CARDIFF BUSINESS SCHOOL
MACROECONOMICS (BS1652)
Spring Semester 2005-06
Maths Macro Lecture 2
Simultaneous equation systems in macroeconomic analysis
I.
The ‘Keynesian Cross’ model
In a model of the macro-economy, one will normally find both exogenous variables
and endogenous variables. The ‘Keynesian Cross’ model is no exception. The model
comprises two basic equations – a national income identity (or equilibrium condition)
and a consumption function. The precise nature of the equations making up this model
depends upon the number of sectors which the economy is considered to have – all
economies must have a minimum of two sectors (households and firms), but an
economy can also comprise a government sector (government expenditure and
taxation) and a trade sector (exports and imports). Assuming that the economy is a
closed one (i.e. no trade) and has no government sector (i.e. no government
expenditure or taxation), then the model is as follows:
(i)
y = c + i
(equilibrium condition)
(ii)
c = a + b. y
(consumption function)
where y represents national income, c is consumption and i is the exogenous
level of investment expenditure, while a and b are constants of the consumption
function ( a > 0 and 0 < b < 1 ).
Since the level of investment expenditure is determined outside of the model (it must
be since there is no equation within the model which explains its level), the variable i
is said to be exogenous (or autonomous). [Similarly, the values of the constants in the
consumption function, i.e. a and b , are also exogenous since they are determined
‘outside’ of the model.]
The ‘Keynesian Cross’ model above comprises only two endogenous variables: y
and c . Note that y depends on c (from the equilibrium condition) but also that c
depends on y (from the consumption function). Hence, if either of c or y changes,
then so too will the other variable (but this, of course, is not the case for the
exogenous variable i , which remains at its pre-determined value – changes in the
value of the endogenous variables can not cause changes in the values of the
exogenous variables). Note, however, that if either the exogenous variable i , or the
constant terms in the consumption function, were to experience a change in their
value (due to an ‘outside’ or exogenous shock), then the values of both c and y will
change, that is, changes in the exogenous values can induce changes in the
equilibrium values of the endogenous variables.
The ‘reduced form’ of a model
The ‘reduced form’ of a model is a series of equations which express the equilibrium
values of the endogenous variables in a model as functions simply of the exogenous
variables/constants. The number of ‘reduced form’ equations will therefore
correspond to the number of the endogenous variables in the model. In the ‘Keynesian
Cross’ model depicted above, there will be two ‘reduced form’ equations, one for c
and one for y .
To obtain the ‘reduced form’ equation for y , substitute for c in equation (i) :
y = a + b. y + i
y - b. y = a + i
y(1- b) = a + i
y = (a + i)/(1- b)
To obtain the ‘reduced form’ equation for c , substitute for y in equation (ii) :
c = a + b. ( c + i )
c - b. c = a + b. i
c ( 1 - b ) = a + b. i
c = ( a + b. i ) / ( 1 - b )
How can we be sure that the equations we have determined are actually the ‘reduced
form’ equations?
Well, it can be seen from looking at the right hand sides of each equation that they
only contain pre-determined values, that is, either the value of the exogenous variable
i and/or the pre-determined constants ( a and b ) from the consumption function.
Thus, if we knew the values of i , a and b , then by substituting these into the
‘reduced form’ equations, we would be able to determine the equilibrium solution
values of c and y . For example, if i = 200 , and a and b are respectively 50
and 0.8, then the equilibrium values for c and y are:
y = ( 50 + 200 ) / ( 1 - 0.8 ) = 250 / 0.2 = 1250
c = [ 50 + (0.8) (200) ] / ( 1 - 0.8 ) = 210 / 0.2 = 1050
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[Note that, since i = 200 , c + i = 1050 + 200 = 1250 , which is the value of y.
Hence the answers are consistent and you have an in-built check to see whether or not
you have worked out the answer correctly.]
Note, however, that we got definitive answers for the values of y and c because it
was assumed that i was exogenous (and hence its level was known). But what if the
level of i depended upon a variable such as the interest rate r ? In this situation the
model becomes:
(i)
y = c + i
(equilibrium condition)
(ii)
c = a + b. y
(consumption function)
(iii)
i = i0 - h. r
(investment function)
where i0 is the exogenous component of investment expenditure, h > 0 , and r is
determined exogenously.
To obtain the ‘reduced form’ equation for y , substitute for c and i in equation (i) :
y = a + b. y + i0 - h. r
y - b. y = a + i0 - h. r
y ( 1 - b ) = a + i0 - h. r
y =
( a + i0 ) _
h_ .r
(1- b)
(1- b)
(A)
To obtain the ‘reduced form’ equation for c , substitute for y and i in equation (ii) :
c = a + b. ( c + i )
c - b. c = a + b. i
c ( 1 - b ) = a + b. ( i0 - h. r )
c = ( a + b. i0 ) _
bh . r
(1- b)
(1- b)
Clearly the solution values for both y and c depend upon the value of r , the rate of
interest. The other important thing to note is that equation (A) , which is derived from
the equilibrium condition (i) , provides us with a relationship between y and r
which describes equilibrium in the real sector of the economy. Equation (A) is also
known as the IS curve, and forms a key part of the IS/LM model (see notes for
Maths Macro Lecture 3).
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Turning equation (A) around to express r as a function of y , we get:
y =
( a + i0 ) _
h_ .r
(1- b)
(1- b)
( 1 – b ). Y = ( a + i0 ) - h. r
h. r = ( a + i0 ) - ( 1 – b ). Y
r = ( a + i0 ) - ( 1 – b ). Y
h
h
The slope of the IS curve is therefore given by:
dr/dy = - ( 1 – b ) / h
Diagrammatically, then, we have the following:
r
Slope = - ( 1 – b ) / h
IS
y
Given that the slope of the IS curve is - ( 1 – b ) / h , then, if b (the marginal
propensity to consume) increases, or h (the interest sensitivity of investment)
increases, then the slope of the IS curve will become flatter; if b or h should
decrease, then the slope of the IS curve will become steeper.
Having examined the IS curve, in Maths Macro Lecture 3 we will investigate the LM
curve and the IS/LM model.
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