In my Advanced Macro script, the professor says take TSA1 of the following equation:

$(1+g)(1+n)k_{t+1} = sk^\alpha_{t+1} + (1-\delta)k_t$

where $g$ is technological progress, $n$ population growth, $k$ capital per effective labour (AL).

In the class we work with discrete time. The results don't seem to make sense to me - could anyone guide me how to take the TSA in this particular case?

The $k_{t+1}$ means subscript and thus capital per AL at time $t+1$.

EDIT: TSA around the steady state.

Many thanks!


1 Answer 1


The steady-state value $k^*$ must be a fixed point :

$(1+g)(1+n)k^* = s(k^*)^{\alpha} +(1 - \delta) k^*$

Taking the difference between this equation and the dynamic one :

$(1+g)(1+n)(k_{t+1} - k^*) = s((k_{t+1})^{\alpha} - (k^*)^{\alpha}) +(1 - \delta) (k_t - k^*)$

Now if you denote by $d_t = k_t - k^*$ the distance to steady-state, this gives :

$(1+g)(1+n)d_{t+1} = s((k_{t+1})^{\alpha} - (k^*)^{\alpha}) +(1 - \delta) d_t$

But :

$(k_{t+1})^{\alpha} = (k^* + d_{t+1})^\alpha = (k^*)^\alpha + \alpha d_{t+1} (k^*)^{\alpha - 1} + o (d_{t+1})$

(Here was the Taylor development.) So that finally, this equality holds approximately around steady state ($d_t$ small):

$(1+g)(1+n)d_{t+1} = \alpha s (k^*)^{\alpha - 1} d_{t+1} +(1 - \delta) d_t$

From the steady state condition you can also compute (assuming that $ng << n+g$):

$(k^*)^{\alpha -1} = \frac{n+g+\delta}{s}$


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