Here is (one of the ways to state) social planner's problem: statement

Eric Sims' notes then immediately gives the solution:


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

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

  • 1
    $\begingroup$ 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$=0$". $\endgroup$ Jan 14, 2018 at 0:57

1 Answer 1


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

$$c_t^{-\sigma} = \lambda_t.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.