<p>In p. 334 of the book (2009 edition), I read:</p>

<blockquote>
  <p><strong>Proposition 9.4</strong> In the overlapping-generations model with two-period lived households, Cobb-Douglas technology and CRRA
  preferences, there exists a unique steady-state equilibrium with the
  capital-labor ratio $k^*$ given by (9.15) and <strong>for any $\theta &gt;0$</strong>
  this steady-state equilibrium is globally
  stable for all k (0) > 0.</p>
</blockquote>

<p>...it does not say "...as long as $\theta \geq 1$".</p>

<p><strong>ADDENDUM :</strong> The case $\theta =0$</p>

<p>The OP elaborated on the case $\theta=0$. So let's attempt to solve this.</p>

<p>Here, intertemporal utility becomes linear in consumption. Also, we will allow consumption to be zero -assume that there is an exogenous endowment that takes care of the biological needs of the individuals here and cannot be used for any other purpose, so the consumption choice is not about survival. Still, we are interested in economically interesting cases, so we exclude the case $c_2=0$ because this would mean that the capital stock will be zero, and the economy will stop. So we allow only for $c_1=0$. Then, incorporating the equality constraint in the objective function we solve</p>

<p>$$\max U = c_1(t) + \beta R(t+1)[w(t)-c_1(t)]\;\;\; s.t.\;\; 0\leq c_1(t) &lt; w(t)$$</p>

<p>So the Lagrangian is </p>

<p>$$\Lambda = c_1(t) + \beta R(t+1)[w(t)-c_1(t)] + \lambda c_1(t)$$</p>

<p>and the f.o.c. is</p>

<p>$$1 - \beta R(t+1) +\lambda \leq 0,\;\;\;\; (1 - \beta R(t+1) +\lambda)\cdot c_1(t) = 0$$</p>

<p>I will ignore issues of second-order conditions for a maximum.</p>

<p><strong>CASE 1.</strong> Assume 
$$c_1^*(t) &gt;0 \implies \lambda=0 \implies 1 = \beta R(t+1)$$</p>

<p>$$\implies 1 = \beta ak(t+1)^{a-1} \implies k(t+1) = (a\beta)^{1/(1-a)}$$ </p>

<p>But this must hold for <em>all</em> time periods, so it can hold only for a specific value of initial capital endowment and not for all $k(0) &gt;0$. So already, we see that this case does not satisfy the conditions for which Proposition 9.4 is stated.</p>

<p><strong>CASE 2.</strong> Assume 
$$c_1^*(t) =0 \implies \lambda&gt;0 \implies 1 - \beta R(t+1) +\lambda \leq 0 \implies \beta R(t+1) &gt; 1$$</p>

<p>$$\implies \beta ak(t+1)^{a-1} &gt;1 \implies k(t+1) &lt; (a\beta) ^{1/(1-a)}$$</p>

<p>Here the capital stock must remain below a certain fixed level at all time periods. Again, this imposes constraints on the value of the initial capital stock $k(0)$.</p>

<p>So in both cases, if a steady-state equilibrium exists, it cannot be "<em>globally</em> stable, for <em>all</em> $k(0)&gt;0$".</p>

CC BY-SA 3.0
2017-04-06T00:54:08.090Z
It seems that I am reading the 2007 edition of the book, which is the draft version. But, even if $\theta = 0$, equation 1 becomes $k(t) = [\frac{1+n}{1-\alpha}k(t+1)]^{\frac{1}{\alpha}}$, the graph with equation 1 and the 45-degrees line will be similar to those shown in my post, the steady-state equilibrium will still be globally stable. Am I correct?
Chris Cheung
CC BY-SA 3.0
2017-04-06T01:14:56.007Z
@ChrisCheung I would suggest to go back at the beginning of this utility optimization problem and to solve it with $\theta =0$, looking also at what happens to the savings rate.
Alecos Papadopoulos
CC BY-SA 3.0
2017-04-06T19:31:02.610Z
I did the following computation: $$
\text{max } U(t) = c_1(t) + \beta(w(t) - c_1(t))R(t+1) \\ = w(t) - s(t) + \frac{\alpha\beta}{(1+n)^{\alpha - 1}}  s(t)^\alpha $$ First-order condition with respect to $s(t)$ gives: $s(t)^* = (1+n)(\alpha^2 \beta)^{\frac{1}{1 - \alpha}}$ Then, $
k(t+1) = \frac{s(t)}{1+n} = (\alpha^2 \beta)^{\frac{1}{1 - \alpha}}\text{ ,a constant}$. This means that the steady-state equilibrium where $k^* = (\alpha^2 \beta)^{\frac{1}{1 - \alpha}}$ is globally stable as even k deviates from $k^*$ for one period, next period k will return to steady state k^*. Am I correct?
Chris Cheung
CC BY-SA 3.0
2017-04-07T10:03:16.857Z
@ChrisCheung The issue with the case $\theta =0$ is that makes utillity linear in consumption, (both contemporaneous and intertemporal), which introduces issues of uniqueness. Also I get the impression that you have differentiated the Gross Interest rate with respect to savings, but this is the "central-planner" solution, not the competitive one where factor rewards are treated by individuals as market-determined and so exogenous with respect to their optimization problems.
Alecos Papadopoulos
CC BY-SA 3.0
2017-04-19T01:49:20.440Z
Thank you for reminding me this is not a central planner problem. I have added my derivation for the case $\theta = 0$ at the bottom of my question. But, I do not see this case introduces issues of uniqueness. I think $\theta = 0$ can also be included as the condition where the steady-state equilibrium is globally stable. Can you explain why the case $\theta=0$ introduces uniqueness problem?
Chris Cheung
CC BY-SA 3.0
2017-04-19T10:05:57.677Z
@ChrisCheung I updated my answer with some elaboration on the case $\theta =0$
Alecos Papadopoulos
CC BY-SA 3.0
2017-04-05T22:20:21.797Z
2017-04-19T10:05:21.537Z
2017-04-19T10:05:21.537Z
Alecos Papadopoulos
<p>From Daron Acemoglu's Introduction to Modern Economic Growth, proposition 9.4 is that:</p>

<blockquote>
  <p>In the overlapping-generations model with two-period lived
  households, Cobb-Douglas technology and CRRA preferences, there exists a unique steady-state equilibrium with the capital-labor ratio k* given by (9.15) and as long as $\theta \geq 1$, this steady-state equilibrium is globally stable for all k (0) > 0.</p>
</blockquote>

<p>where (9.15) is:
$$(1+n)[1+\beta^{-\frac{1}{\theta}}(\alpha(k^*)^{\alpha-1})^{\frac{\theta-1}{\theta}}] = (1-\alpha)(k^*)^{\alpha - 1}$$</p>

<p>My question is why $\theta$ has to be greater than or equal to 1 for the steady-state equilibrium to be globally stable?</p>

<p>As the textbook derives in (9.17):
$$k(t+1)=\frac{(1-\alpha)k(t)^\alpha}{(1+n)[1+\beta^{-\frac{1}{\theta}}(\alpha k(t+1)^{\alpha-1})^{\frac{\theta-1}{\theta}}]}$$</p>

<p>We can rearrange to get:
$$\begin{align}
k(t)&amp;=\big[  \frac{1+n}{1-\alpha}  [k(t+1)+\beta^{-\frac{1}{\theta}}\alpha^{\frac{\theta-1}{\theta}}k(t+1)^{(\alpha-1)(1-\frac{1}{\theta})+1}]   \big]^\frac{1}{\alpha} \text{  .....(1)}
\end{align} 
$$</p>

<p>Let $n = 0.01$, $\alpha = 0.25$, $\beta = 0.75$.</p>

<p>If $\theta = 1$, we can plot the graph:
<a href="https://i.sstatic.net/K6pb1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/K6pb1.png" alt="enter image description here"></a></p>

<p>The blue line is equation (1) where $\theta = 1$  and the red line is 45-degrees line. It can be seen that for all k > 0, k will converge to steady-state k*. The steady-state equilibrium is globally stable.</p>

<p>The case is similar for $\theta &gt; 1$, in which the steady-state equilibrium is globally stable.</p>

<p>If $\theta &lt; 1$, like $\theta = 0.5$, we can plot the graph similar to:
<a href="https://i.sstatic.net/oFXRD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oFXRD.png" alt="enter image description here"></a></p>

<p>The graph is similar to graphs for the case that $\theta \geq 1$. The steady-state equilibrium is still globally stable.</p>

<p>I can't find a case where $\theta &lt; 1$, but the steady-state equilibrium is not globally stable. It seems that $\frac{1}{\alpha} &gt; 1$ for $\alpha \in (0,1)$ determines the shape of equation (1), making the steady-state equilibrium globally stable. It would be good if someone could show me a counter-example where $\theta &lt; 1$, but the steady-state equilibrium is not globally stable. It would be better if someone could show me how to prove proposition 9.4 formally.</p>

<p>Acknowledgement: The graphs are modified from those generated by Wolframalpha.</p>

<p><strong>Edit (Apr 19, 2017)</strong>:
Case $\theta = 0$:
Note that when the textbook derives (9.17), it implicitly assumes that $\theta \neq 0$ (for derivation of Euler equation for consumption in P.333 of 2009 edition of the textbook). When $\theta = 0$, equation (1) no longer applies. Returning to the utility maximization problem with $\theta = 0$:</p>

<p>$$
\text{max } U(t) = c_1(t) + \beta(c_2(t+1)) \text{ such that } c_1(t) + \frac{c_2(t+1)}{R(t+1)} = w(t) \\
\Leftrightarrow 
\text{max } U(t) = c_1(t) + \beta(w(t) - c_1(t))R(t+1) = c_1(t) (1 - \beta R(t+1)) + \beta R(t+1)w(t) \\ \text{   ...Should treat R(t+1) as given as consumer's own optimization problem}
$$</p>

<p>s(t) has to be non-negative for $k(t+1) = \frac{s(t)}{1+n}$ and k(t+1) is non-negative.
$$
c_1(t)^* = \begin{cases}
w(t)\text{, for }\beta R(t+1) &lt;1 \\
[0, w(t)]\text{, for }\beta R(t+1) = 1 \\
0\text{, for }\beta R(t+1) &gt; 1 \\
\end{cases}
$$
$$
s(t)^* = \begin{cases}
0\text{, for }\beta R(t+1) &lt;1 \\
w(t) - c_1(t)^* \in [0, w(t)]\text{, for }\beta R(t+1) = 1 \\
w(t)\text{, for }\beta R(t+1) &gt; 1 \\
\end{cases}
$$
For $R(t+1) = f'(k(t+1)) = \alpha k(t+1)^{\alpha - 1}$,
$$
k(t+1) = \frac{s(t)}{1+n}  = \begin{cases}
0\text{, for }\beta R(t+1) &lt;1 \Leftrightarrow  k(t+1) &lt; (\alpha \beta)^{\frac{1}{1 - \alpha}} \\
\frac{w(t) - c_1(t)}{1+n} \in [0, \frac{w(t)}{1+n}]\text{, for }\beta R(t+1) = 1 \Leftrightarrow  k(t+1) = (\alpha \beta)^{\frac{1}{1 - \alpha}}\\
\frac{w(t)}{1+n} = \frac{k(t)^\alpha - k(t)\alpha k(t)^{\alpha -1}}{1+n} = \frac{1-\alpha}{1+n}k(t)^\alpha\text{, otherwise} \Leftrightarrow k(t) &gt; [\frac{1+n}{1 - \alpha} (\alpha \beta)^{\frac{1}{1 - \alpha}}]^{\frac{1}{\alpha}} \\
\end{cases}
$$</p>

<p>Cases:<br>
Case 1: $\beta R(t+1) &lt; 1 \Leftrightarrow R(t+1) &lt; \frac{1}{\beta}$:
As the production function $f(k)$ is Cobb-Douglas, it satisfies the Inada condition: $lim_{k(t) \to 0} f'(k(t)) = \infty$. But as $f'(k(t)) = R(t)$, $lim_{k(t) \to 0} R(t) &lt; \infty$ for $R(t) &lt; \frac{1}{\beta} &lt; \infty$ as $\beta \in (0,1)$, violating the Inada condition. This contradiction means this case is impossible.</p>

<p>Case 2: $\beta R(t+1) = 1 \Leftrightarrow \beta \alpha k(t+1)^{\alpha-1} = 1 \Leftrightarrow  k(t+1)^{\alpha-1} = \frac{1}{\alpha \beta} \Leftrightarrow k(t+1)^{1-\alpha} = \alpha \beta$:
Denote saving rate at t as $\mathcal{S}(t) = \frac{s(t)}{w(t)}$. $k(t+1) = \frac{s(t)}{1+n} = \frac{\mathcal{S}(t)w(t)}{1+n} = \frac{\mathcal{S}(t)(1-\alpha)k(t)^\alpha}{1+n}$. At steady state, $k^* = \frac{\mathcal{S}^*(1-\alpha){k^*}^\alpha}{1+n}$, meaning $\mathcal{S}^* = \frac{1+n}{1-\alpha}{k^*}^{1-\alpha} = \frac{1+n}{1-\alpha} \alpha \beta$. For $\mathcal{S}^* &gt; 1 \Leftrightarrow (1+n)\alpha\beta &gt; 1 - \alpha \Leftrightarrow \beta &gt; \frac{1 - \alpha}{\alpha (1+n)}$, which is possible. As saving rate cannot be larger than 1, this contradiction means this case is impossible.</p>

<p>Case 3: $\beta R(t+1) &gt; 1$:
This case is possible.
$$
k(t+1) = \frac{1-\alpha}{1+n} k(t)^\alpha
$$
We can draw the graph:<a href="https://i.sstatic.net/t6Q94.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/t6Q94.png" alt="enter image description here"></a><br>
The red line is 45-degrees line. The blue line is $k(t+1) = \frac{1-\alpha}{1+n} k(t)^\alpha$ where $0 &lt; \alpha &lt; 1$. For all k(0) > 0, k will converge to steady state $k^* = \frac{1-\alpha}{1+n} {k^*}^\alpha \Leftrightarrow k^* = [\frac{1-\alpha}{1+n}]^{\frac{1}{1-\alpha}}$. The steady-state equilibrium is globally stable.</p>

<macroeconomics><economic-growth>
Stability of steady-state equilibrium for overlapping generation model
Chris Cheung
# Stability of steady-state equilibrium for overlapping generation model

(Asked by: username_0 on 2017-04-05)

From Daron Acemoglu's Introduction to Modern Economic Growth, proposition 9.4 is that:

> In the overlapping-generations model with two-period lived
> households, Cobb-Douglas technology and CRRA preferences, there exists a unique steady-state equilibrium with the capital-labor ratio k\* given by (9.15) and as long as $\theta \geq 1$, this steady-state equilibrium is globally stable for all k (0) > 0.

where (9.15) is:
$$(1+n)[1+\beta^{-\frac{1}{\theta}}(\alpha(k^\*)^{\alpha-1})^{\frac{\theta-1}{\theta}}] = (1-\alpha)(k^\*)^{\alpha - 1}$$

My question is why $\theta$ has to be greater than or equal to 1 for the steady-state equilibrium to be globally stable?

As the textbook derives in (9.17):
$$k(t+1)=\frac{(1-\alpha)k(t)^\alpha}{(1+n)[1+\beta^{-\frac{1}{\theta}}(\alpha k(t+1)^{\alpha-1})^{\frac{\theta-1}{\theta}}]}$$

We can rearrange to get:
$$\begin{align}
k(t)&=\big[ \frac{1+n}{1-\alpha} [k(t+1)+\beta^{-\frac{1}{\theta}}\alpha^{\frac{\theta-1}{\theta}}k(t+1)^{(\alpha-1)(1-\frac{1}{\theta})+1}] \big]^\frac{1}{\alpha} \text{ .....(1)}
\end{align}
$$

Let $n = 0.01$, $\alpha = 0.25$, $\beta = 0.75$.

If $\theta = 1$, we can plot the graph:
[![enter image description here](https://i.sstatic.net/K6pb1.png)](https://i.sstatic.net/K6pb1.png)

The blue line is equation (1) where $\theta = 1$ and the red line is 45-degrees line. It can be seen that for all k > 0, k will converge to steady-state k\*. The steady-state equilibrium is globally stable.

The case is similar for $\theta > 1$, in which the steady-state equilibrium is globally stable.

If $\theta < 1$, like $\theta = 0.5$, we can plot the graph similar to:
[![enter image description here](https://i.sstatic.net/oFXRD.png)](https://i.sstatic.net/oFXRD.png)

The graph is similar to graphs for the case that $\theta \geq 1$. The steady-state equilibrium is still globally stable.

I can't find a case where $\theta < 1$, but the steady-state equilibrium is not globally stable. It seems that $\frac{1}{\alpha} > 1$ for $\alpha \in (0,1)$ determines the shape of equation (1), making the steady-state equilibrium globally stable. It would be good if someone could show me a counter-example where $\theta < 1$, but the steady-state equilibrium is not globally stable. It would be better if someone could show me how to prove proposition 9.4 formally.

Acknowledgement: The graphs are modified from those generated by Wolframalpha.

**Edit (Apr 19, 2017)**:
Case $\theta = 0$:
Note that when the textbook derives (9.17), it implicitly assumes that $\theta \neq 0$ (for derivation of Euler equation for consumption in P.333 of 2009 edition of the textbook). When $\theta = 0$, equation (1) no longer applies. Returning to the utility maximization problem with $\theta = 0$:

$$
\text{max } U(t) = c\_1(t) + \beta(c\_2(t+1)) \text{ such that } c\_1(t) + \frac{c\_2(t+1)}{R(t+1)} = w(t) \\
\Leftrightarrow
\text{max } U(t) = c\_1(t) + \beta(w(t) - c\_1(t))R(t+1) = c\_1(t) (1 - \beta R(t+1)) + \beta R(t+1)w(t) \\ \text{ ...Should treat R(t+1) as given as consumer's own optimization problem}
$$

s(t) has to be non-negative for $k(t+1) = \frac{s(t)}{1+n}$ and k(t+1) is non-negative.
$$
c\_1(t)^\* = \begin{cases}
w(t)\text{, for }\beta R(t+1) <1 \\
[0, w(t)]\text{, for }\beta R(t+1) = 1 \\
0\text{, for }\beta R(t+1) > 1 \\
\end{cases}
$$
$$
s(t)^\* = \begin{cases}
0\text{, for }\beta R(t+1) <1 \\
w(t) - c\_1(t)^\* \in [0, w(t)]\text{, for }\beta R(t+1) = 1 \\
w(t)\text{, for }\beta R(t+1) > 1 \\
\end{cases}
$$
For $R(t+1) = f'(k(t+1)) = \alpha k(t+1)^{\alpha - 1}$,
$$
k(t+1) = \frac{s(t)}{1+n} = \begin{cases}
0\text{, for }\beta R(t+1) <1 \Leftrightarrow k(t+1) < (\alpha \beta)^{\frac{1}{1 - \alpha}} \\
\frac{w(t) - c\_1(t)}{1+n} \in [0, \frac{w(t)}{1+n}]\text{, for }\beta R(t+1) = 1 \Leftrightarrow k(t+1) = (\alpha \beta)^{\frac{1}{1 - \alpha}}\\
\frac{w(t)}{1+n} = \frac{k(t)^\alpha - k(t)\alpha k(t)^{\alpha -1}}{1+n} = \frac{1-\alpha}{1+n}k(t)^\alpha\text{, otherwise} \Leftrightarrow k(t) > [\frac{1+n}{1 - \alpha} (\alpha \beta)^{\frac{1}{1 - \alpha}}]^{\frac{1}{\alpha}} \\
\end{cases}
$$

Cases:  
Case 1: $\beta R(t+1) < 1 \Leftrightarrow R(t+1) < \frac{1}{\beta}$:
As the production function $f(k)$ is Cobb-Douglas, it satisfies the Inada condition: $lim\_{k(t) \to 0} f'(k(t)) = \infty$. But as $f'(k(t)) = R(t)$, $lim\_{k(t) \to 0} R(t) < \infty$ for $R(t) < \frac{1}{\beta} < \infty$ as $\beta \in (0,1)$, violating the Inada condition. This contradiction means this case is impossible.

Case 2: $\beta R(t+1) = 1 \Leftrightarrow \beta \alpha k(t+1)^{\alpha-1} = 1 \Leftrightarrow k(t+1)^{\alpha-1} = \frac{1}{\alpha \beta} \Leftrightarrow k(t+1)^{1-\alpha} = \alpha \beta$:
Denote saving rate at t as $\mathcal{S}(t) = \frac{s(t)}{w(t)}$. $k(t+1) = \frac{s(t)}{1+n} = \frac{\mathcal{S}(t)w(t)}{1+n} = \frac{\mathcal{S}(t)(1-\alpha)k(t)^\alpha}{1+n}$. At steady state, $k^\* = \frac{\mathcal{S}^\*(1-\alpha){k^\*}^\alpha}{1+n}$, meaning $\mathcal{S}^\* = \frac{1+n}{1-\alpha}{k^\*}^{1-\alpha} = \frac{1+n}{1-\alpha} \alpha \beta$. For $\mathcal{S}^\* > 1 \Leftrightarrow (1+n)\alpha\beta > 1 - \alpha \Leftrightarrow \beta > \frac{1 - \alpha}{\alpha (1+n)}$, which is possible. As saving rate cannot be larger than 1, this contradiction means this case is impossible.

Case 3: $\beta R(t+1) > 1$:
This case is possible.
$$
k(t+1) = \frac{1-\alpha}{1+n} k(t)^\alpha
$$
We can draw the graph:[![enter image description here](https://i.sstatic.net/t6Q94.png)](https://i.sstatic.net/t6Q94.png)  
The red line is 45-degrees line. The blue line is $k(t+1) = \frac{1-\alpha}{1+n} k(t)^\alpha$ where $0 < \alpha < 1$. For all k(0) > 0, k will converge to steady state $k^\* = \frac{1-\alpha}{1+n} {k^\*}^\alpha \Leftrightarrow k^\* = [\frac{1-\alpha}{1+n}]^{\frac{1}{1-\alpha}}$. The steady-state equilibrium is globally stable.

**Tags:** macroeconomics, economic-growth


---

## 1 answers

### Answer by username_1 on 2017-04-05. Score: 1

In p. 334 of the book (2009 edition), I read:

> **Proposition 9.4** In the overlapping-generations model with two-period lived households, Cobb-Douglas technology and CRRA
> preferences, there exists a unique steady-state equilibrium with the
> capital-labor ratio $k^\*$ given by (9.15) and **for any $\theta >0$**
> this steady-state equilibrium is globally
> stable for all k (0) > 0.

...it does not say "...as long as $\theta \geq 1$".

**ADDENDUM :** The case $\theta =0$

The OP elaborated on the case $\theta=0$. So let's attempt to solve this.

Here, intertemporal utility becomes linear in consumption. Also, we will allow consumption to be zero -assume that there is an exogenous endowment that takes care of the biological needs of the individuals here and cannot be used for any other purpose, so the consumption choice is not about survival. Still, we are interested in economically interesting cases, so we exclude the case $c\_2=0$ because this would mean that the capital stock will be zero, and the economy will stop. So we allow only for $c\_1=0$. Then, incorporating the equality constraint in the objective function we solve

$$\max U = c\_1(t) + \beta R(t+1)[w(t)-c\_1(t)]\;\;\; s.t.\;\; 0\leq c\_1(t) < w(t)$$

So the Lagrangian is

$$\Lambda = c\_1(t) + \beta R(t+1)[w(t)-c\_1(t)] + \lambda c\_1(t)$$

and the f.o.c. is

$$1 - \beta R(t+1) +\lambda \leq 0,\;\;\;\; (1 - \beta R(t+1) +\lambda)\cdot c\_1(t) = 0$$

I will ignore issues of second-order conditions for a maximum.

**CASE 1.** Assume
$$c\_1^\*(t) >0 \implies \lambda=0 \implies 1 = \beta R(t+1)$$

$$\implies 1 = \beta ak(t+1)^{a-1} \implies k(t+1) = (a\beta)^{1/(1-a)}$$

But this must hold for *all* time periods, so it can hold only for a specific value of initial capital endowment and not for all $k(0) >0$. So already, we see that this case does not satisfy the conditions for which Proposition 9.4 is stated.

**CASE 2.** Assume
$$c\_1^\*(t) =0 \implies \lambda>0 \implies 1 - \beta R(t+1) +\lambda \leq 0 \implies \beta R(t+1) > 1$$

$$\implies \beta ak(t+1)^{a-1} >1 \implies k(t+1) < (a\beta) ^{1/(1-a)}$$

Here the capital stock must remain below a certain fixed level at all time periods. Again, this imposes constraints on the value of the initial capital stock $k(0)$.

So in both cases, if a steady-state equilibrium exists, it cannot be "*globally* stable, for *all* $k(0)>0$".

> **Comments:**
> **username_0**: It seems that I am reading the 2007 edition of the book, which is the draft version. But, even if $\theta = 0$, equation 1 becomes $k(t) = [\frac{1+n}{1-\alpha}k(t+1)]^{\frac{1}{\alpha}}$, the graph with equation 1 and the 45-degrees line will be similar to those shown in my post, the steady-state equilibrium will still be globally stable. Am I correct?  
> **username_1**: @ChrisCheung I would suggest to go back at the beginning of this utility optimization problem and to solve it with $\theta =0$, looking also at what happens to the savings rate.  
> **username_0**: I did the following computation: $$ \text{max } U(t) = c_1(t) + \beta(w(t) - c_1(t))R(t+1) \\ = w(t) - s(t) + \frac{\alpha\beta}{(1+n)^{\alpha - 1}}  s(t)^\alpha $$ First-order condition with respect to $s(t)$ gives: $s(t)^* = (1+n)(\alpha^2 \beta)^{\frac{1}{1 - \alpha}}$ Then, $ k(t+1) = \frac{s(t)}{1+n} = (\alpha^2 \beta)^{\frac{1}{1 - \alpha}}\text{ ,a constant}$. This means that the steady-state equilibrium where $k^* = (\alpha^2 \beta)^{\frac{1}{1 - \alpha}}$ is globally stable as even k deviates from $k^*$ for one period, next period k will return to steady state k^*. Am I correct?  
> **username_1**: @ChrisCheung The issue with the case $\theta =0$ is that makes utillity linear in consumption, (both contemporaneous and intertemporal), which introduces issues of uniqueness. Also I get the impression that you have differentiated the Gross Interest rate with respect to savings, but this is the "central-planner" solution, not the competitive one where factor rewards are treated by individuals as market-determined and so exogenous with respect to their optimization problems.  
> **username_0**: Thank you for reminding me this is not a central planner problem. I have added my derivation for the case $\theta = 0$ at the bottom of my question. But, I do not see this case introduces issues of uniqueness. I think $\theta = 0$ can also be included as the condition where the steady-state equilibrium is globally stable. Can you explain why the case $\theta=0$ introduces uniqueness problem?  
> **username_1**: @ChrisCheung I updated my answer with some elaboration on the case $\theta =0$  

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