How would I find the Euler equation for: $$ U=\sqrt{C_1}+\beta{\sqrt{C_2}} $$, where $$\beta=1/(1+r)$$

  • $\begingroup$ 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 $\endgroup$
    – mynameparv
    May 30, 2023 at 11:50

1 Answer 1


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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