Independent-tail replacement errors are decreasing and summability implies the rate condition

Fix $m$. Write the innovation input as $(A,B)$, with $A$ its first $m$ coordinates and $B$ its remaining tail. Its law is a product by [F2]. The replacement uses $(A,B^{\mathrm{\prime }})$, where $B,B^{\mathrm{\prime }}$ have the same tail law and are independent of one another and of $A$. To construct the conditional mean without assuming a vector conditional-expectation theorem, take finite-valued measurable approximants $f_j(A,B)$ to the filter $f(A,B)$ with $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_{L^2(H)}\le 2^{-2j}$. They exist explicitly: include zero in a fixed dense sequence in $H$, use the first nearest point among its first $N$ points, and let $N$ increase. The error is at most $\mathrm{\parallel }f\mathrm{\parallel }$ and tends pointwise to zero, so [F3] gives $L^2$ convergence; choose the first $N$ achieving the indicated bound. For each finite-valued $f_j$, its tail average $g_j(A)=\int f_j(A,b){\nu }_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}}(db)$ is a finite linear combination of measurable section probabilities, hence is measurable by [F2]. Convexity of the squared Hilbert norm for finite averages, extended by simple scalar integration, gives $\mathbb{E}\mathrm{\parallel }{g}_j-g_k{\mathrm{\parallel }}^2\le \mathbb{E}\mathrm{\parallel }{f}_j-f_k{\mathrm{\parallel }}^2$.

The last inequality implies ${\sum }_j\mathbb{E}\mathrm{\parallel }{g}_{j+1}-g_j\mathrm{\parallel }<\mathrm{\infty }$, by [F3] and the chosen error bounds. Tonelli [F2] makes the sum of these norms finite almost surely. Completeness of $H$ gives a measurable pointwise limit $g_m$, set to zero on the null complement. Fatou [F3], applied to $\mathrm{\parallel }{g}_j-g_k{\mathrm{\parallel }}^2$ as $k\to \mathrm{\infty }$, gives $g_j\to g_m$ in $L^2(H)$. In particular $g_m$ is square integrable. For finite-valued ${\mathcal{F}}_m$-measurable $Y$, finite sums and the section-integral identity give $\mathbb{E}⟨f_j(A,B),Y⟩=\mathbb{E}⟨g_j(A),Y⟩$. Pass to the limit by Cauchy–Schwarz. Approximate any square-integrable ${\mathcal{F}}_m$-measurable $Y$ by its finite nearest-point approximants as in step 1.1, and pass to the limit again. Thus $\mathbb{E}⟨X-g_m,Y⟩=0$ for every such $Y$, which is the conditional-mean identity and implies $\mathbb{E}\mathrm{\parallel }X-Y{\mathrm{\parallel }}^2=\mathbb{E}\mathrm{\parallel }X-g_m{\mathrm{\parallel }}^2+\mathbb{E}\mathrm{\parallel }{g}_m-Y{\mathrm{\parallel }}^2\mathrm{.}$ This proves the stated minimization, without a regular conditional law on the innovation space.

Conditional product integration for the two copies of the finite-valued $f_j$ gives $\mathbb{E}⟨f_j(A,B),f_j(A,B^{\mathrm{\prime }})⟩=\mathbb{E}\mathrm{\parallel }{g}_j(A){\mathrm{\parallel }}^2$. Passing to the $L^2$ limits by [F3] and Hilbert-space Cauchy–Schwarz [F4] gives $\mathbb{E}⟨X,X_{0,m}⟩=\mathbb{E}\mathrm{\parallel }{g}_m{\mathrm{\parallel }}^2$. The two copies have the same second moment by [F1]. Expanding their squared difference therefore yields ${\delta }_m^2=2(\mathbb{E}\mathrm{\parallel }X{\mathrm{\parallel }}^2-\mathbb{E}\mathrm{\parallel }{g}_m{\mathrm{\parallel }}^2)=2\mathbb{E}\mathrm{\parallel }X-g_m{\mathrm{\parallel }}^2$, the final equality also following from step 2.1 with $Y=g_m$. The sigma-fields increase with $m$, so every ${\mathcal{F}}_m$-measurable competitor is an ${\mathcal{F}}_{m+1}$-measurable competitor. The minimum residual squared norm cannot increase; hence ${\delta }_{m+1}\le {\delta }_m$.

For $m\ge 2$, monotonicity gives $\lfloor m\mathrm{/}2\rfloor {\delta }_m\le {\sum }_{j=\lceil m\mathrm{/}2\rceil }^m{\delta }_j$. If the full nonnegative series is summable, its right-hand tail tends to zero. Since $m\mathrm{/}\lfloor m\mathrm{/}2\rfloor \le 3$, this implies $m{\delta }_m\to 0$. Thus the rate-separation example requested in the former design cannot exist under the actual independent-tail replacement convention. This is a proved implication, not an absence of a known counterexample.