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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 $π‘˜^π‘Ž_{𝑑+1}βˆ’π‘˜_{𝑑+2}=π‘Žπ‘π‘˜^{π‘Žβˆ’1}_{𝑑+1}π‘˜^π‘Ž_π‘‘βˆ’π‘Žπ‘π‘˜^π‘Ž_{𝑑+1}$, we divide by π‘˜^π‘Ž_{𝑑+1} and what I get is $1βˆ’π‘”=π‘Žπ‘π‘”βˆ’π‘Žπ‘$. Now based on the results, $g$ should somehow be $π‘Žπ‘$, but I'm stuck here

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    $\begingroup$ Are you sure it's not that $k_{t+1}=gk_t^a$? $\endgroup$
    – Herr K.
    Feb 4 at 19:31
  • $\begingroup$ @HerrK. yes, my mistake $\endgroup$ Feb 4 at 19:34
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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 $g$.

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  • $\begingroup$ I edited it but I'm not sure if what I wrote is correct :/ $\endgroup$ Feb 4 at 21:58
  • $\begingroup$ @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 $g$. $\endgroup$
    – Herr K.
    Feb 4 at 22:32
  • $\begingroup$ 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. $\endgroup$ Sep 16 at 10:19
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    $\begingroup$ @MaybelineLee: Solving $1-g=\frac{ab}{g}-ab$ for $g$ yields $g=ab$ or $g=1$. $\endgroup$
    – Herr K.
    Sep 16 at 12:59

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