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Solution to Homework 7
Problem 1:
Consider the following simple Keynesian macroeconomic model of the U.S. economy.
Yt = Ct + It + Gt + NXt
Ct = β0 + β1YDt + β2Ct-1 + ε1t
YDt = Yt – Tt
It = β3 + β4Yt + β5rt-1 + ε2t
rt = β6 + β7Yt + β8Mt + ε3t
where:
Yt = gross domestic product (GDP) in year t
Ct = total personal consumption in year t
It = total gross private domestic investment in year t
Gt = government purchases of goods and services in year t
NXt = net exports of goods and services (exports - imports) in year t
Tt = taxes in year t
rt = the interest rate in year t
Mt = the money supply in year t
YDt = disposable income in year t
The endogenous variables are Yt, Ct, It, YDt, and rt. The exogenous and predetermined variables
are Gt, NXt, Ct-1, Tt, rt-1, and Mt. Find the reduced form equations for this model.
Answer: Sub the second equation into the first:
Yt = (β0 + β1YDt + β2Ct-1 + ε1t) + It + Gt + NXt
Now sub for It and YDt :
Yt = (β0 + β1 (Yt – Tt) + β2Ct-1 + ε1t) + (β3 + β4Yt + β5rt-1 + ε2t) + Gt + NXt
Now solve for Yt :
Yt = [1/(1-β1-β4)]{(β0 + β3) - β1Tt + β2Ct-1 + β5rt-1 + Gt + NXt + ε1t + ε2t}
And, in the form that would be estimated:
Yt = λ0 + λ 1Tt + λ2Ct-1 + λ3rt-1 + λ4Gt + λ5NXt + εyt
Now substitute this solution into the other equations to find the rest of the reduced form
relationships. Consumption first. Sub for disposable income, then income:
Ct = β0 + β1(Yt – Tt)+ β2Ct-1 + ε1t
Ct = β0 + β1[(λ0 + λ 1Tt + λ2Ct-1 + λ3rt-1 + λ4Gt + λ5NXt + εyt ) – Tt] + β2Ct-1 + ε1t
Ct = (β0+β1λ0) + β1( λ 1-1) Tt + (β1λ2+β2)Ct-1 + β1λ3rt-1 + β1λ4Gt + β1λ5NXt + (β1εyt + ε1t)
Next, YDt:
YDt = Yt - Tt = Yt = λ0 + λ 1Tt + λ2Ct-1 + λ3rt-1 + λ4Gt + λ5NXt + εyt - Tt
YDt = = λ0 + (λ 1-1)Tt + λ2Ct-1 + λ3rt-1 + λ4Gt + λ5NXt + εyt
Investment is next on the list:
It = β3 + β4Yt + β5rt-1 + ε2t
It = β3 + β4(λ0 + λ 1Tt + λ2Ct-1 + λ3rt-1 + λ4Gt + λ5NXt + εyt) + β5rt-1 + ε2t
It = (β3 + β4λ0) + β4λ 1Tt + β4λ2Ct-1 + (β4λ3+ β5)rt-1 + β4λ4Gt + β4λ5NXt + (β4εyt + ε2t)
And finally, the interest rate:
rt = (β6 + β7λ0) + β7λ 1Tt + β7λ2Ct-1 + β7λ3rt-1 + β7λ4Gt + β7λ5NXt + β8Mt + (β7εyt + ε3t)
Problem 2:
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