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Alesina Drazen (1991) - Why Are Stabilizations Delayed (view source)
Revision as of 00:39, 8 March 2011
, 00:39, 8 March 2011→The Model
*The assumption that both groups pay half of the taxes before the stabilization.
The model then solves out for a feasible consumption path, and computes the flow utility before a stabilization (which is the income effect of taxes plus the welfare loss from the tax distortion). With constant consumption after the stabilization (which was shown to be feasible), one can then compute the different in present discounted value of the excess taxes that the loser must pay relative to the winner and loser lifetime utilities from stabilization forward:
:<math>V^W(T) - V^L(T) = (2 \alpha -1)\overline(bar{b) } e^{(1-\gamma)rT}\,</math> This is the present value of the excess taxes that the loser must pay relative to the winner.
This is used to determine the optimum concession time <math>T_i\,</math> of a group with cost <math>\theta_i\,</math>, subject to:
*<math>\underline{\theta} > \alpha - \frac{1}{2}\,</math> : This prevents a groups group's optimum concession time from being infinite, as otherwise a group may prefer to wait indefinatelyindefinitely, as the cost of living in the unstable economy and bearing half of the tax burden is less than the cost of being the loser.
*<math>F(\theta) = 1-H(T(\theta))\,</math> : as <math>T_i\,</math> is monotonic in <math>\theta_i\,</math> this can (apparently) be derived.
*Ignoring the equilibria in which one group concedes immediately, as the paper wants to examine delay.*Looking and thus looking for a symmetric equilibriaNash equilibrium.*Lemma 1 in the paper give <math>t_iT_i'(\theta_i) < 0\,</math> : The optimal concession time is monotonically decreasing in <math>\theta_i \,</math>
Proposition 1 states that there exists a symmetric Nash equilbirum with each group's concession function described by <math>T(\theta)\,</math> where <math>T(\theta)\,</math> is implicitly defined by:
:<math>\underbrace{\underbrace{\left(-\frac{f(\theta)}{F(\theta)}\frac{1}{tT'(\theta)}\right)}_{\mbox{A1}}\underbrace{\frac{2 \alpha -1}{r}}_{\mbox{A2}}}_{\mbox{A}}= \underbrace{\gamma (\theta + \frac{1}{2} - \alpha)}_{\mbox{B}}\,</math> and the initial boundary condition <math>\T(\bar{\theta})=0\,</math>
Where: