Risk decomposition and monotone bias for ordered spectral filters

Put $c_j:={\lambda }_j{q}_{{\alpha }_m}({\lambda }_j){\theta }_j$ and $d_j:=\sigma s_j{q}_{{\alpha }_m}({\lambda }_j)$. By [F3], $\mathrm{\mid }{c}_j\mathrm{\mid }\le \mathrm{\mid }{\theta }_j\mathrm{\mid }$, so $(c_j{)}_j\in {\mathrm{\ell }}^2$. Also $q_{{\alpha }_m}({\lambda }_j)\le C_q^{\mathrm{\prime }}\mathrm{/}{\alpha }_m$ and ${\sum }_j{\lambda }_j<\mathrm{\infty }$ by [F2], so ${\sum }_j{d}_j^2={\sigma }^2{\sum }_j{\lambda }_j{q}_{{\alpha }_m}({\lambda }_j{)}^2<\mathrm{\infty }$. The partial sums of $Q:={\sum }_j{d}_j^2{\xi }_j^2$ increase, and [F5] gives $\mathbb{E}Q={\sum }_j{d}_j^2<\mathrm{\infty }$. For each integer $r\ge 1$, [F6] gives $\mathbb{P}(Q\ge r)\le \mathbb{E}Q\mathrm{/}r$; since $\{Q=+\mathrm{\infty }\}\subseteq \{Q\ge r\}$ for every $r$, $Q<\mathrm{\infty }$ almost surely. Thus on a measurable event of probability one the coordinate series ${\sum }_j(c_j+d_j{\xi }_j)v_j$ converges in $X$; define its pointwise limit there and set it to zero on the null complement. The finite partial sums are measurable, and their pointwise limit on this event, extended by zero, is measurable because $X$ is separable. By Parseval and [F5], for each $N$, $\mathbb{E}\mathrm{\parallel }{\widehat{x}}_m-X_{m,N}{\mathrm{\parallel }}_X^2={\sum }_{j>N}\mathbb{E}(c_j+d_j{\xi }_j{)}^2={\sum }_{j>N}{c}_j^2+{\sum }_{j>N}{d}_j^2$, where [F4] gives $\mathbb{E}{\xi }_j=0$ and $\mathbb{E}{\xi }_j^2=1$. Both deterministic and noise tails tend to zero, so this constructed measurable pointwise limit is the $L^2(\mathrm{\Omega };X)$ limit. In particular it has finite second moment.

For $m_1<m_2$, [F3] gives $0\le 1-{\lambda }_j{q}_{{\alpha }_{m_2}}({\lambda }_j)\le 1-{\lambda }_j{q}_{{\alpha }_{m_1}}({\lambda }_j)$. Squaring these nonnegative terms and summing against ${\theta }_j^2$ gives $B_{m_2}^2\le B_{m_1}^2$. Moreover, $f_{m_2}-f_{m_1}$ has coordinate ${\lambda }_j(q_{{\alpha }_{m_2}}-q_{{\alpha }_{m_1}}){\theta }_j$, whose multiplier equals $(1-{\lambda }_j{q}_{{\alpha }_{m_1}})-(1-{\lambda }_j{q}_{{\alpha }_{m_2}})$ and lies between zero and $1-{\lambda }_j{q}_{{\alpha }_{m_1}}$; coordinatewise comparison therefore yields $\mathrm{\parallel }{f}_{m_2}-f_{m_1}{\mathrm{\parallel }}_X\le B_{m_1}$.

Subtracting $x$ in each singular coordinate gives deterministic bias coefficient $-(1-{\lambda }_j{q}_{{\alpha }_m}({\lambda }_j)){\theta }_j$ and centered noise coefficient $\sigma s_j{q}_{{\alpha }_m}({\lambda }_j){\xi }_j$. Since $\mathbb{E}{\xi }_j=0$ and $\mathbb{E}{\xi }_j^2=1$ by [F1] and [F4], summing the coordinate second moments and using monotone convergence gives $\mathbb{E}\mathrm{\parallel }{\widehat{x}}_m-x{\mathrm{\parallel }}_X^2={\sum }_j(1-{\lambda }_j{q}_{{\alpha }_m}({\lambda }_j){)}^2{\theta }_j^2+{\sigma }^2{\sum }_j{\lambda }_j{q}_{{\alpha }_m}({\lambda }_j{)}^2=B_m^2+v_m$.

The centered pairwise noise difference has coordinate variance ${\sigma }^2{\lambda }_j(q_{{\alpha }_{m_2}}({\lambda }_j)-q_{{\alpha }_{m_1}}({\lambda }_j){)}^2$. By [F3], each difference lies between zero and $q_{{\alpha }_{m_2}}({\lambda }_j)$; summing its squared value against ${\sigma }^2{\lambda }_j$ proves $v_{m_1,m_2}\le v_{m_2}$.