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Consider the growth model with inelastic labor supply, full depreciation, log utility and CRS technology with the Bellman equation be defined as follows: $$V(k)=\max(log(k^\alpha-k')+\beta V(k'))$$ st $$k\geq0\ \text{and}\ \theta k^\alpha-k'\geq0$$

As a guess I have used the usual $V(k)=a+bln(k)$ substituted on the Bellman and have derived $k'=\frac{k^\alpha\beta b}{1+\beta b}$. From this I have found the $k_{ss}=(\frac{1+\beta b}{\beta b})^{1/(\alpha-1)}$ called the non-trivial SS

This is the first SS while the other is $k_{ss}=0$

My question is how can we use the policy function to show that the system converges to the non-trivial steady steady state given any $k_0 > 0$

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  • $\begingroup$ A few remarks: your technology is not CRS unless $\alpha = 1$. Also, you forgot to add $\theta$ in your Bellman equation? An issue here is that for $k_t = 0$ (and therefore $k_{t+1} = 0$), your instantaneous utility function is not defined as $\ln(0)$ does not exist. As such, the usual convergence results for the value function of the Bellman equation are quite tricky. In fact the value of $V(0)$ does not exist, so there is no 'optimal policy function' for $k = 0$. $\endgroup$
    – tdm
    May 11 at 7:09
  • $\begingroup$ @tdm the $\theta$ was confusing me as well that is not included in the Bellman and I wasn't sure if I should have included in the derivation or not. For $k_ss=0$ it makes sense to not have an optimal policy function. But in one of the previous Q. asked it says to derive the policy function $k′(k)$ using the appropriate guess on the value function and to show that there are two SS with one being $k_ss=0$. Maybe the question is not clear? $\endgroup$ May 11 at 8:06
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Let's guess that the value function is of the form $a + b \ln(k)$.

Then substituting for $V(k) = a + b \ln(k)$ in the Bellman equation gives: $$ a + b \ln(k) = \max_{k'}\left(\ln(k^\alpha - k') + \beta(a + b \ln(k')\right) $$ The first order condition is given by: $$ \begin{align*} &\frac{-1}{k^\alpha - k'} + \beta b \frac{1}{k'} = 0,\\ \to & k' = \beta b (k^\alpha - k'),\\ \to & k' = \frac{\beta b}{1+ \beta b} k^\alpha \end{align*} $$ If we plug this into the objective function of the Bellman equation, we obtain the following identity: $$ \begin{align*} a + b \ln(k) &= \ln\left(k^\alpha - \frac{\beta b}{1 + \beta b}k^\alpha\right) + \beta\left(a + b \ln\left(\frac{\beta b}{1 + \beta b}k^\alpha\right)\right),\\ &= (\alpha + \beta b \alpha) \ln(k) + \ln\left(1 - \frac{\beta b}{1 + \beta b}\right) + \beta a + \beta b \ln\left(\frac{\beta b}{1 + \beta b}\right) \end{align*} $$ As this holds for all $k (> 0)$ we can equate coefficients on both sides: $$ \begin{align*} a &= \ln\left(\frac{1}{1 + \beta b}\right) + \beta a + \beta b \ln\left(\frac{\beta b}{1 + \beta b}\right),\\ b & = \alpha + \beta b \alpha \end{align*} $$ The second one gives a closed form expression for $b$: $$ b = \frac{\alpha}{1 - \beta \alpha}. $$ Then substituting this into the first order condition gives: $$ \begin{align*} k_{t+1} &= \frac{\beta \frac{\alpha}{1 - \beta \alpha}}{1 + \beta \frac{\alpha}{1 - \beta \alpha}}k_t^\alpha,\\ &= \beta \alpha k^\alpha_t \tag{1} \end{align*} $$

This shows that: $$ k_{t + 1} > k_t \iff \beta \alpha k_t^\alpha > k_t \iff k_t < (\beta \alpha)^{\frac{1}{1 - \alpha}} $$ So the capital stock will rise as long as $k_t$ is below $(\beta \alpha)^{\frac{1}{1 - \alpha}}$ and it will decrease if $k_t$ it is above this threshold.

According to the dynamic equation (1) above, it would appear that $k = 0$ is also a steady state. However, for $k = 0$, the first order conditions are not satisfied and in fact the value function does not exist. Anyway, for $k_t$ very close, it's value will be below $(\beta \alpha)^{\frac{1}{1 - \alpha}}$ so the stock of capital should increase to the unique steady state.

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  • $\begingroup$ Thank you so much for answering! $\endgroup$ May 12 at 18:47

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