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7

Given the fixed input price $w$, the cost function can be written as $$C(q)=f^{-1}(q)\times w$$ where $f^{-1}$ is the inverse of the production function $f$. From the discussion here, one can conclude that the inverse of a concave strictly increasing function is convex. Thus, $C(q)$ is convex as well. Going back to your approach, you might like to have ...

5

Let $x(w, q)$ denote the solution to the cost minimization problem : \begin{eqnarray*} \min_{x} & \ w\cdot x \\ \text{s.t.} & \ \ f(x) \geq q \end{eqnarray*} where $f$ is the production function. Since $x(w, q)$ minimizes cost at $(w, q)$, following holds for all $w$ and for all $q$ : \begin{eqnarray*} w\cdot x(w, q) \leq w\cdot x(w', q) \ \ \ \...

2

Perhaps I misunderstand something, maybe you do not allow mixed equilibria. That may be strange in games of imperfect information. Consider an asymmetric matching pennies game. Both players show either Heads or Tails. I forgot how to type game matrices in mathjax, but the payoffs look something like this: $\begin{bmatrix} -1,1 & 3,-3 \\ 1,-1 & -3,3 ... 2 Would this (and why or why not) provide an answer to your question? 2 The economic intuition here is clear: if the single-input production function is concave, the marginal product is diminishing. So for equal increases of the single input (and given fixed unit cost), say unit per unit, production increases less and less, while cost increases equally. Flip it, and then for sequential unit increases in output, cost increases ... 1 You can find in many textbooks (for example Mas-Collel, Winston and Green) that the Independence of irrelevant alternatives axiom implies that preferences over lotteries are linear with respect to the probabilities of each event. Then, a compound lottery is of the form:$\mathcal{L}''=\alpha\mathcal{L}+(1-\alpha)\mathcal{L}'$, If$A$and$B$are the ... 1 The household side of this model is pretty standard. Denote$K(t)$be household's assets at time$t$. Then the transversality condition (which is an optimizing condition, not a constraint), is $$\lim_{t \to \infty} [e^{-\rho t}\lambda(t) K(t)] = 0$$ where$\lambda(t)\$ is the current value multiplier on assets in the Hamiltonian. Given the assumed form of ...

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