James Duesenberry (1958) provided a second contribution to the Hicksian growth cycle without ratcheting consumption but merely by modifying the accelerator. Assuming all autonomous terms fall out, Duesenberry (1958) proposed investment to be:
It = b Yt-1 - d Kt-1
where, note, investment is now a function of the past level of income via parameter b and of the past level of capital via d (which is positive and constant).
The relationship with income can be considered akin to the "accelerator". For the relationship between investment and past capital (assumed negative), one can appeal to depreciation or, alternatively, that the greater the stock of capital, the lower its marginal product - implicating a collapse in MEI and thus investment at any point in time. This is not inconsistent with Tobin's (1969) "q-theory" of investment and is, in fact, almost identical to Kaldor's (1940) formulation of an investment function: as investment is positively related to past levels of Y and negatively related to past levels of K, it is more Kaldorian than Keynesian in spirit. Thus, it breaks substantially from our previous Hicks-Samuelson type of multiplier-accelerator cycle models - but nonetheless can still be referred to the same family as it does employ something akin to those relationships. We should also note that the investment ratchet model of Smithies (1957) is similar in spirit to this model. Finally, we should note that Duesenberry's accelerator remains linear. A richer, non-linear accelerator was introduced by Richard Goodwin (1951) and will be reviewed later.
Let us first set up the possibility of fluctuations in the Duesenberry model. Now, keeping a conventional (non-ratcheted) consumption function (also with no autonomous terms), then in goods market equilibrium Yt = Ct + It, so:
Yt = cYt-1 + b Yt-1 - d Kt-1
= (c+b )Yt-1 - d Kt-1
Since, by definition, It = Kt - Kt-1, then:
Kt - Kt-1 = b Yt-1 - d Kt-1
Kt = Kt-1 + b Yt-1 - d Kt-1
We can easily determine the system. Firstly, we need to find Kt-1. We do so by lagging the second equation so as to obtain:
Kt-1 = b Yt-2 + (1-d )Kt-2
Now, we need the term Kt-2. This we find by lagging the first equation:
Yt-1 = (c+b )Yt-2 - d Kt-2
and then solving for Kt-2:
Kt-2 = -(1/d )Yt-1 + ((c+b )/d )Yt-2
and plugging into our previous equation:
Kt-1 = b Yt-2 + (1-d )[-(1/d )Yt-1 + ((c+b )/d )Yt-2]
Kt-1 = - [(1-d )/d ]Yt-1 + [(1-d )(c+b )/d + b ]Yt-2
Plugging this term into our first equation:
Yt = (c+b )Yt-1 - d [-[(1-d )/d ]Yt-1 + [(1-d )(c+b )/d + b ]Yt-2
which, rearranging, yields simply:
Yt - [c + b + 1 - d ]Yt-1 - [c + b - d c]Yt-2 = 0
Note that it is a homogeneous system (thus equilibrium = 0). The characteristic equation is:
r2 - (c + b + 1 - d )r - (c + b - d c) = 0
thus the values of the roots can be determined as follows:
r1, r2 = [(c + b + 1 - d ) ｱ ﾖ [(c + b + 1 - d )2 + 4(c + b - d c)]]/2
Now, again, we obtain monotonicity if the discriminant D > 0 or (c + b + 1 - d )2 > 4(c + b - d c) and explosive or stability if the roots turn out to be greater or less than modulus 1 is value. The crucial parameters, then, are c, b and d - which create a series of range of values to obtain all the six standard cases. For monotonic stability, the Schur criterion requires that:
(i) 1 + (c + b + 1 - d ) - (c + b - d c) > 0
(ii) 1 - (c + b + 1 - d ) - (c + b - d c) > 0
(iii)1 + (c + b - d c) > 0
The first condition (i) reduces to 2 - (1 - c)d > 0 as (1-c) > 0 whereas d > 0, thus this condition may or may not be fulfilled. The second Schur condition (ii) reduces to (1+c)d > 2(c+b ) which may or may not be fulfilled. Finally the third (iii) Schur condition may not be fulfilled. Thus, it is entirely conceivable that we obtain oscillations as in the Hicks model.
So much for fluctuations. What about growth? Well, recalling our goods market equilibrium:
Yt = cYt-1 + b Yt-1 - d Kt-1
then, subtracting Yt-1 from both sides:
Yt - Yt-1 = cYt-1 + b Yt-1 - d Kt-1 - Yt-1
dividing then through by Yt-1, we obtain:
(Yt - Yt-1)/Yt-1 = c + b - d (Kt-1/Yt-1) - 1
Let us define gY as the growth rate of output, gY = (Yt - Yt-1)/(Yt-1) and let us define v as the capital-output ratio, v = Kt-1/Yt-1, then this can be rewritten as:
gY = c + b - dv - 1
Thus the growth rate of output is some negative linear function of v. This can be shown in (g, v) space in Figure 1 as the downward sloping straight line.
Now, recall that investment by definition is It = Kt - Kt-1. Thus, we can write that:
Kt - Kt-1 = b Yt-1 - d Kt-1
dividing through by Kt-1:
(Kt - Kt-1)/Kt-1 = b (Yt-1/Kt-1) - d
so, defining rate of capital growth, gK = (Kt - Kt-1)/Kt-1 and recalling that Yt-1/Kt-1 = 1/v, then we have:
gK = b /v - d
Where, we can easily note that gK is an non-linear function of the capital-output ratio. This is shown in Figure 1 as the curved function. The equilibrium is defined, of course, where gK = gY which, in Figure 1, is at two points: A and B.
Fig. 1 - Duesenberry growth rates
Duesenberry (1958) shows that A is stable whereas B is an unstable equilibrium, as denoted by directional arrows along the horizontal axis in Figure 1. Examining the figure, we see that at the capital-output ratios, left of vA, capital growth is greater than output growth (gK > gY), thus v = K/Y will grow. To the right of vA, capital growth is less than output growth (gK < gY), thus, over time, v will fall, thus we move to the left. However, to the right of vB, again capital growth is greater then output growth, thus v grows and we move to the right. Thus, the dynamics of this simple multiplier-accelerator system is that A is the stable steady-state equilibrium whereas B is an unstable steady-state equilibrium.
However, Luigi Pasinetti (1960) was to show that, indeed, Duesenberry makes a little mistake. Consider that at steady-state equilibrium, gY = gK, so:
c + b - d v - 1 = b /v - d
thus, multiplying through by v and rearranging:
d v2 + v - d v - cv - b v + b = 0
so dividing through by d and rearranging:
v2 + (1/d - 1 - c/d - b /d )v + b /d = 0
thus, by this quadratic equation we obtain two solutions for v. However, Pasinetti subsequently shows that one of the solutions to the system (the greater v) always yields a negative gY and gK. In short, that the proper diagram should be as in Figure 2 below.
Figure 2 - Pasinett's correction
Thus, the unstable solution B is always at negative growth. Thus, the only positive solution (A) is stable. However, this does not exclude the case where the curves gK and gY are at a tangency to each other at point B (drawn by the superimposed gK｢ , the dotted line in the figure above) where the curves gK and gY are at a tangency at point B - in which case the equilibrium would be stable to the left and unstable to the right. Alternatively, as Pasinetti (1960) points out, we cannot rule out a priori the possibility of no equilibrium altogether (i.e. no intersection or even tangency between gK and gY). Nonetheless, the precise value of equilibrium growth of output and equilibrium capital growth depends on lag structure and the particular values of the parameters c, b and d.