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Consider the following version of the Ak model:$$V^*(A_0,K_0)=max\sum\beta^t\sum P(A^t)\frac{c_t(A^t)^{1-\sigma}}{1-\sigma}$$ st$$ k_{t+1}(A^t)+c_t(A^t)\leq Ak_t(A^{t+1})$$ and non-negativities

$A_t$ is an i.i.d. process with mean $1$ and $P(A_t)$ denotes the probability of a sequence $(A_0,A_1, ...,A_t)$. We assume that σ > 0 and β ∈ (0,1). Output is defined as $y_t = A_tk_t$.

This might seem dumb but why is the Euler equation written as this: $$c_t(A^t)^{-\sigma}=\beta E_t[A_{t+1}c_{t+1}{(A^{t+1})}^{-\sigma}]$$

Then there is put another assumption that $k_{t+1}(A^t)=sA_tk_t(A^{t-1})$ Derive the savings rate $s$ in terms of the parameters of the model. The answer given is $s=[\beta E_t(A_{t+1}^{1-\sigma})]^\frac{1}{\sigma}$. Which again, I'm not getting how it came to that :\

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  • $\begingroup$ I am not sure if I understand the question, are you asking how is it derived, or more generally why does Euler equation equates present consumption to the expected future one in these intertemporal problems? $\endgroup$
    – 1muflon1
    Mar 25 at 20:40
  • $\begingroup$ I'm asking for the derivation behind the equation. $\endgroup$ Mar 25 at 20:41
  • $\begingroup$ You can check the answer here economics.stackexchange.com/questions/42971/…. The idea is the same. $\endgroup$ Mar 25 at 22:56
  • $\begingroup$ Well, now I feel stupid :), it's been a long day. Thanks for the link @Alalalalaki I got it now $\endgroup$ Mar 25 at 23:05