<p>Consider the (simplified) Solow equation, which is a first-order non-linear difference equation:
<span class="math-container">\begin{equation}
k_{t+1} = sf(k_t) + (1-\delta)k_t
\end{equation}</span>
where <span class="math-container">$s$</span> and <span class="math-container">$\delta$</span> are two constants and <span class="math-container">$s,\delta \in (0,1)$</span>.
I assume that <span class="math-container">$f: R_+ \to R_+$</span> is twice continuously differentiable and <span class="math-container">$f'(k)&gt;0$</span> and <span class="math-container">$f''(k)&lt;0$</span> for all <span class="math-container">$k&gt;0$</span>. So, <span class="math-container">$f(k)$</span> is assumed to be strictly concave.</p>
<p>At the steady state, <span class="math-container">$k^{\ast}$</span> (fixed point) satsifies the following equation:
<span class="math-container">\begin{equation}
s f(k^{\ast}) = \delta k^{\ast}
\end{equation}</span></p>
<p>Now, pls, tell me if it's correct how I prove global stability:</p>
<p>First,
<span class="math-container">\begin{equation}
k_{t+1} = g(k_t)
\end{equation}</span>
and for <span class="math-container">$k^{\ast}$</span> being a fixed point it must be
<span class="math-container">\begin{equation}
k^{\ast} = g(k^{\ast})
\end{equation}</span></p>
<p>Then, for all <span class="math-container">$k_t \in (0,k^{\ast})$</span>,
<span class="math-container">\begin{equation}
k_{t+1} - k^{\ast} = g(k_t) - g(k^{\ast}) = - \int_{k_t}^{k^{\ast}} g'(k) dk &lt;0
\end{equation}</span>
The inequality holds because <span class="math-container">$g'(k)&gt;0$</span> for all <span class="math-container">$k$</span>.</p>
<p>This result shall ensure that for any <span class="math-container">$k_t \in (0,k^{\ast})$</span>, then <span class="math-container">$k_{t+1}$</span> is always lower than <span class="math-container">$k^{\ast}$</span>, and the convergence of <span class="math-container">$k \to k^{\ast}$</span> is ensured from the fact that
<span class="math-container">\begin{equation}
\gamma(k_t) = \frac{k_{t+1} -k_t}{k_t} = \frac{s f(k_t) + (1-\delta)k_t - k_t}{k_t} = \frac{sf(k_t)}{k_t} - \delta
\end{equation}</span>
and since <span class="math-container">$h(k_t) \equiv k_t/f(k_t)$</span> is strictly increasing in <span class="math-container">$k$</span> and <span class="math-container">$h(k^{\ast})=s/\delta$</span>, then the growth rate of <span class="math-container">$k$</span> is positive but strictly decreasing in <span class="math-container">$k$</span>, with <span class="math-container">$\gamma(k)&gt;0$</span> iff <span class="math-container">$k&lt; k^{\ast}$</span> and <span class="math-container">$\gamma(k)=0$</span> iff <span class="math-container">$k= k^{\ast}$</span>.</p>
<p>A similar argument applies to <span class="math-container">$k_t &gt; k^{\ast}$</span>. Is that correct?</p>
<p>UPDATE:
For my understanding, global stability means that <span class="math-container">$k^{\ast}$</span> is globally asymptotically stable if for all <span class="math-container">$k_0 &gt; 0$</span>, for any solution <span class="math-container">$\{k_t\}_{t=0}^{\infty}$</span>, we have <span class="math-container">$k_t \to k^{\ast}$</span>. In plain English, it does not matter what initial value the capital takes as long as its initial value belongs to the domain of <span class="math-container">$g()$</span>. Then, <span class="math-container">$k_t$</span> converges to the steady state <span class="math-container">$k^{\ast}$</span>.</p>

CC BY-SA 4.0
2024-03-19T21:27:37.610Z
I mean [we already went over this once](https://economics.stackexchange.com/questions/57806/proving-the-global-stability-in-the-solow-swan-model/). Could you please include what you think is the definition of global stability in your question? Because you seem to think it has something to do with convergence.
Giskard
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2024-03-19T21:35:17.327Z
@Giskard you are right, but going through Acemoglu's book gave me this idea. Let me elaborate on what I mean by global stability in the question above
Maximilian
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2024-03-20T04:43:09.530Z
Notice that in your definition you suddenly have an extra word: "*For my understanding, global stability means that $k^{\ast}$ is globally **asymptotically** stable*". This is indeed the definition for **asymptotic** stability. If this (convergence) is what you want, you should edit your question so that you always write asymptotically stable instead of just stable.
Giskard
CC BY-SA 4.0
2024-03-20T05:44:16.013Z
Also, it would be nice to see a clearer version of your argument, and if possible more general version of the question, something like this: In a one dimensional discrete time dynamic system $$x_{t+1} = f(x_t)$$ if, for a certain equilibrium $x^* \in \mathbb{R}$ we have
\begin{array}{lc}
f(x_t)/x_t >1 & \text{if } x_t < x^* \\ 
f(x_t)/x_t <1 & \text{if } x_t > x^* 
\end{array}
then is $x^*$
 asymptotically stable? Your argument may differ, I find it somewhat unclear.
Giskard
CC BY-SA 4.0
2024-03-20T20:20:14.640Z
@Giskard I am a bit confused.  I had understood that stability was a concept typically explored as the time approaches infinity, i.e. $t \to \infty$. My understanding stems from following Acemoglu's book, who has recently shared some presentation slides (pag.40) : https://economics.mit.edu/sites/default/files/inline-files/Economic%20Growth%20Lectures%202%20and%203%202022_0.pdf
Maximilian
CC BY-SA 4.0
2024-03-20T20:24:23.690Z
Note that in all the definitions the word "asymptotically" is included. See also [this question](https://math.stackexchange.com/q/2628567/312777).
Giskard
CC BY-SA 4.0
2024-03-20T20:33:49.590Z
@Giskard, I appreciate your clarification. I thought the two definitions (stable and asymptotically stable) were the same.
Maximilian
CC BY-SA 4.0
2024-03-20T20:37:49.090Z
@Giskard Is the proof of global asymptotic stability correct? Btw, I have amended the title of the question
Maximilian
CC BY-SA 4.0
2024-03-20T20:57:07.077Z
It would be nice to see a clearer version of your argument, I find it somewhat unclear.
Giskard
CC BY-SA 4.0
2024-03-20T21:19:00.203Z
@Giskard I've elaborated on what Acemoglu says in his book (which is also included in the slides). Which part of the discussion isn't clear? If there are any uncertainties, perhaps I'm conveying something incorrectly
Maximilian
CC BY-SA 4.0
2024-03-20T22:16:21.273Z
>>convergence of $k \to k^{\ast}$ is ensured from the fact that 
\begin{equation}
\gamma(k_t) = \frac{k_{t+1} -k_t}{k_t} = \frac{s f(k_t) + (1-\delta)k_t - k_t}{k_t} = \frac{sf(k_t)}{k_t} - \delta
\end{equation}
and since $h(k_t) \equiv k_t/f(k_t)$ is strictly increasing in $k$ and $h(k^{\ast})=s/\delta$, then the growth rate of $k$ is positive but strictly decreasing in $k$, with $\gamma(k)>0$ iff $k< k^{\ast}$ 
<<
Why does the stuff that you say ensures convergence actually ensure convergence?
Giskard
CC BY-SA 4.0
2024-03-21T08:50:37.877Z
@Giskard if I understand correctly, if the level of capital per worker is below the steady-state level, then it keeps growing until $k=k^{\ast}$. Then, $k$ asymptotically converges to $k^{\ast}$. For $k>k^{\ast}$ you have negative growth, and capital per worker de-grows till $k \to k^{\ast}$
Maximilian
CC BY-SA 4.0
2024-03-21T13:17:37.360Z
Do you feel that this is a rigorously laid out argument?
Giskard
CC BY-SA 4.0
2024-03-21T22:40:40.830Z
I am not sure about it but that's why I'm asking. However, that's what was provided in the Acemoglu's book. I'd love to vote your answer providing a rigorous proof of global asymptotic stability
Maximilian
CC BY-SA 4.0
2024-03-22T04:38:43.123Z
The question as currently phrased is not asking for a proof, it is asking if your proof is correct.
Giskard
CC BY-SA 4.0
2024-03-22T07:36:09.203Z
I am being somewhat nitpicky, so I want to clarify: it is not that the proof you convey is incorrect, but it is definitely not rigorous. There is no reference to any theorems, it just intuitively seems true. (See also: the Earth is flat.) IMO one has to think about whether positive growth rates without lower bounds guarantee convergence. A general theorem on this could settle the issue, and then one would only have to prove that the system described in Solow fulfills the conditions of the theorem.
Giskard
CC BY-SA 4.0
2024-03-22T09:40:08.137Z
@Giskard I see. Can you suggest any reference providing a rigorous proof about global asymptotic stability for the Solow model?
Maximilian
CC BY-SA 4.0
2024-03-22T11:25:11.950Z
Nope, my education in economic dynamics was also pretty poor. It is my impression that books focusing on economic aspects are rather handwavy about mathematics, which is unfortunate when proving general statements. Do you need a rigorous proof? And were the [recommendations by Bakerstreet under the other question](https://economics.stackexchange.com/questions/57806/proving-the-global-stability-in-the-solow-swan-model/#comment95884_57806) not to your satisfaction?
Giskard
CC BY-SA 4.0
2024-03-22T13:31:57.753Z
@Maximilan @ Giskard Thank you for mentioning my references. But the book of Braun is only about differential equations, and in the book of Gandolfo, as far as I remember, there is no proof of the stability of the Solow model in discrete time. As far as I remember (now I am not at home and I cant consult books) Acemoglou provides what needed, including also the relevant theorems.
BakerStreet
<macroeconomics><economic-growth><solow>
Global Asymptotic Stability in the Solow Model - Rigorous proof
Maximilian
# Global Asymptotic Stability in the Solow Model - Rigorous proof

(Asked by: username_0 on 2024-03-19)

Consider the (simplified) Solow equation, which is a first-order non-linear difference equation:
\begin{equation}
k\_{t+1} = sf(k\_t) + (1-\delta)k\_t
\end{equation}
where $s$ and $\delta$ are two constants and $s,\delta \in (0,1)$.
I assume that $f: R\_+ \to R\_+$ is twice continuously differentiable and $f'(k)>0$ and $f''(k)<0$ for all $k>0$. So, $f(k)$ is assumed to be strictly concave.

At the steady state, $k^{\ast}$ (fixed point) satsifies the following equation:
\begin{equation}
s f(k^{\ast}) = \delta k^{\ast}
\end{equation}

Now, pls, tell me if it's correct how I prove global stability:

First,
\begin{equation}
k\_{t+1} = g(k\_t)
\end{equation}
and for $k^{\ast}$ being a fixed point it must be
\begin{equation}
k^{\ast} = g(k^{\ast})
\end{equation}

Then, for all $k\_t \in (0,k^{\ast})$,
\begin{equation}
k\_{t+1} - k^{\ast} = g(k\_t) - g(k^{\ast}) = - \int\_{k\_t}^{k^{\ast}} g'(k) dk <0
\end{equation}
The inequality holds because $g'(k)>0$ for all $k$.

This result shall ensure that for any $k\_t \in (0,k^{\ast})$, then $k\_{t+1}$ is always lower than $k^{\ast}$, and the convergence of $k \to k^{\ast}$ is ensured from the fact that
\begin{equation}
\gamma(k\_t) = \frac{k\_{t+1} -k\_t}{k\_t} = \frac{s f(k\_t) + (1-\delta)k\_t - k\_t}{k\_t} = \frac{sf(k\_t)}{k\_t} - \delta
\end{equation}
and since $h(k\_t) \equiv k\_t/f(k\_t)$ is strictly increasing in $k$ and $h(k^{\ast})=s/\delta$, then the growth rate of $k$ is positive but strictly decreasing in $k$, with $\gamma(k)>0$ iff $k< k^{\ast}$ and $\gamma(k)=0$ iff $k= k^{\ast}$.

A similar argument applies to $k\_t > k^{\ast}$. Is that correct?

UPDATE:
For my understanding, global stability means that $k^{\ast}$ is globally asymptotically stable if for all $k\_0 > 0$, for any solution $\{k\_t\}\_{t=0}^{\infty}$, we have $k\_t \to k^{\ast}$. In plain English, it does not matter what initial value the capital takes as long as its initial value belongs to the domain of $g()$. Then, $k\_t$ converges to the steady state $k^{\ast}$.

**Tags:** macroeconomics, economic-growth, solow
> **Comments:**
> **username_1**: I mean [we already went over this once](https://economics.stackexchange.com/questions/57806/proving-the-global-stability-in-the-solow-swan-model/). Could you please include what you think is the definition of global stability in your question? Because you seem to think it has something to do with convergence.  
> **username_0**: @Giskard you are right, but going through Acemoglu's book gave me this idea. Let me elaborate on what I mean by global stability in the question above  
> **username_1**: Notice that in your definition you suddenly have an extra word: "*For my understanding, global stability means that $k^{\ast}$ is globally **asymptotically** stable*". This is indeed the definition for **asymptotic** stability. If this (convergence) is what you want, you should edit your question so that you always write asymptotically stable instead of just stable.  
> **username_1**: Also, it would be nice to see a clearer version of your argument, and if possible more general version of the question, something like this: In a one dimensional discrete time dynamic system $$x_{t+1} = f(x_t)$$ if, for a certain equilibrium $x^* \in \mathbb{R}$ we have \begin{array}{lc} f(x_t)/x_t >1 & \text{if } x_t < x^* \\  f(x_t)/x_t <1 & \text{if } x_t > x^*  \end{array} then is $x^*$  asymptotically stable? Your argument may differ, I find it somewhat unclear.  
> **username_0**: @Giskard I am a bit confused.  I had understood that stability was a concept typically explored as the time approaches infinity, i.e. $t \to \infty$. My understanding stems from following Acemoglu's book, who has recently shared some presentation slides (pag.40) : https://economics.mit.edu/sites/default/files/inline-files/Economic%20Growth%20Lectures%202%20and%203%202022_0.pdf  
> **username_1**: Note that in all the definitions the word "asymptotically" is included. See also [this question](https://math.stackexchange.com/q/2628567/312777).  
> **username_0**: @Giskard, I appreciate your clarification. I thought the two definitions (stable and asymptotically stable) were the same.  
> **username_0**: @Giskard Is the proof of global asymptotic stability correct? Btw, I have amended the title of the question  
> **username_1**: It would be nice to see a clearer version of your argument, I find it somewhat unclear.  
> **username_0**: @Giskard I've elaborated on what Acemoglu says in his book (which is also included in the slides). Which part of the discussion isn't clear? If there are any uncertainties, perhaps I'm conveying something incorrectly  
> **username_1**: >>convergence of $k \to k^{\ast}$ is ensured from the fact that  \begin{equation} \gamma(k_t) = \frac{k_{t+1} -k_t}{k_t} = \frac{s f(k_t) + (1-\delta)k_t - k_t}{k_t} = \frac{sf(k_t)}{k_t} - \delta \end{equation} and since $h(k_t) \equiv k_t/f(k_t)$ is strictly increasing in $k$ and $h(k^{\ast})=s/\delta$, then the growth rate of $k$ is positive but strictly decreasing in $k$, with $\gamma(k)>0$ iff $k< k^{\ast}$  << Why does the stuff that you say ensures convergence actually ensure convergence?  
> **username_0**: @Giskard if I understand correctly, if the level of capital per worker is below the steady-state level, then it keeps growing until $k=k^{\ast}$. Then, $k$ asymptotically converges to $k^{\ast}$. For $k>k^{\ast}$ you have negative growth, and capital per worker de-grows till $k \to k^{\ast}$  
> **username_1**: Do you feel that this is a rigorously laid out argument?  
> **username_0**: I am not sure about it but that's why I'm asking. However, that's what was provided in the Acemoglu's book. I'd love to vote your answer providing a rigorous proof of global asymptotic stability  
> **username_1**: The question as currently phrased is not asking for a proof, it is asking if your proof is correct.  
> **username_1**: I am being somewhat nitpicky, so I want to clarify: it is not that the proof you convey is incorrect, but it is definitely not rigorous. There is no reference to any theorems, it just intuitively seems true. (See also: the Earth is flat.) IMO one has to think about whether positive growth rates without lower bounds guarantee convergence. A general theorem on this could settle the issue, and then one would only have to prove that the system described in Solow fulfills the conditions of the theorem.  
> **username_0**: @Giskard I see. Can you suggest any reference providing a rigorous proof about global asymptotic stability for the Solow model?  
> **username_1**: Nope, my education in economic dynamics was also pretty poor. It is my impression that books focusing on economic aspects are rather handwavy about mathematics, which is unfortunate when proving general statements. Do you need a rigorous proof? And were the [recommendations by Bakerstreet under the other question](https://economics.stackexchange.com/questions/57806/proving-the-global-stability-in-the-solow-swan-model/#comment95884_57806) not to your satisfaction?  
> **username_2**: @Maximilan @ Giskard Thank you for mentioning my references. But the book of Braun is only about differential equations, and in the book of Gandolfo, as far as I remember, there is no proof of the stability of the Solow model in discrete time. As far as I remember (now I am not at home and I cant consult books) Acemoglou provides what needed, including also the relevant theorems.  

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## 0 answers

