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I have read that the convexity of production sets (for instance non-increasing returns to scale) is not a necessary assumption for the first welfare theorem but it is for the second welfare theorem. Maybe I got things wrong and what the author actually tried to say is that convexity is not necessary for the efficiency of the competitive equilibrium but rather for the existence of the competitive equilibrium.

Could you please elaborate where the assumption of convex production sets is needed in the first and second welfare theorem?

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  • $\begingroup$ "the author actually tried to say" Could you please link to the work where you have read these statements? $\endgroup$ – Giskard Jan 4 '18 at 13:17
  • $\begingroup$ I have phrased this poorly. I wanted to say "maybe the author meant to say convexity is not necessary for the efficiency (...)". So this was conjecture on my behalf and not a quote. $\endgroup$ – Fusscreme Jan 4 '18 at 14:25
  • $\begingroup$ That is fine. Who is the author and where can we read what she said? $\endgroup$ – Giskard Jan 4 '18 at 14:31
  • $\begingroup$ For instance Tirole, page 6 and Nicholson et al., page 350. There is no statement that convexity of productions sets is unnecessary but rather the implicit omission of this assumption (check out Wikipedia too). I apologize if my statement was misleading. $\endgroup$ – Fusscreme Jan 4 '18 at 14:45
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Convexity of the production set is indeed not needed for the proof of the first welfare theorem but for the proof of the second welfare theorem. It is not a necessary condition though.

It is possible to interpret this as an existence issue. The first welfare theorem is about all competitive equilibria and holds trivially if there are none. The second welfare theorem, on the other hand, states that for a given Pareto efficient allocation, there is a price system and redistribution of endowments with respect to which it is a (quasi-)equilibrium.

The standard proof of the second welfare theorem uses a result of Minkowski on the separation of nonoverlapping convex sets by a hyperplane, but it is possible to prove a version of the second welfare theorem as a corollary to an existence result by a nice argument due to Maskin and Roberts. The argument is quite easy in the case of an exchange economy: Take a Pareto efficient allocation as the endowment distribution. If a competitive equilibrium exists from these endowments, everyone will end up with something at least as good as their endowment. Since the endowment distribution was Pareto efficient, nobody can end up with something better. So everyone must be indifferent between their demanded commodity bundle and their endowment, so they might as well just demand just their endowment. The argument generalizes to economies with production.

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  • $\begingroup$ Thank you for your answer! May you explain what you mean with "Convexity of the production set is (...) needed for the proof of the second welfare theorem. It is not a necessary condition though." Is this not contradicting? Furthermore, what I don't understand is if convexity of the production set is irrelevant, why are increasing returns not compatible with perfect competition? Shouldn't this issue (increasing returns/convexity) be treated in the assumptions of the welfare theorems? $\endgroup$ – Fusscreme Jan 5 '18 at 10:39
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    $\begingroup$ The proof of the second welfare theorem makes use of the fact that production sets are convex or at least that the aggregate production set is convex (a strictly weaker assumption). But this does not mean that the conclusion of the second welfare theorem fails as soon as there any kind of non-convexity. In particular, convexity is not a logically necessary condition. $\endgroup$ – Michael Greinecker Jan 5 '18 at 11:19
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    $\begingroup$ Globally increasing returns to scale are usually not compatible with the existence of a competitive equilibrium, but the conclusion of the first welfare theorem is trivially true when there is no competitive equilibrium. Every competitive equilibrium must be Pareto efficient when there is none; it's just that "every" is not worth much in that case. $\endgroup$ – Michael Greinecker Jan 5 '18 at 11:19

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