# Euler equation guess and verify

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 $$𝑘^𝑎_{𝑡+1}−𝑘_{𝑡+2}=𝑎𝑏𝑘^{𝑎−1}_{𝑡+1}𝑘^𝑎_𝑡−𝑎𝑏𝑘^𝑎_{𝑡+1}$$, we divide by 𝑘^𝑎_{𝑡+1} and what I get is $$1−𝑔=𝑎𝑏𝑔−𝑎𝑏$$. Now based on the results, $$g$$ should somehow be $$𝑎𝑏$$, but I'm stuck here

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

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 $$g$$.
• @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 $g$. Feb 4 at 22:32
• 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. Sep 16 at 10:19
• @MaybelineLee: Solving $1-g=\frac{ab}{g}-ab$ for $g$ yields $g=ab$ or $g=1$. Sep 16 at 12:59