Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My starting equations are the following:

$V^{e}=w+\beta.((1-s).V^{e}+s.V^{u}(t))\\V^{u}(t)=\begin{matrix} \\max \\a(t) \end{matrix}b(t)-\psi.a(t)+\beta.[\pi(a(t)).V^{e}+(1-\pi(a(t))).V^{u}(t+1)]\\\pi(a(t))=1-exp(-\phi.a(t))\\b(t)=b.t^{-\mu}$

All the parameters are fixed, except from $a$ and $\phi$. When I use the first order condition on $V^{u}(t+1)$ with respect to $a(t)$, I obtain the following condition: $$\beta.\pi'.[V^{e}-V^{u}(t+1)]=\psi$$ My goal is to find, after calibration of the parameters, for which $\phi$ we have $V^{u}(t)>V^{e}$ after 13 periods. I have tried to put $V^{u}(t)$ on the right side of the first expression and $V^{u}(t+1)$ on the right side of the second one. After several substitutions, I get the following expression:$$V^{u}(t)=\frac{1}{\frac{1-\pi-\pi.\beta}{1-\pi}-\frac{s}{1-\beta(1-s)}}.[b(t)-\psi.a(t)+\frac{1-\pi}{2-\pi}.(\frac{\psi.\pi}{\pi'}+\frac{\pi.\beta.w}{(1-\pi).(1-\beta(1-s))})]$$ But this gets me nowhere near my problem. Does anyone have any idea on how I can have an optimization of $a(t)$, and therefore an a value for $\phi$ ? Thanks in advance.

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.