Can a Discount Rate be Inferred within a Solow-Swan Growth Model? I'm trying to understand the implications of a Solow-Swan growth model with, to keep things simple, constant population, and constant Cobb-Douglas technology written in per capita form as $y=k^{\alpha} (0&lt;\alpha&lt;1)$. Given an initial capital stock, depreciation rate $\delta$ and assumptions on the saving ratio s, I understand how to obtain time paths for y, k and per capita consumption c, properties of steady states and of the golden rule steady state.

Question: Is it possible within this model, perhaps with plausible additional assumptions, to infer the path of discount rates $\rho$, and consequently of interest rates r, corresponding to a given feasible time path of y, k and c?
In the Solow-Swan model, the interest rate will be determined by market-equilibrium: under the baseline assumption of competitive markets (and constant-returns-to-scale technology), we get the familiar per-period equilibrium relation

$$f'(k) = r + \delta \tag{1}$$

Regarding the rate of pure time preference, think of the following: in this model, savings is a fixed percentage of output. So consumption is also a fixed percentage of output:

$$c = (1-s)f(k) \implies \dot c = (1-s)f'(k)\dot k \tag{2}$$

while also we have

$$\dot k = sf(k) - (n+\delta)k \tag{3}$$

Combining, we have

$$\dot c = (1-s)f'(k)\cdot [sf(k) - (n+\delta)k]$$

$$= (1-s)f(k)f'(k)s - (1-s)f'(k)(n+\delta)k$$

and using the previous relations, 

$$\dot c = s(r+\delta)c - (1-s)(r+\delta)(n+\delta)k \tag{4}$$

In the "Ramsey" model with consumer intertemporal utility maximization, we get the relation (for log-utility for simplicity)

$$\dot c = (r-\rho(t)) c \tag{5}$$

Note that I made the rate of pure time preference time-varying and this is indeed the case when we examine time-varying preference rates in the Ramsey model (see for example Barro, R. J. (1999). Ramsey meets Laibson in the neoclassical growth model. Quarterly Journal of Economics, 1125-1152).

For the consumers in the Solow-Swan model to behave like the consumer's that optimize intertemporaly equation $(4)$ must be equivalent to equation $(5)$ so (using now the time variable to clearly indicate what is time-varying and what is not), we get after eliminating and re-arranging

$$s(r(t)+\delta)c(t) - (1-s)(r(t)+\delta)(n+\delta)k(t) =[r(t)-\rho(t)] c(t)$$

$$\implies \big[s(r(t)+\delta) - r(t)+\rho(t)]c(t) = (1-s)(r(t)+\delta)(n+\delta)k(t)$$

$$\implies \rho(t) = (1-s)(r(t)+\delta)(n+\delta)\frac {k(t)}{c(t)} +r(t) - s[r(t)+\delta]$$

$$\implies \rho(t) = (r(t)+\delta)(n+\delta)\frac {k(t)}{f(k(t))} +r(t) - s[r(t)+\delta]$$

The consistent rate of pure time preference is endogenous, and time-varying under diminishing marginal product of capital, and becomes constant only in the steady state. The relation tells us to what $\rho(t)$ must be equal so that the optimizing consumer finds optimal to maintain a constant savings rate.

If the production function has constant output elasticities (say, Cobb-Douglas $f(k) = Ak^a$), then the above simplifies to

$$\rho(t) = (n+\delta)a - s\delta + (1-s)r(t) $$

or

$$\rho(t) = (n+\delta)a  - \delta + (1-s)f'(k(t)) $$

so it is still time varying (it would be constant in a model with constant marginal product of capital). If capital increases to its steady state, we see that the consistent rate of pure time preference should fall over time (which is also the result we get in the Ramsey model with time varying $\rho$).

At the steady-state we have $f'(k^*) = a(n+\delta)/s$ so

$$\implies \rho^* = (n+\delta)a  - \delta + \frac{(1-s)a(n+\delta)}{s}$$

$$\implies \rho^* = \frac {an+ (a-s)\delta}{s}$$

That's a nice one since $a$ is the golden-rule savings rate. To mimic the Ramsey model we should set therefore $s&lt;a$ so we will be to the left of the golden rule (and even in the Solow context, setting $s &gt;a$ is dynamically inefficient). Then the steady-state consistent $\rho$ will certainly be positive. 

Just for a taste, set $a=0.45, s=0.35, n=0.01, \delta =0.05$, to get

$$\rho^* \approx 0.027$$

Not far from the benchmark value $\rho =0.02$.