αI| = 0. I do so using the fact that |M| = 0 as follows:

det(M − αI)

= |M| − α{(M_{2,2} − α)[(M_{3,3} − α)(M_{4,4} − α) − M_{3,4}M_{4,3}] − M_{2,3}[M_{3,2}(M_{4,4} − α) − M_{3,4}M_{4,2}] + M_{2,4}[M_{3,2}M_{4,3} − M_{4,2}(M_{3,3} − α)]}

+ M_{1,1}((M_{2,2} − α)(α^{2} − αM_{3,3} − αM_{4,4}) + αM_{2,3}M_{3,2} + αM_{2,4}M_{4,2}) − M_{1,2}(M_{2,1}(α^{2} − αM_{3,3} − αM_{4,4}) + αM_{2,3}M_{3,1} + αM_{2,4}M_{4,1}) + M_{1,3}(−αM_{2,1}M_{3,2} + α(M_{2,2} − α)M_{3,1}) − M_{1,4}(αM_{2,1}M_{4,2} − α(M_{2,2} − α)M_{4,1})

= −α{α^{2} − αM_{2,2}M_{3,3} − αM_{2,2}M_{4,4} + M_{2,2}M_{3,3}M_{4,4} − M_{2,2}M_{3,4}M_{4,3}

− α^{3} + α^{2}M_{3,3} + α^{2}M_{4,4} − αM_{3,3}M_{4,4} + αM_{3,4}M_{4,3} − M_{2,3}M_{3,2}M_{4,4} + αM_{2,3}M_{3,2} + M_{2,3}M_{3,4}M_{4,2} + M_{2,4}M_{3,2}M_{4,3} − M_{2,4}M_{4,2}M_{3,3} + αM_{2,4}M_{4,2}

− α^{2}M_{1,1} + αM_{1,1}M_{2,2} + αM_{1,1}M_{3,3} + αM_{1,1}M_{4,4} − M_{1,1}M_{2,2}M_{3,3}

− M_{1,1}M_{2,2}M_{4,4} + M_{1,1}M_{2,3}M_{3,2} + M_{1,1}M_{2,4}M_{4,2}

− αM_{1,2}M_{2,1} + M_{1,2}M_{2,1}M_{3,3} + M_{1,2}M_{2,1}M_{4,4}

− M_{1,2}M_{2,3}M_{3,1} − M_{1,2}M_{2,4}M_{4,1}

− αM_{1,3}M_{3,1} − M_{1,3}M_{2,1}M_{3,2} + M_{1,3}M_{2,2}M_{3,1} − αM_{1,4}M_{4,1} − M_{1,4}M_{2,1}M_{4,2} + M_{1,4}M_{2,2}M_{4,1}}

= −α{−α^{3} + α^{2} − α^{2}M_{1,1} + α^{2}M_{3,3} + α^{2}M_{4,4} + αM_{1,1}M_{2,2} + αM_{1,1}M_{3,3} + αM_{1,1}M_{4,4} − αM_{2,2}M_{3,3} − αM_{2,2}M_{4,4} − αM_{3,3}M_{4,4} − αM_{1,2}M_{2,1} − αM_{1,3}M_{3,1} − αM_{1,4}M_{4,1} + αM_{2,4}M_{4,2} + αM_{2,3}M_{3,2} + αM_{3,4}M_{4,3} − M_{1,1}M_{2,2}M_{3,3} − M_{1,1}M_{2,2}M_{4,4} + M_{2,2}M_{3,3}M_{4,4} − M_{2,2}M_{3,4}M_{4,3} + M_{1,1}M_{2,3}M_{3,2} + M_{1,1}M_{2,4}M_{4,2} + M_{1,2}M_{2,1}M_{3,3} + M_{1,2}M_{2,1}M_{4,4} − M_{1,2}M_{2,3}M_{3,1} − M_{1,2}M_{2,4}M_{4,1} − M_{1,3}M_{3,2}M_{2,1} + M_{1,3}M_{2,2}M_{3,1} − M_{1,4}M_{4,2}M_{2,1} + M_{1,4}M_{2,2}M_{4,1} − M_{2,3}M_{3,2}M_{4,4} + M_{2,3}M_{3,4}M_{4,2} + M_{2,4}M_{3,2}M_{4,3} − M_{2,4}M_{4,2}M_{3,3}}

= 0