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Write $x\otimes y$ for the rank-one operator $h\mapsto ⟨h,y⟩x$. The estimator is the kernel representation of the empirical covariance operatorSince $\mathbb{E}X=0$,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 givesConsequently consistency for $C_X$ holds exactly when ${\mu }_0=0$. More precisely, if $Y=(X-{\mu }_0)\otimes (X-{\mu }_0)$, thenThe Hilbert-space central limit theorem also yieldswhere $G$ 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 ${\mu }_0\ne 0$.

Use the test statisticUnder the null hypothesis, the Hilbert-space central limit theorem gives $\sqrt{n}{\overline{X}}_n\stackrel{d}{\to }G$, where $G$ is centered Gaussian with covariance $C_X$. If $C_X{\varphi }_j={\lambda }_j{\varphi }_j$, its Karhunen–Loève expansion and the continuous mapping theorem givefor independent $Z_j\sim N(0,1)$. Reject for $T_n$ above the $(1-\alpha )$ quantile of this weighted chi-squared law; replacing the ${\lambda }_j$ by empirical covariance eigenvalues gives a plug-in estimator of the critical value.

Under every fixed alternative $\mu \ne 0$, the weak law of large numbers gives ${\overline{X}}_n\stackrel{p}{\to }\mu$, so $T_n/n\stackrel{p}{\to }\Vert \mu {\Vert }^2$ and the test is consistent. Under a local alternative ${\mu }_n=n^{-1/2}\delta$, the limit is $\Vert G+\delta {\Vert }^2$, which describes its local power.