SABR微笑右翼带吸收边界：Henry-Labordère猜想的解决

中文摘要 AI 辅助

我们确定了SABR模型在$\beta\in(0,1)$且零处吸收条件下，固定到期日右尾的对数渐近行为，适用于所有相关性$\rho\in(-1,1)$。若$P(k)$表示终端远期至少为$f_0e^k$的概率，则当$k\to\infty$时，$-k^{-2}\ln P(k)\to(1-\beta)^2/(2\nu^2T(1-(\rho\wedge0)^2))$。对于$\rho\ge0$，该速率与无限制双曲几何预测一致；对于$\rho<0$，该速率严格更大：在产生尾部的波动率偏移过程中，生存迫使与波动率正交的噪声保持在移动平方根障碍之上，额外成本为同阶$k^2$。因此，Black-Scholes隐含波动率趋于$\nu\sqrt{1-(\rho\wedge0)^2}/(1-\beta)$：Henry-Labordère猜想的翼部极限$\nu/(1-\beta)$在$\rho\ge0$时成立，在$\rho<0$时失败。对于他的第二个猜想（其无限制距离高斯估计在$\rho<0$时不成立），我们也为该模型确定了正确的替代：在尾部概率层面，速率由生存域内从初始状态到终端目标集的平方内蕴黎曼距离给出。证明是直接的，结合了停止的Lamperti变换、条件时间变换、高斯移动障碍界和测度变换构造，且不使用热核估计。

英文摘要

We determine the leading logarithmic asymptotics of the fixed-maturity right tail of the SABR model with $β\in(0,1)$ and absorption at zero, for every correlation $ρ\in(-1,1)$. If $P(k)$ is the probability that the terminal forward is at least $f_0e^k$, then $-k^{-2}\ln P(k)\to(1-β)^2/(2ν^2T(1-(ρ\wedge0)^2))$ as $k\to\infty$. For $ρ\ge0$ this is the rate predicted by the unrestricted hyperbolic geometry. For $ρ<0$ 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 $ν\sqrt{1-(ρ\wedge0)^2}/(1-β)$: Henry-Labordère's conjectured wing limit $ν/(1-β)$ holds for $ρ\ge0$ and fails for every $ρ<0$. For his second conjecture, whose unrestricted-distance Gaussian estimate does not hold for $ρ<0$, 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.

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35 pages, 4 figures