<p>Let's guess that the value function is of the form <span class="math-container">$a + b \ln(k)$</span>.</p>
<p>Then substituting for <span class="math-container">$V(k) = a + b \ln(k)$</span> in the Bellman equation gives:
<span class="math-container">$$
a + b \ln(k) = \max_{k'}\left(\ln(k^\alpha - k') + \beta(a + b \ln(k')\right)
$$</span>
The first order condition is given by:
<span class="math-container">$$
\begin{align*}
&amp;\frac{-1}{k^\alpha - k'} + \beta b \frac{1}{k'} = 0,\\
\to &amp; k' = \beta b (k^\alpha - k'),\\
\to &amp; k' = \frac{\beta b}{1+ \beta b} k^\alpha
\end{align*}
$$</span>
If we plug this  into the objective function of the Bellman equation, we obtain the following identity:
<span class="math-container">$$
\begin{align*}
a + b \ln(k) &amp;= \ln\left(k^\alpha - \frac{\beta b}{1 + \beta b}k^\alpha\right) + \beta\left(a + b \ln\left(\frac{\beta b}{1 + \beta b}k^\alpha\right)\right),\\
&amp;= (\alpha + \beta b \alpha) \ln(k) + \ln\left(1 - \frac{\beta b}{1 + \beta b}\right) + \beta a + \beta b \ln\left(\frac{\beta b}{1 + \beta b}\right)
\end{align*}
$$</span>
As this holds for all <span class="math-container">$k (&gt; 0)$</span> we can equate coefficients on both sides:
<span class="math-container">$$
\begin{align*}
a &amp;= \ln\left(\frac{1}{1 + \beta b}\right) + \beta a + \beta b \ln\left(\frac{\beta b}{1 + \beta b}\right),\\
b &amp; = \alpha + \beta b \alpha
\end{align*}
$$</span>
The second one gives a closed form expression for <span class="math-container">$b$</span>:
<span class="math-container">$$
b = \frac{\alpha}{1 - \beta \alpha}.
$$</span>
Then substituting this into the first order condition gives:
<span class="math-container">$$
\begin{align*}
k_{t+1} &amp;= \frac{\beta \frac{\alpha}{1 - \beta \alpha}}{1 + \beta \frac{\alpha}{1 - \beta \alpha}}k_t^\alpha,\\
&amp;= \beta \alpha k^\alpha_t \tag{1}
\end{align*}
$$</span></p>
<p>This shows that:
<span class="math-container">$$
k_{t + 1} &gt; k_t \iff \beta \alpha k_t^\alpha &gt; k_t \iff k_t &lt; (\beta \alpha)^{\frac{1}{1 - \alpha}}
$$</span>
So the capital stock will rise as long as <span class="math-container">$k_t$</span> is below <span class="math-container">$(\beta \alpha)^{\frac{1}{1 - \alpha}}$</span> and it will decrease if <span class="math-container">$k_t$</span> it is above this threshold.</p>
<p>According to the dynamic equation (1) above, it would appear that <span class="math-container">$k = 0$</span> is also a steady state. However, for <span class="math-container">$k = 0$</span>, the first order conditions are not satisfied and in fact the value function does not exist. Anyway, for <span class="math-container">$k_t$</span> very close, it's value will be below <span class="math-container">$(\beta \alpha)^{\frac{1}{1 - \alpha}}$</span> so the stock of capital should increase to the unique steady state.</p>

CC BY-SA 4.0
2021-05-12T18:47:44.700Z
Thank you so much for answering!
Maybeline Lee
CC BY-SA 4.0
2021-05-12T09:56:38.583Z
2021-05-12T09:56:38.583Z
tdm
<p>Consider the growth model with inelastic labor supply, full depreciation, log utility and CRS technology with the Bellman equation be defined as follows:
<span class="math-container">$$V(k)=\max(log(k^\alpha-k')+\beta V(k'))$$</span>
st <span class="math-container">$$k\geq0\ \text{and}\ \theta k^\alpha-k'\geq0$$</span></p>
<p>As a guess I have used the usual <span class="math-container">$V(k)=a+bln(k)$</span> substituted on the Bellman and have derived <span class="math-container">$k'=\frac{k^\alpha\beta b}{1+\beta b}$</span>. From this I have found the <span class="math-container">$k_{ss}=(\frac{1+\beta b}{\beta b})^{1/(\alpha-1)}$</span> called the non-trivial SS</p>
<p>This is the first SS while the other is <span class="math-container">$k_{ss}=0$</span></p>
<p>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 <span class="math-container">$k_0 &gt; 0$</span></p>

CC BY-SA 4.0
2021-05-11T07:09:21.757Z
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
CC BY-SA 4.0
2021-05-11T08:06:24.777Z
@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
<macroeconomics><steady-state><bellman-equations>
Non-trivial steady state
Maybeline Lee
# Non-trivial steady state

(Asked by: username_0 on 2021-05-10)

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 $k\_0 > 0$

**Tags:** macroeconomics, steady-state, bellman-equations

> **Comments:**
> **username_1**: 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 $k = 0$.  
> **username_0**: @username_1 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?  

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## 1 answers

### Answer by username_1 on 2021-05-12. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Thank you so much for answering!  

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