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I am trying to derive an expression for the optimal level of consumption in the basic problem:

$ max \hspace{1cm} U_t = E_t \left[\sum_{s=t}^{\infty} \beta^{s-t} \left( C_s - a\frac{Cs}{2}^2 \right) \right] $

$ s.t \hspace{1cm} A_{s + 1} = A_s(1+r) + Y_s - C_s $

The problem leads to the Euler Equation:

$ E_t \left[ \frac{u'(C_s)}{u'(C_{s+1})} \right] = \beta (1+r)$

Assuming $\beta = (1+r)^{-1}$ we have:

$ E_t \left[ \frac{u'(C_s)}{u'(C_{s+1})} \right] = 1$

Which leads to:

$ C_s = E_t[C_{s+1}]$.

Iterating forward the BC and using the trasversality condition I derived the intertemporal BC:

$ \sum_{s=t}^{\infty} \frac{C_s}{(1+r)^{s-t}} = A_t(1+r) + \sum_{s=t}^{\infty} \frac{Y_s}{(1+r)^{s-t}} $

Now, how can I find an expression for $C_t$? If time were finite intuitively (I think) I would have:

$ C_t = \frac{1}{T - t}(A_t + \sum_{s=t}^{T} E_t [Y_s]) $

How do I deal with the infinite horizon to find and expression for $C_t$?

I was wondering if I can use the geometric series, something like:

$C_t \sum_{s = t}^{\infty} \left( \frac{1}{1+r} \right)^{s-t} = C_t \frac{1}{1 - \frac{1}{1+r}} = C_t\frac{1+r}{r} $

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  • $\begingroup$ Are you not missing expectations operators in the second equation from the bottom? $\endgroup$
    – BrsG
    Aug 31 at 15:49
  • $\begingroup$ Yes, I could take expectation of both sides and get: $C_t + \sum_{s=t+1}^{\infty}E_t\left[ \frac{C_s}{(1+r)^{s-t}} \right] = A_t(1+r) + Y_t + \sum_{s=t+1}^{\infty}E_t\left[ \frac{Y_s}{(1+r)^{s-t}} \right]$ $\endgroup$
    – Giorgetto
    Aug 31 at 17:58
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    $\begingroup$ hence: $ C_t = A_t(1+r) + Y_t + \sum_{s=t+1}^{\infty} E_t \left[ \frac{Y_s - C_s}{(1+r)^{s-t}}\right]$ $\endgroup$
    – Giorgetto
    Aug 31 at 18:10
  • $\begingroup$ @Giorgetto is right. If you want to proceed, you would need an expression for $Y_s$ (or the difference). An easy way forward is to consider $Y_t$ a constant, typically $Y_t=1$, making $C_t=f(A_t)$ and having all variables in term of the economy's product for the period. $\endgroup$ Aug 31 at 21:25

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