= Solution
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 $\theta\in\Theta$ of
$$
\mathcal L_n(\theta)
=2\langle Y,\theta\rangle-\|\theta\|_2^2
=2\langle\theta_0,\theta\rangle
+\frac2{\sqrt n}W(\theta)-\|\theta\|_2^2,
$$
where $W$ is the <isonormal Gaussian process> on $\ell^2$. The entropy assumption makes $W$ sample-continuous on compact $\Theta$, so a maximizer exists.
Put $\Delta=\widehat\theta-\theta_0$. Comparison with $\theta_0$ gives the basic inequality
$$
\|\Delta\|_2^2
\leq\frac2{\sqrt n}W(\Delta).
$$
For
$$
Z(r)=\sup\{W(\theta-\theta_0):
\theta\in\Theta,\ \|\theta-\theta_0\|_2\leq r\},
$$
the entropy assumption and the <Dudley entropy integral> give
$$
\mathbb EZ(r)
\leq C\int_0^r
\sqrt{\log N(u,\Theta,\|\cdot\|_2)}du
\leq C\int_0^ru^{-1/8}du
\leq C'r^{7/8}.
$$
The <Borell-TIS inequality> further gives
$$
\mathbb P\{Z(r)>\mathbb EZ(r)+rx\}
\leq e^{-x^2/2}.
$$
Set $r_n=cn^{-4/9}$. This is the balance
$$
r_n^2\asymp n^{-1/2}r_n^{7/8}.
$$
On the shell $2^jr_n\leq\|\theta-\theta_0\|_2<2^{j+1}r_n$, the basic inequality would require
$$
Z(2^{j+1}r_n)
\geq\frac{\sqrt n}{2}(2^jr_n)^2.
$$
For $c$ sufficiently large, the expectation bound is at most half this threshold for every $j$. Borell concentration then bounds the shell probability by
$$
\exp(-c_0\,4^j n r_n^2)
=\exp(-c_0c^2\,4^j n^{1/9}).
$$
Summing the geometric sequence of shell bounds gives a quantity tending to zero, uniformly in $\theta_0\in\Theta$. Therefore
$$
\boxed{
\mathbb P_{\theta_0}^Y
\left(\|\widehat\theta-\theta_0\|_{\ell^2}
\geq cn^{-4/9}\right)\longrightarrow0.}
$$
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