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Generalized Stochastic Dominance with Respect to a Function

Generalized Stochastic Dominance with Respect to a Function. Lecture XX. Introduction. Meyer, Jack “Choice among Distributions.” Journal of Economic Theory 14(1977): 326-36.

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Generalized Stochastic Dominance with Respect to a Function

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  1. Generalized Stochastic Dominance with Respect to a Function Lecture XX

  2. Introduction • Meyer, Jack “Choice among Distributions.” Journal of Economic Theory 14(1977): 326-36. • The general idea of the manuscript is to restrict the risk aversion coefficient for stochastic dominance to those risk aversion coefficients in a given interval r1(x)<r(x)<r2(x). Lecture 20

  3. This problem will be solved by finding the utility function u(x) which satisfies: and minimizes Lecture 20

  4. Given that this integral yields the expected value of F(x) minus the expected value of G(x), the minimum will be greater than zero if F(x) is preferred to G(x) by all agents who prefer F(x) to G(x). • If the minimum is less than zero, then the preference is not unanimous for all agents whose risk aversion coefficients are in the state range. • Another problem is that utility is invariant to a linear tranposition. Thus, we must stipulate that u'(0)=1. Lecture 20

  5. The problem is then to use the control variable -u"(x)/u'(x) to maximize the objective function subject to the equation of motion Lecture 20

  6. and the control constraints with the initial condition u'(0)=1. Lecture 20

  7. Rewriting the problem, substituting z(x)=-u"(x)/u'(x) yields Lecture 20

  8. The traditional Hamiltonian for this problem then becomes Following the basic Pontryagin results, a given path for control variable is optimum given three conditions: • Optimality Condition Lecture 20

  9. Multiplier or Costate condition • Equation of Motion Lecture 20

  10. However, in the current scenario the Hamiltonian has to be amended to account for the two inequality constraints. Specifically, Lecture 20

  11. First examining the derivatives of the Lagrangian with respect to z(x), the control variable. We see that Lecture 20

  12. Given the restriction that the Lagrange multipliers must be nonegative at optimum, we can see that Thus, the minimum value of the integral occurs at one of the boundaries. The question is then: What determines which boundary? Lecture 20

  13. To examine this we begin with the Costate condition: Given this expression, we can work backward from the transversality condition. Specifically, the transversality condition for this control problem specifies that Lecture 20

  14. Given this boundary condition, the value of l(x)u'(x) can be defined by the integral Lecture 20

  15. Hence to complete the derivation, we must determine the value of the derivative under the integral Substituting for l'(x) from the Costate condition yields Lecture 20

  16. Substituting for z(x) and canceling like terms yields Yielding the expression Lecture 20

  17. Merging this result with the previous derivations, we have the infamous Theorem 5: An optimal control -u"(x)/u'(x) which maximizes Lecture 20

  18. is given by Lecture 20

  19. The empirical idea is then to determine whether one distribution is dominated by another distribution in the second degree with respect to a particular set of preferences. To do this we want to know if the integral switches from negative to positive within the range of risk aversion coefficients. Lecture 20

  20. Theorem 5 states that the risk aversion which maximizes the integral will be found at one boundary or the other depending on the sign of the integral. Lecture 20

  21. Application: • Assume that G(x)-F(x) is always nonnegative. Then for any u'(x) we consider. Thus, the optimal control for this particular F(x) and G(x) is to choose -u"(x)/u'(x) equal to its maximum possible Lecture 20

  22. If G(x)-F(x) changes sign a finite number of times, then we know that for some x0, G(x)-F(x) does not change sign in the interval [x0,1]. Thus, the optimal solution in [x0,1] is given by the above, and once the solution in [x0,1] is known, the solution of [0,x0] can be calculated by Theorem 5. Lecture 20

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