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Consider Solow growth model. Let $s$ denote savings rate.

Suppose we estimate China invests 40% of it GDP rather than consuming. Assuming China is dynamically inefficient because they are a developing nation, how can I estimate what it's $s_{gold}$ is? Furthermore, how do I decide whether China is at this $s_{gold}$ versus some other $s$?

Furthermore, how does one determine this in general? Like are there distinct characteristic patterns that allow me to understand these rates?

Let our production function be $Y_t = (A_tL_t)^{1-\alpha} K_t^{\alpha}$.

I am asking how one calculates $s$ and $s_{gold}$ in practice, as in, if I had to find realistic numbers for these how would I do it?

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  • $\begingroup$ For the purposes of calculating the golden-rule savings rate, why are you assuming that "China is dynamically inefficient"? $\endgroup$ – Alecos Papadopoulos Mar 16 '16 at 15:09
  • $\begingroup$ I added a justification, see if u think its sufficient $\endgroup$ – Stan Shunpike Mar 16 '16 at 17:00
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To partially answer this question, in Advanced Macroeconomics by Romer, question 1.5(c) asks

"What saving rate is needed to yield the golden-rule capital stock?"

Answering this tells us what $s_{gold}$ is.

But before we can find this we must find the golden rule level of capital stock $(k_{gold})$. Which we shall derive with the following equations: The intensive form of the production function, the dynamics of capital in the Solow Model, and consumption per unit of effective labour

\begin{cases} Y_t = (A_t L_t)^{1-\alpha} K_t^{\alpha} = A_t L_t \left( \frac{K_t}{A_t L_t} \right)^{\alpha} = A L k^{\alpha} \\ \Rightarrow Y/AL = y = k^{\alpha} \\ \dot{k} = s y - (n+g+\delta) k \\ c = (1-s)y \\ \end{cases}

Under the balanced growth path, $\dot{k}=0$ so

$$ \displaystyle s k^{* \alpha} = (n+g+\delta)k^* \Rightarrow k^* = \left( \frac{n+g+\delta}{s} \right)^{\frac{1}{\alpha-1}} $$

We maximize $c^*$ with respect to $k^*$ to implicitly define the golden-rule level of capital per unit of effective labour

$$ \displaystyle \frac{\partial }{\partial k^*} c^* = \frac{\partial }{\partial k^*} (1-s)k^{* \alpha} = (1-s)(\alpha k^{* \alpha - 1}) $$

$$ \displaystyle = \alpha (1 - s) \left( \frac{n+g+\delta}{s} \right)^{\frac{\alpha-1}{\alpha-1}} $$

Next replace $s$ with $(n+g+\delta)k^{* 1-\alpha}$

$$ \displaystyle = \alpha (1 - (n+g+\delta)k^{* 1-\alpha}) \left( \frac{n+g+\delta}{(n+g+\delta)k^{* 1-\alpha}} \right) $$

$$ \displaystyle 0 = \alpha (k^{* (\alpha-1)} - (n+g-\delta)) \Rightarrow k^*_{gold} = \left( \frac{\alpha}{n+g+\delta} \right)^{1/(1-\alpha)} $$

We can finally compute, $s_{gold}$

$$ \displaystyle s_{gold} = (n+g+\delta) k_{gold}^{*1-\alpha} = (n+g+\delta) \left( \frac{\alpha}{n+g+\delta}\right)^{(1-\alpha)/(1-\alpha)} = \alpha $$

So for a Cobb-Douglas production function the golden-rule saving rate should be whatever $\alpha$ is.

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