Solow model response function

Question

Consider a Solow model without technological progress so that the steady state $k^*$ occurs at $sf(k^*) = (n+\delta)k^*$ where $n$ is population growth rate, $\delta$ is capital depreciation rate and $s$ is the proportion of output that is saved and invested. Given that consumption per worker is $c = f(k) - sf(k)$, how does steady state consumption change if $n$ decreases?

Intuitively, output per worker rises so $c = (1-s)f(k)$ should also rise. However, if I try to solve for the change numerically I would do:

$c^* = f(k^*(s, n, \delta)) - (n+\delta)k^*(s, n, \delta)$ at the steady state.

So, $\frac{\partial c^*}{\partial n} = f'(k^*)\frac{\partial k^*}{\partial n} - k^* -(n + \delta)\frac{\partial k^*}{\partial n}$.

Re-arranging:

$\frac{\partial c^*}{\partial n} = [f'(k^*) - (n+\delta)]\frac{\partial k^*}{\partial n} - k^*$.

We know $\frac{\partial k^*}{\partial n}$ < 0 since a higher slope of the breakeven investment causes it to intersect with $sf(k^*)$ at a lower $k^*$. The sign of the term in the square bracket is ambiguous so $\frac{\partial c^*}{\partial n}$ itself is ambiguous.

There seems to be a contradiction between my intuition and the numerical analysis, so where have I gone wrong?

Answer 1

The direction of $\frac{\partial c^{*}}{\partial n}$ is not ambiguous.

An easy way to show this is taking derivative of $c^*=(1-s)f(k^*)$ so that $\frac{\partial c^{*}}{\partial n}=(1-s)f'\frac{\partial k^{*}}{\partial n}$ and because $f'>0$ and we can prove $\frac{\partial k^{*}}{\partial n}<0$ we thus have $\frac{\partial c^{*}}{\partial n}<0$.

Through this you can also easily see that $c^{*}$ will be monotone in $n$ and $\delta$ but it will not be monotone in $s$.

It is somehow harder to prove this from $\frac{\partial c^{*}}{\partial n}=\left[f^{\prime}\left(k^{*}\right)-(n+\delta)\right] \frac{\partial k^{*}}{\partial n}-k^{*}$