Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 210 (Topics in Statistical Theory)

Let $f(x)=\Vert x{\Vert }_{\infty }$. Away from ties and zero coordinates, its gradient is $\nabla f(x)=\pm e_i$ for a maximizing coordinate $i$. Conditional on $X$, the random variable $⟨\nabla f(X),Y⟩$ is centered Gaussian with variance at most one. The supplied Gaussian concentration inequality therefore givesA Chernoff bound, optimized at $\lambda =4u/{\pi }^2$, yieldsIt remains to bound the mean. For $t>0$, Jensen's inequality and the Gaussian moment-generating function giveTaking $t=\sqrt{2\log (2d)}$ gives $\mathbb{E}\Vert X{\Vert }_{\infty }\le \sqrt{2\log (2d)}$. Consequently

Letso $T(\theta )=a^T({\theta }_1,\dots ,{\theta }_4)$. In the Gaussian sequence model, put $h_k=\sqrt{n}({\theta }_k-Y_k)$. For each relevant coordinate, the posterior density of $h_k$ relative to the standard-normal density $\phi$ isThe log-Lipschitz assumption impliesThese bounds provide Gaussian-integrable domination, while the ratio converges pointwise to one. Dominated convergence, coordinate independence, and the same argument after multiplying by $e^{t{a}^{T}h}$ show that, under the posterior,almost surely. The supplied moment-generating-function criterion therefore gives the finite-functional Bernstein-von Mises theoremwith uniform convergence of distribution functions.

If $z_{1-\alpha }={\Phi }^{-1}(1-\alpha )$, the posterior quantile defining $R_n$ consequently satisfiesin probability. Under $P_{{\theta }_0}^Y$,for every $n$. Quantile convergence and the Slutsky theorem now yield

Write $K=K_n$, $ϵ={ϵ}_n$, and chooseTake a $\delta$-net $({\theta }^1,\dots ,{\theta }^N)$ of $H_1$ in ${\ell }^2$, with centers in $H_1$. Since every alternative lies in the $K$-dimensional Euclidean ball of radius $n$, the volumetric covering bound givesfor all large $n$. Put $v_j={\theta }^j-{\theta }_0$ and define the Gaussian net testThe inner products are well-defined Gaussian linear functionals because $v_j\in {\ell }^2$. Under $H_0$,and $\Vert v_j{\Vert }_2\ge Mϵ$. A Gaussian tail bound and a union bound give

For any $\theta \in H_1$, choose $j$ with $\Vert \theta -{\theta }^j{\Vert }_2\le \delta$. ThenAnother Gaussian tail bound givesBecause $n{ϵ}_n^2=K_n\log n$, the entropy term in the type-I bound is dominated by the signal exponent when $M$ is sufficiently large. Also ${ϵ}_n^2\to 0$ implies $K_n<n$ eventually, so $\log n\ge \log K_n$. Given $c_1>0$, choose $M$ large enough to obtainEnlarging $M$ if necessary handles the finitely many initial $n$.

Because white-noise observations are not themselves in ${\ell }^2$, define the least-squares estimator as a minimizer of the Gaussian least-squares contrast, equivalently a maximizer over $\theta \in \Theta$ ofwhere $W$ is the isonormal Gaussian process on ${\ell }^2$. The entropy assumption makes $W$ sample-continuous on compact $\Theta$, so a maximizer exists.

Put $\Delta =\hat{\theta }-{\theta }_0$. Comparison with ${\theta }_0$ gives the basic inequalityForthe entropy assumption and the Dudley entropy integral giveThe Borell-TIS inequality further gives

Set $r_n=c{n}^{-4/9}$. This is the balanceOn the shell $2^j{r}_n\le \Vert \theta -{\theta }_0{\Vert }_2<2^{j+1}{r}_n$, the basic inequality would requireFor $c$ sufficiently large, the expectation bound is at most half this threshold for every $j$. Borell concentration then bounds the shell probability bySumming the geometric sequence of shell bounds gives a quantity tending to zero, uniformly in ${\theta }_0\in \Theta$. Therefore