---
title: "The Right Wing of the SABR Smile with an Absorbing Boundary: Resolution of Conjectures of Henry-Labordère"
authors:
  - name: "Zeyu Cao"
    affiliation: "Independent Researcher, Long Island City, USA"
  - name: "Shaosai Huang"
    affiliation: "Kspectra Research Inc., Toronto, Canada"
date: "2026-10"
status: "Working paper"
url: https://kspectra.ai/papers/sabr-right-wing-absorbing-boundary/
doi: 10.2139/ssrn.7562378
ssrn: https://papers.ssrn.com/sol3/papers.cfm?abstract_id=7562378
arxiv: https://arxiv.org/abs/2610.04209
---

# The Right Wing of the SABR Smile with an Absorbing Boundary: Resolution of Conjectures of Henry-Labordère

Zeyu Cao and Shaosai Huang — Working paper, version of October 2026.

Links: [Web page](https://kspectra.ai/papers/sabr-right-wing-absorbing-boundary/) · [SSRN](https://papers.ssrn.com/sol3/papers.cfm?abstract_id=7562378) · [arXiv](https://arxiv.org/abs/2610.04209)

> Converted by Kspectra Research from the LaTeX of the posted version. Section, theorem, equation and reference numbers match the PDF. Formulas are LaTeX; the document's own macros are defined below.

## How to cite

```bibtex
@misc{cao2026right,
  author        = {Cao, Zeyu and Huang, Shaosai},
  title         = {{The Right Wing of the SABR Smile with an Absorbing Boundary: Resolution of Conjectures of Henry-Labordère}},
  year          = {2026},
  month         = oct,
  note          = {Working paper, version of October 2026},
  doi           = {10.2139/ssrn.7562378},
  eprint        = {2610.04209},
  archiveprefix = {arXiv},
  url           = {https://kspectra.ai/papers/sabr-right-wing-absorbing-boundary/}
}
```

## Macros

The formulas use these definitions from the paper's preamble:

```latex
\renewcommand{\topfraction}{0.85}
\renewcommand{\bottomfraction}{0.6}
\renewcommand{\textfraction}{0.1}
\renewcommand{\floatpagefraction}{0.8}
\newcommand{\E}{\mathbb{E}}
\newcommand{\Prob}{\mathbb{P}}
\newcommand{\orth}{\bar\rho}
\newcommand{\Rb}{R_\partial}
\DeclareMathOperator{\BS}{BS}
```

## Abstract

We determine the leading logarithmic asymptotics of the fixed-maturity right tail of the SABR model with $\beta\in(0,1)$ and absorption at zero, for every correlation $\rho\in(-1,1)$. If $P(k)$ is the probability that the terminal forward is at least $f_0e^k$, then

\begin{equation*} -\frac {\ln P(k)}{k^2}\ \longrightarrow \ \frac {(1-\beta )^2}{2\nu ^2T\bigl (1-(\rho \wedge 0)^2\bigr )} \qquad (k\to \infty ). \end{equation*}

For $\rho\ge0$ this is the rate predicted by the unrestricted hyperbolic geometry. For $\rho<0$ it is strictly larger: along the volatility excursion that produces the tail, survival forces the noise orthogonal to volatility to stay above a moving square-root barrier, at an additional cost of the same order $k^2$. Consequently the Black–Scholes implied volatility tends to $\nu\sqrt{1-(\rho\wedge0)^2}/(1-\beta)$: Henry-Labordère’s conjectured wing limit $\nu/(1-\beta)$ holds for $\rho\ge0$ and fails for every $\rho<0$. For his second conjecture, whose unrestricted-distance Gaussian estimate does not hold for $\rho<0$, we also identify the correct replacement for this model: at the level of tail probabilities, the rate is given by the squared intrinsic Riemannian distance, within the survival domain, from the initial state to the terminal target set. The proofs are direct, combining a stopped Lamperti transform, a conditional time change, Gaussian moving-barrier bounds and change-of-measure constructions, and use no heat-kernel estimates.

**Keywords.** Absorbed diffusions; SABR model; implied-volatility wings; logarithmic tail asymptotics; moving-boundary crossing; intrinsic Riemannian distance; stochastic volatility.

**2020 Mathematics Subject Classification.** Primary 60J60; secondary 60F10, 60H10, 60J50, 91G20.

**JEL Classification.** C02, G13.

## 1 Introduction

Absorption can alter a fixed-time logarithmic tail asymptotic even when the absorbing boundary is disjoint from the terminal target set. We establish this phenomenon for the SABR diffusion absorbed at zero: the volatility excursion that produces a large terminal value can simultaneously drive the forward toward the boundary, so paths contributing to the right tail incur an additional Brownian boundary-crossing cost. The main theorem concerns the terminal right tail of the absorbed diffusion; the implied-volatility wing then follows from a model-free transfer principle.

### 1.1 The Problem and Henry-Labordère’s Conjectures

The SABR model of Hagan, Kumar, Lesniewski and Woodward [HKLW02] is a widely used stochastic-volatility parametrization of implied-volatility smiles in interest-rate markets and beyond:

$$
\begin{equation}\label{eq:sabr}\tag{1.1} dF_t = a_t F_t^{\beta}\, dW_t, \qquad da_t = \nu a_t\, dZ_t, \qquad d\langle W,Z\rangle_t=\rho\,dt,\tag{1.1} \end{equation}
$$

with $F_0=f_0>0$, $a_0=\alpha>0$, $\beta\in(0,1)$, $\nu>0$ and $\rho\in(-1,1)$. Precisely, we work on a complete filtered probability space satisfying the usual conditions and carrying independent standard Brownian motions $(Z,Z^\perp)$. We set $W=\rho Z+\sqrt{1-\rho^2}\,Z^\perp$ and

$$
a_t=\alpha\exp\!\left(\nu Z_t-\tfrac12\nu^2t\right).
$$

On the open state space $(0,\infty)$, $F$ is the pathwise-unique strong solution of the first equation in (1.1) up to the absorption time

$$
\zeta:=\inf\{t\ge0:F_t=0\}\qquad(\inf\varnothing=\infty),
$$

and we take the continuous absorbed extension $F_t=0$ for $t\ge\zeta$. This boundary condition is part of the model: it selects the absorbing continuation among the possible post-hitting extensions of the CEV equation. Throughout, $T>0$ is a *fixed* maturity, $k=\ln(K/f_0)$ is the log-moneyness, and $\sigma_{\BS}(k)$ denotes the Black–Scholes implied volatility of the undiscounted call with strike $K$ and maturity $T$.

At fixed maturity, the extreme right wing probes joint rare excursions of volatility and the forward. The unrestricted hyperbolic geometry of the model suggests a correlation-independent leading coefficient, whereas the right-tail event for the absorbed process requires survival in $(0,\infty)$. The question is whether this survival constraint is negligible on the leading $k^2$ scale.

In Chapter 10 of his monograph [Hen08], Henry-Labordère writes the generator of a stochastic-volatility model as a Schrödinger-type operator on a Riemannian manifold: a Laplace–Beltrami part determined by the diffusion matrix, a connection one-form, and a scalar potential $Q$. Assuming a Gaussian bound of Li–Yau–Davies type [LY86; Dav89] for the associated heat kernel (his (10.21)), he obtains a wing formula, his (10.22), in which the extreme strikes are governed by the geodesic distance, in the *unrestricted* metric, from the initial point to the terminal fibre $\{F=f\}$. For SABR this metric is hyperbolic, and the resulting limit is $\nu/(1-\beta)$. Let $a_{\min}(f)$ denote the volatility coordinate of the point of the fibre $\{F=f\}$ nearest to the initial point, that is, the saddle point of the fibre integral. For SABR with $\beta<1$, the scalar potential along this saddle has the large-strike limit given in [Hen08, p. 306]:

$$
\begin{equation}\label{eq:Q}\tag{1.2} \lim_{f\to\infty}Q(f,a_{\min}(f)) \;=\; \frac{\nu^2}{4(1-\beta)^2}\Bigl(\beta(\beta-1)-\frac{\beta^2}{2(1-\rho^2)}\Bigr).\tag{1.2} \end{equation}
$$

The potential is therefore bounded along the far-right saddle. On the strength of this boundedness, Henry-Labordère poses two conjectures, which we restate in our notation, keeping his numbering.[^1]

**Conjecture 1 ([Hen08, p. 306]).**  For SABR with $\beta<1$, $\sigma_{\BS}(k)\to\nu/(1-\beta)$ as $k\to\infty$, for every $\rho\in(-1,1)$.

The second conjecture concerns the Gaussian estimate itself.

**Conjecture 2 ([Hen08, p. 306]).**  If the potential $Q(a_{\min})$ is bounded, the Gaussian wing estimate (10.22) of [Hen08] is valid.

Conjecture 1 is the proposed right-wing limit; Conjecture 2 would justify the unrestricted-distance Gaussian estimate from the boundedness of the potential at the saddle point. As the book notes, Conjecture 1 was proved for $\rho=0$ by Benaim, Friz and Lee [BFL08]. To our knowledge, the fixed-maturity correlated case $\rho\ne0$, $\beta\in(0,1)$ has remained open (see Section 1.4). The difficulty is the gap between evaluating a coefficient at the unrestricted saddle point and controlling the Dirichlet semigroup of the diffusion killed at the boundary of its state space.

### 1.2 Main Results

We measure the right tail by the probability that the terminal forward ends at or above a large strike, and compare its decay with two quadratic rates:

$$
\begin{equation}\label{eq:PK}\tag{1.3} P(k) := \Prob\bigl(F_T\ge f_0 e^{k}\bigr), \qquad c:=\frac{(1-\beta)^2}{2\nu^2T}, \qquad c_-:=\frac{c}{1-\rho^2}.\tag{1.3} \end{equation}
$$

For $\rho\ne0$ the rate $c_-$ exceeds $c$, and our main result shows which of the two governs the tail.

**Theorem 1.1 (Logarithmic Right-Tail Asymptotics for Absorbed SABR).**  For every $\rho\in(-1,1)$,

$$
\lim_{k\to\infty}\frac{-\ln P(k)}{k^2} = \begin{cases} c_-=\dfrac{(1-\beta)^2}{2\nu^2T(1-\rho^2)},&\rho<0, \\
[6pt] c=\dfrac{(1-\beta)^2}{2\nu^2T},&\rho\ge0. \end{cases}
$$

Through a model-free transfer from tail probabilities to call prices (Lemma 2.3), the decay rate determines the right wing of the implied-volatility smile.

**Corollary 1.2 (Right-Wing Implied-Volatility Limit).**  For every $\rho\in(-1,1)$,

$$
\lim_{k\to\infty}\sigma_{\BS}(k) = \begin{cases} \dfrac{\nu\sqrt{1-\rho^2}}{1-\beta},&\rho<0, \\
[6pt] \dfrac{\nu}{1-\beta},&\rho\ge0. \end{cases}
$$

Comparing this limit with Henry-Labordère’s prediction settles Conjecture 1.

**Corollary 1.3 (Resolution of Conjecture 1).**  For every $\beta\in(0,1)$, the conjecture holds for $\rho\in[0,1)$ and fails for every $\rho\in(-1,0)$. The wing limit

$$
\begin{equation*} \sigma_\infty(\rho)=\frac{\nu\sqrt{1-(\rho\wedge0)^2}}{1-\beta}, \end{equation*}
$$

as a function of $\rho$, is $C^1$ but not $C^2$ at $\rho=0$: its second derivative from the left is $-\nu/(1-\beta)$ and from the right is $0$. It does not depend on $\alpha$, $f_0$ or $T$, and depends on $\beta$ only through the factor $(1-\beta)^{-1}$.

Both corollaries follow from Theorem 1.1 and the integrability $\E F_T\le f_0$ alone. Figure 1 shows the limit as a function of $\rho$, together with the value predicted by Conjecture 1.

![Figure 1](https://kspectra.ai/papers/sabr-right-wing-absorbing-boundary/figs/fig_wing_limit.svg)

*Figure 1. The right-wing limit $\sigma_\infty$ of Corollary 1.2, normalized by $\nu/(1-\beta)$, against the correlation $\rho$. Henry-Labordère’s Conjecture 1 predicts the value $1$ for every $\rho$. It holds for $\rho\ge0$; for $\rho<0$ survival lowers the limit to $\sqrt{1-\rho^2}$ (shaded gap).*

Our third result explains the correction geometrically. In Section 6 we write the model in Fermi coordinates $(R,q)$ of the hyperbolic plane, in which the survival domain is the coordinate half-plane $D=\{R>\Rb\}$, and identify the decay rate with the intrinsic Riemannian distance $d_D$ within $D$: for every $\rho\in(-1,1)$,

$$
\begin{equation}\label{eq:intro-c2}\tag{1.4} -\ln P(k)=\frac{d_D(z_0,\mathcal T_k)^2}{2T}+o(k^2)\tag{1.4} \end{equation}
$$

(Corollary 6.2), where $z_0$ is the initial state and $\mathcal T_k$ the terminal target set. The unrestricted distance yields the rate $c$ for every $\rho$; the restriction to $D$ changes the leading coefficient exactly when $\rho<0$. Since the limit (1.2) is finite for every $\rho$, the model shows that boundedness of the potential at the unrestricted saddle point does not suffice for the Gaussian estimate in Conjecture 2, and (1.4) is its correct replacement for absorbed SABR.

### 1.3 Mechanism and Proof Strategy

The asymmetry between positive and negative correlation is already visible in an exact decomposition of the forward before absorption. Write $X=F^{1-\beta}/(1-\beta)$ for the Lamperti–CEV coordinate and decompose $W=\rho Z+\orth Z^\perp$, where $\orth:=\sqrt{1-\rho^2}$. The correlated stochastic integral satisfies $\int_0^t a\,dZ=(a_t-\alpha)/\nu$ exactly, and conditionally on the volatility path the orthogonal martingale $\int_0^t a\,dZ^\perp$ is a Brownian motion $B$ run on the quadratic-variation clock $V_t=\int_0^ta_s^2\,ds$. Hence, before absorption (Lemmas 2.1 and 2.2),

$$
\begin{equation}\label{eq:decomp0}\tag{1.5} X_t \;=\; X_0+\underbrace{\tfrac{\rho}{\nu}(a_t-\alpha)}_{\xi_t}\;+\;\orth\,B(V_t)\;-\;b\!\int_0^t\!\frac{a_s^2}{X_s}\,ds, \qquad b=\frac{\beta}{2(1-\beta)}.\tag{1.5} \end{equation}
$$

A large terminal value requires a volatility excursion of logarithmic size $n=(1-\beta)k$, which costs $ck^2+o(k^2)$. For $\rho\ge0$ the correlation term is bounded below along such an excursion, survival is cheap, and the unconstrained rate is attained. For $\rho<0$ the correlation term $\xi_t=-\frac{|\rho|}{\nu}(a_t-\alpha)$ becomes increasingly negative along the very excursion that creates the wing, and survival of $F$ forces the orthogonal Brownian motion to stay above the *moving lower barrier*

$$
B(V_t)\;\ge\;\frac{|\rho|}{\nu\orth}\,(a_t-\alpha)-\frac{X_0}{\orth} \qquad\text{for all } t\le T.
$$

Along the exponential volatility path used in the lower bound, this barrier is a shifted square-root profile in the *operational time* $v=V_t$ (Figure 2(a)). In the logarithmic time $s=\ln v$, the process $U(s)=e^{-s/2}B(e^s)$ is a stationary Ornstein–Uhlenbeck process and the barrier becomes an essentially constant level $\ell\asymp\sqrt n$ (Figure 2(b)). Holding $U$ above $\ell$ costs $\ell^2/8$ per unit of logarithmic time, over a window of length about $2n$. This cost, $\rho^2c\,k^2/(1-\rho^2)$ to leading order, is of the same order as that of the volatility excursion, and the two add up to $c_-=c/(1-\rho^2)$. Equivalently, the effective volatility of volatility in the right wing is $\nu\orth$. The absorbing boundary thus acts on the right tail through survival conditioning, not through the mass at zero; the model-free tail-to-wing transfers [Lee04; BF09; GL14] are unaffected.

![Figure 2](https://kspectra.ai/papers/sabr-right-wing-absorbing-boundary/figs/fig_mechanism.svg)

*Figure 2. Survival as a moving-barrier problem ($\rho<0$; schematic simulation). (a) Along the exponential volatility path, survival requires the orthogonal Brownian motion $B$ to stay above a shifted square-root barrier in operational time $v=V_t$; a typical path is killed early. (b) In logarithmic time, $U(s)=e^{-s/2}B(e^s)$ is a stationary Ornstein–Uhlenbeck process and the barrier rises to the constant level $\ell$, which dominates it. The lower-bound construction of Section 4 holds $U$ just above $\ell$, at a cost of about $\ell^2/8$ per unit of $s$.*

Turning this mechanism into a proof requires matching bounds that do not presuppose path concentration. For $\rho<0$, the upper bound on $P(k)$ (Section 3) drops the nonpositive singular drift, discretizes the volatility excursion by level-hitting times, and converts the conditional survival constraints into a finite-dimensional Cameron–Martin energy. A deterministic convex inequality (Lemma 3.7) then bounds the total cost from below; its key step, Lemma 3.6, shows that the part of the Gaussian cost that is linear in the displacement is minimized by deferring the displacement for as long as the barrier allows. The matching lower bound (Section 4) combines a change-of-measure tube for volatility with a Brownian path that enters and stays above the induced barrier, followed by a bootstrap control of the singular drift. For $\rho\ge0$ (Section 5), a terminal-only upper bound is paired with a lower-bound construction based on scale-invariant doubling and $O(\ln n)$ dyadic level-raising steps. These arguments use elementary probability and finite-dimensional convexity rather than a sample-path large-deviation principle, heat-kernel estimates or spectral asymptotics.

### 1.4 Relation to Prior Work

The problem lies at the intersection of several established theories, but not inside any one of their proved regimes. Among the results we located, summarized in Table 1, none treats the fixed-maturity extreme right wing of correlated SABR with $\beta\in(0,1)$. The closest results concern the uncorrelated $\beta<1$ wing, the moment-explosion regime at $\beta=1$, asymptotics of survival probabilities, small-time geodesic expansions, and the left wing generated by the atom at zero. In particular, Paulot [Pau15] cautions that small-time geodesic expansions are valid only as long as the geodesic does not reach the boundary; the present result is a fixed-maturity, extreme-strike manifestation of the same boundary obstruction.

Table 1. Selected results on extreme strikes and boundary effects for SABR-type models.

| Result | Reference | Scope |
|---|---|---|
| Right-wing limit $\nu/(1-\beta)$ | Benaim–Friz–Lee [BFL08] | $\rho=0$ only |
| Moment explosions, smile slopes | Andersen–Piterbarg [AP07]; Lee [Lee04] | $\beta=1$ regime; model-free |
| Sharp correlated right tail | Forde–Zhang [FZ15] | $\beta=1$ only; moment explosion; discontinuity at $\rho=0^-$ |
| Mass at zero and the left wing | Gulisashvili–Horvath–Jacquier [GHJ18]; De Marco–Hillairet–Jacquier [DHJ17]; Gulisashvili [Gul15] | uncorrelated SABR; model-free atom asymptotics; small-strike side |
| Boundary hitting, SABR plane | Gulisashvili–Horvath–Jacquier [GHJ16] | auxiliary geometric process |
| Survival probability | Yang–Wan [YW18]; Chen–Yang [CY19] | perturbative expansion; small-time regime |
| Correlated$\to$uncorrelated map | Antonov–Spector [AS12]; Antonov–Konikov–Spector [AKS13; AKS19] | short-time-matched approximation |
| Geodesic expansions | Paulot [Pau15] | small time; explicit boundary caveat |
| Regularity classification | Döring–Horvath–Teichmann [DHT17] | symmetric Dirichlet forms only if $\rho=0$ or $\beta\in\{0,1\}$ |
| Large maturity | Forde–Pogudin [FP13] | $(\beta=1,\rho\le0)$ or $(\beta<1,\rho=0)$ |

### 1.5 Scope and Organization

Before turning to the proofs, we record precisely what is proved. All limits are taken for fixed $T>0$, $\beta\in(0,1)$ and $\rho\in(-1,1)$. We do not prove uniformity in these parameters, a quantitative remainder, results at $\rho=\pm1$ or $\beta\in\{0,1\}$, or a left-wing theorem. No Dirichlet heat-kernel estimate is used or proved; Remark 6.3 describes what such an estimate would have to provide. For $\rho<0$ we state a conjectural $k^{3/2}$ correction (Conjecture 8.1), supported by formal calculations only; no proved result depends on it. The numerical results of Section 7 illustrate the theorems and play no role in the proofs. Selected deterministic steps have also been formalized in Lean; see the Supplementary Material section.

Section 2 collects the reductions: the Lamperti decomposition, conditioning on volatility, and the transfer to implied volatility. Sections 3 and 4 prove the upper and lower bounds for $\rho<0$; Section 5 treats $\rho\ge0$ and completes the proof of Theorem 1.1. Section 6 proves the intrinsic-distance identification, Section 7 presents numerical illustrations, and Section 8 discusses the subleading correction and neighbouring parameter regimes. Appendix A records measurability facts and Appendix B the numerical methods.

## 2 Preliminaries

The proofs rest on two reductions. The Lamperti–CEV transform isolates the correlation term and lets us represent the orthogonal martingale in its quadratic-variation time; a model-free argument converts logarithmic tail rates into implied-volatility limits.

Throughout, $\beta\in(0,1)$, $\nu>0$, $T>0$ and $\rho\in(-1,1)$ are fixed, $\orth=\sqrt{1-\rho^2}$, and

$$
C:=\frac{|\rho|}{\nu\orth},\qquad b:=\frac{\beta}{2(1-\beta)},\qquad n:=(1-\beta)k,\qquad m:=\frac{e^{(1-\beta)k}f_0^{1-\beta}}{1-\beta},
$$

so that $\ln m=n+O(1)$. Constants $c_1,c_2,\dots$ depend only on the model parameters and on the auxiliary parameters fixed in the current section, never on $k$; their values may change from section to section. Upper and lower bounds always refer to $P(k)$.

### 2.1 The Lamperti Decomposition

The first reduction is a change of coordinates. In the Lamperti coordinate the component driven by the volatility Brownian motion integrates exactly, leaving an orthogonal Gaussian martingale and a nonpositive singular drift.

**Lemma 2.1 (Lamperti Decomposition).**  Let $X=F^{1-\beta}/(1-\beta)$. There is an event $\Omega_{\mathrm{CEV}}$ with $\Prob(\Omega_{\mathrm{CEV}})=1$ such that, for every $\omega\in\Omega_{\mathrm{CEV}}$ and every $t\in[0,T]$ with $t<\zeta(\omega)$,

$$
\begin{equation}\label{eq:decomp}\tag{2.1} X_t \;=\; X_0+\xi_t+\orth\,M_t-b\int_0^t\frac{a_s^2}{X_s}\,ds, \qquad\xi_t=\frac{\rho}{\nu}(a_t-\alpha),\quad M_t=\int_0^t a_s\,dZ^\perp_s,\tag{2.1} \end{equation}
$$

where continuous versions are used throughout. Moreover, $\{F_T\ge f_0e^k\}=\{X_T\ge m,\ \zeta>T\}$ up to a null set; in particular $P(k)=\Prob(X_T\ge m,\ \zeta>T)$.

*Proof.* Apply Itô’s formula to the processes stopped at $\zeta_r:=\inf\{t:X_t\le r\}$ and let $r\downarrow0$. With $g(F)=F^{1-\beta}/(1-\beta)$ this gives $dX=a\,dW-b(a^2/X)\,dt$ before $\zeta$, using $F^{\beta-1}=((1-\beta)X)^{-1}$. Decompose $W=\rho Z+\orth Z^\perp$ and use $da=\nu a\,dZ$, whence $\int_0^t a\,dZ=(a_t-\alpha)/\nu$. Intersecting the full-measure events obtained at rational stopping levels and using continuity yields one event on which the identity holds simultaneously for every admissible $t$. The event identity follows from the strict monotonicity of $F\mapsto F^{1-\beta}/(1-\beta)$ on $(0,\infty)$ and the absorption convention, since $F_T=0<f_0e^k$ on $\{\zeta\le T\}$.∎

We write

$$
\begin{equation}\label{eq:Xbar}\tag{2.2} \bar X_t:=X_0+\xi_t+\orth M_t\tag{2.2} \end{equation}
$$

for the process with the singular drift removed.

### 2.2 Conditioning on Volatility: Operational Time

Conditioning on the volatility path completes the first reduction: the orthogonal martingale then becomes a Brownian motion run on its quadratic-variation clock.

**Lemma 2.2 (Conditional Brownian Time Change).**  Let $\mathsf W$ be Wiener measure on $C([0,\infty))$, let $\Prob_a$ denote the law of the volatility path $a$ on $C([0,T])$, and for a continuous path $y$ put

$$
\begin{equation*} V_t(y):=\int_0^t y_s^2\,ds,\qquad\Theta_y(w)(t):=w(V_t(y)). \end{equation*}
$$

There is a measurable regular conditional distribution $\mathsf K$ of the $C([0,T])$-valued random variable $M$ given $\sigma(a)$ such that, for $\Prob_a$-almost every volatility path $y$,

$$
\mathsf K(y,\,\cdot\,)=\mathsf W\circ\Theta_y^{-1}.
$$

Equivalently, for such $y$, let $t_y:[0,V_T(y)]\to[0,T]$ be the inverse of $t\mapsto V_t(y)$; then $v\mapsto M_{t_y(v)}$ has the standard Wiener law on $[0,V_T(y)]$.

The jointly measurable versions and the kernel construction are recorded in Appendix A. We call $v=V_t$ the *operational time*, write $t(\cdot)$ for the inverse quadratic-variation clock of the volatility path, and set

$$
B(v):=M_{t(v)},\qquad0\le v\le V_T,
$$

so that $M_t=B(V_t)$. By Lemma 2.2, conditionally on $\sigma(a)$, $B$ is a standard Brownian motion on $[0,V_T]$. “Conditionally on the volatility path” always means evaluation under the kernel $\mathsf K$ for $\Prob_a$-almost every path; the operational times, thresholds and events entering such evaluations are measurable functions of the path.

### 2.3 Transfer to Implied Volatility

The second reduction is model-free: once the logarithmic tail rate is known, no further SABR structure is needed to identify the right-wing limit. The forward $F$ is a nonnegative local martingale, hence a supermartingale, so $\E F_T\le f_0$; this is all the integrability we use. Write $\mathcal C(k):=\E(F_T-f_0e^k)^+$ for the undiscounted call price.

**Lemma 2.3 (Tail-to-Implied-Volatility Transfer).**  If $-\ln P(k)/k^2\to\Lambda\in(0,\infty)$ and $P(k)>0$ for all sufficiently large $k$, then $\sigma_{\BS}(k)\to1/\sqrt{2\Lambda T}$.

*Proof.* *(i) Prices.* By the layer-cake identity (Fubini, using $\E F_T<\infty$), $\mathcal C(k)=f_0\int_k^\infty e^{\kappa}P(\kappa)\,d\kappa$. Fix $\varepsilon>0$ and $k_0$ with $e^{-(\Lambda+\varepsilon)\kappa^2}\le P(\kappa)\le e^{-(\Lambda-\varepsilon)\kappa^2}$ for $\kappa\ge k_0$. Substituting $\kappa=k+u$ and using $(k+u)^2\ge k^2+2ku$, for large $k$

$$
\begin{equation*} \begin{aligned} \mathcal C(k)&\le f_0e^{k-(\Lambda-\varepsilon)k^2}\int_0^\infty e^{u(1-2(\Lambda-\varepsilon)k)}\,du = \frac{f_0\,e^{k-(\Lambda-\varepsilon)k^2}}{2(\Lambda-\varepsilon)k-1}, \\
\mathcal C(k)&\ge f_0\int_k^{k+1}e^{\kappa}P(\kappa)\,d\kappa\ge f_0e^{k-(\Lambda+\varepsilon)(k+1)^2}. \end{aligned} \end{equation*}
$$

Hence $-\ln\mathcal C(k)/k^2\to\Lambda$. Since $0<\mathcal C(k)<f_0$, the implied volatility $\sigma_{\BS}(k)\in(0,\infty)$ exists and is unique by strict monotonicity of the Black–Scholes price in $\sigma$.

*(ii) Black–Scholes tail.* With $d_1=(-k+\sigma^2T/2)/(\sigma\sqrt T)$ and $d_2=d_1-\sigma\sqrt T$ one has $d_1^2-d_2^2=-2k$, hence $e^{k}\varphi(d_2)=\varphi(d_1)$, and the Mills bounds $\Phi(-x)=\varphi(x)(x^{-1}-x^{-3}+3\vartheta x^{-5})$, $\vartheta\in[0,1]$, give

$$
C_{\BS}(k,\sigma)=f_0\bigl[\Phi(d_1)-e^k\Phi(d_2)\bigr] = f_0\,\varphi(d_1)\,\frac{\sigma\sqrt T}{|d_1||d_2|}\bigl(1+O(k^{-2})\bigr),
$$

whence, uniformly on compact $\sigma$-sets,

$$
\begin{equation}\label{eq:bs-tail}\tag{2.3} -\ln C_{\BS}(k,\sigma)=\frac{k^2}{2\sigma^2T}-\frac k2+2\ln k +\Bigl[\frac{\sigma^2T}{8}+\ln\sqrt{2\pi}-\ln\bigl(\sigma^3T^{3/2}\bigr)-\ln f_0\Bigr]+o(1).\tag{2.3} \end{equation}
$$

Only the leading term is used below; the bracketed constant follows from the prefactor $f_0$, $d_1^2/2=k^2/(2\sigma^2T)-k/2+\sigma^2T/8$ and $|d_1d_2|=k^2(\sigma^2T)^{-1}(1+O(1/k))$.

*(iii) Identification of the limit.* Suppose $\sigma_{\BS}(k_j)\to\sigma^*\in(0,\infty]$ along a subsequence with $(\sigma^*)^2>1/(2\Lambda T)$. Pick $\sigma_2\in(1/\sqrt{2\Lambda T},\sigma^*)$; eventually $\mathcal C(k_j)\ge C_{\BS}(k_j,\sigma_2)\ge e^{-k_j^2(1+o(1))/(2\sigma_2^2T)}$, contradicting $-\ln\mathcal C/k^2\to\Lambda>1/(2\sigma_2^2T)$. The case $(\sigma^*)^2<1/(2\Lambda T)$ (including $\sigma^*=0$) is symmetric. Hence $\sigma_{\BS}(k)\to1/\sqrt{2\Lambda T}$.∎

By Lemma 2.3, Theorem 1.1 implies Corollary 1.2, and hence Corollary 1.3; the positivity $P(k)>0$ follows from the lower-bound constructions of Sections 4 and 5. Sections 3 to 5 prove Theorem 1.1.

## 3 Upper Bound for Negative Correlation

We begin with the upper bound on $P(k)$ for negative correlation. This half of Theorem 1.1 must rule out every cheaper way of producing the tail, including paths that raise the orthogonal component before the barrier requires it.

**Proposition 3.1 (Upper Bound, $\rho<0$).**  For $\rho\in(-1,0)$,

$$
\begin{equation}\label{eq:upper}\tag{3.1} \liminf_{k\to\infty}\frac{-\ln P(k)}{k^2}\;\ge\;c_-.\tag{3.1} \end{equation}
$$

Conditionally on volatility, removing the nonpositive drift turns survival into a Brownian moving-barrier event. We discretize the volatility excursion by level-hitting times, convert each conditional event into a convex quadratic energy, bound the combined volatility and boundary-crossing cost from below, and sum over the resulting configurations. Throughout this section $\rho\in(-1,0)$.

### 3.1 Drift Removal

The singular drift is nonpositive before absorption. Dropping it enlarges the joint survival-and-terminal event and isolates the Gaussian constraints that determine the upper bound.

**Lemma 3.2 (Drift Removal).**  $\{X_T\ge m,\ \zeta>T\}\subseteq\{\bar X_T\ge m,\ \inf_{t\le T}\bar X_t>0\}$.

*Proof.* On survival $X>0$, so the drift $-b\,a^2/X$ is nonpositive and $\bar X_t=X_t+b\int_0^t a^2/X\,ds\ge X_t$ for all $t\le T$.∎

### 3.2 A Gaussian Convex-Set Bound

At the level-hitting times introduced below, the relaxed survival and terminal requirements become linear inequalities for a Gaussian vector. The following projection estimate converts their conditional probability into a Cameron–Martin energy.

**Lemma 3.3 (Gaussian Convex-Set Bound).**  Let $Y$ be a centred Gaussian vector in $\mathbb R^p$, $p\ge1$, with symmetric positive definite covariance matrix $\Sigma$, and let $K\subset\mathbb R^p$ be nonempty, closed and convex with $0\notin K$. Then $\Prob(Y\in K)\le\overline\Phi\bigl(d_\Sigma(0,K)\bigr)\le e^{-\frac12 d_\Sigma(0,K)^2}$, where $d_\Sigma$ is the Mahalanobis distance and $\overline\Phi:=1-\Phi$.

*Proof.* Write $Y=\Sigma^{1/2}G$ with $G$ standard Gaussian and $K'=\Sigma^{-1/2}K$. Let $y^*$ be the projection of $0$ onto $K'$, so that $|y^*|=d_\Sigma(0,K)$. The supporting hyperplane at $y^*$ gives $K'\subseteq\{x:\langle x,y^*/|y^*|\rangle\ge|y^*|\}$, and $\langle G,y^*/|y^*|\rangle\sim N(0,1)$.∎

Apply the lemma to $B$ at $\sigma(a)$-measurable operational times $0<v_1<\dots<v_J\le V_T$ with thresholds $(h_j)$ and a terminal threshold $H$ at $V_T$. Put $v_0=0$, $v_{J+1}=V_T$, $\ell_j=h_j$ for $1\le j\le J$, and $\ell_{J+1}=H$. For $s\ge0$ define the extended-real increment energy

$$
\mathcal E_s(x):= \begin{cases} x^2/s,&s>0, \\
0,&s=0\text{ and }x=0, \\
+\infty,&s=0\text{ and }x\ne0. \end{cases}
$$

The constraint set is a nonempty intersection of half-spaces, and the Markov structure of the Brownian covariance identifies its squared Mahalanobis distance with the moving-barrier quadratic program

$$
\begin{equation}\label{eq:QP}\tag{3.2} \mathrm{QP}(v,\ell) :=\min_{\substack{x_0,\ldots,x_{J+1}\in\mathbb R \\
x_0=0,\ x_i\ge\ell_i\ (1\le i\le J+1)}} \sum_{i=1}^{J+1}\mathcal E_{v_i-v_{i-1}}(x_i-x_{i-1}),\tag{3.2} \end{equation}
$$

the Cameron–Martin energy of the cheapest interpolation through the constrained nodes. In the applications below $H>0$ for all sufficiently large $k$, so the constraint set does not contain the origin. If $v_J=V_T$, the zero-increment convention forces $x_{J+1}=x_J$; this amounts to merging the coincident nodes with the single threshold $\max(h_J,H)$, and Lemma 3.3 is applied after this merge, so that its covariance matrix is positive definite. The same formula with no intermediate nodes covers the terminal-only case $J=0$.

### 3.3 Level-Hitting Skeleton

To retain sharp costs with finitely many constraints, we discretize the increase of log-volatility by successive level-hitting times. The same skeleton controls the available quadratic variation, the induced survival thresholds, and the probability of completing each level crossing quickly.

Fix $\eta\in(0,\tfrac12)$ and set $A_j:=\alpha e^{j\eta}$, $\tau_j:=\inf\{t: a_t\ge A_j\}$ (so $\tau_0=0$), $J^*:=\max\{j:\tau_j\le T\}$, $\Delta t_j:=\tau_j-\tau_{j-1}$ and $v_j:=V_{\tau_j}$. Along this skeleton the levels control the available quadratic variation, and survival becomes a finite list of threshold conditions at the nodes.

**Lemma 3.4 (Skeleton Constraints).**  (i) For every $t\in[\tau_{j-1},\tau_j)$, $a_t<A_j$; hence

$$
\begin{equation*} \Delta v_j:=v_j-v_{j-1}\le A_j^2\Delta t_j=e^{2\eta}A_{j-1}^2\Delta t_j,\qquad V_T\le\bar V:=2e^{2\eta}A_{J^*}^2 T. \end{equation*}
$$

(ii) On $\{\bar X_T\ge m,\ \inf_{t\le T}\bar X_t>0\}$ one has $B(v_j)>h_j:=C(A_j-\alpha)-X_0/\orth$ for $1\le j\le J^*$, and $B(V_T)\ge H:=(m-X_0)/\orth+C(a_T-\alpha)\ge(m-X_0)/\orth-C\alpha$.

*Proof.* (i) is immediate from the definition of first passage; the cap on $V_T$ holds because

$$
\begin{equation*} a<A_{J^*+1}=e^{\eta}A_{J^*}\qquad\text{on }[0,T]. \end{equation*}
$$

(ii) Since $a_{\tau_j}=A_j$ by continuity and $\xi_t=-C\orth(a_t-\alpha)$, the inequality $\bar X_{\tau_j}>0$ reads $\orth B(v_j)>C\orth(A_j-\alpha)-X_0$, and $\bar X_T\ge m$ reads $\orth B(V_T)\ge m-X_0-\xi_T$.∎

Conversely, crossing a level quickly is costly for the volatility, as the reflection principle shows.

**Lemma 3.5 (Level-Hitting-Time Bound).**  On $\{\tau_{j-1}<\infty\}$, for every $s>0$,

$$
\begin{equation*} \Prob\bigl(\tau_j<\infty,\ \tau_j\le\tau_{j-1}+s\,\big|\,\mathcal F_{\tau_{j-1}}\bigr)\ \le\ 2\exp\Bigl(-\frac{\eta^2}{2\nu^2 s}\Bigr). \end{equation*}
$$

Consequently, for any thresholds $\bar t_1,\dots,\bar t_J>0$, by iterated conditioning,

$$
\begin{equation*} \Prob(\Delta t_j\le\bar t_j\ \text{for all } j\le J)\ \le\ 2^{J}\exp\Bigl(-\sum_{j\le J}\frac{\eta^2}{2\nu^2\bar t_j}\Bigr). \end{equation*}
$$

*Proof.* By the strong Markov property, from $\tau_{j-1}$ the log-volatility runs as $\nu B'_u-\nu^2u/2$ for a fresh Brownian motion $B'$; rising by $\eta$ within time $s$ forces $\sup_{u\le s}\nu B'_u\ge\eta$, and the reflection principle gives the bound.∎

### 3.4 Deferred Displacement and the Energy Inequality

The preceding reductions leave a deterministic question: can a path bring its combined volatility and Gaussian cost below $c_-k^2$, by redistributing the volatility increase or by raising the Gaussian coordinate before the barrier requires it? The next two lemmas show that it cannot. The key input is an elementary summation-by-parts bound for the part of the Gaussian cost that is linear in the displacement.

**Lemma 3.6 (Deferred Displacement).**  Let $w_1\ge w_2\ge\dots\ge w_J\ge0$ and $h_1\le h_2\le\dots\le h_J$, write $h_j^+:=\max(h_j,0)$, and put $h_0^+:=0$. For all $x_0=0,x_1,\dots,x_J\in\mathbb R$ with $x_j\ge h_j$ for $1\le j\le J$,

$$
\sum_{j=1}^J w_j\,|x_j-x_{j-1}|\;\ge\;\sum_{j=1}^J w_j\,\bigl(h_j^+-h_{j-1}^+\bigr),
$$

with equality for the maximally deferred choice $x_j=h_j^+$.

*Proof.* Let $U_0:=0$ and $U_j:=\sum_{i\le j}(x_i-x_{i-1})^+$. Then $U_j\ge x_j\ge h_j$ and $U_j\ge0$, so $U_j\ge h_j^+$. Summation by parts with $w_{J+1}:=0$ gives

$$
\begin{aligned} \sum_{j=1}^J w_j|x_j-x_{j-1}| &\ \ge\ \sum_{j=1}^J w_j(U_j-U_{j-1}) =\sum_{j=1}^J (w_j-w_{j+1})\,U_j \\
&\ \ge\ \sum_{j=1}^J (w_j-w_{j+1})\,h_j^+ =\sum_{j=1}^J w_j\bigl(h_j^+-h_{j-1}^+\bigr), \end{aligned}
$$

since $w_j-w_{j+1}\ge0$. For $x_j=h_j^+$ the increments are nonnegative and both inequalities are equalities.∎

With this bound for the linear part, the combined volatility and Gaussian energy can be bounded below uniformly over all allocations of time and quadratic variation.

**Lemma 3.7 (Energy Inequality).**  For every $\varepsilon_0\in(0,1)$ and $\varepsilon'\in(0,\tfrac14)$ there are $\eta_0\in(0,\tfrac12)$ and $\kappa_0\in(0,1)$, depending only on $\varepsilon_0$ and the model parameters, with the following property. For every $\eta\in(0,\eta_0]$ and $\kappa\in(0,\kappa_0]$ there is $k_0=k_0(\varepsilon_0,\varepsilon',\eta,\kappa)$ such that the estimate below holds for every $k\ge k_0$, every integer $J\ge1$, and every admissible choice of increments.

Let $\Delta t_j>0$ with $\sum_{j=1}^J\Delta t_j\le(1+\kappa)T$, let $\Delta v_j\in(0,e^{2\eta}A_{j-1}^2\Delta t_j]$, and append a terminal increment $\Delta v_\infty\in[0,\bar V]$, where $\bar V:=2e^{2\eta}A_J^2T$. In (3.2) take the operational-time increments $(\Delta v_1,\ldots,\Delta v_J,\Delta v_\infty)$, the thresholds $h_j$ of Lemma 3.4, and a terminal threshold $H$ with

$$
\begin{equation*} H\ge\frac{m-X_0}{\orth}-C\alpha. \end{equation*}
$$

Then

$$
\sum_{j\le J}\frac{\eta^2}{2\nu^2\Delta t_j}\;+\;\tfrac12\,\mathrm{QP} \;\ge\;(1-\varepsilon')^2(1-\varepsilon_0)\,c_-k^2.
$$

*Proof.* $\mathrm{QP}$ is nonincreasing in each operational-time increment, so we may replace $\Delta v_j$ by its upper bound and the terminal increment by $\bar V$; this remains valid when the actual terminal increment is zero, because the convention in (3.2) is the lower-semicontinuous limit as that increment decreases to zero. Write the QP variables through their increments $\Delta x_j:=x_j-x_{j-1}$ ($j\le J$) and $\Delta x_\infty:=x_{J+1}-x_J$, so that $x_j\ge h_j$ and $x_J+\Delta x_\infty\ge H$. Per interval,

$$
\frac{\eta^2}{2\nu^2\Delta t_j}+\frac{(\Delta x_j)^2}{2e^{2\eta}A_{j-1}^2\Delta t_j} =\frac{z_j}{\Delta t_j},\qquad z_j:=\frac{\eta^2}{2\nu^2}+\frac{e^{-2\eta}(\Delta x_j)^2}{2A_{j-1}^2},
$$

and by Cauchy–Schwarz $\sum_j z_j/\Delta t_j\ge(\sum_j\sqrt{z_j})^2/\sum_j\Delta t_j\ge(\sum_j\sqrt{z_j})^2/((1+\kappa)T)$; the terminal term $(\Delta x_\infty)^2/(2\bar V)$ is kept aside as a separate nonnegative summand. Put $r:=J\eta/n$. All $O$-terms below are uniform in $J$ and in the admissible increments; the coefficient of $O(\eta+\kappa)$ is uniform for small $\eta,\kappa$, and once $\eta,\kappa$ are fixed the coefficient of $O(1/n)$ does not depend on $k$.

*Main case* $r\ge1-\varepsilon'$*.* Drop the terminal term. For any $u,v\ge0$ with $u^2+v^2=1$, $\sqrt{z_j}\ge u\,\eta/(\sqrt2\nu)+v\,e^{-\eta}|\Delta x_j|/(\sqrt2 A_{j-1})$. The weights $1/A_{j-1}$ decrease and the thresholds $h_j$ increase, so Lemma 3.6 gives

$$
\sum_{j\le J}\frac{|\Delta x_j|}{A_{j-1}}\ \ge\ \sum_{j\le J}\frac{h_j^+-h_{j-1}^+}{A_{j-1}}\ \ge\ C(e^\eta-1)\,J_{\mathrm{bind}}, \qquad J_{\mathrm{bind}}:=\#\{j\le J:h_{j-1}\ge0\},
$$

where $h_0:=-X_0/\orth$ continues the formula of Lemma 3.4; indeed $(h_j-h_{j-1})/A_{j-1}=C(e^\eta-1)$ whenever $h_{j-1}\ge0$. If $A_b:=\alpha+X_0/(C\orth)$ denotes the level at which $h$ becomes nonnegative, then $J_{\mathrm{bind}}\ge J-\eta^{-1}\ln(A_b/\alpha)-1$. Raising the Gaussian coordinate earlier than the barrier requires can therefore only increase this sum. Hence

$$
\sum_j\sqrt{z_j}\ \ge\ (J\eta)\Bigl[\frac{u}{\sqrt2\,\nu}+\frac{vC}{\sqrt2}\bigl(1-O(\eta)-O(1/n)\bigr)\Bigr],
$$

and optimizing the direction, $\sup_{u^2+v^2=1}[\,\cdot\,]=\sqrt{1/(2\nu^2)+C^2/2}\,(1-O(\eta)-O(1/n))$. With $J\eta=rn\ge(1-\varepsilon')n$ and $1+\nu^2C^2=(1-\rho^2)^{-1}$,

$$
\begin{aligned} \frac{(\sum_j\sqrt{z_j})^2}{(1+\kappa)T} &\ \ge\ (1-\varepsilon')^2\,\frac{n^2}{2\nu^2T(1-\rho^2)}\bigl(1-O(\eta+\kappa)-O(1/n)\bigr) \\
&\ =\ (1-\varepsilon')^2c_-k^2\bigl(1-O(\eta+\kappa)-O(1/n)\bigr). \end{aligned}
$$

*Case* $r<1-\varepsilon'$*.* Here $H\ge m/(2\orth)$ for large $k$. Since $x_J+\Delta x_\infty\ge H$, either $\Delta x_\infty\ge H/2$ or $x_J=\sum_{j\le J}\Delta x_j\ge H/2$. In the former case

$$
\frac{(\Delta x_\infty)^2}{2\bar V}\ \ge\ \frac{m^2}{64\,\orth^2e^{2\eta}A_J^2T}\ \ge\ \frac{e^{2(1-r)n}}{c_3T}\ \ge\ \frac{e^{2\varepsilon' n}}{c_3T};
$$

in the latter, taking $(u,v)=(0,1)$, $\sum_j\sqrt{z_j}\ge(e^{-\eta}/\sqrt2)(H/2)/A_J\ge e^{(1-r)n}/c_4$, and squaring gives at least $e^{2\varepsilon'n}/(c_5T)$. Both dominate $c_-k^2$ for $k\ge k_0$.

Finally choose $\eta_0$ and $\kappa_0$ so that the $O(\eta+\kappa)$ loss in the main case is at most $\varepsilon_0/2$. For each fixed $\eta\le\eta_0$ and $\kappa\le\kappa_0$, enlarge $k_0$ so that the $O(1/n)$ loss is at most $\varepsilon_0/2$, $H\ge m/(2\orth)$, and the exponential bounds of the second case dominate $c_-k^2$. This gives the factor $1-\varepsilon_0$ uniformly over $J$ and all admissible increments.∎

The proof never identifies the minimizer of the aggregated functional, which is not obtained by exact barrier tracking: the directional bound and Lemma 3.6 apply to all feasible allocations simultaneously.

### 3.5 Assembly

It remains to sum the cellwise bounds without losing the quadratic exponent. We first remove exceptionally high volatility levels and very short passage times, then bin the remaining passage times so that Lemma 3.7 applies uniformly at a combinatorial cost of only $e^{o(n^2)}$.

*Proof of Proposition 3.1.*  By Lemmas 2.1 and 3.2, $P(k)\le\Prob(\bar X_T\ge m,\ \inf_{t\le T}\bar X_t>0)$. Fix $\varepsilon_0,\varepsilon',\eta,\kappa$ in the ranges of Lemma 3.7; let $r_{\mathrm{cap}}:=\lceil(1-\rho^2)^{-1/2}\rceil+1$, $J_{\max}:=\lceil r_{\mathrm{cap}}n/\eta\rceil$, $L:=\lceil\ln(n^{10})/\ln(1+\kappa)\rceil$, and define time bins $\mathcal B_i:=(T(1+\kappa)^{-i-1},\,T(1+\kappa)^{-i}]$ for $i<L$ and $\mathcal B_L:=(0,\,T(1+\kappa)^{-L}]$.

*(a) High volatility.* $\{J^*\ge J_{\max}\}\subseteq\{\sup_{t\le T}(\nu Z_t-\nu^2t/2)\ge r_{\mathrm{cap}}n-\eta\}$, so by the reflection principle its probability is at most $\exp(-(r_{\mathrm{cap}}n-\eta)^2/(2\nu^2T))\le e^{-c_-k^2}$ for large $k$, since $r_{\mathrm{cap}}^2\ge(1-\rho^2)^{-1}$.

*(b) No level crossing,* $J^*=0$*.* Then $a<\alpha e^{\eta}$ on $[0,T]$, $V_T\le\alpha^2e^{2\eta}T$, and the one-node case of Lemma 3.3 bounds the conditional probability by $\exp(-c_6 m^2)$.

*(c) Configurations.* For $1\le J^*<J_{\max}$, partition according to the configuration and write the upper end of the bin of $\Delta t_j$ as

$$
\begin{equation*} \mathbf i=(J^*,(i_j)_{j\le J^*}),\qquad\bar t_j:=\sup\mathcal B_{i_j}, \end{equation*}
$$

where $i_j$ is the bin containing $\Delta t_j$. If some $i_j=L$, Lemma 3.5 alone bounds the probability by $2\exp(-\eta^2(1+\kappa)^L/(2\nu^2T))\le e^{-n^9}$ for large $k$. The number of configurations is at most $\sum_{J<J_{\max}}(L+1)^{J}\le\exp\bigl(O\bigl(\tfrac n\eta(\ln\ln n+\ln\tfrac1\kappa)\bigr)\bigr)=e^{o(n^2)}$, so these cells contribute at most $e^{-n^9+o(n^2)}$. Otherwise $\bar t_j\le(1+\kappa)\Delta t_j$, hence $\sum_j\bar t_j\le(1+\kappa)T$, and on each cell the volatility cost is at most $2^{J^*}\exp(-\sum_j\eta^2/(2\nu^2\bar t_j))$ by Lemma 3.5 with thresholds $\bar t_j$, while the conditional Gaussian cost is at most $\exp(-\frac12\mathrm{QP})$ by Lemmas 3.3 and 3.4, with the operational-time increments bounded through $\bar t_j$ (larger increments decrease $\mathrm{QP}$). Their product is bounded by Lemma 3.7 with $\Delta t_j:=\bar t_j$.

*(d) Sum.* Combining,

$$
P(k)\ \le\ e^{-c_-k^2}+e^{-c_6m^2}+e^{-n^{8}}+e^{o(n^2)}\, e^{-(1-\varepsilon')^2(1-\varepsilon_0)c_-k^2}.
$$

Letting $k\to\infty$ and then $\varepsilon_0,\varepsilon'\downarrow0$, with fixed admissible $\eta,\kappa$ for each $\varepsilon_0$, proves the proposition.∎

## 4 Lower Bound for Negative Correlation

We now construct surviving paths that produce the tail at the rate $c_-$, which gives the reverse inequality.

**Proposition 4.1 (Lower Bound, $\rho<0$).**  For $\rho\in(-1,0)$,

$$
\limsup_{k\to\infty}\frac{-\ln P(k)}{k^2}\;\le\;c_-.
$$

The construction separates the two leading costs. A Girsanov change of measure confines volatility to a tube around an exponential path; conditionally on that tube, the orthogonal Brownian motion enters and stays above the induced square-root barrier (Figure 2). A bootstrap then shows that the chosen margins absorb the singular drift and imply the terminal event. Throughout this section $\rho\in(-1,0)$.

Fix $\varepsilon_1,\varepsilon_2,\delta\in(0,\tfrac14)$, set $\theta:=n(1+\varepsilon_1)/(\nu T)$ and $a^*(t):=\alpha e^{\nu\theta t}$, and let $V^*(t):=\int_0^t a^*(s)^2ds$. The exact identity $a^*(t)=\sqrt{2\nu\theta V^*(t)+\alpha^2}$ shows that *the exponential path is a shifted square-root profile in its own quadratic-variation time*.

### 4.1 Volatility Tube

We first confine volatility near the exponential path suggested by the leading variational balance. The logarithmic tube below has the unconstrained quadratic cost and distorts the volatility level and its quadratic-variation clock only by fixed factors.

**Lemma 4.2 (Volatility Tube).**  For $\theta>0$ and $\delta\in(0,1)$ let $a^*(t)=\alpha e^{\nu\theta t}$ and

$$
\begin{equation*} A_{\mathrm{vol}}:=\Bigl\{\sup_{t\le T}|\ln a_t-\ln a^*(t)|\le\delta\Bigr\}. \end{equation*}
$$

Then

$$
\Prob(A_{\mathrm{vol}})\ \ge\ p(\delta)\,\exp\Bigl(-\frac{(\theta+\nu/2)^2T}{2}-\frac{(\theta+\nu/2)\,\delta}{\nu}\Bigr), \qquad p(\delta):=\Prob\Bigl(\sup_{t\le T}|\nu Z_t|\le\delta\Bigr)>0.
$$

On $A_{\mathrm{vol}}$, $e^{-\delta}a^*(t)\le a_t\le e^{\delta}a^*(t)$ and $e^{-2\delta}V^*(t)\le V_t\le e^{2\delta}V^*(t)$ for all $t\le T$.

*Proof.* Let $\mu:=\theta+\nu/2$, and let $\widetilde\Prob$ be the measure, given by Girsanov’s theorem, under which $\widetilde Z_t:=Z_t-\mu t$ is a Brownian motion. The extra $\nu/2$ absorbs the Itô term, so that exactly

$$
\begin{equation*} \ln a_t=\ln\alpha+\nu Z_t-\tfrac12\nu^2t=\ln a^*(t)+\nu\widetilde Z_t. \end{equation*}
$$

Hence $\widetilde\Prob(A_{\mathrm{vol}})=p(\delta)$, and on $A_{\mathrm{vol}}$, where $|\nu\widetilde Z_T|\le\delta$,

$$
\begin{equation*} \frac{d\Prob}{d\widetilde\Prob}=\exp\Bigl(-\mu\widetilde Z_T-\frac{\mu^2T}{2}\Bigr)\ \ge\ \exp\Bigl(-\frac{\mu\delta}{\nu}-\frac{\mu^2T}{2}\Bigr). \end{equation*}
$$

The bounds on $a_t$ and $V_t$ are immediate.∎

With $\theta=n(1+\varepsilon_1)/(\nu T)$ the lemma gives

$$
\begin{equation}\label{eq:tube}\tag{4.1} \Prob(A_{\mathrm{vol}})\ \ge\ \exp\Bigl(-\frac{n^2(1+\varepsilon_1)^2}{2\nu^2T}-O(n)\Bigr).\tag{4.1} \end{equation}
$$

### 4.2 Target Event for the Orthogonal Component

Conditionally on a volatility path in the tube, the orthogonal Brownian motion must keep $\bar X$ positive before the barrier becomes active, stay above the moving barrier afterwards, and leave enough terminal margin to reach $m$ once the singular drift is restored.

Let $\hat a(v):=a_{t(v)}$ denote volatility in operational time. Set $c_*:=\nu X_0/(2(1+\varepsilon_2)|\rho|)$ and let $v_b$ be the first operational time at which $\hat a$ exceeds $\alpha+c_*$; since $\hat a(0)=\alpha$ and $\hat a$ is continuous, $v_b>0$. Define

$$
\begin{aligned} E_B:= {}&\Bigl\{B(v)\ge(1+\varepsilon_2)C\hat a(v)\ \forall v\in[v_b,V_T]\Bigr\} \cap\Bigl\{B\ge-\tfrac{X_0}{2\orth}\text{ on }[0,v_b]\Bigr\} \\
&{}\cap\Bigl\{B(V_T)\ge\tfrac{1+\varepsilon_2}{\orth}\bigl(m+\tfrac{|\rho|}{\nu}a_T\bigr)\Bigr\}. \end{aligned}
$$

On the tube, with $v_\alpha:=\alpha^2/(2\nu\theta)$, the identity $a^*(t)=\sqrt{2\nu\theta V^*(t)+\alpha^2}$ and Lemma 4.2 give the two-sided bound

$$
\begin{equation}\label{eq:ahat-sqrt}\tag{4.2} e^{-2\delta}\sqrt{2\nu\theta\,(v+v_\alpha)}\ \le\ \hat a(v)\ \le\ e^{2\delta}\sqrt{2\nu\theta\,(v+v_\alpha)},\tag{4.2} \end{equation}
$$

so the moving barrier $(1+\varepsilon_2)C\hat a$ is dominated by $\lambda_+\sqrt{v+v_\alpha}$, where $\lambda_+:=(1+\varepsilon_2)e^{2\delta}C\sqrt{2\nu\theta}$. Evaluating (4.2) at $v_b$ shows $c_{\mathrm b,1}/\theta\le v_b\le c_{\mathrm b,2}/\theta$ with $c_{\mathrm b,1}>0$, provided $\delta<\delta_0:=\tfrac12\ln(1+c_*/\alpha)$, which we assume henceforth. The constants of this section may depend on the fixed $\delta$, which is harmless because $\delta\downarrow0$ only after $k\to\infty$.

### 4.3 Bootstrap Control of the Singular Drift

The slack factors in $E_B$ were chosen to dominate the accumulated singular drift. The next lemma turns the tube-and-barrier event into a surviving path that reaches the terminal target.

**Lemma 4.3 (Bootstrap).**  On $A_{\mathrm{vol}}\cap E_B$, for all $k$ large (given $\varepsilon_1,\varepsilon_2,\delta$), $\zeta>T$ and $X_T\ge m$.

*Proof.* Recall $\bar X=X_0+\xi+\orth B(V)$ from (2.2), and work on $A_{\mathrm{vol}}\cap E_B$.

*Positivity of* $\bar X$*.* For $v\ge v_b$, $\orth B\ge(1+\varepsilon_2)\frac{|\rho|}{\nu}\hat a$, so

$$
\bar X\ge X_0-\frac{|\rho|}{\nu}(\hat a-\alpha)+(1+\varepsilon_2)\frac{|\rho|}{\nu}\hat a =X_0+\frac{|\rho|}{\nu}(\varepsilon_2\hat a+\alpha)\ge\varepsilon_2\frac{|\rho|}{\nu}\hat a.
$$

For $v\le v_b$, $\hat a\le\alpha+c_*$ gives $\xi=-\tfrac{|\rho|}{\nu}(\hat a-\alpha)\ge-\tfrac{|\rho|}{\nu}c_*=-\tfrac{X_0}{2(1+\varepsilon_2)}$, and with $B\ge-X_0/(2\orth)$,

$$
\bar X\ge X_0-\frac{X_0}{2(1+\varepsilon_2)}-\frac{X_0}{2}=\frac{\varepsilon_2X_0}{2(1+\varepsilon_2)}>0.
$$

*Drift control.* Let $\tau_{1/2}:=\inf\{t: X_t\le\bar X_t/2\}$. On $[0,\tau_{1/2})$,

$$
X\ \ge\ \frac{\bar X}2\ \ge\ \frac12\min\Bigl(\frac{\varepsilon_2X_0}{2(1+\varepsilon_2)},\ \varepsilon_2\frac{|\rho|}{\nu}\hat a\Bigr);
$$

on the first branch $a\le\alpha+c_*$, so $a^2\le(\alpha+c_*)a$, and on the second $a^2/\hat a=a$. Hence the drift integral satisfies

$$
D_t:=b\int_0^t\frac{a_s^2}{X_s}\,ds\le c_2(\varepsilon_2)\int_0^t a_s\,ds\le c_2(\varepsilon_2)\,\frac{e^{\delta}a^*(t)}{\nu\theta},
$$

so $D_t/\bar X_t\le c_3(\varepsilon_2,\delta)/\theta\to0$ uniformly in $t$. In particular $D<\bar X/2$ throughout for large $k$, which is incompatible with $\tau_{1/2}\le T$. Hence $X\ge\bar X/2>0$ on $[0,T]$, that is, $\zeta>T$.

*Terminal value.* The terminal constraint in $E_B$ and the drift bound give, for $k$ large,

$$
\begin{equation*} \begin{aligned} \bar X_T&\ge(1+\varepsilon_2)m+\varepsilon_2\tfrac{|\rho|}{\nu}a_T+X_0+\tfrac{|\rho|}{\nu}\alpha, \\
D_T&\le c_2(\varepsilon_2)e^{2\delta}a_T/(\nu\theta)=o(a_T), \\
X_T&=\bar X_T-D_T\ge(1+\varepsilon_2)m+\varepsilon_2\tfrac{|\rho|}{\nu}a_T-o(a_T)\ge m. \end{aligned} \end{equation*}
$$

In the first line the $\xi_T$ term cancels all but the $\varepsilon_2$-fraction of the barrier margin; in the second, $D_T=o(a_T)$ because $\theta\to\infty$.∎

### 4.4 Barrier Tracking

It remains to show that $E_B$ has the required conditional probability. Under the logarithmic change of operational time, a square-root Brownian barrier becomes a constant barrier for a stationary Ornstein–Uhlenbeck process (Figure 2(b)). This representation does not control the initial entry, so the construction first follows an explicit Cameron–Martin path segment and then concatenates Ornstein–Uhlenbeck survival blocks.

Fix $\varepsilon_3\in(0,\tfrac14)$ and $L\ge L_0$, where $L_0=L_0(\text{model parameters})$ is chosen so that $v_b\le Lv_\alpha$ on the tube (possible since $v_b\le c_{\mathrm b,2}/\theta$ and $v_\alpha=\alpha^2/(2\nu\theta)$). Set

$$
\begin{aligned} v_L&:=Lv_\alpha, & s_L&:=\ln v_L, & s_T&:=\ln V_T, \\
\lambda_L&:=\lambda_+\sqrt{1+1/L}, & u_0&:=\lambda_L(1+\varepsilon_3), & h_L&:=\lambda_+\sqrt{(L+1)v_\alpha}. \end{aligned}
$$

Here $h_L=(1+\varepsilon_2)e^{2\delta}C\alpha\sqrt{L+1}$ and $u_0\sqrt{v_L}=(1+\varepsilon_3)h_L$ are constants independent of $k$, while $u_0\asymp\lambda_+\asymp\sqrt n$. We use the constant barrier for $U$ only on $[v_L,V_T]$, where $v+v_\alpha\le(1+1/L)v$, and bridge $[0,v_L]$ explicitly. Two elementary lemmas isolate the probabilistic inputs.

**Lemma 4.4 (Brownian Barrier Entry).**  Conditionally on a volatility path in the tube, for all $k\ge k_0(L,\varepsilon_3,\varepsilon_2,\delta,\varepsilon_1)$,

$$
\Prob\Bigl(B\ge-\tfrac{X_0}{2\orth}\ \text{on}\ [0,v_b];\ \ B\ge h_L\ \text{on}\ [v_b,v_L];\ \ B(v_L)\in\bigl[u_0\sqrt{v_L},(u_0+1)\sqrt{v_L}\bigr]\Bigr)\ \ge\ e^{-c_L\lambda_+^2},
$$

with $c_L=c(L,\varepsilon_3,\text{params})$.

*Proof.* Let $\phi$ be piecewise linear: $\phi\equiv0$ on $[0,v_b/2]$, rising linearly to $(u_0+\tfrac12)\sqrt{v_L}$ at $v_b$, and constant on $[v_b,v_L]$. On $\{\sup_{v\le v_L}|B-\phi|\le\tfrac12\sqrt{v_L}\}$ all three constraints hold: on $[0,v_b]$, $B\ge-\tfrac12\sqrt{v_L}\ge-X_0/(2\orth)$ for $k$ large (as $v_L\to0$); on $[v_b,v_L]$, $B\ge u_0\sqrt{v_L}=(1+\varepsilon_3)h_L\ge h_L$; and $B(v_L)\in[u_0\sqrt{v_L},(u_0+1)\sqrt{v_L}]$. The localized Cameron–Martin shift along $\phi$ and Brownian scaling, $\Prob\bigl(\sup_{v\le v_L}|\widetilde B|\le\tfrac12\sqrt{v_L}\bigr)=\Prob\bigl(\sup_{u\le1}|B_u|\le\tfrac12\bigr)=:p_*>0$, give

$$
\Prob(\cdot)\ \ge\ p_*\exp\Bigl(-\frac{(u_0+\tfrac12)^2v_L}{v_b}-\frac{2(u_0+\tfrac12)\,v_L}{v_b}\Bigr)\ \ge\ e^{-c_L\lambda_+^2},
$$

since

$$
\begin{equation*} \frac12\int\dot\phi^2=\frac{(u_0+\frac12)^2v_L}{v_b},\qquad\Bigl|\int\dot\phi\,d\widetilde B\Bigr|\le|\dot\phi|_\infty\sqrt{v_L}=\frac{2(u_0+\frac12)\,v_L}{v_b} \end{equation*}
$$

on the localization event, while $v_L/v_b\le Lv_\alpha\theta/c_{\mathrm b,1}$ is a constant depending only on $L$ and the parameters, and $u_0\asymp\lambda_+$.∎

After the entry, the path is carried through blocks of logarithmic length between $\tfrac12$ and $1$; the next lemma gives the cost of a single block.

**Lemma 4.5 (Ornstein–Uhlenbeck Survival Block).**  Let $U$ solve $dU=-\tfrac12U\,ds+dW$. For each $\varepsilon_3\in(0,\tfrac14)$ there exist $p_1>0$ and $\lambda_0$ such that for all $\lambda\ge\lambda_0$, all durations $\Delta s\in[\tfrac12,1]$, and all $x\in I_\lambda:=[\lambda(1+\varepsilon_3),\lambda(1+\varepsilon_3)+1]$,

$$
\Prob_x\bigl(U_s\ge\lambda\ (0\le s\le\Delta s),\ U_{\Delta s}\in I_\lambda\bigr)\ \ge\ p_1\exp\Bigl(-\frac{\lambda^2(1+\varepsilon_3)^2}{8}\,\Delta s-\lambda^{3/2}\Bigr).
$$

*Proof.* Write $u_\lambda:=\lambda(1+\varepsilon_3)$ and tilt the driving Brownian motion by the constant drift $u_\lambda/2$: under the tilt, $dU=-\tfrac12(U-u_\lambda)\,ds+d\widetilde W$ is an Ornstein–Uhlenbeck process centred at $u_\lambda$. The localized Cameron–Martin inequality $\Prob_x(A)\ge e^{-\frac12\int g^2-K}\bigl(\widetilde\Prob_x(A)-\widetilde\Prob(\int g\,d\widetilde W>K)\bigr)$ applies with $\tfrac12\int g^2=u_\lambda^2\Delta s/8$ and $K:=\lambda^{3/2}$; since $\int g\,d\widetilde W\sim N(0,u_\lambda^2\Delta s/4)$ and $K$ is at least $\lambda^{1/2}$ standard deviations, the subtracted term is at most $e^{-\lambda/2}$. For the main term, the recentred process $\widetilde U:=U-u_\lambda$ is a centred Ornstein–Uhlenbeck process started at $x-u_\lambda\in[0,1]$, and the event contains $\{\widetilde U_s\ge-1\ (s\le\Delta s),\ \widetilde U_{\Delta s}\in[0,1]\}$ because $\lambda-u_\lambda=-\lambda\varepsilon_3\le-1$ for $\lambda\ge1/\varepsilon_3$. That event contains the tube event $\{\sup_{s\le\Delta s}|\widetilde U_s-\psi(s)|\le\tfrac14\}$ with $\psi$ linear from $x-u_\lambda$ to $\tfrac12$. The law of the centred process on $[0,\Delta s]$ is a nondegenerate Gaussian measure with full support on continuous paths from $x-u_\lambda$, so the tube probability is positive and depends continuously on $(x-u_\lambda,\Delta s)\in[0,1]\times[\tfrac12,1]$; hence it is bounded below by some $2p_1>0$ on that compact set. Therefore $\Prob_x(A)\ge e^{-u_\lambda^2\Delta s/8-\lambda^{3/2}}(2p_1-e^{-\lambda/2})\ge p_1e^{-u_\lambda^2\Delta s/8-\lambda^{3/2}}$ for $\lambda\ge\lambda_0$.∎

Concatenating the entry segment with about $\ln(V_T/v_L)$ such blocks yields the conditional probability of $E_B$.

**Lemma 4.6 (Barrier Tracking).**  For every fixed $L\ge L_0$ and $\varepsilon_3\in(0,\tfrac14)$, conditionally on a volatility path in the tube,

$$
\Prob(E_B)\ \ge\ \exp\Bigl(-\bigl(1+\tfrac1L\bigr)(1+\varepsilon_3)^2\, \frac{\lambda_+^2}{8}\,S\,(1+o(1))\Bigr),\qquad S:=\ln(V_T/v_b)=2n(1+\varepsilon_1)(1+o(1)),
$$

with $o(1)\to0$ as $k\to\infty$ for fixed $(L,\varepsilon_3,\varepsilon_2,\delta,\varepsilon_1)$.

*Proof.* *Domination.* For $v\in[v_L,V_T]$ one has $v+v_\alpha\le(1+1/L)v$, hence

$$
\begin{aligned} \bigl\{B(v)\ge\lambda_L\sqrt v\ \ \forall v\in[v_L,V_T]\bigr\} &\ \subseteq\ \bigl\{B(v)\ge\lambda_+\sqrt{v+v_\alpha}\ \ \forall v\in[v_L,V_T]\bigr\} \\
&\ \subseteq\ \bigl\{B\ge(1+\varepsilon_2)C\hat a\ \text{on}\ [v_L,V_T]\bigr\}, \end{aligned}
$$

while on $[v_b,v_L]$ the moving barrier is at most $\lambda_+\sqrt{v_L+v_\alpha}=h_L$. In the coordinate

$$
\begin{equation*} U(s)=e^{-s/2}B(e^s),\qquad dU=-\tfrac12U\,ds+dW, \end{equation*}
$$

a stationary Ornstein–Uhlenbeck process with covariance $e^{-|s-s'|/2}$, the first event reads

$$
\begin{equation*} \bigl\{U(s)\ge\lambda_L\ \text{for all } s\in[s_L,s_T]\bigr\}. \end{equation*}
$$

*Entry and block concatenation.* Apply Lemma 4.4. Then, by the Markov property of $B$ at the $\sigma(a)$-measurable operational time $v_L$ and the Markov property of $U$ at the endpoints $s_L<s_L+\Delta s<\dots<s_T$ of $N$ blocks of equal duration, where

$$
\begin{equation*} N:=\lceil s_T-s_L\rceil,\qquad\Delta s:=\frac{s_T-s_L}{N}\in\bigl(\tfrac12,1\bigr], \end{equation*}
$$

Lemma 4.5—uniform over starting points in $I_{\lambda_L}$ and over $\Delta s$—applies at every block:

$$
\Prob\bigl(\text{entry};\ U\ge\lambda_L\ \text{on}\ [s_L,s_T];\ U(s_T)\in I_{\lambda_L}\bigr) \ \ge\ e^{-c_L\lambda_+^2}\;p_1^N\exp\Bigl(-\frac{u_0^2}{8}(s_T-s_L)-N\lambda_L^{3/2}\Bigr).
$$

Since $N\le S+1$, $\lambda_L^{3/2}=O(n^{3/4})$ and $\lambda_+^2=O(n)$, all terms other than $u_0^2(s_T-s_L)/8$ total $O(n^{7/4})=o(n^2)$.

*The constructed event implies* $E_B$*.* On the intersection above, $B\ge-X_0/(2\orth)$ on $[0,v_b]$ and $B\ge h_L\ge(1+\varepsilon_2)C\hat a$ on $[v_b,v_L]$ (entry); $B\ge(1+\varepsilon_2)C\hat a$ on $[v_L,V_T]$ (domination); and the terminal threshold holds because

$$
B(V_T)\ \ge\ u_0\sqrt{V_T}\ \ge\ \lambda_+\sqrt{V_T+v_\alpha}\;+\;\varepsilon_3\lambda_+\sqrt{V_T}\ \ge\ (1+\varepsilon_2)C\,a_T\;+\;\frac{(1+\varepsilon_2)m}{\orth}
$$

for $k\ge k_0$: the first summand dominates $(1+\varepsilon_2)C\hat a(V_T)=(1+\varepsilon_2)Ca_T$ by (4.2), and $\varepsilon_3\lambda_+\sqrt{V_T}\ge\varepsilon_3(1+\varepsilon_2)C\,a_T\,(1-o(1))\ge(1+\varepsilon_2)m/\orth$ since $a_T\ge e^{-\delta}\alpha e^{n(1+\varepsilon_1)}\gg m$. Hence the constructed event is contained in $E_B$, and $u_0^2(s_T-s_L)\le(1+1/L)(1+\varepsilon_3)^2\lambda_+^2S$ gives the claim.∎

### 4.5 Assembly

The volatility-tube and conditional barrier-tracking costs add at the quadratic scale; all entry, block-concatenation, localization and bootstrap terms are of lower order.

*Proof of Proposition 4.1.*  By Lemmas 2.1 and 4.3, $P(k)\ge\Prob(A_{\mathrm{vol}}\cap E_B)$, and by Lemma 2.2, (4.1) and Lemma 4.6,

$$
-\ln P(k)\ \le\ \frac{n^2}{2\nu^2T}\Bigl[(1+\varepsilon_1)^2+\nu^2C^2\bigl(1+\tfrac1L\bigr)(1+\varepsilon_3)^2(1+\varepsilon_2)^2e^{4\delta}(1+\varepsilon_1)^2\Bigr](1+o(1)),
$$

where we used

$$
\begin{equation*} \Bigl(1+\frac1L\Bigr)(1+\varepsilon_3)^2\,\frac{\lambda_+^2S}{8} =\nu^2C^2\Bigl(1+\frac1L\Bigr)(1+\varepsilon_3)^2(1+\varepsilon_2)^2e^{4\delta}(1+\varepsilon_1)^2\, \frac{n^2}{2\nu^2 T}\,(1+o(1)). \end{equation*}
$$

Let $k\to\infty$, then $\varepsilon_3,\varepsilon_1,\varepsilon_2,\delta\downarrow0$ and $L\to\infty$, and use $1+\nu^2C^2=(1-\rho^2)^{-1}$.∎

## 5 Nonnegative Correlation

For nonnegative correlation the correlation term no longer creates a growing negative barrier, and the rate is the unconstrained one. The upper bound therefore needs only a terminal Gaussian constraint, while the lower bound must still control the singular drift. Throughout this section $\rho\in[0,1)$.

### 5.1 Upper Bound

Even after the survival constraint is dropped, no terminal strategy improves on the unconstrained quadratic rate.

**Proposition 5.1 (Upper Bound, $\rho\ge0$).**  For $\rho\in[0,1)$,

$$
\liminf_{k\to\infty}\frac{-\ln P(k)}{k^2}\;\ge\;c.
$$

Conditioning on volatility reduces the problem to a one-node Gaussian estimate coupled to the level-hitting cost of Section 3. By Lemmas 2.1 and 3.2, dropping the survival constraint as well, $P(k)\le\Prob(\bar X_T\ge m)$, and conditionally on the volatility path the one-node case of Lemma 3.3 gives

$$
P(k)\le\E\exp\Bigl(-\frac{((m-X_0-\xi_T)^+)^2}{2\orth^2V_T}\Bigr),\qquad\xi_T=\frac\rho\nu(a_T-\alpha)\le\frac\rho\nu a_T.
$$

We reuse the level-hitting skeleton, the operational-time bounds, the time bins and the configuration count of Section 3.5, with *no* running constraints and only the terminal node, and take $r_{\mathrm{cap}}=3$. The high-volatility tail then has probability at most $e^{-2ck^2}$ and the $J^*=0$ cell at most $e^{-c_8m^2}$, exactly as in Section 3.5(a)–(b). The per-configuration estimate is the following terminal-only analogue of Lemma 3.7.

**Lemma 5.2 (Terminal-Only Energy).**  Fix $\eta\in(0,\tfrac12)$, $\kappa\in(0,1)$ and $\varepsilon'\in(0,\tfrac14)$. There are constants $c_6,c_7$, depending only on the model parameters, and $k_0=k_0(\eta,\kappa,\varepsilon')$ such that for all $k\ge k_0$, all $J\ge1$, and all $\Delta t_j>0$ with $\sum_{j\le J}\Delta t_j\le(1+\kappa)T$,

$$
\sum_{j\le J}\frac{\eta^2}{2\nu^2\Delta t_j}\ +\ \frac{\bigl((m-X_0-(\rho/\nu)A_Je^{\eta})^+\bigr)^2}{4e^{2\eta}\orth^2A_J^2T}\ \ \ge\ \ \min\Bigl\{(1-\varepsilon')^2\,c\,k^2\,(1-c_7\kappa),\ \ \frac{e^{2\varepsilon' n}}{c_6T}\Bigr\}.
$$

*Proof.* Set $r:=J\eta/n$. *Case 1:* $(\rho/\nu)A_Je^{\eta}\ge m/2$ (possible only for $\rho>0$). Then

$$
\begin{equation*} J\eta=\ln(A_J/\alpha)\ \ge\ \ln\bigl(\nu m/(2\rho\alpha)\bigr)-\eta\ \ge\ n-c_7', \end{equation*}
$$

and Cauchy–Schwarz gives, for $k\ge k_0$,

$$
\begin{equation*} \sum_j\frac{\eta^2}{2\nu^2\Delta t_j}\ \ge\ \frac{(J\eta)^2}{2\nu^2(1+\kappa)T}\ \ge\ (1-\varepsilon')^2ck^2(1-c_7\kappa), \end{equation*}
$$

the factor $(1-\varepsilon')^2$ absorbing the $O(1/n)$ loss from $c_7'$. *Case 2: otherwise.* Then, for $k\ge k_0$,

$$
\begin{equation*} \bigl(m-X_0-(\rho/\nu)A_Je^{\eta}\bigr)^+\ \ge\ \frac m2-X_0\ \ge\ \frac m4, \end{equation*}
$$

so the terminal term is at least

$$
\begin{equation*} \frac{m^2}{64e^{2\eta}\orth^2A_J^2T}\ \ge\ \frac{e^{2(1-r)n}}{c_6T}. \end{equation*}
$$

If $r\le1-\varepsilon'$ this is at least $e^{2\varepsilon'n}/(c_6T)$; if $r>1-\varepsilon'$, Cauchy–Schwarz as in Case 1 gives

$$
\begin{equation*} \sum_j\frac{\eta^2}{2\nu^2\Delta t_j}\ \ge\ \frac{(rn)^2}{2\nu^2(1+\kappa)T}\ \ge\ (1-\varepsilon')^2ck^2(1-c_7\kappa). \end{equation*}
$$

∎

The upper bound now follows by summing over configurations as in Section 3.5.

*Proof of Proposition 5.1.*  Within each configuration, Lemma 3.4(i) gives $a<A_{J^*}e^{\eta}$ on $[0,T]$ and hence

$$
\begin{equation*} \xi_T\le\frac\rho\nu a_T\le\frac\rho\nu A_{J^*}e^{\eta},\qquad V_T\le\bar V=2e^{2\eta}A_{J^*}^2T, \end{equation*}
$$

so the product of the level-hitting cost (Lemma 3.5 at the thresholds $\bar t_j$) and the conditional Gaussian factor is bounded by Lemma 5.2 applied with $\Delta t_j:=\bar t_j$ and $J:=J^*$; the second branch of the minimum is superquadratic in $k$. Summing over the $e^{o(n^2)}$ configurations, letting $k\to\infty$ and then $\varepsilon',\eta,\kappa\downarrow0$ proves the proposition.∎

### 5.2 Lower Bound

For the matching lower bound it suffices that survival and the terminal adjustment cost only $O(n)$ beyond the volatility excursion.

**Proposition 5.3 (Lower Bound, $\rho\ge0$).**  For $\rho\in[0,1)$,

$$
\limsup_{k\to\infty}\frac{-\ln P(k)}{k^2}\;\le\;c.
$$

We confine volatility near an exponential path whose endpoint is a fixed multiple of the target level. The forward is then raised, first by scale-invariant doublings and then by $O(\ln n)$ dyadic level-raising steps; both the orthogonal Brownian cost and the accumulated drift loss remain subquadratic.

*Parameters.* Fix $\delta\in(0,\tfrac14)$ with $\rho\alpha\delta/\nu\le X_0/4$. Let $\kappa_0:=1/(16b)$ and choose a constant $c_9$, depending only on the parameters, such that with

$$
\theta:=\frac{n+c_9}{\nu T},\qquad a^*(t)=\alpha e^{\nu\theta t},\qquad\gamma:=\frac{a^*(T)}{m}=\alpha(1-\beta)e^{c_9}f_0^{\beta-1},
$$

one has $\gamma^2\ge16(\kappa_0+1)e^{2\delta}\max(1,1/T)$. Unlike in Section 4, the terminal volatility is only a fixed multiple of $m$. This keeps $V_T\asymp\gamma^2m^2T/(2n)$ on the tube, so that the singular drift accumulated during the level-raising and holding phases below is only $O(m\ln n/n)$; with an exponential overshoot $a^*(T)=e^{\varepsilon_1n}m$, the drift $b\int a^2/X$ accumulated at height $2m$ would exceed that level.

*Volatility tube.* By Lemma 4.2 with the present $\theta$,

$$
\begin{equation}\label{eq:pos-tube}\tag{5.1} \Prob(A_{\mathrm{vol}})\ge\exp\bigl(-n^2/(2\nu^2T)-O(n)\bigr).\tag{5.1} \end{equation}
$$

On the tube $a_t\ge\alpha e^{-\delta}$; hence, as $\rho\ge0$,

$$
\begin{equation}\label{eq:pos-xi}\tag{5.2} \xi_t=\tfrac\rho\nu(a_t-\alpha)\ge-\tfrac{\rho\alpha}{\nu}(1-e^{-\delta})\ge-\tfrac{\rho\alpha\delta}{\nu}\ge-\tfrac{X_0}{4}.\tag{5.2} \end{equation}
$$

*Comparison process.* Let $Y^-$ denote a solution of

$$
\begin{equation}\label{eq:Yminus}\tag{5.3} dY^-_v=\orth\,dB_v-\frac{b}{Y^-_v}\,dv,\tag{5.3} \end{equation}
$$

absorbed at its first zero. The equation is invariant under the Brownian scaling

$$
\begin{equation*} (Y^-_v)_{v\ge0}\mapsto(y\,Y^-_{v/y^2})_{v\ge0},\qquad y>0, \end{equation*}
$$

and two solutions driven by the same $B$ with ordered initial values remain ordered up to the absorption of the lower one (the argument of Lemma 5.4 below with $\hat\xi\equiv0$). The following lemma reduces survival of $X$ to survival of $Y^-$ on the operational-time horizon.

**Lemma 5.4 (Comparison on the Operational-Time Horizon).**  Fix a volatility path in the tube, and put $Y_v:=X_{t(v)}$ and $\hat\xi_v:=\xi_{t(v)}$, so that

$$
Y_v=X_0+\hat\xi_v+\orth B_v-b\int_0^v\frac{du}{Y_u},\qquad0\le v\le V_T,
$$

up to the first zero of $Y$. Let $Y^-$ solve (5.3) on $[0,V_T]$ with $Y^-_0=X_0/2$ and the same $B$, and let $\zeta^-:=\inf\{v\in[0,V_T]:Y^-_v=0\}$. Then $Y_v\ge Y^-_v$ for $0\le v\le V_T\wedge\zeta^-$. In particular, if $\zeta^-=\infty$ then $\zeta>T$ and $X_T=Y_{V_T}\ge Y^-_{V_T}$.

*Proof.* The path $\hat\xi$ is continuous but need not have finite variation; only the bound $\hat\xi\ge-X_0/4$ from (5.2) is used. For $v<\zeta^-$ and before any first crossing of $D_v:=Y_v-Y^-_v$,

$$
D_v=\frac{X_0}{2}+\hat\xi_v+b\int_0^v\Bigl(\frac1{Y^-_u}-\frac1{Y_u}\Bigr)du\ \ge\ \frac{X_0}{4}>0.
$$

By continuity no crossing occurs, so $Y\ge Y^-+X_0/4>0$ on $[0,V_T\wedge\zeta^-)$, and the claims follow.∎

Starting from $X_0/2$, the comparison process is first raised geometrically. By scale invariance, the success probability of each doubling is bounded below independently of the level.

**Lemma 5.5 (Doubling).**  Let $Y^-$ solve (5.3) with $Y^-_0=y>0$. Then

$$
\Prob\Bigl(\inf_{v\le\kappa_0y^2}Y^-_v\ge\tfrac y2,\ \ Y^-_{\kappa_0y^2}\ge2y\Bigr)\ \ge\ q_1 :=\Prob\Bigl(\inf_{u\le\kappa_0}B_u\ge-\tfrac{1}{4\orth},\ B_{\kappa_0}\ge\tfrac{3}{\orth}\Bigr)>0,
$$

and $q_1$ does not depend on $y$.

*Proof.* By Brownian scaling, the event $\{\inf_{u\le\kappa_0y^2}B_u\ge-\tfrac{y}{4\orth},\ B_{\kappa_0y^2}\ge\tfrac{3y}{\orth}\}$ has probability $q_1$ for every $y$. On it, as long as $Y^-\ge y/2$ the drift integral is at most $2b\kappa_0y^2/y=y/8$, so $Y^-\ge y-\tfrac y4-\tfrac y8=\tfrac{5y}8>\tfrac y2$ throughout $[0,\kappa_0y^2]$; hence there is no first passage below $y/2$, and $Y^-_{\kappa_0y^2}\ge y+3y-\tfrac y8\ge2y$.∎

Doublings at the natural time scale $\kappa_0y^2$ can continue only up to the level $y_{\mathrm{crit}}:=m/\sqrt n$, because the operational-time budget $V_T$ is only of order $m^2/n$. Above that level we force the remaining steps by a Cameron–Martin shift, each certified by the following Gaussian estimate.

**Lemma 5.6 (Brownian One-Step Lower Bound).**  Let $x,g,t>0$ with $x/\sqrt t\ge2$. Then $\Prob\bigl(\inf_{u\le t}B_u\ge-x,\ B_t\ge g\bigr)\ \ge\ \tfrac13\exp\bigl(-g^2/(2t)-2g/\sqrt t\bigr)$.

*Proof.* Tilt by the drift $\mu:=g/t$. The localized Cameron–Martin inequality with $K:=2g/\sqrt t$ gives $\Prob(A)\ge e^{-g^2/(2t)-K}\bigl(\widetilde\Prob(A)-\widetilde\Prob(\mu\widetilde B_t>K)\bigr)$. Under the tilt $A\supseteq\{\inf_{u\le t}\widetilde B_u\ge-x,\ \widetilde B_t\ge0\}$, which has probability at least $\tfrac12-2\Phi(-x/\sqrt t)\ge\tfrac12-2\Phi(-2)$ by the reflection principle; and $\mu\widetilde B_t\sim N(0,g^2/t)$, so the subtracted term equals $\Phi(-2)$. Since $\tfrac12-3\Phi(-2)>\tfrac13$, the claim follows.∎

For the forced steps let $N_B:=\lceil\log_2(2m/y_{\mathrm{crit}})\rceil=O(\ln n)$, and define the *nominal* levels $\hat y_i:=2^iy_{\mathrm{crit}}$ for $0\le i<N_B$ and $\hat y_{N_B}:=2m$, so that the last ratio lies in $(1,2]$. With $V_M:=V_T/4$, allocate the operational-time increments $\Delta v_i:=\hat y_iV_M/(4m)$ to the steps $i=0,\dots,N_B-1$; since $\sum_{i<N_B}\hat y_i\le4m$, they total at most $V_M$. Over these steps the process climbs from $y_{\mathrm{crit}}$ to $2m$ at an exponential cost of order $n$ only.

**Lemma 5.7 (Level Raising).**  There is $k_0=k_0(\delta)$ such that the following holds for $k\ge k_0$, conditionally on a volatility path in the tube. Let $Y^-$ solve (5.3) with $Y^-_0\ge y_{\mathrm{crit}}$, and let $v_+:=\sum_{i<N_B}\Delta v_i\le V_M$. With probability at least $e^{-c_{10}n}$, $Y^-$ stays positive on $[0,v_+]$ and $Y^-_{v_+}\ge2m$; along the way its total drift loss is $O(m\ln n/n)$.

*Proof.* Restart the comparison at the nominal levels: a solution entering step $i$ at a value at least $\hat y_i$ dominates the solution started at $\hat y_i$, so it suffices that the latter reach $\hat y_{i+1}$ within $\Delta v_i$. Apply Lemma 5.6 with $x=\hat y_i/(4\orth)$, $g=3\hat y_i/(2\orth)$ and $t=\Delta v_i$. On the tube $V_M\le e^{2\delta}\gamma^2m^2T/(8n)$, hence, uniformly in $i$,

$$
\frac{x}{\sqrt{\Delta v_i}}\ =\ \frac{\sqrt{\hat y_i\,m}}{2\orth\sqrt{V_M}}\ \ge\ \frac{\sqrt{y_{\mathrm{crit}}\,m}}{2\orth\sqrt{V_M}}\ \ge\ \frac{\sqrt2\,e^{-\delta}}{\orth\gamma\sqrt T}\;n^{1/4}\ \ge\ 2
$$

for $k\ge k_0$, so each step certificate has probability at least

$$
\begin{equation*} \frac13\exp\Bigl(-\frac{(3\hat y_i/(2\orth))^2}{2\Delta v_i}-\frac{3\hat y_i}{\orth\sqrt{\Delta v_i}}\Bigr). \end{equation*}
$$

On the certificate the bootstrap of Lemma 5.5 applies: the drift loss during the step is at most

$$
\begin{equation*} \frac{2b\,\Delta v_i}{\hat y_i}=\frac{bV_M}{2m}=O\Bigl(\frac mn\Bigr)\le\frac{\hat y_i}{8}\qquad\text{for $k$ large,} \end{equation*}
$$

since $\hat y_i\ge m/\sqrt n$. Hence $Y^-\ge\hat y_i/2$ throughout the step, and at its end

$$
\begin{equation*} Y^-\ \ge\ \hat y_i+\tfrac32\hat y_i-\tfrac18\hat y_i\ \ge\ 2\hat y_i\ \ge\ \hat y_{i+1}. \end{equation*}
$$

The certificates concern Brownian increments over disjoint intervals, so their probabilities multiply. The costs are

$$
\begin{equation*} \begin{aligned} &\text{Gaussian:}&&\sum_i\frac{(3\hat y_i/(2\orth))^2}{2\Delta v_i}=\frac{9m}{2\orth^2V_M}\sum_i\hat y_i\le\frac{18m^2}{\orth^2V_M}=O(n), \\
&\text{localization:}&&\sum_i\frac{3\hat y_i}{\orth\sqrt{\Delta v_i}}=O\Bigl(\frac{m}{\sqrt{V_M}}\Bigr)=O(\sqrt n), \\
&\text{per-step constants:}&&N_B\ln3=O(\ln n), \end{aligned} \end{equation*}
$$

and the total drift loss is at most $N_B\,bV_M/(2m)=O(m\ln n/n)$.∎

Once the level $2m$ is reached, keeping the process above $m$ until the horizon is nearly free.

**Lemma 5.8 (Holding).**  Fix a volatility path in the tube and a $\sigma(a)$-measurable operational time $v_u\le V_T/2$, and suppose $Y^-_{v_u}\ge2m$. Let

$$
A_{\mathrm{hold}}:=\Bigl\{\inf_{0\le s\le V_T-v_u}\bigl(B(v_u+s)-B(v_u)\bigr)\ge-\frac{m}{2\orth}\Bigr\}.
$$

Then $\Prob(A_{\mathrm{hold}}\mid\sigma(a),\mathcal F^B_{v_u})=2\Phi\bigl(m/(2\orth\sqrt{V_T-v_u})\bigr)-1$, which tends to one uniformly over the tube and over $v_u$; and on $A_{\mathrm{hold}}$, for $k$ large, $Y^-$ stays positive on $[v_u,V_T]$ and $Y^-_{V_T}\ge m$.

*Proof.* The formula is the reflection principle, and it is at least $2\Phi(m/(2\orth\sqrt{V_T}))-1$. On the tube,

$$
V_T\le e^{2\delta}\int_0^T(a^*(t))^2\,dt =e^{2\delta}\frac{\gamma^2m^2-\alpha^2}{2\nu\theta}, \qquad\frac{m}{2\orth\sqrt{V_T}} \ge\frac{e^{-\delta}}{2\orth\gamma}\sqrt{\frac{2n}{T}}\,(1+o(1))\longrightarrow\infty.
$$

For the drift, bootstrap once more: let $\tau^*:=\inf\{v\ge v_u:\ Y^-_v\le\tfrac{5m}4\}$. On $[v_u,\tau^*]$ the drift integral is at most $bV_T/(5m/4)=O(m/n)$, while on $A_{\mathrm{hold}}$ the decrease of $\orth B$ is at most $m/2$; hence $Y^-\ge2m-\tfrac m2-O(m/n)>\tfrac{5m}4$ up to $\tau^*$, so $\tau^*>V_T$, and the same bounds give $Y^-_{V_T}\ge\tfrac{3m}2-o(m)\ge m$.∎

The three phases, doubling, level raising and holding, combine to give the lower bound.

*Proof of Proposition 5.3.*  Fix a volatility path in the tube. By Lemmas 2.1 and 5.4, it suffices to show that, with conditional probability at least $e^{-O(n)}$, the solution $Y^-$ of (5.3) with $Y^-_0=X_0/2$ stays positive on $[0,V_T]$ and satisfies $Y^-_{V_T}\ge m$. We combine three phases by the Markov property at their $\sigma(a)$-measurable endpoints.

*Doubling.* Apply Lemma 5.5 at the dyadic levels from $X_0/2$ up to $y_{\mathrm{crit}}$: there are $\log_2(2y_{\mathrm{crit}}/X_0)=n/\ln2+O(\ln n)$ stages, each starting no lower than its nominal level by the ordering of solutions of (5.3). They succeed together with probability at least $q_1^{\,O(n)}=e^{-O(n)}$ and use operational time at most $2\kappa_0y_{\mathrm{crit}}^2=2\kappa_0m^2/n\le V_T/4$, by the choice of $\gamma$.

*Level raising.* Lemma 5.7 raises $Y^-$ from at least $y_{\mathrm{crit}}$ to at least $2m$ within further operational time $V_M=V_T/4$, with probability at least $e^{-c_{10}n}$.

*Holding.* The level-raising phase ends at an operational time $v_u\le V_T/2$; by Lemma 5.8 the conditional probability that $Y^-$ then stays positive and ends above $m$ tends to one.

Multiplying the three conditional bounds and integrating over the tube with (5.1) gives

$$
-\ln P(k)\ \le\ \frac{n^2}{2\nu^2T}+O(n),
$$

which proves the proposition.∎

### 5.3 Proof of Theorem 1.1

The four propositions combine to give the main theorem and its corollaries.

*Proof of Theorem 1.1.*  For $\rho<0$ the theorem is Propositions 3.1 and 4.1; for $\rho\ge0$ it is Propositions 5.1 and 5.3. The lower-bound constructions show $P(k)>0$ for all large $k$, so Lemma 2.3 applies and yields Corollaries 1.2 and 1.3. At $\rho=0$ this recovers [BFL08].∎

## 6 The Intrinsic Distance in the Survival Domain

We now prove the geometric identification (1.4). Henry-Labordère’s derivation of (10.22) in [Hen08] represents the SABR generator by a Riemannian metric, a connection and a scalar potential on a patch of the Poincaré half-plane, and then uses a Gaussian estimate based on the unrestricted geodesic distance. For the absorbed model the correct leading quantity is instead the intrinsic distance within the open survival domain, the state space of the process killed at the boundary. The calculation below is elementary and uses no heat-kernel theorem.

### 6.1 Fermi Coordinates and the Survival Domain

The hyperbolic geometry of SABR is most transparent in Fermi coordinates around a geodesic, in which the survival domain becomes the coordinate half-plane $\{R>\Rb\}$, bounded by a curve at constant distance from that geodesic. Recall $X=F^{1-\beta}/(1-\beta)$, and set

$$
y:=\frac a\nu,\qquad x:=\frac{X-\rho y}{\orth},\qquad R:=\operatorname{arsinh}(x/y),\qquad q:=\tfrac12\ln(x^2+y^2), \qquad\Rb:=-\operatorname{artanh}\rho.
$$

Before absorption, $x$ and $y$ are driven by the independent Brownian motions $Z^\perp$ and $Z$ with the common volatility $a=\nu y$, so the principal symbol of the generator is that of the hyperbolic metric $(dx^2+dy^2)/(\nu y)^2$. In the coordinates $(R,q)$,

$$
x=e^q\tanh R,\qquad y=e^q\operatorname{sech}R, \qquad X=e^q\Psi(R),\qquad\Psi(R):=\orth\tanh R+\rho\operatorname{sech}R,
$$

and the open survival domain $\{X>0\}$ and the metric are

$$
\begin{equation}\label{eq:killed-fermi-geometry}\tag{6.1} D=\{R>\Rb\}, \qquad g=\nu^{-2}\bigl(dR^2+\cosh^2R\,dq^2\bigr).\tag{6.1} \end{equation}
$$

Thus $R/\nu$ is the signed distance from the geodesic $\{x=0\}$, and motion in $q$ at fixed $R$ costs $\cosh R/\nu$ per unit of $q$.

Distances in $D$ are measured along paths that stay in $D$. For $z,z'\in D$, let $\mathcal A_D(z,z')$ be the class of finite chains $\gamma=(\gamma_1,\ldots,\gamma_N)$, $N\ge1$, of $C^1$ legs $\gamma_j=(R_j,q_j):[0,1]\to D$ with $\gamma_1(0)=z$, $\gamma_N(1)=z'$ and $\gamma_j(1)=\gamma_{j+1}(0)$. Every leg respects the domain constraint, while minimizing sequences may approach the boundary. For $\gamma\in\mathcal A_D(z,z')$ set

$$
L_g(\gamma):=\frac1\nu\sum_{j=1}^N\int_0^1 \sqrt{\dot R_j(s)^2+\cosh^2(R_j(s))\,\dot q_j(s)^2}\,ds, \qquad d_D(z,z'):=\inf_{\gamma\in\mathcal A_D(z,z')}L_g(\gamma),
$$

and $d_D(z,A):=\inf_{z'\in A}d_D(z,z')$ for $A\subset D$. Let $z_0=(R_0,q_0)$ be the initial point, so that $X_0=e^{q_0}\Psi(R_0)$ and $m=X_0e^n$, and let

$$
\mathcal T_k:=\{(R,q)\in D:e^q\Psi(R)\ge m\}
$$

be the terminal target set.

### 6.2 The Distance to the Target Set

Reaching $\mathcal T_k$ requires a displacement of about $n$ in $q$, and such motion is cheapest where $\cosh R$ is smallest. The distance is therefore governed by the infimum of $\cosh R$ over $D$.

**Proposition 6.1 (Intrinsic Distance to the Target Set).**  Let $s_\rho:=\sqrt{1-(\rho\wedge0)^2}$, that is, $s_\rho=1$ for $\rho\ge0$ and $s_\rho=\orth$ for $\rho<0$. Then, as $k\to\infty$,

$$
\begin{equation}\label{eq:killed-distance}\tag{6.2} d_D(z_0,\mathcal T_k) =\frac{n}{\nu s_\rho}+O(1) =\frac{(1-\beta)k}{\nu s_\rho}+O(1).\tag{6.2} \end{equation}
$$

For $\rho\le0$ the bound is approached by curves whose long tangential segment tends to the absorbing boundary; for $\rho>0$ the minimizing tangential line $R=0$ lies in the interior.

*Proof.* Since $\Psi(R)=(\orth\sinh R+\rho)/\cosh R$, one has $\Psi>0$ precisely when $R>\Rb$. As $(\orth,\rho)$ and $(\tanh R,\operatorname{sech}R)$ are unit vectors, $0<\Psi\le1$ on $D$. Hence every $(R,q)\in\mathcal T_k$ satisfies

$$
q-q_0\ge n+\ln\Psi(R_0)-\ln\Psi(R)\ge n+\ln\Psi(R_0).
$$

Moreover $\inf_{R>\Rb}\cosh R=1/s_\rho$: if $\rho\ge0$, then $\Rb\le0$ and the infimum is $\cosh0=1$ (approached from within $D$ when $\rho=0$); if $\rho<0$, then $\Rb>0$ and the infimum is $\cosh\Rb=1/\orth$. Therefore every admissible chain from $z_0$ to $\mathcal T_k$ satisfies, for all sufficiently large $k$,

$$
L_g(\gamma)\ge\frac1{\nu s_\rho}\sum_{j=1}^N\int_0^1|\dot q_j(s)|\,ds\ge\frac{n+\ln\Psi(R_0)}{\nu s_\rho}.
$$

For the reverse bound, first let $\rho>0$. The three-segment path

$$
(R_0,q_0)\longrightarrow(0,q_0)\longrightarrow(0,q_0+n)\longrightarrow(R_0,q_0+n)
$$

stays in $D$, ends in $\mathcal T_k$, and has length $n/\nu+2|R_0|/\nu$. If $\rho\le0$, set $R_k:=\Rb+n^{-1}$. For all sufficiently large $k$ the analogous path through $R_k$ stays in $D$, has the same endpoint $(R_0,q_0+n)\in\mathcal T_k$, and has length

$$
\frac{2(R_0-R_k)}{\nu}+\frac{n\cosh R_k}{\nu} =\frac{n\cosh\Rb}{\nu}+O(1) =\frac{n}{\nu s_\rho}+O(1).
$$

Returning to the fixed level $R_0$ makes the terminal constraint exact and avoids the displacement $-\ln\Psi(R_k)$ that would arise if the endpoint stayed in the boundary layer.∎

![Figure 3](https://kspectra.ai/papers/sabr-right-wing-absorbing-boundary/figs/fig_geometry.svg)

*Figure 3. Fermi coordinates for the reference parameters of Section 7 with $\beta=0.5$ and $n=3$. The survival domain is $D=\{R>R_\partial\}$; the shaded strip lies outside $D$, and $\mathcal T_k$ is the target set. Motion in $q$ costs $\cosh R/\nu$ per unit, least on the line $R=0$. (a) For $\rho>0$ the line $R=0$ lies in $D$, and the paths of the proof of Proposition 6.1 have length $n/\nu+O(1)$, as in the unrestricted metric. (b) For $\rho<0$ the line $R=0$ lies outside $D$; admissible paths can only approach the boundary, where the cost is $\cosh R_\partial/\nu=1/(\nu\bar\rho)$, so the distance grows like $n/(\nu\bar\rho)$.*

Combined with Theorem 1.1, the proposition identifies the tail rate with the intrinsic distance. This is the corrected form of Conjecture 2 for the absorbed model.

**Corollary 6.2 (Corrected Conjecture 2 for Absorbed SABR).**  For every $\rho\in(-1,1)$,

$$
\begin{equation}\label{eq:corrected-c2}\tag{6.3} -\ln P(k)=\frac{d_D(z_0,\mathcal T_k)^2}{2T}+o(k^2).\tag{6.3} \end{equation}
$$

*Proof.* By Proposition 6.1, $d_D(z_0,\mathcal T_k)^2/(2T)=(1-\beta)^2k^2/(2\nu^2Ts_\rho^2)+o(k^2)$, which equals $-\ln P(k)+o(k^2)$ by Theorem 1.1.∎

### 6.3 Unrestricted versus Intrinsic Distance

The comparison with the unrestricted distance shows exactly where the Gaussian estimate behind Conjecture 2 breaks down. Let $d_{\mathrm{free}}$ denote the distance to the same set $\mathcal T_k$ when connecting curves may range over the whole Fermi plane. The lower-bound and three-segment arguments above give $d_{\mathrm{free}}(z_0,\mathcal T_k)=n/\nu+O(1)$ for every $\rho$: the unrestricted plane always permits the long segment on $R=0$. If $\rho<0$, however, $\Rb>0$ and the line $R=0$ lies outside $D$; admissible curves can only approach $R=\Rb$, where the tangential cost is $\cosh\Rb/\nu=1/(\nu\orth)$ (Figure 3). If $\rho>0$ the line $R=0$ is interior, and if $\rho=0$ it is the boundary but can be approached without changing the leading coefficient. Thus the intrinsic and unrestricted distances agree to leading order exactly when the excluded region is irrelevant on the $k^2$ scale. Paulot [Pau15] pointed out the corresponding boundary restriction on small-time geodesic expansions; Proposition 6.1 is its fixed-maturity, extreme-strike analogue.

### 6.4 The Potential Is Unbounded on the Survival Domain

The endpoint value in Conjecture 2 does not control the lower-order part of the operator. In the Lamperti–CEV coordinate the full scalar potential is

$$
Q(F,a)=-\Gamma_{\beta,\rho}\left(\frac aX\right)^2, \qquad\Gamma_{\beta,\rho} =\frac{\beta(1-\beta)+\beta^2/(2\orth^2)}{4(1-\beta)^2}>0,
$$

which has an inverse-square singularity at the absorbing boundary. For the unrestricted saddle on the fibre $\{F=f\}$, with $X=f^{1-\beta}/(1-\beta)$ and $X_0=f_0^{1-\beta}/(1-\beta)$, the exact formula [Hen08, (6.45)] is

$$
a_{\min}(f)^2=\alpha^2+2\rho\alpha\nu(X-X_0)+\nu^2(X-X_0)^2.
$$

Thus $a_{\min}(f)/X\to\nu$ as $f\to\infty$, which gives (1.2) and shows that the potential remains bounded along the far-right saddle, although its finite-strike value is generally not constant. Moreover, in Henry-Labordère’s convention the connection one-form is

$$
A=-\frac{b}{\orth^2X}\,dX+\frac{\rho b}{\nu\orth^2X}\,da, \qquad dA=-\frac{\rho b}{\nu\orth^2X^2}\,dX\wedge da,
$$

which is singular and, for $\rho\ne0$, not globally gauge-removable. These terms do not affect Corollary 6.2, but the formal calculation of Section 8.1 indicates that they contribute at order $k^{3/2}$. They are also the obstacle to a heat-kernel proof of Corollary 6.2.

**Remark 6.3 (A Heat-Kernel Route).**  A Dirichlet heat-kernel theorem would give a second route to (6.3), closer to Henry-Labordère’s method. Let $K_D$ be the Dirichlet heat kernel of the generator on $D$—the transition density of the killed process—with respect to a reference measure $\mu$, so that $P(k)=\int_{\mathcal T_k}K_D(T;z_0,z)\,\mu(dz)$. It would suffice to prove

$$
\ln K_D(T;z_0,z)=-\frac{d_D(z_0,z)^2}{2T}+o(k^2)
$$

uniformly on a subset $\mathcal N_k\subset\mathcal T_k$ with

$$
\begin{equation*} \sup_{z\in\mathcal N_k}d_D(z_0,z)^2=d_D(z_0,\mathcal T_k)^2+o(k^2),\qquad-\ln\mu(\mathcal N_k)=o(k^2), \end{equation*}
$$

together with an exponentially matching bound for the integral over $\mathcal T_k\setminus\mathcal N_k$, for instance a global upper estimate with subquadratic logarithmic volume factors. Gaussian bounds of Li–Yau type [LY86; Dav89; Stu92] do not apply directly, because the potential and the connection are singular at $\partial D$; whether a suitable singular-potential framework applies is open. The proofs in this paper establish both tail inequalities directly.

## 7 Numerical Illustrations

The theorems concern leading $k^2$ rates, which finite-strike computations can illustrate but not establish. We report Monte Carlo estimates of $P(k)$ at moderate strikes, which display the correlation asymmetry, and solutions of the deterministic variational problems behind the two rates up to $k=160$. Methods are summarized in Appendix B. All computations use $f_0=1$, $\alpha=0.3$, $\nu=0.6$, $T=1$, $\beta\in\{0.5,0.7\}$ and $\rho\in\{0,\pm0.6\}$, for which

$$
\beta=0.5:\quad c=0.3472,\quad c_-=0.5425;\qquad\beta=0.7:\quad c=0.1250,\quad c_-=0.1953.
$$

Table 2. Monte Carlo estimates of $-\ln P(k)$ with delta-method standard errors. $^\dagger$: effective sample size below $100$; these errors should be read with caution.

|  | $\beta=0.5$ | $\beta=0.7$ |  |  |  |
|---|---|---|---|---|---|
| $k$ | $\rho=+0.6$ | $\rho=0$ | $\rho=-0.6$ | $\rho=0$ | $\rho=-0.6$ |
| 2 | $10.72\pm0.01$ | $13.67\pm0.01$ | $22.26\pm0.02$ | — | — |
| 3 | $16.13\pm0.01$ | $20.37\pm0.01$ | $33.47\pm0.02$ | — | — |
| 4 | $21.90\pm0.01$ | $27.34\pm0.01$ | $45.81\pm0.05$ | $22.21\pm0.02$ | $36.62\pm0.04$ |
| 5 | $28.14\pm0.01$ | $34.76\pm0.01$ | $59.50\pm0.11^\dagger$ | — | — |
| 6 | $34.84\pm0.02$ | $42.79\pm0.01$ | $74.52\pm0.24^\dagger$ | $32.07\pm0.01$ | $54.43\pm0.05$ |
| 8 | — | $60.77\pm0.02$ | — | $42.43\pm0.02$ | $74.44\pm0.20^\dagger$ |
| 10 | — | $81.50\pm0.02$ | — | $53.51\pm0.02$ | — |
| 12 | — | $105.02\pm0.02$ | — | $65.63\pm0.03$ | — |

At these strikes the asymptotic regime is still distant: at $\beta=0.5$, $\rho=0$ the implied volatility is only about $0.87$ at $k=12$, against the limit $1.2$. Table 2 therefore cannot be read as a rate. The correlation asymmetry is nevertheless visible in the spreads

$$
\Delta_\rho(k):=\ln P_0(k)-\ln P_\rho(k),
$$

for which Theorem 1.1 gives $\Delta_{+0.6}(k)=o(k^2)$ and $\Delta_{-0.6}(k)=(c_--c)k^2+o(k^2)$. In Figure 4(a), $\Delta_{+0.6}$ is close to linear in $k$ (its second differences at unit spacing are within $0.15$ of zero), while $\Delta_{-0.6}$ is convex, with second differences between $0.7$ and $0.9$. These exceed the value $2(c_--c)=0.39$ contributed by the $k^2$ term alone, consistent with the positive $k^{3/2}$ correction of Conjecture 8.1; the window is too short to separate the two, and we draw no quantitative conclusion from it.

![Figure 4](https://kspectra.ai/papers/sabr-right-wing-absorbing-boundary/figs/fig_numerics.svg)

*Figure 4. (a) Monte Carlo spreads $\Delta_\rho(k)=\ln P_0(k)-\ln P_\rho(k)$ at $\beta=0.5$ from Table 2, with $\pm1$ standard-error bars (marginal errors combined in quadrature; mostly smaller than the markers). The spread for $\rho=+0.6$ is close to linear, consistent with equal $k^2$ coefficients for $\rho=+0.6$ and $\rho=0$ (Theorem 1.1 gives $\Delta_{+0.6}=o(k^2)$); the spread for $\rho=-0.6$ is convex. (b) Eliminated coefficients $\hat c(k)$ of Table 3, divided by their predictions: $c_-$ for the survival problems ($\rho=-0.6$) and $c$ for the terminal problems ($\rho\in\{0,\pm0.6\}$).*

**Variational Rates.** The logarithmic rates are governed by deterministic control problems: minimize the Cameron–Martin energy of the volatility driver plus the Gaussian energy of the orthogonal component, subject to the terminal constraint (the *terminal* problem) or to the terminal and survival constraints (the *survival* problem); see Appendix B. Their values $R(k)$, normalized by $k^2$, are predicted to approach $c$, and $c_-$ for the survival problem with $\rho<0$. We solve them for $k=20,40,80,160$ and remove a term linear in $k$ by the elimination

$$
\hat c(k):=\frac1k\Bigl(\frac{R(2k)}{2k}-\frac{R(k)}{k}\Bigr).
$$

At the pair $(80,160)$ every estimate lies within $1.5\%$ of its prediction (Table 3 and Figure 4(b)). The individual coefficients depend on $\beta$ through $(1-\beta)^2$; the $\beta$-independent prediction is the ratio $c_-/c=(1-\rho^2)^{-1}=1.5625$, against which the survival-to-terminal ratios at $(80,160)$ are $1.55$ for $\beta=0.5$ and $1.57$ for $\beta=0.7$.

Table 3. Eliminated quadratic coefficients $\hat c(k)$ of the variational problems, from the pairs $(k,2k)$ with $k=20,40,80$.

| problem | $\beta$ | $\rho$ | $\hat c(20)$ | $\hat c(40)$ | $\hat c(80)$ | prediction |
|---|---|---|---|---|---|---|
| survival | 0.5 | $-0.6$ | 0.5381 | 0.5498 | 0.5458 | $c_-=0.5425$ |
| terminal | 0.5 | $-0.6$ | 0.3347 | 0.3488 | 0.3522 | $c=0.3472$ |
| terminal | 0.5 | $0$ | 0.3412 | 0.3470 | 0.3474 | $c=0.3472$ |
| terminal | 0.5 | $+0.6$ | 0.3430 | 0.3463 | 0.3470 | $c=0.3472$ |
| survival | 0.7 | $-0.6$ | 0.1825 | 0.1940 | 0.1981 | $c_-=0.1953$ |
| terminal | 0.7 | $-0.6$ | 0.1047 | 0.1210 | 0.1260 | $c=0.1250$ |
| terminal | 0.7 | $0$ | 0.1139 | 0.1227 | 0.1248 | $c=0.1250$ |

**Checks of Analytic Ingredients.** We also tested several formulas and implementations numerically: the Gaussian convex-set bound of Lemma 3.3 by $10^7$-path Monte Carlo on a three-node instance; the functional of Lemma 3.7 in the limit $\kappa\downarrow0$ by exhaustive minimization over $J$ at $\eta=0.1$, which gives normalized minima $0.927$, $0.917$ and $0.926$ at $k=20,40,80$, consistent with the slack factor $(1-\varepsilon')^2(1-\varepsilon_0)$ of that lemma, and starts with early Gaussian displacement converge to the same minima; the block probabilities of Lemma 4.5 at $\lambda=5,20,80$; the doubling and level-raising certificates of Section 5.2 on a time grid; the expansion (2.3) up to $k=200$; and the Ornstein–Uhlenbeck eigenvalue computations of Section 8.1, including the exact value $\Theta(1)=1$. These checks are not part of the proofs.

## 8 Discussion

We close with two complementary perspectives on the main theorem: what lies beyond the leading order when $\rho<0$, and how the absorbed model sits between the neighbouring regimes $\beta=1$ and $\beta=0$.

### 8.1 Beyond the Leading Order

Theorem 1.1 settles only the leading $k^2$ scale. For $\rho<0$ four statements of different status must be kept apart: the square-root-boundary eigenvalue expansion below is classical for the drift-free comparison process; the Fourier–Laplace reduction of the full generator with Dirichlet boundary condition is exact; the ensuing boundary-layer localization and Legendre inversion are formal; and the resulting $k^{3/2}$ correction is conjectural.

Along the leading exponential volatility excursion, dropping the singular drift from the survival constraint gives the moving-barrier comparison problem $B(v)\ge\ell\sqrt v$. Under $v=e^s$, $U(s)=e^{-s/2}B(e^s)$ is a stationary Ornstein–Uhlenbeck process constrained to stay above $\ell$, with survival exponent given by the principal Dirichlet eigenvalue $\Theta(\ell)$ on $(\ell,\infty)$; this is the square-root-boundary theory of Breiman [Bre67] and Uchiyama [Uch80]. Its classical large-$\ell$ expansion is

$$
\begin{equation}\label{eq:theta-exp}\tag{8.1} \Theta(\ell)=\frac{\ell^2}{8}+\frac{2^{1/3}|a_1|}{4}\,\ell^{2/3}-\frac14+o(1), \qquad|a_1|=2.33811\ldots,\tag{8.1} \end{equation}
$$

where $a_1$ is the first zero of the Airy function. The leading term reproduces $c_-$. The Airy term is valid for the comparison problem, but it cannot be transferred to the full SABR diffusion without restoring the singular drift.

The exact lower-order obstruction appears after a transform in the tangential variable. In the coordinates $(R,q)$ of Section 6, with

$$
\chi(R)=\frac{1}{\orth\cosh R\,(\orth\sinh R+\rho)},
$$

the backward generator on $D$, with Dirichlet boundary condition at $R=\Rb$, is

$$
\mathcal L =\frac{\nu^2}{2}\left(\partial_R^2+\tanh R\,\partial_R+\operatorname{sech}^2R\,\partial_q^2\right) -b\nu^2\chi(R)\left(\partial_R+\tanh R\,\partial_q\right).
$$

All coefficients and the boundary are invariant under translation in $q$. A real Fourier mode yields a non-self-adjoint fibre; positive $q$-displacement leads instead to the bilateral Laplace mode $e^{\lambda q}$, equivalently the complex Fourier parameter $\xi=-i\lambda$. After conjugation by the radial speed measure, the analytically continued fibre is a real one-dimensional differential expression with the intended Friedrichs realization. For $\rho=-r<0$ and $Y:=R-\Rb$, its boundary expansion, shifted by the quadratic maximum, is

$$
\begin{equation}\label{eq:fiber-layer}\tag{8.2} H_\lambda+\frac{\nu^2\orth^2\lambda^2}{2} =\nu^2\left[-\frac12\partial_Y^2 +r\orth^2\lambda^2Y+\frac{br\lambda}{Y} +\frac{b(b+1)}{2Y^2} +O(\lambda^2Y^2+\lambda+Y^{-1}+1)\right].\tag{8.2} \end{equation}
$$

The term $br\lambda/Y$ is the singular tangential component of the drift. Balancing it with $r\orth^2\lambda^2Y$ gives

$$
Y_\lambda=\frac{\sqrt b}{\orth\sqrt\lambda}, \qquad r\orth^2\lambda^2Y_\lambda+\frac{br\lambda}{Y_\lambda} =2r\orth\sqrt b\,\lambda^{3/2}.
$$

A formal one-dimensional min–max localization, with a bump of width $\lambda^{-5/8}$ about $Y_\lambda$, predicts the tilted principal-eigenvalue scaling

$$
\begin{equation}\label{eq:fiber-eigen}\tag{8.3} \Lambda(\lambda) =\frac{\nu^2\orth^2}{2}\lambda^2 -2\nu^2r\orth\sqrt b\,\lambda^{3/2} +O(\lambda^{5/4}).\tag{8.3} \end{equation}
$$

We do not prove (8.3): a proof must specify the quadratic-form domain, establish the relevant discrete eigenvalue, and supply a global localization upper bound. The inverse-square term is only $O(\lambda)$ at this scale but formally selects the Dirichlet Friedrichs realization. At the ordinary Airy scale $Y\asymp\lambda^{-2/3}$ the term $br\lambda/Y$ is of order $\lambda^{5/3}$ and dominates the $\lambda^{4/3}$ Airy balance; within this formal balance the Airy regime survives only in the driftless case $b=0$, or for the comparison process without the singular drift.

Formal Legendre inversion of (8.3), and a second, likewise formal, calculation that conditions on the autonomous angular path and optimizes its exact Gaussian tail, give the same prediction.

**Conjecture 8.1 (Singular-Drift Subleading Correction).**  For $\beta\in(0,1)$ and $\rho\in(-1,0)$,

$$
-\ln P(k) =c_-k^2+\kappa_{3/2}k^{3/2}+o(k^{3/2}), \qquad\kappa_{3/2} =\frac{\sqrt{2\beta}(1-\beta)|\rho|}{\nu(1-\rho^2)\sqrt T}.
$$

A proof would require a uniform shrinking-tube lower bound or, alternatively, complex saddle-point estimates controlling endpoint overlaps and transverse frequencies. The numerical window $k\le12$ of Section 7 cannot distinguish the $k^{3/2}$ term reliably from other $o(k^2)$ corrections.

### 8.2 Neighbouring Parameter Regimes

The geometry above identifies absorption, rather than correlation alone, as the source of the negative-correlation correction. The neighbouring regimes are therefore informative: the correlation structure persists, but the state space, and with it the tail mechanism, changes.

At $\beta=1$ there is no absorption. Forde and Zhang [FZ15] show that the right tail for $\rho<0$ is governed by the critical moment $m^*=(1-\rho^2)^{-1}$ [LM07], with a one-sided discontinuity at $\rho=0^-$; the implied-volatility smile grows rather than approaching a finite right-wing limit, consistently with Lee’s moment formula [Lee04]. For $0<\beta<1$, by contrast, moments of every positive order are finite as a consequence of the quadratic log-moneyness tail proved here, and the right-wing limit is finite; its value is $C^1$ but not $C^2$ at $\rho=0$ (Corollary 1.3). The same factor $(1-\rho^2)^{-1}$ thus appears in the two regimes through distinct mechanisms: moment explosion at $\beta=1$, absorption at $0<\beta<1$.

The normal SABR model $\beta=0$ supplies the complementary boundary case. Its state space is all of $\mathbb{R}$ and it has no absorbing boundary; its right-wing limit is $\nu$ for every $\rho$ [Hen08, Example 10.4]. The limit $\beta\downarrow0$ in Corollary 1.2, however, retains the factor $\sqrt{1-\rho^2}$, because it approaches $\beta=0$ through models that still have an absorbing boundary at $F=0$. The difference is caused by the change of state space, not by the CEV exponent alone.

Finally, the mass-at-zero literature [GHJ18; DHJ17; Gul15] describes how the atom at the origin shapes the left wing. Theorem 1.1 and Corollary 1.2 show that for negative correlation the same boundary also changes the right-wing limit, acting through the survival constraint rather than through the size of the atom. Taken together, these regimes show that the diffusion metric alone does not determine the extreme wing: its interaction with the survival domain is decisive.

## A Measurability, Versions and Uniqueness

This appendix records the measurability and well-posedness facts used when conditioning on a volatility path, passing to operational time, and summing level-hitting configurations; it introduces no additional probabilistic estimate.

(i) On every compact subset of $(0,\infty)\times(0,\infty)$ the coefficients of the $(F,a)$ equation are locally Lipschitz. Standard localization therefore gives a pathwise-unique strong solution up to $\zeta$, and the model convention fixes its continuation by $F_t=0$ after $\zeta$. We use continuous adapted versions throughout. The stopped Itô identities at levels $X\ge r$, $r\in\mathbb Q_{>0}$, hold simultaneously outside one null set; continuity then gives the all-times formulation of Lemma 2.1.

(ii) The volatility path space $C([0,T])$ is Polish. For a fixed continuous positive path $y$, the Itô integral $M(y)=\int_0^\cdot y_s\,dZ_s^\perp$ with deterministic integrand is a centred continuous Gaussian process with covariance $V_{s\wedge t}(y)$. Since $a$ is $\sigma(Z)$-measurable, independence of $Z^\perp$ and $Z$ implies independence of $Z^\perp$ and $a$, and the deterministic-integrand stochastic-integral theorem gives, for $\Prob_a$-almost every $y$, the conditional path law

$$
\mathcal L(M\mid a=y)=\mathsf W\circ\Theta_y^{-1}, \qquad\Theta_y(w)(t)=w(V_t(y)),
$$

as stated in Lemma 2.2 [KS91, Ch. 3]. The map $(y,w)\mapsto w\circ V(y)$ is Borel measurable for the compact-open topologies, so $\mathsf K(y,A):=\mathsf W(\Theta_y^{-1}(A))$ is a measurable probability kernel. Regular conditional distributions are unique only $\Prob_a$-almost everywhere, and every conditional assertion in the paper is made with that qualification.

(iii) Since $y>0$ for $\Prob_a$-almost every path, $V(y)$ is $C^1$ and strictly increasing; its inverse $t_y:[0,V_T(y)]\to[0,T]$ is $C^1$, and under the kernel above $(M_{t_y(v)})_{0\le v\le V_T(y)}$ has Wiener law on the finite random interval $[0,V_T(y)]$. In particular, the comparison statements of Section 5.2 are confined to this operational-time horizon.

(iv) The level-hitting times $\tau_j$ are stopping times of the continuous adapted process $a$, and all skeleton data ($v_j$, $h_j$, $H$ and bin memberships) are $\sigma(a)$-measurable. After coincident operational-time nodes are merged, the QP in (3.2) is the minimum of a finite-dimensional coercive convex functional over a closed polyhedron and is a Borel function of its operational-time and threshold data. The conditional applications of Lemma 3.3 and the tower property are therefore legitimate.

## B Numerical Methods

This appendix summarizes the methods behind Section 7.

*Monte Carlo estimators.* We write $P(k)=\E[q(a)]$, where $q(a)$ is the conditional probability of the target event given the volatility path. Volatility paths are sampled from a finite mixture of exponentially tilted path measures and reweighted by the mixture likelihood ratio. Except for $\beta=\tfrac12$, $\rho=0$, $q(a)$ is estimated by an Euler scheme in operational time with $128$ antithetic inner draws and the drift-free Brownian-bridge correction

$$
p_{\mathrm{hit}}=\exp\!\Bigl(-\frac{2X_sX_{s+1}}{\orth^2\Delta v}\Bigr)
$$

when both endpoints of a step are positive. The inner Brownian increments are tilted toward the terminal target, by a constant tilt or by an equal mixture of a constant and a variationally scheduled tilt, on grids of $500$ to $600$ steps. For $\beta=\tfrac12$, $\rho=0$ the inner probability is exact: conditionally on the volatility path, $4F$ is a squared Bessel process of dimension $0$ in operational time, absorbed at $0$, so that with $u:=2f_0/V_T$ and $w:=2K/V_T$,

$$
q(a)=\sum_{j\ge1}e^{-u}\frac{u^j}{j!}\,G(j,w),
$$

where $G$ is the regularized upper incomplete gamma function. The series is truncated adaptively, with analytic bounds on the omitted Poisson tail that are negligible relative to the estimate and its Monte Carlo error; independent Bessel-density quadrature agrees to relative error $10^{-15}$, and the remaining discretization is the trapezoidal evaluation of $V_T$ on a $2000$-step grid.

The reported errors are first-order delta-method standard errors $\mathrm{SE}(\hat P)/\hat P$ of $-\ln\hat P$; the second-order correction $\tfrac12(\mathrm{SE}/\hat P)^2$ is at most $0.03$. Effective sample sizes are $(\sum_iW_i)^2/\sum_iW_i^2$ for importance weights $W_i$; the threshold $100$ for $^\dagger$ was chosen after the runs. The cell $(\beta,\rho,k)=(\tfrac12,-0.6,6)$ pools two batches of $36\,000$ paths. Grid-refinement ladders with $60$, $125$, $250$ and $500$ steps at the representative cells $(\tfrac12,+0.6,2)$, $(\tfrac12,-0.6,4)$ and $(0.7,-0.6,4)$ bound the production-grid bias in $-\ln P$ by $0.003$, $0.013$ and $0.024$ (two standard errors). They do not bound the bias of the other cells; the sources of discretization error are the trapezoidal integrated variance, the Euler treatment of the singular drift, and the drift-free bridge correction on steps with drift.

*Variational problems.* With the volatility driven by a control $\varphi$, $a(t)=\alpha e^{\nu\varphi(t)}$, operational time $V(t)=\int_0^ta(s)^2ds$ and correlation term $\xi(t)=(\rho/\nu)(a(t)-\alpha)$, the survival problem is

$$
R(k)=\min_{\varphi(0)=0}\Bigl\{\frac12\int_0^T\dot\varphi(t)^2\,dt +\min_{\psi}\frac1{2\orth^2}\int_0^{V(T)}\dot\psi(v)^2\,dv\Bigr\},
$$

where the inner minimum is over paths $\psi$ with $\psi(0)=0$, $\psi(V(T))\ge m-X_0-\xi(T)$, and $\psi(V(t))\ge-X_0-\xi(t)$ for all $t\le T$; the terminal problem omits the last constraint. (The Itô correction in $\ln a$ is of lower order and is omitted.) We take $\varphi$ piecewise linear on $20$ control nodes, interpolated on a $400$-step grid with exact operational-time increments on each step. With the survival constraint imposed at the grid nodes, the inner problem is solved exactly: its minimizer is the least nondecreasing concave majorant of the constraint points, anchored at the origin. The free terminal value imposes a nonnegative final slope. We compute the ordinary concave upper hull and replace its nonpositive-slope terminal portion by a constant continuation; equivalently, only positive-slope segments contribute to the energy. The outer minimization uses L-BFGS-B followed by Nelder–Mead from several initializations, with continuation in $k$; the computation is deterministic. An earlier implementation, which imposed the inner constraints by penalties, gives consistent but less regular elimination sequences; both are included in the supplementary material.

*Reproducibility.* The supplementary material contains the scripts, the stored outputs and a cell-by-cell provenance record. Not every displayed Monte Carlo value can be regenerated from a recorded seed; the record states which can. For $\beta=\tfrac12$, $\rho=+0.6$, Table 2 reports a run with recorded seeds; an earlier production run agrees with it to within $0.032$ in $-\ln P$.

## Supplementary Material

The supplementary material, available from the authors on request, contains the code and data for Section 7 and the figures, and a Lean development. The Lean development formalizes the fixed-volatility Black–Scholes call-price rate, the identification of the implied-volatility limit from a call-price rate, and the intrinsic-distance calculation of Proposition 6.1, and derives Corollary 6.2 from that proposition with the two tail bounds as hypotheses. It states the stochastic inputs of the main proof as explicit interfaces and is not a formal proof of Theorem 1.1. A claim manifest links statements of this paper to Lean declarations.

**Acknowledgments and Tool Disclosure.** The authors used OpenAI Codex and Anthropic’s Claude for language editing, restructuring of the exposition, preparation of figures, and assistance with the supplementary Lean development and computational checks. The authors take full responsibility for the mathematical claims, exposition, code and numerical results.

## Notes

[^1]: The book writes the right-hand side of (1.2) as a constant value; the finite-point potential and saddle formula in Section 6.4 give the large-strike limit stated here. Both conjectures appear on p. 306 of [Hen08]. Conjecture 1 is stated there without restriction on $\rho$ and is immediately followed by the attribution of the case $\rho=0$ to [BFL08] (the book’s reference [55]), which fixes its intended generality.

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