<p><strong>Hint:</strong> The problem is quite simple and straightforward, so instead I'll try to explain how we derive the non-stochastic Euler equation. Using the derivation you can try to do this problem on your own and post a solution yourself.</p>
<p>Consider a standard utility maximization problem of the form:
<span class="math-container">$$\begin{align}
\max_{\{C_t\}_{t=1}^{\infty}} \quad &amp; \sum_{t=1}^{\infty}\beta ^tu(C_t) \\
\textrm{s.t.} \quad &amp; \sum_{t=1}^{\infty}\frac{C_t}{(1+r)^{t-1}}\leq w
\end{align}$$</span>
where, <span class="math-container">$\beta \in (0,1)$</span> , <span class="math-container">$u$</span> is strictly increasing and convex i.e., <span class="math-container">$u'(.)&gt;0, \; u''(.)&lt;0$</span> , and <span class="math-container">$w$</span> denotes the lifetime wealth</p>
<p>Note that the constraint will bind at optimum given our assumption about <span class="math-container">$u(.)$</span>, therefore we can use the Lagrangian Method to solve the above.  The Lagrangian function is given by:
<span class="math-container">$$\mathcal{L}=\sum_{t=1}^{\infty}\beta ^tu(C_t)-\lambda\left (\sum_{t=1}^{\infty}\frac{C_t}{(1+r)^{t-1}} - w \right)$$</span></p>
<p>The FOC tangency condition for some arbitrary <span class="math-container">$C_t, C_{t+1}$</span> are:
<span class="math-container">$$\begin{align}
 &amp; \frac{\partial \mathcal{L}}{\partial C_t} = \beta^tu'(C_t)-\lambda \frac{1}{(1+r)^{t-1}} \overset{set}=0 \tag1\\
&amp; \frac{\partial \mathcal{L}}{\partial C_{t+1}}=\beta^{t+1}u'(C_{t+1})-\lambda \frac{1}{(1+r)^{t}} \overset{set}=0 \tag2
\end{align}$$</span></p>
<p>From <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>: <span class="math-container">$$\frac{\beta^{t}u'(C_{t})}{\beta^{t+1}u'(C_{t+1})}=\frac{(1+r)^t}{(1+r)^{t-1}}$$</span></p>
<p>The above gives us the Euler Equation as: <span class="math-container">$\boxed{u'(C_t)=\beta(1+r) u'(C_{t+1})} \quad \forall \; t\geq 1$</span></p>

CC BY-SA 4.0
2023-05-30T11:48:17.163Z
2023-05-30T11:48:17.163Z
mynameparv
<p>How would I find the Euler equation for:
<span class="math-container">$$ U=\sqrt{C_1}+\beta{\sqrt{C_2}} $$</span>, where <span class="math-container">$$\beta=1/(1+r)$$</span></p>

CC BY-SA 4.0
2023-05-30T11:50:13.290Z
try to be more specific in your question instead of just asking for the solution and try to provide your own attempt to the problem as well
mynameparv
<macroeconomics><utility><international-economics>
Euler Equation - small open economy
f.Greening
# Euler Equation - small open economy

(Asked by: username_0 on 2023-05-30)

How would I find the Euler equation for:
$$ U=\sqrt{C\_1}+\beta{\sqrt{C\_2}} $$, where $$\beta=1/(1+r)$$

**Tags:** macroeconomics, utility, international-economics
> **Comments:**
> **username_1**: try to be more specific in your question instead of just asking for the solution and try to provide your own attempt to the problem as well  

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## 1 answers

### Answer by username_1 on 2023-05-30. Score: 2

**Hint:** The problem is quite simple and straightforward, so instead I'll try to explain how we derive the non-stochastic Euler equation. Using the derivation you can try to do this problem on your own and post a solution yourself.

Consider a standard utility maximization problem of the form:
$$\begin{align}
\max\_{\{C\_t\}\_{t=1}^{\infty}} \quad & \sum\_{t=1}^{\infty}\beta ^tu(C\_t) \\
\textrm{s.t.} \quad & \sum\_{t=1}^{\infty}\frac{C\_t}{(1+r)^{t-1}}\leq w
\end{align}$$
where, $\beta \in (0,1)$ , $u$ is strictly increasing and convex i.e., $u'(.)>0, \; u''(.)<0$ , and $w$ denotes the lifetime wealth

Note that the constraint will bind at optimum given our assumption about $u(.)$, therefore we can use the Lagrangian Method to solve the above. The Lagrangian function is given by:
$$\mathcal{L}=\sum\_{t=1}^{\infty}\beta ^tu(C\_t)-\lambda\left (\sum\_{t=1}^{\infty}\frac{C\_t}{(1+r)^{t-1}} - w \right)$$

The FOC tangency condition for some arbitrary $C\_t, C\_{t+1}$ are:
$$\begin{align}
& \frac{\partial \mathcal{L}}{\partial C\_t} = \beta^tu'(C\_t)-\lambda \frac{1}{(1+r)^{t-1}} \overset{set}=0 \tag1\\
& \frac{\partial \mathcal{L}}{\partial C\_{t+1}}=\beta^{t+1}u'(C\_{t+1})-\lambda \frac{1}{(1+r)^{t}} \overset{set}=0 \tag2
\end{align}$$

From $(1)$ and $(2)$: $$\frac{\beta^{t}u'(C\_{t})}{\beta^{t+1}u'(C\_{t+1})}=\frac{(1+r)^t}{(1+r)^{t-1}}$$

The above gives us the Euler Equation as: $\boxed{u'(C\_t)=\beta(1+r) u'(C\_{t+1})} \quad \forall \; t\geq 1$


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