3
$\begingroup$

Is it enough to assume that agents have perfect foresight, or have 'rational' expectations for the economy to always - except in few cases - be in the stable saddle path?

With rational expectations, I think with shocks, there can be a temporary dislocation from the stable path, but then the agents will reoptimize, shifting the economy back to the stable path.

Are there more situations where this may happen?

Edit: Looking for conditions in the deterministic and stochastic case.

$\endgroup$
  • $\begingroup$ Have you realized that in a stochastic environment there is no saddle-path; $\endgroup$ – Alecos Papadopoulos Oct 3 '16 at 19:12
  • $\begingroup$ @AlecosPapadopoulos , I didn't know that... but in Wickens' book, in the mathematical appendix, when he solves stochastic 2nd order difference equation there's the case for a saddle path... Is this the case of another mistake by the author? $\endgroup$ – An old man in the sea. Oct 3 '16 at 20:56
  • 1
    $\begingroup$ We use the same nomenclature for convenience. The solution describing the saddle "path" includes random variables and so it is a random variable itself. Meaning, we have managed to describe it, but the actual realizations, the points where the economy will find itself will jump around according to the unknown realizations of the random variables. What we try to obtain in such settings is an eventually unchanging statistical distribution for the economic variables of interest, it's the closest to a "steady-state" that we can have in stochastic environments. $\endgroup$ – Alecos Papadopoulos Oct 3 '16 at 21:07
  • $\begingroup$ @AlecosPapadopoulos So, could I salvage my question, by changing "saddle path" to "stable/saddle optimal solution"? Would it make sense then? $\endgroup$ – An old man in the sea. Oct 4 '16 at 7:19
  • $\begingroup$ I wonder what essentially you try to find out here. Conditions that can guarantee that the economy will not diverge to a corner solution? $\endgroup$ – Alecos Papadopoulos Oct 4 '16 at 8:41
1
$\begingroup$

To stick to the simpler deterministic case, it is perfect foresight and the Transvarsality condition (TVC)that guarantees that an optimizing agent will stay on the saddle-path, because all other paths lead to the violation of TVC (corner solutions violate the TVC). In a deterministic environment and with perfect foresight the agent knows this and so he sticks to the saddle path. The chapter on the basic Ramsey model in Barrow and Sala-i-Martin's Growth book has a good discussion on the matter.

$\endgroup$
  • $\begingroup$ Not trying to be a nuisance, what about in the stochastic case? ;) $\endgroup$ – An old man in the sea. Oct 4 '16 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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