Simple partial derivative with optimization function

Question

I have the following equation $$L= \sum_{t=0}^{\infty}E_{0}\beta^{t}\left( \log c_t +\phi \log \left( 1-h_{t}\right)\right)+\sum_{t=0}^{\infty}E_{0}\beta^{t}\left[\lambda_{t}\left( (1+r_{t})a_{t}+w_{t}h_{t}+div_{t}-a_{t+1}-c_{t}\right) \right]$$

I want to find partial derivative which is equal to:

$$\frac{\partial L}{\partial a_{t+1}}=-E_{0}\lambda_{t}+\beta^{t}E_{0}\left[ \lambda_{t+1}(1+r_{t+1})\right]$$

Question is how did we derive this result? I can not see e.g. $r_{t+1}$ in original function. Thus my question is what transformations were applied to achieve such a result in a simplified form?

Answer 1

Think that given

$$ S(a) = \sum_{k=0}^{\infty}(\lambda a_k-\mu a_{k+1}) $$

we have

$$ S(a) = \lambda a_0 + \sum_{k=1}^{\infty}(\lambda-\mu)a_k $$

then

$$ \frac{\partial S}{\partial a_k} = \left\{\begin{array}{rcl}\lambda & & k = 0\\ \lambda-\mu& & k > 0\end{array}\right. $$

Answer 2

Take first order condition of $a_{t+1}$ in the objective function. Then flip the time index from t to t+1 in the objective function for all variable then take FOC. again for the $a_{t+1}$. Then you will get two equations for the same FOC. Then rest is simple algebra. Clearly you will have $-E_0β^tλt=E_0β^{t+1}λ_{t+1}[1+r_{t+1}]$ equality, in which left hand side is FOC of given Lagrange, right hande side is FOC of equation after flipped from t to $t_{t+1}$.