Paper 4, Section II, J

Statistical Modelling | Part II, 2020

(a) Define a generalised linear model (GLM)(\mathrm{GLM}) with design matrix XRn×pX \in \mathbb{R}^{n \times p}, output variables Y:=(Y1,,Yn)TY:=\left(Y_{1}, \ldots, Y_{n}\right)^{T} and parameters μ:=(μ1,,μn)T,βRp\mu:=\left(\mu_{1}, \ldots, \mu_{n}\right)^{T}, \beta \in \mathbb{R}^{p} and σi2:=aiσ2(0,),i=1,,n\sigma_{i}^{2}:=a_{i} \sigma^{2} \in(0, \infty), i=1, \ldots, n. Derive the moment generating function of YY, i.e. give an expression for E[exp(tTY)],tRn\mathbb{E}\left[\exp \left(t^{T} Y\right)\right], t \in \mathbb{R}^{n}, wherever it is well-defined.

Assume from now on that the GLM satisfies the usual regularity assumptions, XX has full column rank, and σ2\sigma^{2} is known and satisfies 1/σ2N1 / \sigma^{2} \in \mathbb{N}.

(b) Let Y~:=(Y~1,,Y~n/σ2)T\tilde{Y}:=\left(\tilde{Y}_{1}, \ldots, \tilde{Y}_{n / \sigma^{2}}\right)^{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\tilde{\mu}:=\left(\tilde{\mu}_{1}, \ldots, \tilde{\mu}_{n / \sigma^{2}}\right)^{T} and σ~2:=(σ~12,,σ~n/σ22)\tilde{\sigma}^{2}:=\left(\tilde{\sigma}_{1}^{2}, \ldots, \tilde{\sigma}_{n / \sigma^{2}}^{2}\right). Suppose the output variables may be split into nn blocks of size 1/σ21 / \sigma^{2} with constant parameters. To be precise, for each block i=1,,ni=1, \ldots, n, if j{(i1)/σ2+1,,i/σ2}j \in\left\{(i-1) / \sigma^{2}+1, \ldots, i / \sigma^{2}\right\} then

μ~j=μi and σ~j2=ai\tilde{\mu}_{j}=\mu_{i} \quad \text { and } \quad \tilde{\sigma}_{j}^{2}=a_{i}

with μi=μi(β)\mu_{i}=\mu_{i}(\beta) and aia_{i} defined as in part (( a )). Let Yˉ:=(Yˉ1,,Yˉn)T\bar{Y}:=\left(\bar{Y}_{1}, \ldots, \bar{Y}_{n}\right)^{T}, where Yˉi:=\bar{Y}_{i}:= σ2k=11/σ2Y~(i1)/σ2+k\sigma^{2} \sum_{k=1}^{1 / \sigma^{2}} \tilde{Y}_{(i-1) / \sigma^{2}+k}.

(i) Show that Yˉ\bar{Y} is equal to YY 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\tilde{y} \in \mathbb{R}^{n / \sigma^{2}}, let yˉ=(yˉ1,,yˉn)T\bar{y}=\left(\bar{y}_{1}, \ldots, \bar{y}_{n}\right)^{T}, where yˉi:=σ2k=11/σ2y~(i1)/σ2+k\bar{y}_{i}:=\sigma^{2} \sum_{k=1}^{1 / \sigma^{2}} \tilde{y}_{(i-1) / \sigma^{2}+k}. Show that the model function of Y~\tilde{Y} satisfies

f(y~;μ~,σ~2)=g1(yˉ;μ~,σ~2)×g2(y~;σ~2)f\left(\tilde{y} ; \tilde{\mu}, \tilde{\sigma}^{2}\right)=g_{1}\left(\bar{y} ; \tilde{\mu}, \tilde{\sigma}^{2}\right) \times g_{2}\left(\tilde{y} ; \tilde{\sigma}^{2}\right)

for some functions g1,g2g_{1}, g_{2}, and conclude that Yˉ\bar{Y} is a sufficient statistic for β\beta from Y~\tilde{Y}.

(iii) For the model and data from part (a), let μ^\hat{\mu} be the maximum likelihood estimator for μ\mu and let D(Y;μ)D(Y ; \mu) be the deviance at μ\mu. Using (i) and (ii), show that

D(Y;μ^)σ2=d2log{supμ~M~1f(Y~;μ~,σ~2)supμ~M~0f(Y~;μ~,σ~2)}\frac{D(Y ; \hat{\mu})}{\sigma^{2}}={ }^{d} 2 \log \left\{\frac{\sup _{\tilde{\mu}^{\prime} \in \widetilde{\mathcal{M}}_{1}} f\left(\tilde{Y} ; \tilde{\mu}^{\prime}, \tilde{\sigma}^{2}\right)}{\sup _{\tilde{\mu}^{\prime} \in \widetilde{\mathcal{M}}_{0}} f\left(\tilde{Y} ; \tilde{\mu}^{\prime}, \tilde{\sigma}^{2}\right)}\right\}

where =d=^{d} means equality in distribution and M~0\widetilde{\mathcal{M}}_{0} and M~1\widetilde{\mathcal{M}}_{1} are nested subspaces of Rn/σ2\mathbb{R}^{n / \sigma^{2}} which you should specify. Argue that dim(M~1)=n\operatorname{dim}\left(\widetilde{\mathcal{M}}_{1}\right)=n and dim(M~0)=p\operatorname{dim}\left(\widetilde{\mathcal{M}}_{0}\right)=p, and, assuming the usual regularity assumptions, conclude that

D(Y;μ^)σ2dχnp2 as σ20\frac{D(Y ; \hat{\mu})}{\sigma^{2}} \rightarrow^{d} \chi_{n-p}^{2} \quad \text { as } \sigma^{2} \rightarrow 0

stating the name of the result from class that you use.

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