This means that inflation remains constant, uh at zero. So downward nominal rigidity and inflation obsessed monetary policy imply inflation is always zero and nominal wages and prices never change. I will normalize the price p to 1. So the model allows two regimes. One a liquidity trap with unemployment over the natural rate, inflation exactly zero and nominal and real interest rates exactly zero. The other with unemployment equal to the natural rate and the real interest rate equal to the corresponding marginal product of capital (net of depreciation).
When not in the liquidity trap, the economy acts like a Ramsey Cass Koopmans totally standard growth model. Employment is kept at an exogenous level not by flexible wages which make unemployment zero, but by a monetary authority which targets a fixed unemployment level. However, the economy can also be in a bad situation in which unemployment is higher than the natural rate (so inflation is zero) even if the nominal and real interest rates are zero. Production is standard 1) Y = (K^alpha)L^(1-alpha) There is no technological progress or population growth. The labor force is constant. Employment (L) is that constant times one minus unemployment. So when the economy is not in the liquidity trap, it is constant corresponding to the natural rate of unemployment and when the economy is in the liquidity trap it can take any positive value less than that natural level of employment.
I will normalize the natural level of employment to be 1 so L is less than or equal to 1. Whether or not output is demand constrained cost minimization implies 2) (K/L) = (alpha/(1-alpha))[w/(r+delta)p]= (alpha/(1-alpha))[w/(r+delta)] This means that the share of capital in GDP is alpha
3) (r+delta)Kt = (alpha)Y = (alpha)(K^alpha)L^(1-alpha)
So Y = (r+delta)Kt/alpha The social budget constraint is
4) Ktdot = (K^alpha)L^(1-alpha) - Ct -deltaKt
Where Ct is consumption at time t. Consumers maximize
5) Integral e^-(Rho t) Ct^(1-sigma)/(1-sigma)
6) Ctdot/Ct = (r-rho)/sigma
NOW a crazy assumption ! Sigma = alpha !!!! This gives the Ramsey Cass Koopmans regime a simple closed form solution with consumption proportional to capital. It is crazy.
All estimates of sigma are greater than one (often much greater than one). I will explore this crazy assumption and then relax it below. OK RCK part. The economy has a steady state with r = rho Cdot = Kdot = 0
7) K/L = K= (alpha/(1-alpha))[w/(rho+delta)]
8) C = Y –(Kt)delta = ((rho+ddelta)/alpha – delta)K
I guess and check that there is a solution with
9) Ct = ((rho+delta)/alpha – delta)Kt
All along this path which goes straight up to the steady state. This occurs if and only if ln(Ct/Kt) is fixed so Ctdot/CT = Ktdot/Kt So check
10) Ctdot/Ct – Ktdot/Kt = (r-rho)/alpha - (r+d)/alpha-delta + ((rho+d)/alpha – delta)=0
Yes this is a solution for everything which happens from 0 on as a function of K0. More generally, the RCK natural unemployment rate growth can start at any time T with all variables a function of capital at time T. Well that was pointless. Now how about dynamics during the bad times when the economy is in the liquidity trap ? Here r = 0 so everything is fairly simple.
11) Ctdot/Ct =-rho/sigma = -rho/alpha
Initial capital K0 is given. I am going to treat initial consumption C0 as given, because the equations are much simpler. I will be able in principal to solve for it at the end. So
12) Ct = C0e^(-t)rho
There are going to be no surprises, so this must be true in the liquidity trap. This is true whether or not there is fiscal stimulus in the form of wasteful government spending. Also note that employment is determined by labor demand at the rigid wage and price. This means that more capital implies more employment and the same capital employment ratio. Because of the excess supply of labor, output is proportional to capital. First assume Gt = 0 for all t.
13) (r+delta)Kt = (delta)Kt = alphaYt
14) Ktdot = (delta/alpha)Kt – (delta)Kt – Ct
There is a balanced shrinking case with C/K fixed so long as (delta/alpha) – (delta) – Ct/Kt = -rho/alpha so balanced if
15) C/K = (delta+rho)/alpha)-delta
Oh look that is exactly the same ratio as in the growth with natural unemployment case ! (this algebra surprised me honestly I hadn’t noticed that the sigma = alpha stunt means that r doesn’t matter for balanced growth C/K). This means that the economy can switch from growth with unemployment at the natural rate and positive r to shrinking with r = 0 without consumption jumping.
This means that such switches can occur in equilibrium. They can be predictable from time 0 or can depend on an unpredictable sunspot. Equilibrium with market clearing or downward nominal rigidity requires that C/K = (delta+rho)/alpha)-delta , but the economy can move up or down that path unpredictably or predictably. Well that’s silly. The problem is that this “result” really depends on the crazy assumption that sigma = alpha. I assumed that because it made C/K constant in the RCK equilibrium. The alternative is consumption is some function of K, but I don’t know what function and sure can’t find another case with a closed form solution. However the data are firm that Sigma > 1 > alpha. Also the “result” that C/K is fixed doesn’t fit the facts. Consumption moves less than investment but more than the capital stock.
Now the crazy example is useful in characterizing the less crazy case with sigma>alpha. Balanced shrinking doesn’t change much. Now C and K shrink at rate –rho/sigma (that was easy). This means 15 becomes
16) C/K = (delta/alpha)-delta + rho/sigma
But dyanmics with positive r and fixed L change a lot. In a graph of C on K, the saddle path must lie above the line with slope (delta+rho)/alpha)-delta from the origin to the steady state. Along that line, Kdot/K > Cdot/C so if the economy touches that line it will stay below it. This means that K will rise until Y = (delta)K and consumption will fall to zero. This would be a stupid mistake (violating the transversality condition). In contrast balanced growth shrinking lies below that line.
Now there is a set of sunspot equilibria in which the economy starts out growing with natural rate L = 1 and a bad sunspot arrives at a low rate lambda. If the sunspot arrives C falls sharply and the economy drops to the balanced shrinking path and goes back to the origin. The Euler equation for Cdot in case the sunspot doesn’t arrive will have a term of extra growth of the order lambda so that the Euler equation holds in expected value. Similarly there is a path close to the balanced shrinking path with consumption shrinking a little faster except if a good sunspot arrives and it jumps up. But it is also possible to have deterministic cycles.
I have extensive algebra for an economy which shrinks from zero to T then predictably starts growing. Here the idea is that off the balanced shrinking path there are paths which deviate with the gap increasing exponentially. One can be chosen which hits the positive r L = 1 saddle path exactly at T. Similarly the economy can have L = 1 until tau then shrink. A path below the natural employment rate saddle path will diverge and can hit the balanced shrinking path exactly at tau.
Finally the economy can switch to shrinking when (K,C) is just above the balanced growth path then switch to growing when (K,C) is just below the L=1 saddle path then etc forever.
Now I will not do the algebra of dynamics for L=1 and sigma>alpha. It is hard to the extent which I can’t find an example with a closed form solution. But dynamics with r = 0 is simple. I will consider an solution in which the economy is in the liquidity trap until t= T and from then on has natural rate employment L=1.
The challenge is to get liquidity trap dynamics so that (K_T,C_T) is on the L=1 saddle path to the L=1 steady state. I am going to define a made up term
17) Ktdif = Kt – [alpha/((1-alpha)delta+rho)]Ct
This is how much more capital there is than on the balanced shrinking path.
It is easy to understand the dynamics of Ktdif. Ktdif more capital means gross production is higher by (Ktdif)delta/alpha and depreciation higher by (Ktdif)delta
18) Ktdifdot = (delta)((alpha-1)/alpha)Ktdif C_T = C_0e^(-rhoT) .
K_t must be the K on the saddle path which corresponds to C_T. Inventing notation again, I call this K(C_T) = K(C_00e^(-rhoT))
19) KTdif = K(C_0e^(-rhoT)) - [alpha/((1-alpha)delta+rho)]C_0e^(-rhoT)
20) K0dif = [(K(C_0e^(-rhoT)) - [alpha/((1-alpha)delta+rho)]C_0e^(-rhoT))]e^[-T(delta)(alpha-1)/alpha]
21) C_0= (K_0-K0dif) ((1-alpha)delta+rho)/alpha = (K_0- [(K[C_0e^(-rhoT)] - [alpha/((1-alpha)delta+rho)]C_0e^(-rhoT))]e^[-T(delta)(alpha-1)/alpha] ) ((1-alpha)delta+rho)/alpha
That sure is an ugly equation, but the right hand side is K_0 >0 if C_0=0 and decreases in C_0 so it has a unique solution for some positive C_0. I claim to have proven that the economy can be in the liquidity trap until T and then never again. This is a third solution in addition to the always in the liquidity trap balanced shrinking solution and the never in the liquidity trap perfectly conventional Ramsey Cass Koopmans solution.
It should be clear that the economy can switch from not in the liquidity trap r >0 L=1 to in the liquidity trap and can indeed switch back and forth in a cycle. Basically downward nominal rigidity makes equilibrium indeterminate. Pretty much anything can happen. Things can happen for any reason or for no reason. Even the confidence fairy can matter in this model.