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:
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:
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:
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.
