Conditional quantile couplings for Gaussian anchored sums

Work with a fixed prefix of length $j\le 4$ and write its dimension as $d$ during the proof. Its law satisfies [F1]–[F2]. Set $\sigma =\sqrt{s}$, $ϵ=\tau \mathrm{/}\sigma$, and normalize coordinates by $\sigma$. Its cumulant $f$ has mean zero, covariance $I$, and third derivatives bounded by $ϵ\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v{\mathrm{\mid }}^2$. Polarization and [F6], integrating along a real segment, give $\mathrm{\parallel }{f}^{\mathrm{\prime }\mathrm{\prime }}(a)-I\mathrm{\parallel }\le ϵ\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }{f}^{\mathrm{\prime }}(a)-a\mathrm{\mid }\le ϵ\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}2$. Fix a prefix $x^{\mathrm{\prime }}$, put $b=x^{\mathrm{\prime }}\mathrm{/}\sigma$, $\delta =\mathrm{\mid }b\mathrm{\mid }$, and assume $ϵ\delta$ sufficiently small. The map $a^{\mathrm{\prime }}\mapsto b-({\mathrm{\nabla }}^{\mathrm{\prime }}f(a^{\mathrm{\prime }},0)-a^{\mathrm{\prime }})$ sends $\mathrm{\mid }{a}^{\mathrm{\prime }}\mathrm{\mid }\le 2\delta$ into itself and has Lipschitz constant at most $2ϵ\delta <1\mathrm{/}2$. Its iterates from zero have geometrically decreasing consecutive distances, hence converge by [F6] to a solution of ${\mathrm{\nabla }}^{\mathrm{\prime }}f(a^{\mathrm{\prime }},0)=b$ with $\mathrm{\mid }{a}^{\mathrm{\prime }}\mathrm{\mid }\le 2\delta$. For the empty prefix use $a^{\mathrm{\prime }}=0$. Put $a=(a^{\mathrm{\prime }},0)$ and let $y_0=\sigma {\mathrm{\partial }}_{d}f(a)$ be the physical last mean under this prefix tilt. The last mixed covariance row at zero vanishes; integrating its Hessian deviation yields $\mathrm{\mid }{y}_0\mathrm{\mid }\le C\tau {\delta }^2$. The physical tilt parameter is $h=a\mathrm{/}\sigma$, so $\tau \mathrm{\mid }h\mathrm{\mid }\le 2ϵ\delta$ can be kept below $1\mathrm{/}8$.

Let $p,p^{\mathrm{\prime }}$ be the joint and prefix densities. The Gaussian convolution representations in [F2] and [F4] give ${\int }_{\mathbb{R}}p(x^{\mathrm{\prime }},y)dy=p^{\mathrm{\prime }}(x^{\mathrm{\prime }})$ at every $x^{\mathrm{\prime }}$: integrate the Gaussian joint density over its last coordinate and then integrate the bounded-summand law. Both densities are everywhere positive. Thus $p(x^{\mathrm{\prime }},y)\mathrm{/}{p}^{\mathrm{\prime }}(x^{\mathrm{\prime }})$ defines a probability density for each $x^{\mathrm{\prime }}$ and, by integration, a version of the conditional kernel. Denote tilted densities by $\bar{p},{\bar{p}}^{\mathrm{\prime }}$. Since $h_d=0$, their exponential factors cancel exactly, and the density $v$ of $S_d-y_0$ conditional on $S^{\mathrm{\prime }}=x^{\mathrm{\prime }}$ is $v(u)=p(x^{\mathrm{\prime }},u+y_0)\mathrm{/}{p}^{\mathrm{\prime }}(x^{\mathrm{\prime }})=\bar{p}(x^{\mathrm{\prime }},u+y_0)\mathrm{/}{\bar{p}}^{\mathrm{\prime }}(x^{\mathrm{\prime }})$. The centered tilted covariance $C$ satisfies $\mathrm{\parallel }C\mathrm{/}s-I\mathrm{\parallel }\le Cϵ\delta$ by step 1.1. Apply [F2] to its full density at $(0,u)$ and prefix density at zero. The ratio of the matching Gaussian densities is a centered normal density with variance $\det C\mathrm{/}\det C^{\mathrm{\prime }}=1\mathrm{/}(C^{-1}{)}_{dd}=s(1+O(ϵ\delta ))$; the determinant identity follows by eliminating the off-diagonal blocks in $C$. Consequently $\mid \log \frac{v(u)}{{\varphi }_s(u)}\mid \le Cϵ(1+\delta (1+\frac{u^2}{s})+\frac{\mathrm{\mid }u{\mathrm{\mid }}^3}{s^{3\mathrm{/}2}})$ for $\mathrm{\mid }u\mathrm{\mid }\le cs\mathrm{/}\tau$. Here ${\varphi }_s$ is the density of $N(0,s)$, and the small constants also ensure all [F2] moderate bounds for the full centered tilt.

In standardized units set $g(t)=\sigma v(\sigma t)$ and let $\varphi$ be the standard normal density. Step 2.1 implies, on $\mathrm{\mid }t\mathrm{\mid }\le c\mathrm{/}ϵ$, $\mathrm{\mid }\log (g(t)\mathrm{/}\varphi (t))\mathrm{\mid }\le Cϵ(1+\delta +\delta t^2+\mathrm{\mid }t{\mathrm{\mid }}^3)$ and hence $c_1{e}^{-C_1{t}^2}\le g(t)\le C_1{e}^{-c_1{t}^2}$ after reducing $ϵ$ and $ϵ\delta$. There is also a global bound $g(t)\le C\exp [-c\min (t^2,\mathrm{\mid }t\mathrm{\mid }\mathrm{/}ϵ)]$. Indeed the standardized prefix-tilted law has prefix density at its mean bounded below by a positive constant by [F2]. Tilt its centered full law further by $q{e}_d$, where $\mathrm{\mid }q\mathrm{\mid }\le c\mathrm{/}ϵ$. Its full density is bounded above by a constant by the global part of [F2], while integration of its bounded Hessian gives a centered log mgf at most $C{q}^2$. The density tilt identity therefore gives $g(t)\le C{e}^{C{q}^2-qt}$ at every real $t$. Choosing $q=\mathrm{sign}(t)\min (\mathrm{\mid }t\mathrm{\mid }\mathrm{/}(4C),c\mathrm{/}(2ϵ))$ proves the bound. The combined real tilt remains in a fixed smaller analytic ball, where the Hessian bound from step 1.1 applies; the global density bound in [F2] itself permits every real tilt.

We next convert the density estimate into a CDF displacement without a coupling theorem. Write $R(t)=ϵ(1+\delta +\delta t+t^2)$ for $t\ge 0$. Choose a fixed large $T\ge 10$ and then a sufficiently large fixed $K$. Put $r(t)=KR(t)$ and $b=c_0\mathrm{/}ϵ$, with fixed $c_0$ sufficiently small. Reduce the allowed bounds on $ϵ$ and $ϵ\delta$ so that for $T\mathrm{/}2\le t\le 2b$ one has $0\le r(t)\le t\mathrm{/}8$ and $0\le r^{\mathrm{\prime }}(t)\le 1\mathrm{/}2$; this is possible because $R^{\mathrm{\prime }}(t)\le 2R(t)\mathrm{/}t$ for $t\ge 1$ and $r(t)\mathrm{/}t\le CKϵ(1+\delta +t)$. Also choose $3b$ within the moderate interval of step 3.1. For $t\ge T$, the maps $f_{\pm }(t)=t\pm r(t)$ are increasing and their arguments are comparable with $t$. Since $(\log \varphi {)}^{\mathrm{\prime }}(t)=-t$, step 3.1 bounds the logarithmic density error at either perturbed argument by $CtR(t)$, whereas the Gaussian log-density changes by at least a fixed positive multiple of $tr(t)$ in the favorable direction. The log Jacobians are at most $r^{\mathrm{\prime }}(t)$ and at least $-2r^{\mathrm{\prime }}(t)$, respectively. Since $r^{\mathrm{\prime }}(t)\le 2r(t)\mathrm{/}t$, first taking $T$ large and then $K$ large gives $g(f_{+}(t))f_{+}^{\mathrm{\prime }}(t)\le \varphi (t)$ and $g(f_{-}(t))f_{-}^{\mathrm{\prime }}(t)\ge \varphi (t)$. These comparisons follow by adding the three logarithmic contributions; the Jacobian contribution divided by $tr(t)$ is at most $4\mathrm{/}{t}^2$, and the density-error contribution is at most $C\mathrm{/}K$.

The inverse equations and monotonicity of $r$ imply $f_{+}^{-1}(z)\ge z-r(z)$ and $f_{-}^{-1}(z)\le z+r(2z)$ in the range below. Substituting $u=f_{+}(t)$ and $u=f_{-}(t)$ in the inequalities of step 4.1 gives, for $2T\le z\le b$, ${\int }_z^{b}g(u)du\le {\int }_{z-r(z)}^{\mathrm{\infty }}\varphi (u)du,{\int }_z^{b}g(u)du\ge {\int }_{z+r(2z)}^b\varphi (u)du\mathrm{.}$ Only a smaller range $z\le c_1\mathrm{/}ϵ$, with $c_1\ll c_0$, will be retained, so the lower integral has its endpoints in the stated order. The global tail estimate in step 3.1 and the Gaussian tail give ${\int }_b^{\mathrm{\infty }}g+{\int }_b^{\mathrm{\infty }}\varphi \le C{e}^{-c{b}^2}$: split the first integral at a constant multiple of $1\mathrm{/}ϵ$ and integrate the Gaussian and exponential parts. Enlarge the shifts in the displayed inequalities to $3r(2z)$. Each extra Gaussian interval has mass at least $cR(z)e^{-C{z}^2}\ge cϵe^{-C{z}^2}$. Choose $c_1$ so small that $C{c}_1^2<c{c}_0^2\mathrm{/}2$. For sufficiently small $ϵ$, uniformly for $2T\le z\le c_1\mathrm{/}ϵ$, that interval mass exceeds the discarded tail, since ${ϵ}^{-1}\exp (-(c{c}_0^2-C{c}_1^2)\mathrm{/}{ϵ}^2)\to 0$. Thus the full tails have the corresponding inequalities with shift $CR(z)$; here $R(2z)\le 4R(z)$. Reflecting $g,\varphi$ proves the negative-tail version as well.

At $z=2T$ step 5.1 bounds the tail-probability difference by $CR(2T)$. For $\mathrm{\mid }z\mathrm{\mid }\le 2T$, step 3.1 bounds the density-ratio error by $Ca$, where $a=ϵ(1+\delta )$; integrating over this fixed interval gives a CDF difference at most $Ca$. The Gaussian density is bounded below by a fixed positive constant on $[-2T-1,2T+1]$, so a shift of length $Ca$ in either direction dominates that difference, after reducing the allowed bound on $a$ to keep the shift within this compact interval. Since $R(\mathrm{\mid }z\mathrm{\mid })\ge a$, this proves together with step 5.1, for the standardized conditional CDF $F_g$, $\mathrm{\Phi }(z-CR(\mathrm{\mid }z\mathrm{\mid }))\le F_g(z)\le \mathrm{\Phi }(z+CR(\mathrm{\mid }z\mathrm{\mid }))(\mathrm{\mid }z\mathrm{\mid }\le c\mathrm{/}ϵ)\mathrm{.}$ All constant choices are finite and made in the indicated order; reducing the common cutoff enforces every smallness requirement. Returning to physical units, the CDF $F_v$ of the centered conditional coordinate satisfies ${\mathrm{\Phi }}_s(u-\rho (u))\le F_v(u)\le {\mathrm{\Phi }}_s(u+\rho (u))$, where $\rho (u)=C\tau (1+\delta (1+\mathrm{\mid }u\mathrm{\mid }\mathrm{/}\sigma )+u^2\mathrm{/}s)$.

Every conditional density is positive and continuous by step 2.1, so its CDF is continuous, strictly increasing and has limits zero and one. Its inverse on $(0,1)$ is therefore unique. For a fresh $N(0,s)$ variable $Z$, [F3] implies that ${\mathrm{\Phi }}_s(Z)$ is uniform: for $0<u<1$, the event ${\mathrm{\Phi }}_s(Z)\le u$ is $Z\le {\mathrm{\Phi }}_s^{-1}(u)$, of probability $u$. Thus the inverse-CDF map has the asserted conditional law. If $Y=T_j(x^{\mathrm{\prime }},Z)$ and $U=Y-y_0$, then $F_v(U)={\mathrm{\Phi }}_s(Z)$. Step 6.1 and strict monotonicity imply $\mathrm{\mid }U-Z\mathrm{\mid }\le \rho (U)$ whenever $U$ is moderate. Step 1.1 gives $\mathrm{\mid }{y}_0\mathrm{\mid }\mathrm{/}\sigma \le Cϵ{\delta }^2\le Cc\delta$, so $\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }\le cs\mathrm{/}\tau$ implies the required moderate bounds on $U$ and $\delta$. The inequalities $2\delta \mathrm{\mid }U\mathrm{\mid }\mathrm{/}\sigma \le {\delta }^2+U^2\mathrm{/}s$ and $\delta \le 1+{\delta }^2$ now give $\mathrm{\mid }Y-Z\mathrm{\mid }\le C\tau (1+(\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2+Y^2)\mathrm{/}s)$.

The bound in Gaussian rather than target values follows by bracketing, not by substituting $Z$ for $Y$ in step 7.1. In standardized units fix a Gaussian coordinate $z$ and put $A=1+{\delta }^2+z^2$, $L=K_1ϵA$, with fixed large $K_1$. If $ϵ\delta$ and $ϵ\mathrm{\mid }z\mathrm{\mid }$ are small enough, $z\pm L$ remain in the moderate region and the standardized version of $\rho (z\pm L)$ is at most $L\mathrm{/}2$. Indeed $\rho (t)\le Cϵ(1+{\delta }^2+t^2)$ and $(z\pm L{)}^2\le 2z^2+2L^2$; the extra $CϵL^2$ is absorbed into $L$ because ${ϵ}^{2}A$ is small. Step 6.1 therefore gives $F_g(z-L)<\mathrm{\Phi }(z)<F_g(z+L)$, so its inverse at $\mathrm{\Phi }(z)$ lies in $[z-L,z+L]$. Adding the shift $y_0$ from step 1.1 and restoring units gives $\mathrm{\mid }{T}_j(x^{\mathrm{\prime }},Z)-Z\mathrm{\mid }\le C\tau (1+(\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2+Z^2)\mathrm{/}s)$ under the stated Gaussian-value cutoff.

The maps just constructed are jointly measurable, also under the additional parameter in the Statement. On each fixed row length the product of the bounded input-law kernels exists by [F5]. The Gaussian density is jointly Borel in its positive definite covariance and its argument: its determinant and inverse are continuous functions of the matrix entries by their polynomial and cofactor formulas. Consequently $p_{\theta }(x)=\int q_{G_{\theta }}(x-{\sum }_i{t}_i){⨂}_i{K}_i(\theta ,d{t}_i)$ is jointly measurable by [F5], as are the prefix densities, their positive ratios and their CDF integrals. Lebesgue integration here uses its common exhaustion by bounded intervals. The conditional inverse can be written $\inf \{r\in \mathbb{Q}:H_{j,\theta }(x^{\mathrm{\prime }},r)\ge u\}$; by [F7] this equals the continuous strictly increasing inverse and is jointly measurable, since strict sublevel sets are countable unions of measurable sets indexed by $r\in \mathbb{Q}$. Compose with the continuous ${\mathrm{\Phi }}_s(z)=\mathrm{\Phi }(z\mathrm{/}\sqrt{s})$. Different finite row lengths are handled on their countable measurable disjoint union. This constructs the kernel versions explicitly at every prefix.

Starting with independent Gaussian coordinates of variance $s$, apply the conditional maps successively to coordinates $1,\dots ,d$. Conditional integration of the densities in step 2.1 shows inductively that the first $j$ constructed coordinates have density $p_j$; their joint density is the product of the successive density ratios, which telescopes to $p_j$. The full map $R$ therefore has the law of $S$. On $\mathrm{\mid }R(g)\mathrm{\mid }\le cs\mathrm{/}\tau$, every coordinate and prefix meets the target-value cutoff in step 7.1 after reducing $c$ by a fixed factor. Summing at most four coordinate bounds gives $\mathrm{\mid }R(g)-g\mathrm{\mid }\le C\tau (1+\mathrm{\mid }R(g){\mathrm{\mid }}^2\mathrm{/}s)$. For $S=(P,D)$ apply only its last $a$ conditional maps with the specified prefix $P=p$ and fresh independent Gaussian coordinates; the same telescoping gives the conditional law of $D$ and the same coordinate estimates give $\mathrm{\mid }{T}_p(e)-e\mathrm{\mid }\le C\tau (1+(\mathrm{\mid }p{\mathrm{\mid }}^2+\mathrm{\mid }{T}_p(e){\mathrm{\mid }}^2)\mathrm{/}s)$. Separate bounds on the two norms imply the joint cutoff by reducing $c$. The Gaussian reference has covariance $sI$, so its parent and bridge coordinate blocks are independent by the product Gaussian density in [F3]. Joint measurability follows by composing the maps of step 8.2.

For the scalar extension put $a=\sqrt{s\mathrm{/}W}$, $b=\sqrt{s\mathrm{/}v}$ and $Q=(aP,bD)$. Independence gives $\mathrm{Var}(P)=W$, $\mathrm{Cov}(P,D)=V_L-wW=0$ and $\mathrm{Var}(D)=V_L{V}_R\mathrm{/}W=v$, hence $\mathrm{cov}(Q)=s{I}_2$. The stated ranges follow from $V_L,V_R\in [s\mathrm{/}2,s]$; the function $uv\mathrm{/}(u+v)$ is increasing in either positive argument, giving $v\in [s\mathrm{/}4,s\mathrm{/}2]$. Thus $a\in [1\mathrm{/}\sqrt{2},1]$, $b\in [\sqrt{2},2]$ and $w\in [1\mathrm{/}3,2\mathrm{/}3]$. A bounded left increment $X$ contributes $(aX,b(1-w)X)$ to $Q$, whose norm is at most $2B$; a bounded right increment contributes $(aX,-bwX)$ with the same bound. The joint Gaussian component is obtained from independent child Gaussians by the same linear map. Since their variances are at least $\gamma V_L,\gamma V_R$, its covariance, by congruence, is at least $\gamma \mathrm{cov}(Q)=\gamma s{I}_2$. Therefore $Q$ satisfies the original hypotheses with bound $2B$ and parameter $2\tau$. Apply step 9.1 conditionally on $aP=ap$, feeding $be\sim N(0,s)$, and divide the output by $b$. The uniform bounds on $a,b$ convert its estimate and its cutoffs into the asserted scalar bridge bound after changing $c_{\gamma },C_{\gamma }$ by fixed factors. Applying the scalar root construction to $aP$ and rescaling gives the root bound. Finally $D=L-wP$ rearranges to the two stated child formulas. Every coefficient is a continuous function of the positive variances, so the parameter measurability in step 8.2 is preserved. All conclusions follow.