1
$\begingroup$

Consumer's problem \begin{equation} \max \sum_{t}\beta^{t}[c_{t}-1/2(1-x_{t})^{2}], \end{equation} \begin{equation} \ s.t. c_{t}+q_{t}b_{t+1} \leq (1-\tau_{t})(1-x_{t})+b_{t}, \end{equation} where c=consumption, x=leisure(1-x=labor), $\tau$=labor income tax, b_{t+1}=the gov's bond, which is sold in period t at price q_{t}. Time endowment is normalized to 1.

\begin{equation} \ L=\sum_{t}\beta^{t}[c_{t}-1/2(1-x_{t})^{2}]+\lambda[c^{t}+q_{t}b_{t+1}-b_{t}-(1-\tau_{t})(1-x_{t})] \end{equation} First order conditions \begin{equation} w.r.t. c_{t}: \beta^{t}-\lambda=0 \end{equation} \begin{equation} w.r.t. x_{t}: -\beta^{t}(1-x_{t})(-1)-\lambda(1-\tau_{t})(-1)=0 \end{equation} \begin{equation} w.r.t. b_{t}: ? \end{equation}

$\endgroup$
  • $\begingroup$ This seems trivial. What exactly is causing you trouble? Is it the $\sum$ sign? $\endgroup$ – Giskard Oct 25 '18 at 13:18
  • $\begingroup$ Also $c^t$ seems to be a misspelled version of $c_t$. $\endgroup$ – Giskard Oct 25 '18 at 13:18
  • $\begingroup$ Just take the derivative of $L$ w.r.t. $b_t$, but don't forget that in period $t-1$, $b_t$ shows up with a coefficient of $\lambda q_{t-1}$. $\endgroup$ – Kenneth Rios Oct 25 '18 at 15:09
  • $\begingroup$ Two things 1. I think that signs inside the brackets after lambda should be reversed. 2. Although taking derivative with $b_t$ works, technically it should be with $b_(t+1)$, since this period’s bond is observed and next period’s bond is to be selected. $\endgroup$ – erik Oct 26 '18 at 0:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.