<p>Assuming that <span class="math-container">$f$</span> is the Cobb-Douglas production function in per capita terms, then is the steady-state equilibrium consumption <span class="math-container">$c^*$</span> and capital <span class="math-container">$k^*$</span>  uniquely determined by
<span class="math-container">$$f'(k^*) = \delta + \rho \quad \text{and} \quad c^* = f (k^*) - (n + \delta) k^*$$</span>
with depreciation rate <span class="math-container">$\delta$</span>, time preference <span class="math-container">$\rho$</span>, and growth rate <span class="math-container">$n$</span>.</p>
<p>Since <span class="math-container">$f$</span> is assumed to be Cobb-Douglas it holds <span class="math-container">$f' &gt; 0$</span> and <span class="math-container">$f'' &lt; 0$</span>. Hence is <span class="math-container">$f$</span> strictly monotone increasing and <span class="math-container">$f'$</span> strictly monotone decreasing.
Thus, for <span class="math-container">$\tilde \rho$</span> with <span class="math-container">$\tilde \rho &lt; \rho$</span>,
we have
<span class="math-container">$$f'(k^*) = \delta + \rho \quad \text{and} \quad f'(\tilde k^*) = \delta + \tilde \rho,$$</span>
and since <span class="math-container">$f'$</span> is strictly monotone decreasing we have that <span class="math-container">$\tilde k^* &gt; k^*$</span> holds.
By usind that <span class="math-container">$f$</span> is strictly monotone increasing we obtain by
<span class="math-container">$$c^* = f (k^*) - (n + \delta) k^* \quad \text{and} \quad \tilde c^* = f (k^*) - (n + \delta) k^*$$</span>
that <span class="math-container">$\tilde c^* &gt; c^*$</span> holds, too.</p>
<p>In <a href="https://press.princeton.edu/books/hardcover/9780691132921/introduction-to-modern-economic-growth" rel="nofollow noreferrer">Introduction to Modern Economic Growth by Daron Acemoglu</a> does Proposition 8.3 yield the sensitivity analysis for all parameters.</p>

CC BY-SA 4.0
2022-07-01T07:43:49.200Z
Thank you! To tell the truth I still don’t get why ̃∗ is greater than the other if its first derivative is lower.
tryingtogetsmth
CC BY-SA 4.0
2022-07-01T09:22:12.273Z
@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., $k(\rho) = f'^{-1}(\delta + \rho)$
maximilian43
CC BY-SA 4.0
2022-07-01T09:34:54.133Z
Ok now it makes sense, thank you indeed!
tryingtogetsmth
CC BY-SA 4.0
2022-06-30T15:04:46.583Z
2022-07-01T09:12:14.827Z
2022-07-01T09:12:14.827Z
maximilian43
<p>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.</p>

<macroeconomics>
Effect of an increase/decrease of ρ or θ on consumption per effective labour and on capital per effective labour? And on their steady-state values?
tryingtogetsmth
# Effect of an increase/decrease of ρ or θ on consumption per effective labour and on capital per effective labour? And on their steady-state values?

(Asked by: username_0 on 2022-06-30)

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.

**Tags:** macroeconomics


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## 1 answers

### Answer by username_1 on 2022-06-30. Score: 2

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](https://press.princeton.edu/books/hardcover/9780691132921/introduction-to-modern-economic-growth) does Proposition 8.3 yield the sensitivity analysis for all parameters.

> **Comments:**
> **username_0**: Thank you! To tell the truth I still don’t get why ̃∗ is greater than the other if its first derivative is lower.  
> **username_1**: @username_0 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., $k(\rho) = f'^{-1}(\delta + \rho)$  
> **username_0**: Ok now it makes sense, thank you indeed!  

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