(a) Define a generalised linear model (GLM) with design matrix X∈Rn×p, output variables Y:=(Y1,…,Yn)T and parameters μ:=(μ1,…,μn)T,β∈Rp and σi2:=aiσ2∈(0,∞),i=1,…,n. Derive the moment generating function of Y, i.e. give an expression for E[exp(tTY)],t∈Rn, wherever it is well-defined.
Assume from now on that the GLM satisfies the usual regularity assumptions, X has full column rank, and σ2 is known and satisfies 1/σ2∈N.
(b) Let Y~:=(Y~1,…,Y~n/σ2)T be the output variables of a GLM from the same family as that of part (a) and parameters μ~:=(μ~1,…,μ~n/σ2)T and σ~2:=(σ~12,…,σ~n/σ22). Suppose the output variables may be split into n blocks of size 1/σ2 with constant parameters. To be precise, for each block i=1,…,n, if j∈{(i−1)/σ2+1,…,i/σ2} then
μ~j=μi and σ~j2=ai
with μi=μi(β) and ai defined as in part ( a ). Let Yˉ:=(Yˉ1,…,Yˉn)T, where Yˉi:=σ2∑k=11/σ2Y~(i−1)/σ2+k.
(i) Show that Yˉ is equal to Y in distribution. [Hint: you may use without proof that moment generating functions uniquely determine distributions from exponential dispersion families.]
(ii) For any y~∈Rn/σ2, let yˉ=(yˉ1,…,yˉn)T, where yˉi:=σ2∑k=11/σ2y~(i−1)/σ2+k. Show that the model function of Y~ satisfies
f(y~;μ~,σ~2)=g1(yˉ;μ~,σ~2)×g2(y~;σ~2)
for some functions g1,g2, and conclude that Yˉ is a sufficient statistic for β from Y~.
(iii) For the model and data from part (a), let μ^ be the maximum likelihood estimator for μ and let D(Y;μ) be the deviance at μ. Using (i) and (ii), show that
where =d means equality in distribution and M0 and M1 are nested subspaces of Rn/σ2 which you should specify. Argue that dim(M1)=n and dim(M0)=p, and, assuming the usual regularity assumptions, conclude that
σ2D(Y;μ^)→dχn−p2 as σ2→0
stating the name of the result from class that you use.