<p>The Lagrangian is also a function of the multipliers <span class="math-container">$(\lambda_t, \mu_t, \omega_t, \alpha_t)$</span> 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, <span class="math-container">$-u_{H}=u_l$</span> since <span class="math-container">$l=1-H$</span> (I would re-write the second FOC, in terms of <span class="math-container">$u_l$</span> 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 <span class="math-container">$\pi_t=0$</span> as required by your second-to-last target equation, and in turn implies <span class="math-container">$C_t=Y_t$</span> (from your first constraint). Second, using the second constraint you can replace the integral in your last FOC with <span class="math-container">$Y_t$</span>.</p>

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

CC BY-SA 4.0
2019-04-27T21:48:46.560Z
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
CC BY-SA 4.0
2019-04-28T08:42:01.717Z
I see it now! I was missing the fact that $y_{t}(j)$ doesn't depend on $j$. Thank you for that comment!
Ali
CC BY-SA 4.0
2019-04-28T20:18:43.860Z
I'm happy it helped :D
Regio
CC BY-SA 4.0
2019-04-27T21:31:09.433Z
2019-04-27T21:31:09.433Z
Regio
<p>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 <span class="math-container">$\{ C_{t}, H_{t}, Y_{t}, \pi_{t}, \{h_{t}(j)_{j=0}^{\infty} \}, \{y_{t}(j)_{j=0}^{\infty} \} \}_{t=0}^{\infty}$</span> to maximize <span class="math-container">$$ U_{0} = E_{0} \sum_{t=0}^{\infty} \beta^{t}u(C_{t}, 1-H_{t}) $$</span></p>

<p>subject to: <span class="math-container">$$ C_{t} = (1-\frac{\emptyset}{2}\pi_{t}^{2})Y_{t} $$</span> <span class="math-container">$$ Y_{t} = \ [\int_{0}^{1}y_{t}^{\frac{\epsilon -1}{\epsilon}}(j)\;dj ] ^{\frac{\epsilon}{\epsilon-1}} $$</span> <span class="math-container">$$ y_{t}(j) = A_{t}h_{t}(j) $$</span> <span class="math-container">$$ H_{t} = \int_{0}^{1} h_{t}(j) \; dj $$</span></p>

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

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

<p>to obtain the FOCs:</p>

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

<p>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!</p>

<macroeconomics><optimization><new-keynesian-economics>
Pareto efficiency (optimality conditions) in simple New Keynesian model
Ali
# Pareto efficiency (optimality conditions) in simple New Keynesian model

(Asked by: username_0 on 2019-04-18)

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!

**Tags:** macroeconomics, optimization, new-keynesian-economics


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## 1 answers

### Answer by username_1 on 2019-04-27. Score: 0 | Accepted 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.

> **Comments:**
> **username_1**: 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.  
> **username_0**: I see it now! I was missing the fact that $y_{t}(j)$ doesn't depend on $j$. Thank you for that comment!  
> **username_1**: I'm happy it helped :D  

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