ST 762, HOMEWORK 4 EXTRA PROBLEMS SOLUTIONS, FALL 20091. Here, there are no covariates.(a) Using the notation in the notes, we h ave fβj= mhβ(β) = meβ/(1 + eβ)2, and g2j=meβ/(1 + eβ)2. Thus, these terms cancel in the GLS estimating equation, and the GLSestimatorˆβ satisfies at C = ∞nXj=1{Yj− meˆβ/(1 + eˆβ)} = 0.Letting¯Y = n−1= n−1Pnj=1Yj, we haveˆβ = log ¯Ym −¯Y!.By the weak law of large numbers,¯Yp−→ meβ0/(1+eβ0). Because log{η/(m−η)} is continuousin η, we thus haveˆβp−→ log(meβ0m(1 + eβ0) − meβ0)= β0.(b) We have n1/2(ˆβ − β0) = n1/2[log{¯Y /(m −¯Y )} − β0]. Because¯Yp−→ meβ0/(1 + eβ0), wecan approximate by a first-order Taylor serieslog{¯Y /(m −¯Y )} ≈ log(mh(β0)m − mh(β)))+"{(¯Y /(m −¯Y )}−1(1m −¯Y+¯Y(m −¯Y )2)#¯Y =mh(β0)× ¯Y −meβ01 + eβ0!= β0+(meβ0(1 + eβ0)2)−1 ¯Y −meβ01 + eβ0!.Thus,n1/2(ˆβ − β0) ≈ n1/2{n−1nXj=1Yj− meβ0/(1 + eβ0)}(meβ0(1 + eβ0)2).Nown1/2{n−1nXj=1Yj− meβ0/(1 + eβ0)}L−→ N(0,meβ0(1 + eβ0)2).Thus, the full expression satisfiesn1/2(ˆβ − β0)L−→ N0, meβ0(1 + eβ0)2!−1.From the folklore theorem, the limiting variance should be, using shorthand notation,σ20n−1nXj=1g−20jfβ0jfTβ0j−1=n−1nXj=1 meβ0(1 + eβ0)2!−1 meβ0(1 + eβ0)2!2−1,1which reduces to the above (σ0= 1 here). So the above result coincides with the generalfolklore result.(c) The GLS estimator is still the same as above. However, now the variance is different. Wenow haven−1nXj=1varYj= n−1nXj=1(meβ0(1 + eβ0)2){1 + θ(m − 1)},which is a constant, so that n−2Pnj=1var(Yj) → 0. Thus, the weak law of large numbersstill applies as in (a), and we have the result that, despite the misspecification of variance,ˆβp−→ β0.(d) We may make the same approximation as in (b) to arrive atn1/2(ˆβ − β0) ≈ n1/2{n−1nXj=1Yj− meβ0/(1 + eβ0)}(meβ0(1 + eβ0)2).Now, however, by the central limit theorem, we haven1/2{n−1nXj=1Yj− meβ0/(1 + eβ0)}L−→ N"0,meβ0(1 + eβ0)2{1 + θ(m − 1)}#,so that, after simplification, we haven1/2(ˆβ − β0)L−→ N0, meβ0(1 + eβ0)2!−1{1 + θ(m − 1)}.From page 224, we expect that the covariance matrix should be of the formσ20Σ−11Σ2Σ−11,whereΣ−11≈ n−1nXj=1k−20jfβ0jfTβ0jand k0jis the misspecified variance function. It is straightforward to obtainΣ1≈meβ0(1 + eβ0)2,andΣ2≈meβ0(1 + eβ0)2{1 + θ(m − 1)}.Thus,Σ−11Σ2Σ−11=(meβ0(1 + eβ0)2)−1{1 + θ(m − 1)},as required.22. (a) This is the s ame argument as in the notes, only messier. You should have ended up withC∗n= n−1/2nXj=1(2+κj−ζ2j)−1 {(2 + κj)g−10jfβ0j− 2ζjσ0νβ0j}ǫj+ {−ζjg−10jfβ0j+ 2σ0νβ0j}(ǫ2j− 1){−2ζjǫj+ 2(ǫ2j− 1)τθ0j!andA∗n= XTW X + (XT∗W1/2− 2σ0RT∗)(W1/2X∗− 2σ0R∗) −2(XT∗− 2σ0RT∗)Q∗−2QT∗(X∗− 2σ0R∗) 4QT∗Q∗!.(b) We could proceed to evaluate the RHS ofn1/2(ˆβ − β0)/σ0(ˆσ − σ0)/σ0ˆθ − θ0≈ A∗ −1nC∗n.However, we know from the usual M-estimator argument that when the “covariance matrix”Vjin the original estimating equation is correctly specified, we have immediately that the LHSconverges in distribution to a normal with mean 0 and covariance matrix {lim n−1A∗ −1n}−1.So all we really need to do is find the upper left-hand (p × p) submatrix of A∗ −1n. From theresults for inversion of partitioned matrices on page 252, we can find that this matrix is{XTW X + (XT∗W1/2− 2σ0RT∗)(W1/2X∗− 2σ0R∗)−(XT∗W1/2− 2σ0RT∗)Q∗(QT∗Q∗)−1QT∗(W1/2X∗− 2σ0R∗)}−1which is equal to{XTW X + (XT∗W1/2− 2σ0RT∗)P∗(W1/2X∗− 2σ0R∗)}−1.ThusΣ ≈ {n−1XTW X + n−1(XT∗W1/2− 2σ0RT∗)P∗(W1/2X∗− 2σ0R∗)}−1.(c) We know that the GLS estimator has approximate large sample covariance matrix{n−1XTW X}−1. So we want to show that (XTW X)−1− {XTW X + (XT∗W1/2−2σ0RT∗)P∗(W1/2X∗−2σ0R∗)}−1is nonnegative definite Because P∗is symmetric and idem-potent, we can write this second term, letting G∗= (W1/2X∗− 2σ0R∗), as{XTW X + GT∗PT∗P∗G∗}−1= {XTW X + KT∗K∗}−1,say. Now from the partitioned inverse results, we have that this may be written equivalentlyas(XTW X)−1− (XTW X)−1{KT∗K∗+ (XTW X)−1}−1(XTW X)−1,which is clearly ≤ (XTW X)−1, assuming all matrices in question are positive defin ite. Thus,the quadratic estimator is at least as efficient as GLS.(d) When ζj= 0 and κj= κ, then it is straightforward to see that Z = 0, H = (2 + κ)I, andthus X∗= 0, R∗= (2 + κ)−1/2R, Q∗= (2 + κ)−1/2Q, and P∗= P . Under these conditions,the general result reduces to the one for (c) on page 256 of the notes.3(e) When gj= fθj, log gj= θ log fj, and thus νβ0j= θ0fβ0j/fj. Now when ζj= ζ = aσ0andκj= κ, X∗= ζX/(2 + κ − ζ2)1/2, Q∗= Q/(2 + κ − ζ2)1/2, and R∗= R/(2 + κ − ζ2)1/2, sothat P∗= P . Moreover, defining F = diag(f−10j), R = θ0F X. Now for ζ = aσ0and κ = bσ20,2 + κ − ζ2= 2 + (b − a2)σ20. Substituting all of this into the approximation to Σ we found in(b), we obtainΣ ≈ {n−1XTW X + n−1(2 + (b − a2)σ20)−1σ20XT(aW1/2− 2F )P (aW1/2− 2F )X}−1.(f) Note that if in fact θ0= 1, W1/2= F , so if we furth er take a = 2, the second term in Σis zero, as long as 2 + (b − a2)σ206= 0, Σ ≈ (n−1XTW x)−1, as required. Thus, with a = 2,b must satisfy 2 + (b − 4)σ206= 0. The only way this can equal zero is if σ20= 2/(4 − b).Note that if b = 6, then these choices of a and b with θ0correspond second, third, and fourthmoments of a gamma distribution (and certainly this value of b does not violate the above,as σ20> 0). This shows that if we set up the estimating equations using th e third and fourthmoments of the gamma distribution, and in fact the variance is identical to that of a gamma(power model with θ0= 1), then there is no benefit to trying to incorporate information fromthe variance about β to gain efficiency; the ML estimator under these conditions solves thelinear GLS equations, so this would seem to be the best we can do.3. (a) The optimal estimating equation isnXj=1(fβj, 2σ2g2jνβj) σ2g2j00 (2 + κ)σ4g4j!−1 Yj− fj(Yj− fj)2− σ2g2j!= 0.Here fj= β, gj= f1/2j= β1/2, fβj= 1, νβj= 1/(2β) for j = 1, . . . , m, and fj= 1/β,gj= f1/2j=