<p>Consumer's problem 
<span class="math-container">\begin{equation}
\max \sum_{t}\beta^{t}[c_{t}-1/2(1-x_{t})^{2}],
\end{equation}</span>
<span class="math-container">\begin{equation}
\ s.t. c_{t}+q_{t}b_{t+1} \leq (1-\tau_{t})(1-x_{t})+b_{t},
\end{equation}</span>
where c=consumption, x=leisure(1-x=labor), <span class="math-container">$\tau$</span>=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. </p>

<p><span class="math-container">\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}</span>
First order conditions
<span class="math-container">\begin{equation}
w.r.t. c_{t}: \beta^{t}-\lambda=0
\end{equation}</span>
<span class="math-container">\begin{equation}
w.r.t. x_{t}: -\beta^{t}(1-x_{t})(-1)-\lambda(1-\tau_{t})(-1)=0 
\end{equation}</span>
<span class="math-container">\begin{equation}
w.r.t. b_{t}: ?
\end{equation}</span></p>

CC BY-SA 4.0
2018-10-25T13:18:03.103Z
This seems trivial. What exactly is causing you trouble? Is it the $\sum$ sign?
Giskard
CC BY-SA 4.0
2018-10-25T13:18:42.003Z
Also $c^t$ seems to be a misspelled version of $c_t$.
Giskard
CC BY-SA 4.0
2018-10-25T15:09:11.310Z
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 $\lambda q_{t-1}$.
Kenneth Rios
CC BY-SA 4.0
2018-10-26T00:53:14.300Z
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
<macroeconomics><microeconomics>
Lagrangian multiplier
user19912
# Lagrangian multiplier

(Asked by: username_0 on 2018-10-25)

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}

**Tags:** macroeconomics, microeconomics

> **Comments:**
> **username_1**: This seems trivial. What exactly is causing you trouble? Is it the $\sum$ sign?  
> **username_1**: Also $c^t$ seems to be a misspelled version of $c_t$.  
> **username_2**: 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 $\lambda q_{t-1}$.  
> **username_3**: 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.  

---

## 0 answers

