<p>In solow model, you can find steady-state of transformed variable <span class="math-container">$\hat{y}^{*} =\left(\frac{s}{n+g+\delta+n g}\right)^{\frac{a}{1-a}}$</span>. If <span class="math-container">$s$</span> goes up, <span class="math-container">$\hat{y^*}$</span> goes up, and we can say a higher <span class="math-container">$s$</span> is better by just comparing old and new steady-states. Maybe, because <span class="math-container">$s$</span> does not affect the balanced growth path of the steady-state, comparing <span class="math-container">$\hat{y^*}$</span> for different values of <span class="math-container">$s$</span> is enough even though what we care about is non-transformed variables.</p>
<p>However, the same cannot be said for an increase in <span class="math-container">$g$</span>, right? Thus, we need to compare the entire equilibrium path of the steady-state of the non-transformed variables. For example, <span class="math-container">$y_{t}^{*}=A_{0}(1+g)^{t}\left(\frac{s}{n+g+\delta+n g}\right)^{\frac{a}{1-a}}$</span> because <span class="math-container">$g$</span> affects the entire balanced growth path, and not just <span class="math-container">$\hat{y^*}$</span>.</p>
<p>But the problem is the choice of <span class="math-container">$t$</span> affects whether <span class="math-container">$y_{t}^{*}$</span> under a higher <span class="math-container">$g$</span> is better or worse. For example, for <span class="math-container">$t&lt;n$</span>, <span class="math-container">$y_{t}^{*}$</span> under a higher <span class="math-container">$g$</span> is worse. And for <span class="math-container">$t&gt;n+i, \; i = (0,1,2,3...)$</span>, <span class="math-container">$y_{t}^{*}$</span> under a higher <span class="math-container">$g$</span> is better.</p>
<p>So I guess a higher <span class="math-container">$g$</span> is better for an economy in the long-run <span class="math-container">$t&gt;n+i, \; i = (0,1,2,3...)$</span>. If we wanna compare <span class="math-container">$y_{t}^{*}$</span> under old and new <span class="math-container">$g$</span>, what <span class="math-container">$t$</span> to choose?</p>

CC BY-SA 4.0
2022-02-24T05:42:38.050Z
Can you elaborate on what you mean by "*better*"? In terms of consumption, or utility? Or just raw output, though most people would argue that $s = 100\%$ is a bad way to live?
Giskard
CC BY-SA 4.0
2022-02-24T05:44:03.957Z
Also, if you really want to have a normative "*better*" as in one path is better than the other then why are you assuming the economy is always on the equilibrium path instead of just approaching it from a specific capital level? An analogy is examining if a KIA or a Ferrari would be a better car to have, instead of thinking about what to do in your current financial situation.
Giskard
CC BY-SA 4.0
2022-02-24T05:48:31.533Z
Finally, I don't understand what you mean by "*what $t$ to choose"*. For what purpose? You can compare "*old and new*" $y_t^*$ for any $t$.
Giskard
CC BY-SA 4.0
2022-02-24T15:37:08.603Z
What is $y^*$ and what is $\hat{y}^*$ ?
erik
CC BY-SA 4.0
2022-02-24T15:40:11.883Z
Also, shouldn’t there be a 1 in the denominator of the steady state fraction?
erik
CC BY-SA 4.0
2022-02-24T15:43:37.733Z
@Giskard  Solow model has no utility. And there is already a well defined optimum consumption for this model. “Better” in the context of this type of model is generally limited to asking which parameter changes generate a higher level effect or, trivially, a higher rate effect.
erik
CC BY-SA 4.0
2022-02-24T19:01:57.247Z
@erik I am not sure I understand - do you have an interpretation of what OP means when they ask which $\hat{y}^*$ steady state level is better? Will you please share it?
Giskard
CC BY-SA 4.0
2022-02-24T20:13:13.530Z
I think the OP means to say “higher” when they write “better”.
erik
<solow>
Steady-state VS equilibrium path
Emmanuel Ameyaw
# Steady-state VS equilibrium path

(Asked by: username_0 on 2022-02-24)

In solow model, you can find steady-state of transformed variable $\hat{y}^{\*} =\left(\frac{s}{n+g+\delta+n g}\right)^{\frac{a}{1-a}}$. If $s$ goes up, $\hat{y^\*}$ goes up, and we can say a higher $s$ is better by just comparing old and new steady-states. Maybe, because $s$ does not affect the balanced growth path of the steady-state, comparing $\hat{y^\*}$ for different values of $s$ is enough even though what we care about is non-transformed variables.

However, the same cannot be said for an increase in $g$, right? Thus, we need to compare the entire equilibrium path of the steady-state of the non-transformed variables. For example, $y\_{t}^{\*}=A\_{0}(1+g)^{t}\left(\frac{s}{n+g+\delta+n g}\right)^{\frac{a}{1-a}}$ because $g$ affects the entire balanced growth path, and not just $\hat{y^\*}$.

But the problem is the choice of $t$ affects whether $y\_{t}^{\*}$ under a higher $g$ is better or worse. For example, for $t, $y\_{t}^{\*}$ under a higher $g$ is worse. And for $t>n+i, \; i = (0,1,2,3...)$, $y\_{t}^{\*}$ under a higher $g$ is better.

So I guess a higher $g$ is better for an economy in the long-run $t>n+i, \; i = (0,1,2,3...)$. If we wanna compare $y\_{t}^{\*}$ under old and new $g$, what $t$ to choose?

**Tags:** solow
> **Comments:**
> **username_1**: Can you elaborate on what you mean by "*better*"? In terms of consumption, or utility? Or just raw output, though most people would argue that $s = 100\%$ is a bad way to live?  
> **username_1**: Also, if you really want to have a normative "*better*" as in one path is better than the other then why are you assuming the economy is always on the equilibrium path instead of just approaching it from a specific capital level? An analogy is examining if a KIA or a Ferrari would be a better car to have, instead of thinking about what to do in your current financial situation.  
> **username_1**: Finally, I don't understand what you mean by "*what $t$ to choose"*. For what purpose? You can compare "*old and new*" $y_t^*$ for any $t$.  
> **username_2**: What is $y^*$ and what is $\hat{y}^*$ ?  
> **username_2**: Also, shouldn’t there be a 1 in the denominator of the steady state fraction?  
> **username_2**: @Giskard  Solow model has no utility. And there is already a well defined optimum consumption for this model. “Better” in the context of this type of model is generally limited to asking which parameter changes generate a higher level effect or, trivially, a higher rate effect.  
> **username_1**: @erik I am not sure I understand - do you have an interpretation of what OP means when they ask which $\hat{y}^*$ steady state level is better? Will you please share it?  
> **username_2**: I think the OP means to say “higher” when they write “better”.  

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## 0 answers

