<p><img src="https://i.sstatic.net/lv6kh.png" alt=""></p>

<p>At $t=0$, consumption in future periods is indeed uncertain because output is uncertain since $A_t$ is stochastic. But the maximization we are doing here is taking place <strong>in each period</strong> $t$. In each period $t$, the realization of $c_t$ is observed; that is, $E_t[c_t] = c_t$. By the same logic, $E_t[\lambda_t] = \lambda_t$. Since we are maximizing in each period $t$, the first-order condition of $\mathcal{L}$ with respect to $c_t$ is not a summation. Instead, $\frac{\partial \mathcal{L}}{\partial c_t} = 0$ implies</p>

<p>$$ 
\begin{align}
\beta^t E_t[c_t^{-\sigma}] - \beta^tE_t[\lambda_t] &amp;= 0 \\
\beta^t c_t^{-\sigma} - \beta^t \lambda_t = 0.
\end{align}
$$</p>

<p>Note that if the coefficient of relative risk aversion were not assumed to be constant across time; that is, if $\sigma_t \neq \sigma$, then the result we wish to show does not follow. Just some subtleties about expectations to keep in mind.</p>

<p>But since indeed $\sigma_t = \sigma$, it then follows that</p>

<p>$$c_t^{-\sigma} = \lambda_t.$$</p>

CC BY-SA 3.0
2018-01-14T02:56:50.353Z
2018-01-14T03:48:57.073Z
2018-01-14T03:48:57.073Z
Kenneth Rios
<p>Here is (one of the ways to state) social planner's problem:
<a href="https://i.sstatic.net/lv6kh.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lv6kh.png" alt="statement"></a></p>

<p><a href="https://www3.nd.edu/~esims1/neoclassical_growth_11.pdf" rel="nofollow noreferrer">Eric Sims'</a> notes then immediately gives the solution:</p>

<p><a href="https://i.sstatic.net/MhtMB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MhtMB.png" alt="solution"></a></p>

<p>I am trying to connect these two lines. This is what I get after taking a derivative with respect to consumption.</p>

<p><a href="https://i.sstatic.net/N4EA3.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/N4EA3.png" alt="enter image description here"></a></p>

<p>Question is: how did he get rid of expectation and discount factor?</p>

CC BY-SA 3.0
2018-01-14T00:57:42.920Z
What does it mean to take the "derivative with respect to consumption"? Consumption is given by infinitely many variables, a whole sequence, and you only take the derivative with respect to $c_t$. There is no summation remaining.The expectation there is irrelevant because only $A_t$ is stochastic. You will end up with something of the form "$\beta^t$ blah$=0$", which is equivalent to "blah$=0$".
Michael Greinecker
<macroeconomics><optimization>
Social planner's first order condition (current-value Lagrangian)
user3349993
# Social planner's first order condition (current-value Lagrangian)

(Asked by: username_0 on 2018-01-13)

Here is (one of the ways to state) social planner's problem:
[![statement](https://i.sstatic.net/lv6kh.png)](https://i.sstatic.net/lv6kh.png)

[Eric Sims'](https://www3.nd.edu/~esims1/neoclassical_growth_11.pdf) notes then immediately gives the solution:

[![solution](https://i.sstatic.net/MhtMB.png)](https://i.sstatic.net/MhtMB.png)

I am trying to connect these two lines. This is what I get after taking a derivative with respect to consumption.

[![enter image description here](https://i.sstatic.net/N4EA3.png)](https://i.sstatic.net/N4EA3.png)

Question is: how did he get rid of expectation and discount factor?

**Tags:** macroeconomics, optimization

> **Comments:**
> **username_1**: What does it mean to take the "derivative with respect to consumption"? Consumption is given by infinitely many variables, a whole sequence, and you only take the derivative with respect to $c_t$. There is no summation remaining.The expectation there is irrelevant because only $A_t$ is stochastic. You will end up with something of the form "$\beta^t$ blah$=0$", which is equivalent to "blah$=0$".  

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## 1 answers

### Answer by username_2 on 2018-01-14. Score: 1 | Accepted answer

![](https://i.sstatic.net/lv6kh.png)

At $t=0$, consumption in future periods is indeed uncertain because output is uncertain since $A\_t$ is stochastic. But the maximization we are doing here is taking place **in each period** $t$. In each period $t$, the realization of $c\_t$ is observed; that is, $E\_t[c\_t] = c\_t$. By the same logic, $E\_t[\lambda\_t] = \lambda\_t$. Since we are maximizing in each period $t$, the first-order condition of $\mathcal{L}$ with respect to $c\_t$ is not a summation. Instead, $\frac{\partial \mathcal{L}}{\partial c\_t} = 0$ implies

$$
\begin{align}
\beta^t E\_t[c\_t^{-\sigma}] - \beta^tE\_t[\lambda\_t] &= 0 \\
\beta^t c\_t^{-\sigma} - \beta^t \lambda\_t = 0.
\end{align}
$$

Note that if the coefficient of relative risk aversion were not assumed to be constant across time; that is, if $\sigma\_t \neq \sigma$, then the result we wish to show does not follow. Just some subtleties about expectations to keep in mind.

But since indeed $\sigma\_t = \sigma$, it then follows that

$$c\_t^{-\sigma} = \lambda\_t.$$


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