<p>A system is explosive if its coefficients are non-stationary. </p>

<p>Stationary is an important property to have in dynamic models as it tells us that an equilibrium value is obtainable (which is important in finding the BGP). If your system is explosive no equilibrium (and BGP) exists.</p>

<p>In your example where you have the equality:
$$k_t-k^* \simeq \lambda^t (k_0-k^*)$$</p>

<p>If $\lambda\le1$, then as time progresses ($t=1,2,...,n$) the impact of a policy change approaches a new equilibrium and BGP.</p>

<p>However if $\lambda&gt;1$ then your system is explosive and equilibrium will not be reached, thus making the system unsolvable as $t$ tends to $\infty$.</p>

CC BY-SA 3.0
2018-03-11T17:13:36.450Z
2018-03-11T17:13:36.450Z
EconJohn
<p>Eq. (2.64) can be written as (at first order)</p>

<p>$$
k_{t+1} - k^* = \lambda (k_t - k^*) \tag{1a}
$$ </p>

<p>Define the quantity $\kappa_t$ as </p>

<p>$$
\kappa_t \stackrel{\rm def}{=} k_t - k^* \tag{2}
$$</p>

<p>So that</p>

<p>$$
\kappa_{t+1} = \lambda \kappa_t \tag{1b}
$$</p>

<p>This can be solved very easy by realizing that</p>

<p>\begin{eqnarray}
\kappa_1 &amp;=&amp;\lambda \kappa_0 \\
\kappa_2 &amp;=&amp;\lambda \kappa_1 = \lambda(\lambda\kappa_0) = \lambda^2\kappa_0 \\
&amp;\vdots&amp; \\
\kappa_t &amp;=&amp; \lambda^t \kappa_0 \tag{3}
\end{eqnarray}</p>

<p>So the dynamics is $\kappa$ is driven by the value $\lambda^t$. You can see that if $\lambda = 1$ then $\lambda^t=\lambda$ for each $t$ and $\kappa$ remains constant. However, if $\lambda &gt; 1$ the values $k^t$ will become increasingly large (<em>explodes</em>) as $t$ increases, that is </p>

<p>$$
\lim_{t\to \infty} \kappa_t \stackrel{\lambda &gt; 1}{=} \infty
$$</p>

<p>Similarly</p>

<p>$$
\lim_{t\to \infty} \kappa_t \stackrel{0&lt; \lambda &lt; 1}{=} 0
$$</p>

CC BY-SA 3.0
2018-03-11T15:26:02.437Z
@AlecosPapadopoulos Thank you for catching that, I just fixed it
caverac
CC BY-SA 3.0
2018-03-07T22:48:34.460Z
2018-03-11T15:25:34.940Z
2018-03-11T15:25:34.940Z
caverac
<p><a href="https://en.wikipedia.org/wiki/David_Romer" rel="noreferrer">David Romer</a> in his textbook <a href="https://rads.stackoverflow.com/amzn/click/com/B004HP3V5U" rel="noreferrer" rel="nofollow noreferrer">Advanced Macroeconomics (<em>Third Edition</em>)</a> writes regarding the speed of convergence of the <a href="https://en.wikipedia.org/wiki/Overlapping_generations_model" rel="noreferrer">Diamond model</a> the following:</p>
<p>(<em>Pg. 83</em>)</p>
<blockquote>
<p><em>Equation (2.60) [<span class="math-container">$k_{t+1}={1\over{(1+n)(1+g)}}{1\over{2+\rho}}(1-\alpha)k_t^\alpha$</span>] gives <span class="math-container">$k_{t+1}$</span> as a function of <span class="math-container">$k_t$</span>. The economy is on its balanced growth path when these two are equal, that is <span class="math-container">$k^*$</span> is defined by</em></p>
<p><span class="math-container">$k^*={1\over{(1+n)(1+g)}}{1\over{2+\rho}}(1-\alpha)k^{* \alpha}\space\space\space\space\space\space\space\space\space\space\space\space$</span> (2.61)</p>
<p><em>Solving this expression for <span class="math-container">$k^*$</span> yields</em></p>
<p><span class="math-container">$k^*=\left[{(1-\alpha)\over{(1+n)(1+g)(2+\rho)}}\right]^{1 \over{1-\alpha}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$</span> (2.62)</p>
<p><em>[...] We can also find out how quickly the economy converges to the balanced growth path. To do this, we again linearize around the balanced growth path. [...] Thus:</em></p>
<p><span class="math-container">$k_{t+1} \simeq k^*+\left( {{dk_{t+1}}\over{dk_t}}\Big{|}_{k_t=k^*}\right) \left(k_t-k^*\right)\space\space\space\space$</span> (2.64)</p>
<p><em>Let <span class="math-container">$\lambda$</span> denote <span class="math-container">$dk_{t+1}/dk_t$</span> evaluated at <span class="math-container">$k_t=k^*$</span>. With this definition, we can rewrite (2.64) [...]</em></p>
<p><span class="math-container">$k_t-k^* \simeq \lambda^t (k_0-k^*)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$</span> (2.65)</p>
<p><em>Where <span class="math-container">$k_0$</span> is the initial value of <span class="math-container">$k$</span></em></p>
</blockquote>
<h1>My Question</h1>
<p>(<em>Pg 84</em>)</p>
<blockquote>
<p><em>[...] if <span class="math-container">$\lambda$</span> is greater than 1, the system explodes</em></p>
</blockquote>
<p>What does it mean the system explodes when <span class="math-container">$\lambda$</span> is greater than 1? Why does it explode when <span class="math-container">$\lambda$</span> is greater than 1?</p>

<macroeconomics><steady-state><olg>
"If $\lambda$ is greater than than 1, the system explodes." Why does the system explode?
Bensstats
# "If $\lambda$ is greater than than 1, the system explodes." Why does the system explode?

(Asked by: username_0 on 2018-03-07)

[David Romer](https://en.wikipedia.org/wiki/David_Romer) in his textbook [Advanced Macroeconomics (*Third Edition*)](https://rads.stackoverflow.com/amzn/click/com/B004HP3V5U) writes regarding the speed of convergence of the [Diamond model](https://en.wikipedia.org/wiki/Overlapping_generations_model) the following:

(*Pg. 83*)

> *Equation (2.60) [$k\_{t+1}={1\over{(1+n)(1+g)}}{1\over{2+\rho}}(1-\alpha)k\_t^\alpha$] gives $k\_{t+1}$ as a function of $k\_t$. The economy is on its balanced growth path when these two are equal, that is $k^\*$ is defined by*
>
> $k^\*={1\over{(1+n)(1+g)}}{1\over{2+\rho}}(1-\alpha)k^{\* \alpha}\space\space\space\space\space\space\space\space\space\space\space\space$ (2.61)
>
> *Solving this expression for $k^\*$ yields*
>
> $k^\*=\left[{(1-\alpha)\over{(1+n)(1+g)(2+\rho)}}\right]^{1 \over{1-\alpha}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ (2.62)
>
> *[...] We can also find out how quickly the economy converges to the balanced growth path. To do this, we again linearize around the balanced growth path. [...] Thus:*
>
> $k\_{t+1} \simeq k^\*+\left( {{dk\_{t+1}}\over{dk\_t}}\Big{|}\_{k\_t=k^\*}\right) \left(k\_t-k^\*\right)\space\space\space\space$ (2.64)
>
> *Let $\lambda$ denote $dk\_{t+1}/dk\_t$ evaluated at $k\_t=k^\*$. With this definition, we can rewrite (2.64) [...]*
>
> $k\_t-k^\* \simeq \lambda^t (k\_0-k^\*)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ (2.65)
>
> *Where $k\_0$ is the initial value of $k$*

# My Question

(*Pg 84*)

> *[...] if $\lambda$ is greater than 1, the system explodes*

What does it mean the system explodes when $\lambda$ is greater than 1? Why does it explode when $\lambda$ is greater than 1?

**Tags:** macroeconomics, steady-state, olg


---

## 2 answers

### Answer by username_2 on 2018-03-07. Score: 10 | Accepted answer

Eq. (2.64) can be written as (at first order)

$$
k\_{t+1} - k^\* = \lambda (k\_t - k^\*) \tag{1a}
$$

Define the quantity $\kappa\_t$ as

$$
\kappa\_t \stackrel{\rm def}{=} k\_t - k^\* \tag{2}
$$

So that

$$
\kappa\_{t+1} = \lambda \kappa\_t \tag{1b}
$$

This can be solved very easy by realizing that

\begin{eqnarray}
\kappa\_1 &=&\lambda \kappa\_0 \\
\kappa\_2 &=&\lambda \kappa\_1 = \lambda(\lambda\kappa\_0) = \lambda^2\kappa\_0 \\
&\vdots& \\
\kappa\_t &=& \lambda^t \kappa\_0 \tag{3}
\end{eqnarray}

So the dynamics is $\kappa$ is driven by the value $\lambda^t$. You can see that if $\lambda = 1$ then $\lambda^t=\lambda$ for each $t$ and $\kappa$ remains constant. However, if $\lambda > 1$ the values $k^t$ will become increasingly large (*explodes*) as $t$ increases, that is

$$
\lim\_{t\to \infty} \kappa\_t \stackrel{\lambda > 1}{=} \infty
$$

Similarly

$$
\lim\_{t\to \infty} \kappa\_t \stackrel{0< \lambda < 1}{=} 0
$$

> **Comments:**
> **username_2**: @AlecosPapadopoulos Thank you for catching that, I just fixed it  

---

### Answer by username_1 on 2018-03-11. Score: 5

A system is explosive if its coefficients are non-stationary.

Stationary is an important property to have in dynamic models as it tells us that an equilibrium value is obtainable (which is important in finding the BGP). If your system is explosive no equilibrium (and BGP) exists.

In your example where you have the equality:
$$k\_t-k^\* \simeq \lambda^t (k\_0-k^\*)$$

If $\lambda\le1$, then as time progresses ($t=1,2,...,n$) the impact of a policy change approaches a new equilibrium and BGP.

However if $\lambda>1$ then your system is explosive and equilibrium will not be reached, thus making the system unsolvable as $t$ tends to $\infty$.


---

