<h2>Hint:</h2>
<p>Note that
<span class="math-container">\begin{equation}
k_{t+1}=gk_t^a \quad\Rightarrow\quad g=\frac{k_{t+1}}{k_t^a}=\frac{k_{t+2}}{k_{t+1}^a}.
\end{equation}</span></p>
<p>Rearrange your Euler condition into a form such that you can use <span class="math-container">$g$</span> to sub-out the above ratios. Then it should be straightforward to solve for <span class="math-container">$g$</span>.</p>

CC BY-SA 4.0
2021-02-04T21:58:02.167Z
I edited it but I'm not sure if what I wrote is correct :/
Maybeline Lee
CC BY-SA 4.0
2021-02-04T22:32:44.210Z
@MaybelineLee: Try cross-multiplying the denominators in the original equation, and then divide both sides by $k_{t+1}^a$. Simplify using the hint to get an equation that only involves $g,a,\beta$. Lastly solve for $g$.
Herr K.
CC BY-SA 4.0
2021-09-16T10:19:03.013Z
So I'm going through the derivations again on this one and it's still confusing me. We have $k_{t+1}^a-k_{t+2}=abk_{t+1}^{a-1}k_t^a-abk_{t+1}^a$, we divide by $k_{t+1}^a$ and what I get is $1-g=\frac{ab}{g}-ab$. Now based on the results, g should somehow be $ab$ but I'm just not getting it.
Maybeline Lee
CC BY-SA 4.0
2021-09-16T12:59:41.163Z
@MaybelineLee: Solving $1-g=\frac{ab}{g}-ab$ for $g$ yields $g=ab$ or $g=1$.
Herr K.
CC BY-SA 4.0
2021-02-04T20:04:47.850Z
2021-02-04T20:04:47.850Z
Herr K.
<p>So I have the Euler condition written as follows: <span class="math-container">$$\frac{1}{k_t^a-k_{t+1}}=\frac{a\beta k_{t+1}^{a-1}}{k_{t+1}^a-k_{t+2}}$$</span>
and it says that <span class="math-container">$k_{t+1}$</span> takes the form <span class="math-container">$gk_t^a$</span>, where g is an unknown to be determined. I know the result is <span class="math-container">$k_{t+1}=a\beta k_t^a$</span>, but don't know how the derivation goes to reach this result.</p>
<p>From the hint, I got that <span class="math-container">$^_{+1}−_{+2}=^{−1}_{+1}^_−^_{+1}$</span>, we divide by ^_{+1} and what I get is <span class="math-container">$1−=−$</span>. Now based on the results, <span class="math-container">$g$</span> should somehow be <span class="math-container">$$</span>, but I'm stuck here</p>

CC BY-SA 4.0
2021-02-04T19:31:06.423Z
Are you sure it's not that $k_{t+1}=gk_t^a$?
Herr K.
CC BY-SA 4.0
2021-02-04T19:34:17.010Z
@HerrK. yes, my mistake
Maybeline Lee
<macroeconomics>
Euler equation guess and verify
Maybeline Lee
# Euler equation guess and verify

(Asked by: username_0 on 2021-02-04)

So I have the Euler condition written as follows: $$\frac{1}{k\_t^a-k\_{t+1}}=\frac{a\beta k\_{t+1}^{a-1}}{k\_{t+1}^a-k\_{t+2}}$$
and it says that $k\_{t+1}$ takes the form $gk\_t^a$, where g is an unknown to be determined. I know the result is $k\_{t+1}=a\beta k\_t^a$, but don't know how the derivation goes to reach this result.

From the hint, I got that $^\_{+1}−\_{+2}=^{−1}\_{+1}^\_−^\_{+1}$, we divide by ^\_{+1} and what I get is $1−=−$. Now based on the results, $g$ should somehow be $$, but I'm stuck here

**Tags:** macroeconomics
> **Comments:**
> **username_1**: Are you sure it's not that $k_{t+1}=gk_t^a$?  
> **username_0**: @HerrK. yes, my mistake  

---

## 1 answers

### Answer by username_1 on 2021-02-04. Score: 5 | Accepted answer

## Hint:

Note that
\begin{equation}
k\_{t+1}=gk\_t^a \quad\Rightarrow\quad g=\frac{k\_{t+1}}{k\_t^a}=\frac{k\_{t+2}}{k\_{t+1}^a}.
\end{equation}

Rearrange your Euler condition into a form such that you can use $g$ to sub-out the above ratios. Then it should be straightforward to solve for $g$.

> **Comments:**
> **username_0**: I edited it but I'm not sure if what I wrote is correct :/  
> **username_1**: @MaybelineLee: Try cross-multiplying the denominators in the original equation, and then divide both sides by $k_{t+1}^a$. Simplify using the hint to get an equation that only involves $g,a,\beta$. Lastly solve for $g$.  
> **username_0**: So I'm going through the derivations again on this one and it's still confusing me. We have $k_{t+1}^a-k_{t+2}=abk_{t+1}^{a-1}k_t^a-abk_{t+1}^a$, we divide by $k_{t+1}^a$ and what I get is $1-g=\frac{ab}{g}-ab$. Now based on the results, g should somehow be $ab$ but I'm just not getting it.  
> **username_1**: @MaybelineLee: Solving $1-g=\frac{ab}{g}-ab$ for $g$ yields $g=ab$ or $g=1$.  

---

