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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. EC6012 Lecture 5
Stephen Kinsella
Lecture Outline
Notation
The Model
Derivation
Problems
Steady States
Figure: Table 3.4 of Godley/Lavoie.