68 and 六∑1(r-川） To compute thethe solution to (1)and(2),we start with initial values for a and B,say a(0)and B(0),then update as follows: -）aw ∑=1(ri-p(a),gG) ∑1x(r:-p(a6),3) We iterate until convergence
68 and Jn2(µ, σ2) = ⎡⎣ nσ2 1σ4 ni=1(xi − µ) 1σ4 ni=1(xi − µ) − n2σ4 + 1σ6 ni=1(xi − µ)2 ⎤⎦ To compute the the solution to (1) and (2), we start with initial values for α and β, say α(0) and β(0), then update as follows: α(j+1) β(j+1) = α(j) β(j) +{Jn1(α(j), β(j))}−1· ni=1(ri − pi(α(j), β(j)) ni=1 xi(ri − pi(α(j), β(j)) We iterate until convergence.
69 Assume that la≤K1,ll≤K2,luW≤K3,andK4≤o2≤K5, where K1,...,K5 are finite positive quantities.Is the parameter space,e,compact?YES!Assume that 0o (truth)belongs to the interior of c.Verify all of the regularity conditions that are necessary to show that the MLE is both consistent and asymptotically normal. Refer to Section 8c.Condition i)is satisfied by assumption. Condition ii)is satisfied since logp(x,r;0)is continuous at each for all -oo<x<oo and r =0,1.Now,we need to show condition ii)that|logp(c,r;0)川≤d(x,r)for all0∈日and Eo[d(X,R)l<o. IogDz.r;0)--jlog(2z)-log(a)-z(-p)+r(@+Ox)-log(1+exp(a+0) By the triangle inequality,we know that Ilogp(z,r:0)I<>log(2z)+llog(o)l+lzc2(-p)2|+lr(@+Bz)l+llog(1+exp(@+Bz)l
69 Assume that |α| ≤ K1, |β| ≤ K2, |µ| ≤ K3, and K4 ≤ σ2 ≤ K5, where K1,...,K5 are finite positive quantities. Is the parameter space, Θ, compact? YES! Assume that θ0 (truth) belongs to the interior of Θ. c. Verify all of the regularity conditions that are necessary to show that the MLE is both consistent and asymptotically normal. Refer to Section 8c. Condition i) is satisfied by assumption. Condition ii) is satisfied since log p(x, r; θ) is continuous at each θ for all −∞ <x< ∞ and r = 0, 1. Now, we need to show condition iii) that | log p(x, r; θ)| ≤ d(x, r) for all θ ∈ Θ and Eθ0 [d(X, R)] < ∞. log p(x, r; θ) = −12 log(2π)−log(σ)− 12σ2 (x−µ)2+r(α+βx)−log(1+exp(α+βx)) By the triangle inequality, we know that | log p(x, r; θ)| ≤ 12 log(2π)+| log(σ)|+| 12σ2 (x−µ)2|+|r(α+βx)|+| log(1+exp(α+βx))|