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EC6012 Lecture 5

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EC6012 Lecture 5

  1. 1. EC6012 Lecture 5 Stephen Kinsella Lecture Outline Notation EC6012 Lecture 5 The Model Numerical Examples Derivation Problems Steady States Stephen Kinsella Dept. Economics, University of Limerick. stephen.kinsella@ul.ie January 20, 2008
  2. 2. EC6012 Lecture 5 Stephen Kinsella Lecture Outline Notation EC6012 Lecture 5 The Model Numerical Examples Derivation Problems Steady States Stephen Kinsella Dept. Economics, University of Limerick. stephen.kinsella@ul.ie January 20, 2008
  3. 3. Outline EC6012 Lecture 5 Stephen Kinsella Lecture Outline Lecture Outline Notation The Model Derivation Notation Problems Steady States The Model Derivation Problems Steady States
  4. 4. EC6012 Lecture 5 Notation Stephen Kinsella Lecture Outline Symbol Meaning Notation G Pure government expenditures in nominal terms The Model Y National Income in Nominal Terms Derivation C Consumption of goods supply by households, in nominal terms Problems T Taxes Steady States θ Personal Income Tax Rate YD Disposable Income of Households α1 Propensity to consume out of regular (present) income α2 Propensity to consume out of past wealth ∆Hs Change in cash money supplied by the central bank ∆Hh Cash money held by households H, H−1 High Powered cash money today, and yesterday (−1 )
  5. 5. The Model EC6012 Lecture 5 Stephen Kinsella Lecture Outline Notation G (1) The Model Y = G +C (2) Derivation T = θ×Y (3) Problems Steady States YD = Y − T (4) C = α1 × YD + α2 × H1 (5) ∆Hs = G −T (6) δHh = YD − C (7) H = ∆H + H−1 (8)
  6. 6. Derivation EC6012 Lecture 5 Stephen Kinsella If we start by solving the model for Y , everything will Lecture Outline become clear. Thus Y = G + C and T = θY , and by Notation substituting in for T and factoring, we get The Model Derivation Problems YD = Y − T (9) Steady States = Y × (1 − θ). (10) By similar logic, C = α1 × YD + α2 × H−1 .
  7. 7. Derivation, continued EC6012 Lecture 5 Stephen Kinsella Since, in period 2, H−1 = 0, we can say that Lecture Outline C = α1 × Y (1 − θ). Substitute this into Y = G + C Notation and we get The Model Derivation Problems Y = G + α1 Y (1 − θ), (11) Steady States Y − α1 (Y )(1 − θ)) = G , (12) Y [1 − α1 × (1 − θ)] = G , (13) G Y = (14) 1 − α1 + α1 θ
  8. 8. Derivation, continued EC6012 Lecture 5 Stephen Kinsella We have numbers for α1 , G [Period1], and θ—0.6, 20, Lecture Outline and 0.2. Plugging these into equation (14), we can Notation calculate Y for period 2. We obtain The Model Derivation 20 Problems Y = = 38.462 38.5. 1 − 0.6 + 0.6 × 0.2 Steady States
  9. 9. EC6012 Lecture 5 As soon as you have solved for Y , you can fill in all the Stephen Kinsella remaining numbers in column 2 including ∆H and therefore H. You now have all the material you need to Lecture Outline solve for Y in period 3 (H−1 = 12.3) and the whole Notation column in period 3. And so on. The Model Derivation The system reaches a steady state when ∆H = 0 and Problems hence YD = C . Steady States
  10. 10. Problems EC6012 Lecture 5 Stephen Kinsella Fill in all the values for column 2 of table 3.4 and show Lecture Outline your workings. Ask me if you get stuck. Notation What happens to this model if θ changes from 20% to The Model 30%? Work out the first period and then give and Derivation economic explanation for the figures you see. Problems Steady States
  11. 11. EC6012 Lecture 5 Steady States Stephen Kinsella G = T∗ = θ × W × N∗ Lecture Outline Notation θ × W × N∗ = θ × Y The Model Derivation G Y∗ = . (15) Problems θ Steady States
  12. 12. Stock-Flow Consistency EC6012 Lecture 5 Stephen Kinsella Lecture Outline Notation C = YD − ∆Hh (16) The Model = α1 × YD + α2 × Hh−1 (17) Derivation δHh = (1 − α1 ) × YD − α2 × Hh−1 (18) Problems Steady States 1 − α1 ∆Hh = α2 × ( × YD − Hh−1 ) (19) α2
  13. 13. Expectations EC6012 Lecture 5 Stephen Kinsella Lecture Outline Cd = α1 × YD e + α2 × Hh−1 . (20) Notation The Model e Derivation ∆Hd = Hd − Hh−1 = YD − Cd . (21) Problems Steady States Hh − Hd = YD − YD e . (22)
  14. 14. Dynamics EC6012 Lecture 5 Stephen Kinsella Lecture Outline G + α2 × H1 Y = . (23) Notation 1 − α1 × (1 − θ) The Model Household’s demand for money is Derivation Problems Steady States Hh = (1 − α1 ) × (1 − θ) × Y + (1 − α2 ) × H−1 . (24)
  15. 15. For Next Week EC6012 Lecture 5 Stephen Kinsella What do you think will happen to the steady state Lecture Outline value(s) of output when θ changes? Why does this Notation happen? Post the answers on your blogs by next The Model Monday. Derivation Read Godley and Lavoie, Chapter 4. Problems Steady States
  16. 16. EC6012 Lecture 5 Stephen Kinsella Lecture Outline Notation The Model Derivation Problems Steady States Figure: Table 3.4 of Godley/Lavoie.

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