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So I was replicating the results obtained in section 4 of Prescott's original paper, which derives optimality conditions in steady state without shock. I hope to solve the social planner's maximization problem by using the method of Lagrange.

An important feature in Prescott's model is that capital is built in multiple periods and each project costs resources, i.e., a fraction of output, in each intermediate period. My question is that how to formulate this constraint into the Lagrangian so I can solve the system of equations of FOC? Here is what I have now: \begin{multline} \mathcal{L}=\sum_{t=0}^\infty\beta^tu(c_t,\alpha(L)l_t) -\sum_{t=0}^\infty\lambda_{1t} \left( k_{t+1} - (1-\delta)k_t-s_{1t}\right) -\sum_{t=0}^\infty\lambda_{2t}\left(i_t-\sum_{j=1}^J \varphi_j s_{jt}+y_{t+1}-y_{t}\right) \\ -\sum_{t=0}^\infty\lambda_{3t}\left(c_t+i_t- f(\lambda_t,k_t,n_t,y_t)\right) -\sum_{t=0}^\infty\sum_{j=1}^{J-1}\left[\lambda_{j+3,t}\left(\sum_{t=0}^\infty s_{j,t+1}-s_{j+1,t}\right)\right]\\ \end{multline} where $s_{j,t}$ is the investment project that takes $j$ more periods to complete as of time $t$ and $y_t$ is the stock of inventory at time $t$. For the purpose of this question, let's ignore the lag polynomial $\alpha(L)$ and assume it to be have effect on $l_t$. Note that $l_t$ is leisure and $1-l_t=n_t$ is labor input where time is normalized to 1. In addition, output goes to inventory, consumption, and capital projects. I have tried to use the following substitutions: \begin{align} k_{t+1} &= (1-\delta)k_t + s_{J,t-(J-1)}\\ i_t&=\left(\sum_{j=1}^J \varphi_j s_{J,t-(J-j)}\right)+y_{t+1}-y_{t} \end{align} which allows me to get rid to the last term in $\mathcal{L}$. However, I still do not seem to be making any progress. It will be great if you can give me suggestions on formulation of my lagrangian and what FOCs will give me the steady state consumption, capital stock.

EDIT: I tried many online lecture notes, but none of them seem to want to deal with Prescott's time to build formulation and alike...

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  • $\begingroup$ Is there a reason why you are not using the Bellman method to solve for this dynamic programming problem? A Lagrangian seems prohibitively time consuming... $\endgroup$ – Kitsune Cavalry Nov 15 '16 at 6:08
  • $\begingroup$ @KitsuneCavalry in fact, i don't know how to use bellman method... $\endgroup$ – Kun Nov 15 '16 at 12:50
  • $\begingroup$ I see. I may come back to this question if I have a lot of time then. $\endgroup$ – Kitsune Cavalry Nov 15 '16 at 17:05
  • $\begingroup$ @KitsuneCavalry I think I solved the lagrangian by reindexing all the investment projects. Would you mind just check my work please? Thanks! $\endgroup$ – Kun Nov 15 '16 at 17:09
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By re-indexing all investment projects in the form of $s_{J,t}$, I get the following two equivalent constraints \begin{align} c_t+y_{t+1}-y_t&\leq f(k_t)-\sum_{i=1}^\infty\varphi_j s_{J,t-(J-j)} % Resource Constraint \\ k_{t+1}&=(1-\delta)k_t+S_{J,t-(J-1)} %Law of motion for capital formation \end{align} Hence, the Lagrangian for the maximization problem is \begin{multline} \mathcal{L} = \sum_{t=0}^\infty\beta^tu\left(c_t,1- \alpha_0n_t - \eta(1-\alpha_0) a_t\right) %Objective Function - \sum_{t=0}^\infty\lambda_{1t}\left(c_t+y_{t+1}-y_t+\sum_{i=1}^\infty\varphi_j s_{J,t-(J-j)}-f(k_t)\right) % Resource Constraint \\ - \sum_{t=0}^\infty\lambda_{2t}\left(k_{t+1}-(1-\delta)k_t-S_{J,t-(J-1)}\right) %Law of motion for capital formation - \sum_{t=0}^\infty\lambda_{3t} \left(a_{t+1}-(1-\eta)a_t-n_{t}\right) % Law of motion for past leisure \end{multline} where the state variables at time $t$ are $a_t, k_t, y_t$ and the decision variables at time $t$ are $c_t, n_t, s_{Jt}, y_{t+1}$. It is easy to see that all constraints must bind. The necessary conditions are \begin{align} &\frac{\partial \mathcal{L}}{\partial c_t} =0\implies \beta^t \frac{\partial u}{\partial c_t}=\lambda_{1t} \\ &\frac{\partial \mathcal{L}}{\partial s_{J,t}} =0\implies \sum_{i=1}^J\lambda_{1,t+J-i}\varphi_i=\lambda_{2,t+J-1} \\ &\frac{\partial \mathcal{L}}{\partial k_t}=0\implies \lambda_{1t}\frac{\partial f}{\partial k_t}+\lambda_{2t}(1-\delta)=\lambda_{2,t-1}, \end{align} which holds for all $t$. Combining 14,15,16 and simplifying, we have \begin{equation} \frac{\partial u}{\partial c_t}\frac{\partial f}{\partial k_t} + (1-\delta)\sum_{i=1}^J\beta^{1-i} \frac{\partial u}{\partial c_{t+1-i}}\varphi_i = \beta^{-1}\sum_{i=1}^J\beta^{1-i} \frac{\partial u}{\partial c_{t-i}}\varphi_i \end{equation} In steady state, $\frac{\partial u}{\partial c_t}=\frac{\partial u}{\partial c_{t'}}$, so we have \begin{equation} \frac{\partial f}{\partial k_t} = \sum_{i=0}^{J-1}\beta^{J-i-1} \varphi_{J-i} \left(\frac{1-\beta}{\beta}+\delta\right) \end{equation} One observation is that since in steady state, no borrowing can occur between household and firm, the real interest rate $r=\frac{1-\beta}{\beta}$. We see that $\sum_{i=0}^{J-1}\beta^{J-i-1} \varphi_{J-i} $ is the price of the capital in steady state.

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  • $\begingroup$ I don't...think there are any mistakes in the algebra. So have my +1 $\endgroup$ – Kitsune Cavalry Nov 15 '16 at 18:23

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