A reporting rule and its prediction error

Let \(m\ge3\), let \(I_1,\ldots,I_m\) be intervals, and let \(\iota:\mathbb R\to\{1,\ldots,m\}\) be a selector with finitely many cut points. All endpoint and tie conventions are part of the rule. Write its inputs as \((u,z_1,\ldots,z_m)\) and their sum as \(L=u+\sum_{j=1}^m z_j\). The coordinate \(u\) selects the leaf \(z_{\iota(u)}\). Define \[B=\{z_j\in I_j\text{ for every }j\},\qquad P=\{z_j\in I_j\text{ for every }j\ne\iota(u)\},\qquad A=P\setminus B.\] Thus \(A\) is the event that the selected leaf alone fails its interval test. Since \(B\subseteq P\), the sets \(A\), \(B\), and \(P^c\) partition the input space. Define the tagged observation \(\mathcal W\) by \[\begin{array}{c|l} \text{Event}&\text{Information reported}\\ \hline A&\text{the tag }A\text{ only}\\ B&(B,z_1,\ldots,z_m)\\ P^c&(P^c,u,\iota(u),(z_j)_{j\ne\iota(u)}). \end{array}\] On \(B\), the central input is omitted. On \(P^c\), the selected leaf is omitted. Although the exceptional report on \(A\) can depend on all \(m+1\) inputs, its indicator has the pointwise identity \[ \mathbf1_A= \sum_{i=1}^m\mathbf1_{\{\iota(u)=i\}} \prod_{j\ne i}\mathbf1_{I_j}(z_j) -\prod_{j=1}^m\mathbf1_{I_j}(z_j). \tag{3}\] Each summand uses only \(m\) coordinates. This cancellation is what allows the exceptional report to have the same degree cost as the ordinary reports.

The remaining task is now precise: choose the intervals, selector, and constant \(b\) so that \(\Delta_b>0\) for Gaussian inputs. A strict positive margin will survive the normal approximation used in amplification. The next calculation produces that margin by subtracting the contribution of \(B\) from that of \(P\) in (3).

Choosing the intervals in the Gaussian model

We will obtain a positive predictor deficit and also record the conditional variance on \(A\). The latter describes the event’s gain without specifying a predictor: on \(A\), the sum varies less than one unobserved variance-one input.

Transfer to finite inputs and amplification

Fix all the Gaussian choices from Lemma 6, including \(m\), the intervals, the selector, \(b\), and \(p_*>0\). Set \(\eta=p_*/2\). If centered variance-one laws \(Z_n\) converge weakly to a standard normal and have uniformly bounded fourth moments, apply Lemma 2 to the fixed function \[g(u,z_1,\ldots,z_m)=\mathbf1_A(1-(L-b)^2).\] It has quadratic growth and discontinuities only on the finitely many hyperplanes from Lemma 6; these have Gaussian measure zero. For independent copies of \(Z_n\), therefore, \(\mathbb Eg\to d_A>3p_*/4\), and eventually \(\mathbb Eg>\eta\). Lemma 5 then gives \[\operatorname{Var}\!\left( \mathbb E\!\left[\sum_{i=1}^{m+1} Z_{n,i} \,\middle|\,\mathcal W(\mathbf Z_n)\right]\right) >m+\eta.\] Together with the pointwise output-indicator decomposition in that lemma, this verifies both hypotheses of Proposition 4 with \((r,d,\eta)=(m+1,m,p_*/2)\).

The Gaussian parameters and \(\eta\) remain fixed throughout the iteration. Proposition 4 therefore increases the ratio of retained variance to cell degree bound by at least \(1+\eta/m>1\) at each stage, using a sufficiently large finite copy count chosen for the current observation. Starting from one revealed sign, finitely many stages make this ratio exceed \(4C^2\) for any prescribed \(C>0\). Lemma 3 then supplies a nonconstant Boolean function with signed singleton sum greater than \(C\sqrt{\deg(f)}\), proving Theorem 1.

Merging the all-leaves-pass reports

Start with any integer \(m\ge3\), intervals \(I_1,\ldots,I_m\), and a selector \(\iota:\mathbb R\to\{1,\ldots,m\}\) with finitely many cut points. Fix all endpoint and tie conventions. As before, on inputs \((u,z_1,\ldots,z_m)\) put \[L=u+\sum_{j=1}^m z_j,\qquad B=\{z_j\in I_j\text{ for every }j\},\] \[P=\{z_j\in I_j\text{ for }j\ne\iota(u)\},\qquad A=P\setminus B.\] Define the coarse report \(\mathcal O\) by \[\begin{array}{c|l} \text{Event}&\text{Information reported}\\ \hline A&\text{the mark }X\\ B&\text{the mark }Y\\ P^c&(O,u,\iota(u),(z_j)_{j\ne\iota(u)}). \end{array}\] The marks \(X,Y\) and the ordinary-report tag \(O\) are distinct. The rule differs from the refined report only on \(B\). The next lemma separates its pointwise cost from the choice of an input law, so that other interval constructions can use the same reporting rule.

Now fix the Gaussian intervals and selector from Lemma 6. Write \[M_\mu=\sum_{j=1}^m\mu_j,\qquad b=M_\mu-1,\qquad p_*=\mathbb P(B)>0,\qquad S=\sum_{j=1}^m s_j^2.\] The Gaussian calculation gave \(d_A=\mathbb E[\mathbf 1_A(1-(L-b)^2)]=p_*(1+S-J)\) with \(J<1/4\). In Lemma 7 take \(b_0=b\) and \(b_1=M_\mu\). On \(B\), the central Gaussian remains unrestricted and the leaves are independently restricted to their intervals. Thus \(L-M_\mu\) has conditional mean zero and variance \(1+S\), giving \[ \mathbb Eg_{b,M_\mu}=d_A-p_*S=p_*(1-J)>\frac34p_*. \tag{9}\] The term \(p_*S\) is precisely the added error on \(B\): the refined predictor had error \(u\) there, while merging its reports leaves the conditional leaf variance \(S\) unobserved. Equation (9) is an identity for this predictor’s gain; (8) is the resulting lower bound for optimal retained variance.

All these Gaussian choices are now fixed. Let centered variance-one laws \(Z_n\) converge weakly to a standard normal with uniformly bounded fourth moments, and take \(m+1\) independent copies as inputs. The fixed function \(g_{b,M_\mu}\) has quadratic growth; its discontinuities lie on the finitely many interval and selector hyperplanes. Lemma 2 and (9) imply that eventually \(\mathbb Eg_{b,M_\mu}>\eta\), where \(\eta=p_*/2>0\). Lemma 7 then gives retained variance greater than \(m+\eta\) and the required pointwise cell cost. Therefore Proposition 4 applies with \[(r,d,\eta)=(m+1,m,p_*/2).\] This criterion and Lemma 3 give Theorem 1 by the coarse rule as well.

Switching to an independent score

Keep the fixed Gaussian event \(A\) and predictor \(b\) above, and put \(q=m+1\). Add one independent centered variance-one input \(w\) to the original tuple \(\mathbf z=(u,z_1,\ldots,z_m)\). Define \[\mathcal O_+(\mathbf z,w)= \begin{cases} (\mathrm{off},\mathbf z),&\mathbf z\notin A,\\ (\mathrm{on},w),&\mathbf z\in A. \end{cases}\] Off \(A\) the original sum \(L\) is known and \(w\) is omitted. On \(A\) the extra input is known and the original sum must be predicted. The rule gains variance whenever the conditional variance of \(L\) on \(A\) is smaller than the one unit of residual variance of \(w\).

This rule has \(q+1\) inputs but cell cost at most \(q\). Indeed, an off-\(A\) output fixes the first \(q\) inputs and places no condition on \(w\). The indicator of an on-\(A\) output \((\mathrm{on},w_0)\) is \[\mathbf 1_A(\mathbf z)\mathbf 1_{\{w=w_0\}}.\] Multiplying (3) by the last value test expresses this indicator as a sum of functions of at most \(m+1=q\) inputs. The tags keep the two types of report separate, and unattained labels have zero indicators. These pointwise expansions also cover the prescribed endpoints and ties. The rule has finite range on every finite product alphabet.

To establish a gain for the approximating laws, first transfer the unconditional deficit, before conditioning on \(A\). For independent copies of any centered variance-one law \(Z_n\) tending to a standard normal with uniformly bounded fourth moments, Lemma 2 gives \[\Delta_{b,n}:=\mathbb E[\mathbf 1_A(1-(L-b)^2)] \longrightarrow d_A>\frac34p_*.\] Thus, for all sufficiently large \(n\), \(\Delta_{b,n}>\eta=p_*/2\). Since \(\Delta_{b,n}\le\mathbb P(A)\), these laws have positive probability of \(A\). Only now fix such an \(n\), write \(\Delta_b=\Delta_{b,n}\), and define \[p=\mathbb P(A)>0,\qquad c=\mathbb E[L\mid A].\] Independence of \(w\) from the original tuple shows that \(\mathbb E[L+w\mid\mathcal O_+]\) equals \(L\) off \(A\) and \(c+w\) on \(A\). The sum \(L+w\) is centered, so its conditional mean is centered too. Using \(\mathbb Ew=0\), \(\mathbb Ew^2=1\), and \(\mathbb EL^2=q\) gives the exact identity \[\begin{align*} \mathop{\mathrm{Var}}(\mathbb E[L+w\mid\mathcal O_+]) &=\mathbb E[\mathbf 1_{A^c}L^2]+p(c^2+1)\\ &=q+p\bigl(1-\mathop{\mathrm{Var}}(L\mid A)\bigr). \tag{10}\end{align*}\] The same input law satisfies \[ p\bigl(1-\mathop{\mathrm{Var}}(L\mid A)\bigr) =\Delta_b+p(c-b)^2>\eta. \tag{11}\] The identities in (10) and the equality in (11) hold more generally for any independent centered variance-one original inputs and an independent centered variance-one \(w\), whenever \(p>0\), with \(\Delta_b=\mathbb E[\mathbf 1_A(1-(L-b)^2)]\). Their marginal laws need not agree. The nonnegative correction \(p(c-b)^2\) records the improvement from the fixed Gaussian predictor \(b\) to the actual conditional mean \(c\). For the fixed approximating law, (10) therefore exceeds \(q+\eta\). This argument uses convergence of one fixed predictor error, not continuity of conditional variances.

The pointwise cost and the transferred gain verify Proposition 4 with the distinct triple \[(r,d,\eta)=(m+2,m+1,p_*/2).\] The criterion again proves Theorem 1, with ratio multiplier at least \(1+\eta/(m+1)>1\). For both rules in this section, the Gaussian parameters and \(\eta\) stay fixed throughout. Each stage starts from one fixed finite observation and chooses its own finite averaging count; the fourth-moment bound is uniform in that count, not across all stages.

From a variance saving to an average cost saving

Let \(q\ge2\) be an integer, and let \(\mathcal P:\mathbb R^q\to\mathcal L\) be a measurable map to a finite label set. Give label \(l\) an integer cost \(\ell(l)\in\{1,\ldots,q\}\). The relevant algebraic requirement is that, on every product of finite input alphabets, the cell indicator for \(l\) is a finite linear combination of functions each depending on at most \(\ell(l)\) input coordinates. The indicator itself need not depend on only that many coordinates. When the inputs are scores of observations with cell degree bound \(D\), this requirement bounds the degree of the output cell by \(D\ell(l)\).

Suppose now that a Gaussian construction supplies an event \(A\) on which one coordinate of degree cost can be saved, while the sum has conditional variance less than one. The next lemma turns this event into an observation whose retained variance exceeds its average cost. We keep \(A\) as one cell and refine its complement finely enough that the extra prediction error uses less than the available variance saving.

The event is fixed before the complement grid is chosen. The resulting partition is then fixed before any normal approximation or iteration. This gives the concrete inequality \(V>c\) that the amplification theorem below needs.

Amplifying the average cost saving

Only restricted first moments need to pass to the Gaussian limit in this criterion. The next lemma isolates that fact; it also explains why zero limiting cell probabilities cause no difficulty.

Applying the criterion to the reporting event

Apply the criterion first with \(q=m+1\) and the reporting event \(A\) of Lemma 6. That lemma gives positive probability, conditional variance of the coordinate sum below one, and Gaussian-null boundary. Identity (3) gives the required pointwise expansion into functions of at most \(m\) coordinates. Lemma 8 therefore supplies a finite partition with costs \(m\) and \(m+1\) and with \(V>c=m+1-\mathbb P(A)\). Theorem 9 gives the signed Boolean conclusion. Appendix 6 supplies a second event with these same four properties and uses the same refinement and amplification.

The event and its conditional bin weights

For \(R>0\), put \[ m=1+\lceil e^{R^2/4}\rceil,\qquad \Delta=\frac{2R}{m-1},\qquad h=e^{-3R^2/8},\qquad q=m+1. \tag{16}\] Let \(G=(G_0,\ldots,G_m)\) have independent standard normal coordinates and set \(L=\sum_{j=0}^mG_j\). Partition \([-R,R)\) into \(m-1\) half-open intervals \(B_i\) of length \(\Delta\), and let \(B_m=\mathbb R\setminus[-R,R)\). Define \[\pi_i=\mathbb P(G_0\in B_i),\qquad v_i=\mathbb E[G_0\mid G_0\in B_i],\qquad u_i=\mathop{\mathrm{Var}}(G_0\mid G_0\in B_i).\] Every \(\pi_i\) is positive. For each leaf \(j\in\{1,\ldots,m\}\), take the half-open interval \[I_j=[v_j+1-h/2,v_j+1+h/2),\qquad \chi_j=\mathbf 1_{I_j}.\] For any real input \(g=(g_0,\ldots,g_m)\), define \(g\in A\) when, for the unique \(i\) such that \(g_0\in B_i\), all leaves other than \(i\) belong to their intervals and leaf \(i\) does not. The half-open conventions assign every endpoint a definite outcome. All unqualified bin and leaf sums below range from \(1\) to \(m\).

Application to mean-cost amplification

Fix one finite \(R\) for which Lemma 11 holds. Thus the integer \(q=m+1\), all intervals and bins, and the positive numbers \[ p_A=\mathbb P(G\in A)>0,\qquad 1-r_A>0 \tag{27}\] are fixed before further approximation. The pointwise expansion (17) uses at most \(q-1\) coordinates in each term, and the finite hyperplane boundary is Gaussian-null. These are precisely the hypotheses of Lemma 8. That lemma refines \(A^c\) into a finite collection of cells, with total squared approximation error below the actual positive quantity \(p_A(1-r_A)\). It assigns cost \(m\) to \(A\) and cost \(q\) to every other cell, giving a fixed partition \(\mathcal P\) for which \[ V:=\mathop{\mathrm{Var}}(\mathbb E[L\mid\mathcal P(G)])> c:=\mathbb E\ell(\mathcal P(G))=q-p_A. \tag{28}\] The partition includes its Gaussian-null cells as pointwise sets; later discrete input laws may give those cells positive probability.

With \(R\) and this partition fixed, Lemma 10 supplies the finite-input transfer used in Theorem 9. The costs \(m\) and \(q\) are positive integers, and the pointwise coordinate bounds, Gaussian-null boundaries, and strict inequality \(V>c\) verify all its hypotheses. The theorem therefore gives arbitrarily large ratios of retained variance to cell degree and the signed Boolean conclusion of Theorem 1. Neither the conditional variance saving nor the required batch sizes are asserted to be uniform in \(R\).

The parameters and a law-dependent error bound

For \(0<w<1\), put \[ h=w^{2/3},\qquad T=\sqrt{4\log(1/w)}. \tag{29}\] List the points of \(h\mathbb Z\cap[-T,T]\) in increasing order as \(t_1,\ldots,t_m\), and let \(i_*\) index zero. Thus \(m=2\lfloor T/h\rfloor+1=O(T/h)\) as \(w\downarrow0\); in particular, \(m\ge3\) for sufficiently small \(w\). On \(|t|\le T\), let \(I(t)\) be the index of a nearest grid point, breaking ties by the smaller index. On \(|t|>T\), set \(I(t)=i_*\). The grid endpoints need not equal \(\pm T\), but \[|t-t_{I(t)}|\le h\qquad (|t|\le T).\] Give leaf \(j\) the closed interval \[A_j=[t_j+1-w/2,t_j+1+w/2].\] These endpoint and tie conventions define the rule on every real input. Restrict henceforth to sufficiently small \(w\) that \(m\ge3\).

The choice \(h=w^{2/3}\) balances two power scales: \(h^2/w\) from rounding the central input to a grid point, and \(w/h\) from adding the leaf-interval variances. The accompanying cutoff factors grow more slowly than every power of \(1/w\). The choice of \(T\) makes \(e^{-T^2/2}=w^2\), so the Gaussian tail second moment is \(o(w)\). The estimates below retain all these factors.

Apply the coarse report of Lemma 7 with central input \(t\), leaf inputs \(u_1,\ldots,u_m\), selector \(I\), and leaf intervals \(A_j\). Denote it by \(\mathcal O_w\). Thus it returns \(Y\) when all leaves pass, \(X\) when only the selected leaf fails, and \((O,t,I(t),(u_j)_{j\ne I(t)})\) otherwise. In the notation of that lemma, these events are \(B\), \(A=P\setminus B\), and \(P^c\). For clarity, the special-cell identity here is \[ \mathbf 1_A(t,u)= \sum_{i=1}^m\mathbf 1_{\{I(t)=i\}}\prod_{j\ne i}\mathbf 1_{A_j}(u_j) -\prod_{j=1}^m\mathbf 1_{A_j}(u_j). \tag{30}\] The lemma proves that every cell has coordinate cost at most \(m\): the displayed summands use \(m\) coordinates, the \(Y\) cell ignores \(t\), and an ordinary report imposes no condition on its omitted leaf. This holds pointwise on every finite product alphabet, including unused labels. The boundaries of \(A\), \(B\), and \(P\) lie in finitely many coordinate hyperplanes.

Let \(U\) be centered with variance one, and take independent copies \(U_0,\ldots,U_m\). Write \[L=\sum_{j=0}^mU_j,\qquad \mathcal R_w(U)=\mathbb E[(L-\mathbb E[L\mid\mathcal O_w(\mathbf U)])^2].\] For the moment assume \[q_j=\mathbb P(U\in A_j)\in(0,1),\qquad Q=\prod_{j=1}^m q_j.\] Let \(\mu_j,a_j\) be the conditional mean and variance on \(A_j\), and put \[M_\mu=\sum_j\mu_j,\qquad S_a=\sum_j a_j.\] The interval length gives \(0\le a_j\le w^2\). Use the predictor \(M_\mu-1\) on \(X\), the predictor \(M_\mu\) on \(Y\), and the sum of reported coordinates on ordinary outputs. Denote this function of the report by \(\widehat L\). We will compute its deficit \(1-\mathbb E[(L-\widehat L)^2]\) exactly.

The ordinary reports contribute \(\mathbb P(P^c)\) to the squared error, by Lemma 7. For the exceptional reports, fix \(U_0=t\) and put \(i=I(t)\). The event \(P\) has conditional probability \(Q/q_i\), restricts every unselected leaf to its interval, and leaves \(U_i\) unrestricted. Under these restrictions, \(L-(M_\mu-1)\) has mean \(t+1-\mu_i\) and variance \(1+S_a-a_i\). Hence \[\mathbb E\bigl[\mathbf 1_P(1-(L-(M_\mu-1))^2)\bigr] =-Q\mathbb E_t\left[\frac{(t+1-\mu_i)^2+S_a-a_i}{q_i}\right].\] On \(B\), the central input remains unrestricted. The two respective predictor deficits on \(B\) are therefore \[\begin{align*} \mathbb E\bigl[\mathbf 1_B(1-(L-(M_\mu-1))^2)\bigr]&=-Q(1+S_a),\\ \mathbb E\bigl[\mathbf 1_B(1-(L-M_\mu)^2)\bigr]&=-QS_a. \end{align*}\] Subtracting the first of these from the \(P\) contribution gives the \(X\) contribution, because \(A=P\setminus B\). Adding the \(Y\) contribution cancels \(S_a\), exactly as in the earlier Gaussian calculation. Define \[ H_w(U)=1-\mathbb E_t\left[ \frac{(t+1-\mu_{I(t)})^2+S_a-a_{I(t)}}{q_{I(t)}}\right]. \tag{31}\] We have proved \[ 1-\mathcal R_w(U)\ge 1-\mathbb E[(L-\widehat L)^2]=QH_w(U). \tag{32}\] The equality is the gain of the specified predictor; the inequality uses optimality of conditional expectation. The calculations integrate against the law of \(t\), so they require no positive probability for any individual central value. We next show that both subtracted errors in (31) tend to zero for Gaussian inputs. The product \(Q\) may be extremely small; only its strict positivity will matter after \(w\) is fixed.

Gaussian estimates, including the rare-probability factors

In this subsection \(U=G\) is standard normal and \(w\downarrow0\). Write \(\phi\) for its density. Uniformly over the grid, \[q_j\le\frac{w}{\sqrt{2\pi}},\qquad |\mu_j-(t_j+1)|\le w/2,\qquad \sum_ja_j\le mw^2=O(Tw^{4/3})\longrightarrow0.\] We must control the two errors in (31) after division by the selected interval probability. A uniform lower bound of order \(w\) for all \(q_j\) is unavailable, because the grid extends into the Gaussian tails. Instead we compare each interval’s density with the density of the central coordinate that selects it.

For \(|t|\le T\) and \(z\in A_{I(t)}\), we have \(|z-t|\le1+h+w/2\). Hence, for sufficiently small \(w\), \[q_{I(t)}\ge w\inf_{z\in A_{I(t)}}\phi(z),\qquad \frac{\phi(t)}{q_{I(t)}} \le w^{-1}\exp(C(1+T)).\] The constant \(C\) is absolute: the logarithm of the density ratio is \((z^2-t^2)/2\le |t||z-t|+|z-t|^2/2\). Since \(T=2\sqrt{\log(1/w)}\), for each fixed \(\varepsilon>0\), \[ \frac{\phi(t)}{q_{I(t)}}=O(w^{-1-\varepsilon}) \qquad (|t|\le T). \tag{33}\] The fallback interval is centered at \(1\), so \(q_{i_*}\ge cw\) for an absolute \(c>0\) and small \(w\). Integrating the central bound and using this fallback bound on the tails gives \[ \mathbb E\frac1{q_{I(G)}}=O(Tw^{-1-\varepsilon}). \tag{34}\] Take \(\varepsilon=1/6\). Since \(0\le S_a-a_{I(G)}\le mw^2\), the variance contribution in (31) is at most \[mw^2\mathbb E\frac1{q_{I(G)}} =O(T^2w^{1/6})\longrightarrow0.\]

For the squared mismatch, first suppose \(|t|\le T\). Then \[|t+1-\mu_{I(t)}| \le |t-t_{I(t)}|+w/2\le h+w/2.\] Its central contribution is therefore at most \[O\bigl(Tw^{-7/6}(h+w)^2\bigr) =O(Tw^{1/6})\longrightarrow0.\] On \(|t|>T\), the fallback mean is bounded, so the contribution is \[O\left(w^{-1}\int_{|t|>T}(1+t^2)\phi(t)\,dt\right) =O\bigl(w^{-1}(1+T)e^{-T^2/2}\bigr) =O((1+T)w)\longrightarrow0.\] For the integral estimate, integration by parts gives \(\int_T^\infty t^2\phi(t)\,dt =T\phi(T)+\int_T^\infty\phi(t)\,dt\), and \(\int_T^\infty\phi(t)\,dt\le\phi(T)/T\) for \(T>0\). The last equality uses \(e^{-T^2/2}=w^2\). Both subtracted errors therefore vanish, proving \[ H_w(G)\longrightarrow1. \tag{35}\] Thus the same coarse report has a strictly positive Gaussian improvement for this independent choice of parameters. To use it repeatedly on finite cubes, we now fix those parameters and transfer the particular lower bound just proved.

Restricted moments and the finite-score transfer

Fix a sufficiently small \(w>0\) with \(m\ge3\) and \(H_w(G)>1/2\). All intervals, grid points, selector cells, and \(m\) are fixed from now on. Put \[Q_G=\prod_{j=1}^m\mathbb P(G\in A_j)>0,\qquad \eta=Q_G/4.\] Let \(U_n\) be centered variance-one variables such that \(U_n\Rightarrow G\) and \(\sup_n\mathbb EU_n^4<\infty\). Write \(U_{n,0},\ldots,U_{n,m}\) for independent copies. Use a subscript \(n\) for the interval moments of \(U_n\) and a superscript \(G\) for their Gaussian values, and put \(Q_n=\prod_j q_{j,n}\). For each bounded interval \(A_j\), weak convergence off its Gaussian-null boundary gives convergence of its probability and its restricted first and second moments. Thus \[q_{j,n}\to q_j^G\in(0,1),\qquad \mu_{j,n}\to\mu_j^G,\qquad a_{j,n}\to a_j^G.\] All these conditional moments are defined for sufficiently large \(n\). There are finitely many intervals, so their probabilities are then simultaneously bounded away from zero and one. In particular, every coefficient in (31) converges, including the reciprocal-probability factors.

Coefficient convergence alone does not yet transfer its expectation. Partition the line into the fixed selector cells \(E_i=\{t:I(t)=i\}\), whose boundaries are finite. The fallback cell includes both tails as well as the central cell for the grid point zero; it still has a finite boundary. For \(k=0,1,2\), \[ \mathbb E[U_n^k\mathbf 1_{\{U_n\in E_i\}}] \longrightarrow\mathbb E[G^k\mathbf 1_{\{G\in E_i\}}]. \tag{36}\] To see this, multiply \(t^k\mathbf 1_{E_i}(t)\) by a continuous cutoff that is one on \([-K,K]\) and zero outside \([-K-1,K+1]\). The resulting function is bounded and has only Gaussian-null discontinuities, so weak convergence applies. The cutoff errors for \(k=1,2\) vanish uniformly as \(K\to\infty\), by \[\mathbb E[|U_n|\mathbf 1_{\{|U_n|>K\}}]\le K^{-1},\qquad \mathbb E[U_n^2\mathbf 1_{\{|U_n|>K\}}] \le K^{-2}\sup_j\mathbb EU_j^4.\] The corresponding Gaussian errors also vanish. The case \(k=0\) follows directly from the continuity-set criterion. These are the same weak-convergence and uniform-integrability principles used in Lemma 2; see [2].

On a fixed selector cell \(E_i\), the integrand in (31) is a polynomial of degree at most two in \(t\). Expanding that polynomial expresses its expectation as a finite sum of convergent coefficients times the restricted moments in (36). Therefore \[Q_nH_w(U_n)\longrightarrow Q_GH_w(G)>Q_G/2.\] This proves convergence despite the moving predictor coefficients; we have not assumed continuity of conditional variances. For all sufficiently large \(n\), (32) now gives \(1-\mathcal R_w(U_n)>\eta\). Since the sum of the \(m+1\) independent inputs is centered and has variance \(m+1\), \[\mathop{\mathrm{Var}}\!\left(\mathbb E\!\left[\sum_{j=0}^mU_{n,j} \,\middle|\,\mathcal O_w(\mathbf U_n)\right]\right) =m+1-\mathcal R_w(U_n)>m+\eta.\] The pointwise coordinate-cost bound from Lemma 7, with the substitution specified above, and this transferred gain verify both hypotheses of Proposition 4 with \((r,d,\eta)=(m+1,m,Q_G/4)\). The criterion gives the signed conclusion of Theorem 1. Each averaging limit uses one fixed finite observation and fixed parameters. At that stage, the normalized batch scores have a fourth-moment bound uniform in the batch size, by (2); no uniformity across subsequent stages is required.