(Note: the other posts do not cover this part of the derivation) enter image description here I have tried to compute the FOC of $k_{t+1}(s^t)$. I get that $0 = -\lambda_t(s^t)$; I can't see why the sigma remains for the FOC of $k_{t+1}(s^t)$ and not $c_t(s^t)$.

[SOLVED: had to use the fact that we can consider $s^{t+1}$ that realizes $s^t$]

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    $\begingroup$ Try to explicitly expand the time summation, and try differentiating a specific time, say $t = 1$ and it should be clear. $\endgroup$ May 10 '21 at 4:39
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    $\begingroup$ Please use TeX to write equations in your posts rather than images. Images are non-searchable decreasing the usefulness of this site. $\endgroup$ May 10 '21 at 12:38
  • $\begingroup$ @WalrasianAuctioneer Here's what I've tried. I have expanded it out. The first conceptual thing is that I have a sum over all these possible $s^t$ paths, but the FOC is taking the derivative w.r.t one path. That means that the innersum over all these paths can be reduced to one path. Moreover, because I am taking the derivative with respect to a specific $t$, that means that the left sum over time can be reduced to just one time. In other words, I am taking the derivative with respect to one line with no sigmas- just like consumption FOC has no sigma, so would $k_{t+1}$? $\endgroup$
    – JustJ
    May 10 '21 at 13:17
  • $\begingroup$ @Justin your second statement is not true, when you take derivative with respect to $k_{t+1}$, it appears in two time periods ($t$ and $t+1$)! Try again. $\endgroup$ May 10 '21 at 16:01
  • $\begingroup$ @WalrasianAuctioneer Great, I ended up figuring it out. The work was definitely not trivial but good to finally understand it $\endgroup$
    – JustJ
    May 11 '21 at 2:54

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