# Lagrangian multiplier

Consumer's problem $$$$\max \sum_{t}\beta^{t}[c_{t}-1/2(1-x_{t})^{2}],$$$$ $$$$\ s.t. c_{t}+q_{t}b_{t+1} \leq (1-\tau_{t})(1-x_{t})+b_{t},$$$$ where c=consumption, x=leisure(1-x=labor), $$\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.

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

• This seems trivial. What exactly is causing you trouble? Is it the $\sum$ sign? Oct 25, 2018 at 13:18
• Also $c^t$ seems to be a misspelled version of $c_t$. Oct 25, 2018 at 13:18
• 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 $\lambda q_{t-1}$. Oct 25, 2018 at 15:09
• 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.
– erik
Oct 26, 2018 at 0:53