I am studying a paper with the following optimization problem:
$\max U_{t-1}=u(c_{m,t})+\beta [u(c_{o,t+1})+\phi u(x_{t+1})]$ ,$\beta>0$, $\phi>0$
subject to
$b_{t-1}\leq\bar{b}_{t-1}$
$-s_{t}\leq \bar{s}_{t}$
$c_{m,t}=\omega_{m}+f(b_{t-1})-b_{t-1}R+x_{t}-s_{t}$
$c_{o,t+1}=\omega_{o}+s_{t}R-x_{t+1}$,
where $f'(b_{t-1})>0$, $f(0)=0$, and $f''(b_{t-1})<0$.
The authors obtain the following Kuhn-Tucker conditions:
$f'(\hat{b}_{t-1})\geq R, = \text{if } \hat{b}_{t-1}<\bar{b}_{t-1}$,
$u'(\hat{c}_{m,t})\geq \beta R u'(\hat{c}_{o,t+1}), = \text{if } -\hat{s}_{t}<\bar{s}_{t}$,
$u'(\hat{c}_{o,t+1})=\phi u'(\hat{x}_{t+1})$,
where $\hat{b}_{t-1}$, $\hat{c}_{m,t}$, $\hat{c}_{o,t+1}$, $x_{t+1}$, and $\hat{s}_{t}$ is the optimal allocation.
I'm struggling to see how they get to these conditions. Specifically, I cannot get the first condition. Any help is appreciated. Here's my attempt:
$\ell=u(c_{m,t})+\beta u(c_{o,t+1})+\phi \beta u(x_{t+1})+\lambda_{0}[\omega_{m}+f(b_{t-1})-b_{t-1}R+x_{t}-s_{t}-c_{m,t}]+\lambda_{1}[\omega_{o}+s_{t}R-x_{t+1}-c_{o,t+1}]+\lambda_{2}[\bar{b}_{t-1}-b_{t-1}]+\lambda_{3}[\bar{s}_{t}+s_{t}]$
$\frac{\partial \ell}{\partial c_{m,t}}=u'(c_{m,t})-\lambda_{0}=0$
$\frac{\partial \ell}{\partial c_{o,t+1}}=\beta u'(c_{o,t+1})-\lambda_{1}=0$
$\frac{\partial \ell}{\partial x_{t+1}}=\phi \beta u'(x_{t+1})-\lambda_{1}=0$
$\frac{\partial \ell}{\partial s_{t}}=-\lambda_{0}+\lambda_{1}R+\lambda_{3}=0$
$\frac{\partial \ell}{\partial b_{t-1}}=\lambda_{0}[f'(b_{t-1})-R]-\lambda_{2}=0$
$\frac{\partial \ell}{\partial \lambda_{0}}=\omega_{m}+f(b_{t-1})-b_{t-1}R+x_{t}-s_{t}-c_{m,t}=0$
$\frac{\partial \ell}{\partial \lambda_{1}}=\omega_{o}+s_{t}R-x_{t+1}-c_{o,t+1}$
$\frac{\partial \ell}{\partial \lambda_{2}}=\bar{b}_{t-1}-b_{t-1}=0$
$\frac{\partial \ell}{\lambda_{3}}=\bar{s}_{t}+s_{t}=0$.
From here we can consider the case where $\lambda_{0}$, $\lambda_{1}$, $\lambda_{2}$ $\lambda_{3}$ $\geq 0$. Using $\lambda_{0}=u'(\hat{c}_{m,t})$,$-\lambda_{0}+\lambda_{1}R+\lambda_{3}=0$, and $\beta u'(\hat{c}_{o,t+1})-\lambda_{1}=0$ $\phi \beta u'(x_{t+1})-\lambda_{1}=0$, we obtain
$u'(\hat{c}_{m,t})\geq R \beta u'(\hat{c}_{o,t+1})$,
which will be equal if $\lambda_{3}=0$, or $-\hat{s}_{t}<\bar{s}_{t}$ (i.e., constraint not binding), since this implies $\lambda_{1}R=\lambda_{0}$, $u'(\hat{c}_{m,t})=\lambda_{1}$, $\beta u'(\hat{c}_{o,t+1})=\lambda_{1}$. We also clearly have
and
$u'(\hat{c}_{o,t+1})=\phi u'(\hat{x}_{t+1})$.
Finally, from the first order conditions, it must be the case that $\lambda_{0}[f'(\hat{b}_{t-1})-R]-\lambda_{2}=0$. Since $\lambda_{2} \geq 0$, this implies $f'(\hat{b}_{t-1})-R \geq 0$.
I don't see how to show $f'(\hat{b}_{t-1})=R$ if $\hat{b}_{t-1}<\bar{b}_{t-1}$ (i.e. $\lambda_{2}=0$) though.
Best Answer
I wrote the Lagrangian of the problem (with Lagrange multipliers for the constraints and bounds on them for the inequalities), then set all partial derivatives to 0. After a bit of manipulation, I get exactly the same result.
What did you try?