Solow model, time and steady state

Question

Suppose we have a Solow model:

$$ Y(t)=C(t)+I(t) $$

$$ I(t)=sY(t) $$

$$ \dot K=I(t)-δK(t) $$

With a given Cobb-Douglas:

$$ Y(t)=Z(t)K^aL^{1-a} $$

$$ y(t)=Y(t)/L(t) $$

$$ k(t)=K(t)/L(t) $$

$$ y=Zk^a $$

We also know these:

$$ L=L(t), \dot L/L=n $$

$$ Z=Z(t), \dot Z/Z=g $$

Therefore we have:

$$ L(t)=L(0)e^{nt} $$

$$ Z(t)=Z(0)e^{gt} $$

We get to this point:

$$ \dot k=sZk^a-(n+δ)k $$

In the steady state:

$$ k=k^* $$

$$ k^*=\left(\frac{sZ}{n+δ}\right)^{1/(1-a)} $$

The question has to do with the time dimension in Z. Time enters the final expression if we substitute Z with what we found above, but I'm not sure what to make of this.

$$ k^*=\left[\frac{sZ(0)e^{gt}}{n+δ}\right]^{1/(1-a)} $$

Surely there's something this neophyte is mixing up. I thought that in a steady state capital per worker remains the same.

Answer 1

In the model with technological progress the capital per effective worker remains constant, implies that capital per worker grows at the rate of exogenous rate of technological progress. See Barro and Martin book, Chapter 1.

Answer 2

Kshitiz is right. If you define effective capital per worker as $\hat k(t) \equiv \dfrac{K(t)}{L(t)Z(t)}$, the equation for the dynamic of $\hat k(t)$ is:

$$ \dot {\hat k} (t) = s {\hat k}^a - (n+g+\delta){\hat k} $$

from where you get the constant effective capital per worker in the steady state:

$$ {\hat k} = \left(\frac{s}{n+g+\delta}\right)^{1/(1-a)} $$

Intuitively, every period the marginal product of capital is increasing because of the exogenous expansion in $Z(t)$. In the standard Solow diagram with capital per worker in the horizontal axis, this exogenous technical change means an outward expansion of the $y$ and $sy$ curves. This higher return to capital induces more investment and an expansion of capital per worker. As technical change occurs every period, this process continues forever.