I am reading through McCandless "The ABCs of RBCs" this summer to get a preview of what I need to know for the coming Fall semester. It did not take long to find a statement that I can easily accept but cannot prove. On Page 9 after deriving the steady state condition of $(\delta + n)\bar{k} = \sigma A_0 f(\bar{k})$ in a zero-technological growth regime (where $\delta$ is depreciation and $n$ is growth rate of the labor force), the author says that "the stability of the positive stationary state can be seen from the equation $$ k_{t+1} = g(k_t) = \frac{(1-\delta)k_t + \sigma A_0 f(k_t)}{1+n}$$ Notice that between 0 and the positive $\bar{k}$, the function $g(k_t)$ is above the 45 degree line, so that $k_{t+1}$ is greater than k_t." He provides a standard-looking Solow model state diagram where I can verify this graphically, but not analytically.

I'm trying to prove every statement in the book in order to better familiarize myself with the details of macroeconomic theory before diving deeper, but I am absolutely stumped as to how I would prove that $k_{t+1} > k_t$ when $0 < k_t < \bar{k}$ and vice versa. I first tried manipulating the capital motion equation to prove it directly, and I could not reach a proof. The next strategy I have been trying to employ is to differentiate the capital motion equation and prove that its derivative is less than 1 at the steady state point, but it fails:

$$ \frac{\partial k_{t+1}}{\partial k_t} = \frac{(1-\delta) + \sigma A_0 f'(k_t)}{1+n} > \frac{(1-\delta) + \sigma A_0 f'(\bar{k})}{1+n} = \frac{(1-\delta)+(\delta+n)}{1+n} = 1 $$

Does anyone have an alternative strategy I can follow? It's been driving me nuts for two days.

  • $\begingroup$ I also have a handwaving "proof:" $f(k_t)$ is increasing and concave, so $g(k_t)$ is also increasing and concave. Therefore if there exists a $\bar{k}$ such that $(\delta + n)\bar{k} = \sigma A_0 f(\bar{k})$, $g(k_t)$ must cross the 45-degree line from above to below. This just doesn't seem like a "real proof" to me. $\endgroup$ May 12, 2017 at 20:41
  • $\begingroup$ What a great question! Glad to see the high quality answers too $\endgroup$ Apr 12, 2019 at 19:07

2 Answers 2


For stability, we want $$\frac{\partial k_{t+1}}{\partial k_t}\Big|_{\bar k} <1 \implies \frac{(1-\delta) + \sigma A_0 f'(\bar k)}{1+n} <1$$

$$ \implies f'(\bar k) < \frac {\delta+n}{\sigma A_0 } = \frac {f(\bar k)}{\bar k}$$

So we need the marginal product of capital to be smaller than the average product at the steady state.

Equivalently, we need $\bar k f'(\bar k) < f(\bar k) \implies f(\bar k)-\bar k f'(\bar k)>0$. And this holds, doesn't it? Otherwise...

  • $\begingroup$ Ah, yes. I keep forgetting that I can and should employ economic intuition in an economics course. Since all the connections work both ways, it looks like your proof works even better if you start with the MPK < APK rule and work back from there. Thanks! $\endgroup$ May 13, 2017 at 16:21

For completeness, let me illustrate this in the continuous time framework. The Solow equation, in the simplest of cases, is

$\dot{k} = s f(k) - \delta k = \phi(k)$

Then we have

$\frac{\partial \phi}{\partial k} = s f'(k) - \delta = \frac{sf'(k)k - \delta k }{k}$.

In steady state (i.e., $\dot{k} = \phi(k^{\ast}) = 0$), we have $\delta k = s f(k)$, hence

$\left. \frac{\partial \phi}{\partial k} \right|_{k=k^{\ast}} = \frac{s f' k - s f(k)}{k} = \frac{s}{k} f(k) \left[ \frac{k \cdot f'}{f(k)} - 1 \right]$

Since both $\frac{s}{k}$ and $f(k)$ are always positive, the sign of that expression hinges on the term in brackets. But remember that $\frac{k \cdot f'}{f(k)}$ is the output elasticity of capital, which is smaller than 1 if $f$ is concave in $k$, which means $\frac{\partial \phi}{\partial k}$ at $k^{\ast}$ is negative. Therefore by theorem 3.2.1 in Zhang (2005), the steady state in the Solow model is stable.


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