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In a Ramsey-Cass-Koopmans framework, having defined ρ as the rate of time preference, I was wondering what is the effect of a variation in the two parameters on consumption per effective labour and on capital per effective labour and on their steady state values.

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Assuming that $f$ is the Cobb-Douglas production function in per capita terms, then is the steady-state equilibrium consumption $c^*$ and capital $k^*$ uniquely determined by $$f'(k^*) = \delta + \rho \quad \text{and} \quad c^* = f (k^*) - (n + \delta) k^*$$ with depreciation rate $\delta$, time preference $\rho$, and growth rate $n$.

Since $f$ is assumed to be Cobb-Douglas it holds $f' > 0$ and $f'' < 0$. Hence is $f$ strictly monotone increasing and $f'$ strictly monotone decreasing. Thus, for $\tilde \rho$ with $\tilde \rho < \rho$, we have $$f'(k^*) = \delta + \rho \quad \text{and} \quad f'(\tilde k^*) = \delta + \tilde \rho,$$ and since $f'$ is strictly monotone decreasing we have that $\tilde k^* > k^*$ holds. By usind that $f$ is strictly monotone increasing we obtain by $$c^* = f (k^*) - (n + \delta) k^* \quad \text{and} \quad \tilde c^* = f (k^*) - (n + \delta) k^*$$ that $\tilde c^* > c^*$ holds, too.

In Introduction to Modern Economic Growth by Daron Acemoglu does Proposition 8.3 yield the sensitivity analysis for all parameters.

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  • $\begingroup$ Thank you! To tell the truth I still don’t get why 𝑘̃∗ is greater than the other if its first derivative is lower. $\endgroup$ Jul 1, 2022 at 7:43
  • $\begingroup$ @tryingtogetsmth I accidentally described $f'$ as monotone increasing at the very end which as been fixed. Clearly $f'$ is strictly decreasing, and, thus, the inverse is strictly decreasing, too, i.e., for $\tilde \rho < \rho$ we have that $\tilde k^* > k^*$. You have to see $k^*$ as function in dependency of $\rho$, i.e., $k(\rho) = f'^{-1}(\delta + \rho)$ $\endgroup$ Jul 1, 2022 at 9:22
  • $\begingroup$ Ok now it makes sense, thank you indeed! $\endgroup$ Jul 1, 2022 at 9:34

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