= Solution
Write $x\otimes y$ for the <rank-one operator> $h\mapsto\langle h,y\rangle x$. The estimator is the kernel representation of the <empirical covariance operator>
$$
\widehat C_{\mu_0}=\frac1n\sum_{i=1}^n(X_i-\mu_0)\otimes(X_i-\mu_0).
$$
Since $\mathbb EX=0$,
$$
\mathbb E\widehat C_{\mu_0}=C_X+\mu_0\otimes\mu_0.
$$
Thus the estimator has the fixed <bias of an estimator> $\mu_0\otimes\mu_0$ for $C_X$.
The fourth-moment assumption makes $(X_i-\mu_0)\otimes(X_i-\mu_0)$ square-integrable in the Hilbert space of <Hilbert-Schmidt operators>. The <weak law of large numbers> therefore gives
$$
\widehat C_{\mu_0}\xrightarrow{p}C_X+\mu_0\otimes\mu_0.
$$
Consequently <consistency (statistics)> for $C_X$ holds exactly when $\mu_0=0$. More precisely, if $Y=(X-\mu_0)\otimes(X-\mu_0)$, then
$$
\mathbb E\lVert\widehat C_{\mu_0}-C_X\rVert_{\mathrm{HS}}^2
=\lVert\mu_0\otimes\mu_0\rVert_{\mathrm{HS}}^2
+\frac1n\mathbb E\lVert Y-\mathbb EY\rVert_{\mathrm{HS}}^2
=\lVert\mu_0\rVert^4+O(n^{-1}).
$$
The <Hilbert-space central limit theorem> also yields
$$
\sqrt n\{\widehat C_{\mu_0}-(C_X+\mu_0\otimes\mu_0)\}
\xrightarrow dG,
$$
where $G$ is a centered <Gaussian random element> in the Hilbert-Schmidt operator space with covariance determined by $Y$. Relative to $C_X$, the same fluctuation is displaced by $\sqrt n(\mu_0\otimes\mu_0)$ and hence does not have a finite centered limit when $\mu_0\ne0$.
Back to article page
