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 $$k_0 > 0$$

• 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 $k = 0$. – tdm May 11 at 7:09
• @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? – Maybeline Lee May 11 at 8:06

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.