I'm currently reading Romer's advanced macroeconomics, and encounter a question of linear stability of the model at the steady state.

We have the key equation $k'(t)=sf(k(t))-(n+g+\delta)k(t)$, where $k$ is capital per effective labour, $n,g,\delta$ each represents constant growth rate of labour, knowledge and depreciation rate of capital, respectively.

Use linear stability analysis, steady state is where $k'(t)$ is zero, denoted by $\bar k$. We can check the nature of the steady state by using $sf'(\bar k)-(n+g+\delta)$. We want to prove that the sign of it is negative.

We know $f'(k) >0$ and $f''(k)<0$, also other assumptions of $f$ given in the Solow model. Is it possible to prove it mathematically?


1 Answer 1


You outline the assumption that $f$ is strictly concave in $k$, so the average slope $$ \frac{f(k) - f(0)}{k - 0} $$ is larger than $f'(k)$.

I will also use the frequently made assumption that $f(0) = 0$, though $f(0) \geq 0$ would be sufficient.

Taking these two together, we get $$ f'(k) < \frac{f(k) - f(0)}{k - 0} = \frac{f(k)}{k} $$ or $$ f'(k)k < f(k). $$ In the steady state $$ sf( \bar k ) - (n+g+\delta) \bar k = 0. $$ Substituting in our inequality $$ sf'( \bar k ) \bar k - (n+g+\delta) \bar k < 0. $$ if $\bar k \neq 0$, so we are not in the 'trivial' steady state, we can divide by $\bar k$ and we get $$ sf'( \bar k ) - (n+g+\delta) < 0. $$

In the case of $f(0) < 0$ you would have a steady state where $f(k)$ intersects the line $(n+g+\delta)k$ from below, with a higher slope. This steady state would indeed be unstable. The same is true for $\bar k = 0$ when $f(0) = 0$.


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