<p>My Lagrangian is:</p>

<p><span class="math-container">$L=\sum\limits_{t=0}^T \beta^tU(f(k_t)-k_{t+1})+\sum\limits_{t=0}^T\lambda_t(f(k_t)-k_{t+1}).$</span></p>

<p>My FOC for <span class="math-container">$[k_{t+1}]$</span> is:</p>

<p><span class="math-container">$\beta^tU'(f(k_t)-k_{t_1}^*)(-1)-\lambda_t^*+\beta^{t+1}U'(f(k_{t+1}^*-k_{t+2})+\lambda^*_{t+1}f'(k_{t+1})^*=0$</span>.</p>

<p><span class="math-container">$\textbf{My Question:}$</span>
I try to recover from their FOC and backtrack, but how did they get (5) as the Euler Equation on page 11?</p>

<p>Their EE is:</p>

<p><span class="math-container">$\beta f'(k_t)U'[f(k_t-k_{t+1})]=U'[f(k_{t-1})-k_t]$</span> for <span class="math-container">$t=1,...,T$</span>.</p>

<p>Reference:</p>

<p>Stokey, N., R. Lucas, and E. Prescott, Recursive Methods in Economic Dynamics, Harvard Univ. Press, 1989</p>

<macroeconomics><optimization>
Stokey, Lucas (1989) p 11 FOC
Frank Swanton
# Stokey, Lucas (1989) p 11 FOC

(Asked by: username_0 on 2019-06-08)

My Lagrangian is:

$L=\sum\limits\_{t=0}^T \beta^tU(f(k\_t)-k\_{t+1})+\sum\limits\_{t=0}^T\lambda\_t(f(k\_t)-k\_{t+1}).$

My FOC for $[k\_{t+1}]$ is:

$\beta^tU'(f(k\_t)-k\_{t\_1}^\*)(-1)-\lambda\_t^\*+\beta^{t+1}U'(f(k\_{t+1}^\*-k\_{t+2})+\lambda^\*\_{t+1}f'(k\_{t+1})^\*=0$.

$\textbf{My Question:}$
I try to recover from their FOC and backtrack, but how did they get (5) as the Euler Equation on page 11?

Their EE is:

$\beta f'(k\_t)U'[f(k\_t-k\_{t+1})]=U'[f(k\_{t-1})-k\_t]$ for $t=1,...,T$.

Reference:

Stokey, N., R. Lucas, and E. Prescott, Recursive Methods in Economic Dynamics, Harvard Univ. Press, 1989

**Tags:** macroeconomics, optimization

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## 0 answers

