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6. (Bayesian Hardy-Weinberg model) Let again X1 = 3:1,X2 = 3:2,X3 = 333 be the observed counts having jointly the trinomial distribution with n trials

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6. (Bayesian Hardy-Weinberg model) Let again X1 = 3:1,X2 = 3:2,X3 = 333 be the observed counts having jointly the trinomial distribution with n trials and correspond ing probabilities 191, p2, p3. According to the Hardy Weinberg equilibrium law, for some parameter 6 E [0, 1], 191(9) = (1 - (9)2: 192(0) = 20(1- 9), 193(0) = 92- The unknown parameter 0 is population specic and needs to be estimated based on the data. The maximum likelihood estimator (MLE) of the parameter 6 was found earlier in the course to be 2333 + 51:2 ($1,$2,$3) = 2\" Suppose that the unknown parameter 6 has the uniform prior distribution on the interval [0, 1]. Consider the Bayesian estimator 0*(w1, 3:2, $3) of 6, with respect to the weighted Bayesian mean squared error n 6(1_6)(5(X1,X2,X3) 9)2 . Ex,0 a) Prove that the Bayesian estimator 6*(311, :32, 3:3) coincides with the MLE 9(azl, $2,313). Hint: Use the beta integral fol 6\"'1(1 60131 d6 = W. b) Is the MLE 9 an admissible estimator? Explain your

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