# Pareto efficiency (optimality conditions) in simple New Keynesian model

I am looking for the pareto-optimal equilibrium for a central planner's problem in a simple New Keynesian model. The planner's problem is to choose $$\{ C_{t}, H_{t}, Y_{t}, \pi_{t}, \{h_{t}(j)_{j=0}^{\infty} \}, \{y_{t}(j)_{j=0}^{\infty} \} \}_{t=0}^{\infty}$$ to maximize $$U_{0} = E_{0} \sum_{t=0}^{\infty} \beta^{t}u(C_{t}, 1-H_{t})$$

subject to: $$C_{t} = (1-\frac{\emptyset}{2}\pi_{t}^{2})Y_{t}$$ $$Y_{t} = \ [\int_{0}^{1}y_{t}^{\frac{\epsilon -1}{\epsilon}}(j)\;dj ] ^{\frac{\epsilon}{\epsilon-1}}$$ $$y_{t}(j) = A_{t}h_{t}(j)$$ $$H_{t} = \int_{0}^{1} h_{t}(j) \; dj$$

The first-order conditions for this problem imply that for all $$j$$ and $$t$$: $$\frac{u_{l}(C_{t}, 1-H_{t})}{u_{c}(C_{t}, 1-H_{t})} = A_{t}$$ $$y_{t}(j) = Y_{t}$$ $$h_{t}(j) = H_{t}$$ $$Y_{t} = A_{t}H_{t}$$ $$\pi_{t} = 0$$ $$C_{t} = Y_{t}$$

To arrive to the above implications I tried setting up the following Lagrangian: $$L_{0}(j) = E_{0} \sum_{t=0}^{\infty} \beta^{t} u(C_{t}, 1-H_{t}) + \lambda_{t}[(1-\frac{\emptyset}{2}\pi_{t}^{2})Y_{t} -C_{t}] + \mu_{t}[\ [\int_{0}^{1}y_{t}^{\frac{\epsilon -1}{\epsilon}}(j)\;dj ] ^{\frac{\epsilon}{\epsilon-1}} - Y_{t}] + \omega_{t}[A_{t}h_{t}(j)- y_{t}(j)] + \alpha_{t}[H_{t} - \int_{0}^{1} h_{t}(j) \; dj ]$$

to obtain the FOCs:

$$\frac{\partial L_{0}(j)}{\partial C_{t}}: u_{c}(C_{t}, 1 - H_{t}) - \lambda_{t} = 0$$ $$\frac{\partial L_{0}(j)}{\partial H_{t}}: -u_{H}(C_{t}, 1 - H_{t}) + \alpha_{t} = 0$$ $$\frac{\partial L_{0}(j)}{\partial Y_{t}}: \lambda_{t}(1 - \frac{\emptyset}{2}\pi_{t}^{2}) - \mu_{t} = 0$$ $$\frac{\partial L_{0}(j)}{\partial \pi_{t}}: -\emptyset \pi_{t} Y_{t} \lambda_{t} = 0$$ $$\frac{\partial L_{0}(j)}{\partial h_{t}(j)}: \omega_{t}A_{t} - \alpha_{t} = 0$$ $$\frac{\partial L_{0}(j)}{\partial y_{t}(j)}: \mu_{t}[\int_{0}^{1} y_{t}^{\frac{\epsilon -1}{\epsilon}}(j)]^{\frac{1}{\epsilon -1}} y_{t}^{\frac{-1}{\epsilon}}(j) - \omega_{t} = 0$$

Now, I have no idea what I am doing wrong or how to get to the implications above. Any help would be very much appreciated!

## 1 Answer

The Lagrangian is also a function of the multipliers $$(\lambda_t, \mu_t, \omega_t, \alpha_t)$$ so you are missing their FOC's (these will give you the constraints as FOC's so if you were already taking them into account, omit my comment, please). Also, $$-u_{H}=u_l$$ since $$l=1-H$$ (I would re-write the second FOC, in terms of $$u_l$$ in order to construct the first "target" equation that has the MRS=MRT. Finally, there are a couple of useful facts that should help you. First, note that in your 4th FOC, you have a product equal to zero, so at least one of the elements is equal to zero. Try to convince yourself, that the only possibility (so that you don't run into a contradiction with the other FOC's) is for $$\pi_t=0$$ as required by your second-to-last target equation, and in turn implies $$C_t=Y_t$$ (from your first constraint). Second, using the second constraint you can replace the integral in your last FOC with $$Y_t$$.

This should help for further simplifying the 10 FOC's you should have, into the 6 target equations you want.

• Oh, also, after you do the substitution I mentioned since you will find that $y(j)$ is implicitly determined by an equation that does not depend on $j$, so it must be a constant, and from the second constraint, it must be that $y_t(j)=Y_t$, and from this and the third constraint $h_t(j)$ should also be a constant. – Regio Apr 27 '19 at 21:48
• I see it now! I was missing the fact that $y_{t}(j)$ doesn't depend on $j$. Thank you for that comment! – user20105 Apr 28 '19 at 8:42
• I'm happy it helped :D – Regio Apr 28 '19 at 20:18