On Small-Noise Equations with Degenerate Limiting System Arising from Volatility Models

Abstract

The one-dimensional SDE with non Lipschitz diffusion coefficient

$$\begin{aligned} \textit{dX}_{t} = b(X_{t})\textit{dt} + \sigma X_{t}^{\gamma } \textit{dB}_{t}, \quad X_{0}=x, \quad \gamma <1 \end{aligned}$$

(1)

  is widely studied in mathematical finance. Several works have proposed asymptotic analysis of densities and implied volatilities in models involving instances of (1), based on a careful implementation of saddle-point methods and (essentially) the explicit knowledge of Fourier transforms. Recent research on tail asymptotics for heat kernels (Deuschel et al. Comm. in Pure and Applied Math., 67(1):40–82, 2014, [11]) suggests to work with the rescaled variable $X^{\varepsilon }:=\varepsilon ^{1/(1-\gamma )} X$: while allowing to turn a space asymptotic problem into a small-$\varepsilon$ problem, the process $X^{\varepsilon }$ satisfies a SDE in Wentzell–Freidlin form (i.e. with driving noise $\varepsilon \textit{dB}$). We prove a pathwise large deviation principle for the process $X^{\varepsilon }$ as $\varepsilon \rightarrow 0$. As it will be seen, the limiting ODE governing the large deviations admits infinitely many solutions, a non-standard situation in the Wentzell–Freidlin theory. As for applications, the $\varepsilon$-scaling allows to derive leading order asymptotics for path functionals: while on the one hand the resulting formulae are confirmed by the CIR-CEV benchmarks, on the other hand the large deviation approach (i) applies to equations with a more general drift term and (ii) potentially opens the way to heat kernel analysis for higher-dimensional diffusions involving (1) as a component.

Appendix A

We complete the proof of Proposition 3.2 here.

Proof of Proposition 3.2 Let us define an auxiliary process $\overline{X}$ by

$$\begin{aligned} d\overline{X}_{t} = |\alpha |_{\infty }\textit{dt} +\sigma \exp (-(1-\gamma ) |\beta | t ) \overline{X}_{t}^{\gamma }\textit{dB}_{t}, \quad \overline{X}_{0}=x; \end{aligned}$$

after a simple application of the product rule, one has that the process $Z_{t} := \exp (|\beta | t)\overline{X}_{t}$ is a solution to

$$\begin{aligned} dZ_{t} = \bigl (|\alpha |_{\infty } \exp (|\beta | t) + |\beta | Z_{t}\bigr ) \textit{dt} + \sigma Z_{t}^{\gamma }\textit{dB}_{t}, \quad Z_{0} = x. \end{aligned}$$

Since $|\alpha |_{\infty } \exp (|\beta | t) \ge |\alpha |_{\infty }$, an application of the comparison principle for SDE’s [17, Proposition 5.2.18] yields $Z_{t} \ge \tilde{X}_{t}$, for all $t \ge 0$. Therefore, if $\overline{X}^{2(1-\gamma )}$ admits (some) exponential moments, so does $Z^{2(1-\gamma )}_{t}$ and by comparison $\tilde{X}^{2(1-\gamma )}_{t}$. In this sense, the process $\overline{X}$ is not covered by Proposition 3.3 in [9], since the latter deals with the case of a diffusion coefficient that does not depend on time (see [9, Eq. (3.1)]); nonetheless, the essential condition that [9, Proposition 3.3] relies on is the presence of a non-strictly positive slope coefficient, say b in the drift term $a+bX$ (cf. [9, Eq. (3.3)]). Since this is the case for the process $\overline{X}$ (which has zero slope coefficient b), it is straightforward to extend the proof to the present setting: in particular, in the spirit of Lamperti’s change-of-variable argument, one still defines the function $\varphi (x)=\int _0^{x}\frac{1}{\sigma x^{\gamma }}=\frac{1}{\sigma (1-\gamma )}x^{1-\gamma }$ and studies the process $\tilde{\varphi }(X_t)$, where the function $\tilde{\varphi }$ is a modification of $\varphi$ identically null around zero. Itô’s formula shows that $\tilde{\varphi }(X_t)$ is an Itô process with bounded quadratic variation and a bounded drift term; the existence of quadratic exponential moments for $\tilde{\varphi }(X_t)$, then, is a consequence of Dubins–Schwarz time-change argument and Fernique’s theorem. As a consequence, there exist $c',C > 0$ such that $\sup _{t \le T} \mathbb {E}[\exp (c' \overline{X}^{2(1-\gamma )}_{t})] \le C$; it follows $\sup _{t \le T} \mathbb {E}[\exp (c \tilde{X}^{2(1-\gamma )}_{t})] \le \sup _{t \le T} \mathbb {E}[\exp (c Z^{2(1-\gamma )}_{t})] \le C$ with $c:=c' \exp (-2|\beta |(1-\gamma )T)$, and the claim is proved.$\blacksquare$

We report the statement given in [26, Chap. 2, Theorem 2.13].

Lemma A.1

(Garsia-Rodemich-Rumsey’s Lemma) Let p and $\Psi$ be continuous, strictly increasing functions on $[0, +\infty )$ such that $p(0) = \Psi (0)=0$ and $\lim _{t \rightarrow + \infty } \Psi (t)= + \infty$. If $\omega \in \Omega$ is such that:

$$\begin{aligned} \int _{0}^{T} \int _{0}^{T} \Psi \left( \frac{|\omega _{t} - \omega _{s} |}{p(|t-s|)}\right) \textit{ds} \textit{dt} \le K, \end{aligned}$$

(A.1)

then

$$\begin{aligned} |\omega _{t} - \omega _{s} | \le 8 \int _{0}^{|t-s|} \Psi ^{-1} \left( \frac{4K}{u^2}\right) dp(u). \end{aligned}$$

(A.2)

Lemma A.1 allows us to prove Proposition 3.3:

Proof of Proposition 3.3 Assume that (3.3) holds true with the left hand side replaced by $K>0$. Applying Lemma A.1 with the choice of functions $\Psi (y)=\exp (\varepsilon ^{-2} y)-1, p(y) = \sqrt{y}$, one has for all s, t

$$\begin{aligned} \left| \omega _{t}- \omega _{s} \right| \le 8\int _{0}^{|t-s|} \Psi ^{-1} \left( \frac{4K}{u^2} \right) dp(u)&= 8\varepsilon ^{2}\int _{0}^{|t-s|} \log \left( \frac{4K}{u^2} +1 \right) dp(u) \\&\le 8 \varepsilon ^{2} \left[ \int _{0}^{|t-s|} \log \left( 4K + T^2 \right) dp(u)\right. \\&\qquad \qquad \left. + \int _{0}^{|t-s|} \log \left( u^{-2} \right) dp(u) \right] \\&\le 8 \varepsilon ^{2}\left[ \sqrt{|t-s|} \log \left( 4K+T^{2} \right) \right. \\&\qquad \qquad \left. +\,\sqrt{|t-s|} \left( 4-2\log \left( |t-s|\right) \right) \right] . \end{aligned}$$

Dividing on both sides by $(t-s)^{\eta }$ and taking suprema we obtain

$$\begin{aligned} \left\| \omega \right\| _{\eta } \le 8 \varepsilon ^{2} \left( \log \left( 4K+T^{2} \right) T^{1/2-\eta } +4T^{1/2-\eta } + K_{\eta } \right) . \end{aligned}$$

Since the right hand side in the last estimate is $K^{-1}_{\varepsilon , \eta } \left( K \right)$, (3.3) yields (3.4).$\blacksquare$

Finally, we prove Lemma 3.10.

Proof of Lemma 3.10 Denote $T^{\varepsilon }$ the stopping time

$$\begin{aligned} T^{\varepsilon } (\omega ) = \inf \left\{ t \ge 0 : \omega _{t} \le \frac{1}{2} \varepsilon x^{1-\gamma } \right\} . \end{aligned}$$

(A.3)

We can apply Itô formula to the function $f(x) = x^{1-\gamma }$ up to time $T^{\varepsilon }(Y^{\varepsilon ,h})$, and obtain

$$\begin{aligned} Y^{\varepsilon ,h}_{t}\,-\,\varepsilon x^{1-\gamma }= \int _{0}^{t} \tilde{b}^{\varepsilon } (Y^{\varepsilon ,h}_{s})\textit{ds}\,+\,\sigma (1\,-\,\gamma ) h_{t}\,+\,\varepsilon \sigma (1\,-\,\gamma ) B_t, \quad \forall \, t \le T^{\varepsilon }(Y^{\varepsilon ,h}), \quad a.s. \end{aligned}$$

(A.4)

where $\tilde{b}^{\varepsilon }$ is given by

$$\begin{aligned} \tilde{b}_{\varepsilon }(y) := (1-\gamma )\varepsilon ^{\frac{1}{1-\gamma }}\alpha (\varepsilon ^{-\frac{1}{(1-\gamma )}}y^{\frac{1}{(1-\gamma )}})\frac{1}{y^{\frac{\gamma }{1-\gamma }}} - \frac{\sigma ^{2} \gamma (1-\gamma )}{2}\varepsilon ^{2} \frac{1}{y} + \beta (1-\gamma ) y\quad \end{aligned}$$

(A.5)

We need to prove

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0 } W \left( \sup _{t \in [0,T]} | Y^{\varepsilon ,h}_{t}- \mathcal {S}_{0}(h)_{t}| \le R \right) = 1 \qquad \forall R>0. \end{aligned}$$

(A.6)

In order to simplify the notation, there is no ambiguity in writing Y instead of $Y^{\varepsilon ,h}$ inside this proof.

Step 1. We first prove (A.6) under the assumption

$$\begin{aligned} k:=\inf _{t \in [0,T]}\dot{h}_{t} >0 \end{aligned}$$

(A.7)

Let us fist show that

$$\begin{aligned} \lim _{ \varepsilon \rightarrow 0} W \left( T^{\varepsilon } \left( Y^{\varepsilon ,h} \right) \le T \right) =0 \end{aligned}$$

(A.8)

A direct computation shows that there exist a constant $c>0$ depending on $x,\sigma ,\alpha (\cdot )$ such that:

$$\begin{aligned} \inf _{y \ge \frac{1}{2} \varepsilon x^{1-\gamma }} \Bigl \{ \tilde{b}^{\varepsilon } (y)-\beta (1-\gamma ) y \Bigr \} \ge - c \varepsilon . \end{aligned}$$

(A.9)

Define $(Z_{t})_{t \in [0,T]}$ by

$$\begin{aligned} Z_{t} = \varepsilon x^{1-\gamma } + \left( -c\varepsilon +\sigma (1-\gamma ) k \right) t + \beta (1-\gamma ) \int _{0}^{t} Z _{s}\textit{ds} + \varepsilon \sigma (1-\gamma ) B_{t} \end{aligned}$$

(A.10)

Using (A.9), it follows from the comparison principle for SDEs that

$$\begin{aligned} \quad Y_{t} \ge Z_{t} \quad \forall \, t \le T^{\varepsilon }(Y), \quad a.s. \end{aligned}$$

(A.11)

We claim that

$$\begin{aligned} W \left( T^{\varepsilon } \left( Z\right) \le T \right) \rightarrow 0 \end{aligned}$$

(A.12)

holds true. Since $W \left( T^{\varepsilon } \left( Y \right) \le T \right) \le W \left( T^{\varepsilon } \left( Z\right) \le T \right)$ by (A.11), then (A.8) holds. We prove (A.12) later on. Now, it follows from the definition of $S_0(h)_t$ and an application of Gronwall’s Lemma that

$$\begin{aligned} |Y_{t} -\mathcal {S}_{0}(h)_{t}| \le \varepsilon \biggl ( c+ \sigma (1-\gamma ) \sup _{t \in [0,T]}| B_{t}| \biggr ) e^{|\beta |(1-\gamma )T} =:\Theta _T \qquad \forall t\le T^{\varepsilon } \left( Y \right) , \end{aligned}$$

therefore, for any $R >0$ and $\varepsilon$ small enough

$$\begin{aligned} W \biggl ( \sup _{t \in [0,T^{\varepsilon }] } |Y_{t} -\mathcal {S}_{0}(h)_{t}| \le R \biggr )&\ge W \biggl (\biggl \{ \sup _{t \in [0,T^{\varepsilon }(Y)] }|Y_{t} -\mathcal {S}_{0} (h)_{t}| \le \Theta _T \biggr \} \cap \left\{ \Theta ^{\varepsilon }_T \le R \right\} \biggr ) \\&\ge W \left( \left\{ T^{\varepsilon } (Y) \ge T \right\} \cap \left\{ \Theta ^{\varepsilon }_T \le R \right\} \right) . \end{aligned}$$

Since both the events in the right hand side of the last inequality have probability converging to 1, (A.6) follows, and Lemma 3.10 is proved under condition (A.7).

Step 2. We assume that (A.7) holds only on the time interval $[0,\rho ]$, that is $\dot{h}_{t} \ge k$ for every $t \le \rho$, for some $k, \rho >0$. Repeating the argument of Step 1 with $T= \rho$, we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} W\biggl (\sup _{t \in [0,\rho ]} | Y_{t}-S_{0}(h)_{t} | \le R^{'} \biggr ) =1, \quad \forall R^{'}>0 \end{aligned}$$

(A.13)

We apply estimate (A.13) together with a localization argument. Define a time-shift operator $\tau _{\rho } \omega$, for every $\omega \in \Omega$, by $(\tau _{ \rho } \omega )_{t} = \omega _{\rho +t}$ for all $t \in [0, T-\rho ]$. For any fixed $y>0$, denote $X^{y,\rho }$ the strong solution of the SDE:

$$\begin{aligned} X^{y,\rho }_{t} = y^{\frac{1}{(1-\gamma )}} + \int _{0}^{t} b^{\varepsilon }(X^{y,\rho }_{s}) + \sigma |X^{y,\rho }_{s}|^{\gamma } \dot{h}_{\rho +s} \textit{ds} + \varepsilon \sigma \int _{0}^{t} |X^{y,\rho }_{s}|^{\gamma } \textit{dB}_{s} \end{aligned}$$

and set

$$\begin{aligned} Y^{y,\rho }:= (X^{y,\rho })^{1-\gamma }. \end{aligned}$$

Note that $Y^{y,\rho }$ is well defined since $X^{y,\rho } \ge 0$ for all $t \in [0,T]$, W-almost surely. If $h=0$ the non negativity of the trajectories of $X^{y,\rho }$ follows from an application Proposition 3.1 in [9] and extends to $h \in H$ by an application of the Girsanov theorem. By definition of Y and $Y^{y,\rho }$, the Markov property yields

$$\begin{aligned} \mathbb {E} ( f(\tau _{\rho }Y) | \mathcal {F}_{\rho }) = \mathbb {E} ( f(Y^{Y_{\rho },\rho }) ) \end{aligned}$$

By the continuity of the map $(h,y) \mapsto \mathcal {S}_{y}(h)$ we can choose $R'>0$ such that

$$\begin{aligned} \sup _{y \in B(S_{0}(h)_{\rho },R')} \sup _{t \in [0,T-\rho ]} | \mathcal {S}_{y}(\tau _{\rho } h)_{t} - \mathcal {S}_{\mathcal {S}_{0}(h)_{\rho }}(\tau _{\rho }h)_{t} | \le \frac{R}{2} \end{aligned}$$

(A.14)

Therefore, using (A.14) the following inclusion of events holds (assume w.lo.g $R' \le \frac{R}{2}$):

$$\begin{aligned} \biggl \{ \sup _{t \in [0,T]} |Y_{t} - \mathcal {S}_{0} (h)_{t} | \le R \biggr \}&\supseteq \biggl \{ \sup _{[0,\rho ]}|Y_{t} - \mathcal {S}_{0}(h)_{t} | \le R' \biggr \}\\&\cap \biggl \{ \sup _{t\in [0,T-\rho ]} |\tau _{\rho }(Y)_{t} - \mathcal {S}_{Y_{\rho }}(\tau _{\rho }h)_{t} | \le \frac{R}{2} \biggr \} \end{aligned}$$

Applying the Markov property

$$\begin{aligned} W \biggl ( \sup _{t \in [0,T]}&|Y_{t} - \mathcal {S}_{0}(h)_{t} | \le R \biggr )\nonumber \\&\ge \mathbb {E} \biggl ( \mathbb {1}_{ \{ \sup _{t \in [0,\rho ]}|Y_{t} - \mathcal {S}_{0}(h)_{t} | \le R' \}} W \biggl ( \sup _{t \in [0,T-\rho ]} |Y^{ Y_{\rho }, \rho }_{t} - S_{Y_{\rho }}(\tau _{\rho }h)_{t} | \le \frac{R}{2} \biggr ) \biggr ) \nonumber \\&\ge W \biggl ( \sup _{t \in [0,\rho ]}|Y_{t} - \mathcal {S}_{0}(h)_{t} | \le R' \biggr )\\&\quad \times \inf _{y \in B(\mathcal {S}_{0} (h)_{\rho } ,R') } W \biggl ( \sup _{t \in [0,T-\rho ]} | Y^{y,\rho }_{t} - \mathcal {S}_{y}(\tau _{ \rho }h)_{t}| \le \frac{R}{2} \biggr ) \end{aligned}$$

(A.15)

We want to show that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0 }\inf _{ y \in B \left( S_{0} \left( h \right) _{\rho },R' \right) } W \left( \sup _{t \in [0,T-\rho ]} | Y^{y,\rho }_{t} - \mathcal {S}_{y} (\tau _{ \rho } h)_{t} | \le \frac{R}{2} \right) = 1 \end{aligned}$$

(A.16)

It follows from the hypothesis $\mathcal {S}_{0}(h)_{t} >0 \ \forall t>0$ and the continuity of the map $(y,h) \mapsto \mathcal {S}_{y}(h)$ that, if $R',R$ are small enough

$$\begin{aligned} y^{*}:=\inf _{y \in B\left( \mathcal {S}_{0} \left( h \right) _{\rho },R' \right) } \inf _{t \in [0,T-\rho ]} \mathcal {S}_{y} ( \tau _{\rho } h )_{t} -\frac{R}{2} >0. \end{aligned}$$

(A.17)

Define $U^{y,\rho }$ as the unique strong solution of the SDE:

$$\begin{aligned} U^{y,\rho }_{t} = y + \int _{0}^{t} \bigl ( \tilde{b}^{\varepsilon }_{u} (U^{y,\rho }_{s}) + \sigma (1-\gamma ) \dot{h}_{s+\rho } \bigr )\textit{ds} + \varepsilon \sigma (1-\gamma ) B_{t}, \end{aligned}$$

where

$$\begin{aligned} \tilde{b}^{\varepsilon }_{u}(y) = {\left\{ \begin{array}{ll} \tilde{b}^{\varepsilon }(y) &{} \text {if}\, y \ge y^* \\ \beta (1-\gamma )y + (1-\gamma )\varepsilon ^{\frac{1}{1-\gamma }}\alpha (\varepsilon ^{-\frac{1}{(1-\gamma )}}(y^*)^{\frac{1}{(1-\gamma )}}) \frac{1}{(y^*)^{\frac{\gamma }{1-\gamma }}} - \frac{\sigma ^{2} \gamma (1-\gamma )}{2}\varepsilon ^{2} \frac{1}{y^{*}} &{} \text {if}\, y < y^{*}. \end{array}\right. } \end{aligned}$$

Then one has

$$\begin{aligned} W \biggl ( \sup _{t \in [0,T - \rho ]} | Y^{y,\rho }_{t} - S_{y}(\tau _{ \rho }h)_{t} | \le \frac{R}{2} \biggr ) = W \biggl ( \sup _{t \in [0, T-\rho ]} | U^{y,\rho }_{t} - S_{y}(\tau _{\rho }h )_{t}| \le \frac{R}{2} \biggr ). \end{aligned}$$

(A.18)

Now observing that $\tilde{b}^{u}_{\varepsilon }$ is globally Lipschitz continuous $\forall \varepsilon >0$ and $C^{\varepsilon } := \sup _{y \in \mathbb {R}}|\tilde{b}^{\varepsilon }_{u} (y) -\beta (1-\gamma )y| \rightarrow 0$, an application of Gronwall’s lemma gives

$$\begin{aligned} \mathbb {E} \biggl ( \sup _{t \in [0,T-\rho ]} \left| U^{y,\rho }_{t} - \mathcal {S}_{y}(\tau _{\rho } h)_{t} \right| \biggr ) \le (C^{\varepsilon }T + 2\varepsilon \sigma (1-\gamma )\sqrt{T} )\exp ( |\beta (1-\gamma )| T). \end{aligned}$$

(A.19)

By letting $\varepsilon \rightarrow 0$ and applying the Markov inequality, observing that the right hand side of (A.19) does not depend on y, we have proven (A.16). By letting $\varepsilon \rightarrow 0$ in (A.15) and applying (A.13) and (A.16), the proof of Lemma 3.10 is complete.$\blacksquare$

Proof of (A.12). Observe that $\tilde{Z}:=\frac{1}{\varepsilon }Z$ is an Ornstein-Uhlenbeck process,

$$\begin{aligned} \tilde{Z}_t = x^{1-\gamma } + \mu _{\varepsilon } t + \beta (1-\gamma )\int _{0}^{t}\tilde{Z}_{s}\textit{ds} + \sigma (1-\gamma ) B_t \end{aligned}$$

(A.20)

where $\mu _{\varepsilon }:= \frac{1}{\varepsilon }(-c \varepsilon \,+\,\sigma (1\,-\,\gamma )k) = -c\,+\,\frac{\sigma (1-\gamma )k}{\varepsilon }$. It is immediate by the definition of $\tilde{Z}$ that $W \left( T^{\varepsilon }(Z) \le T \right) = W \left( \inf _{t \in [0,T]} \tilde{Z} \le \frac{x^{1-\gamma }}{2} \right)$. The explicit representation of $\tilde{Z}$ reads

$$\begin{aligned} \tilde{Z}_t := x^{1-\gamma }e^{\beta (1-\gamma ) t} + f_{\varepsilon }(t) + \sigma (1-\gamma )\exp (\beta (1-\gamma ) t) \int _{0}^{t} \exp (- \beta (1-\gamma ) s) \textit{dB}_s \end{aligned}$$

(A.21)

with $f_{\varepsilon }(t) = -\frac{\mu _{\varepsilon } (1-\exp (\beta (1-\gamma ) t) )}{\beta (1-\gamma )}$. Consider a deterministic time $\tau _{\varepsilon }$ with $\tau _{\varepsilon } \rightarrow 0$ as $\varepsilon \rightarrow 0$, to be chosen precisely later on. Noting that $f_{\varepsilon }$ is a decreasing function, for $\tau _{\varepsilon } \le t \le T$ one has

$$\begin{aligned} \tilde{Z}_t \ge f_{\varepsilon }(\tau _{\varepsilon }) - \sigma (1-\gamma ) \Bigl | \int _{0}^{t} \exp (-\beta (1-\gamma ) s ) \textit{dB}_s \Bigr |; \end{aligned}$$

(A.22)

hence, using Markov’s inequality and Doob’s inequality

Now, the choice $\tau _{\varepsilon } = \sqrt{\varepsilon }$ gives $f_{\varepsilon }(\tau _{\varepsilon }) \sim \mu _{\varepsilon } \tau _{\varepsilon } \rightarrow \infty$ as $\varepsilon \rightarrow 0$, so that $\left( f_{\varepsilon }(\tau _{\varepsilon }) - x^{1-\gamma }/2 \right) ^{-1} \rightarrow 0$. On the other hand, $\inf _{t \in [0,\tau _{\varepsilon }]} \tilde{Z}_t \rightarrow x^{1-\gamma }$ a.s. as $\varepsilon \rightarrow 0$, hence $W \Bigl ( \inf _{t \in [0,\tau _{\varepsilon }]} \tilde{Z}_t \le x/2 \Bigr ) \rightarrow 0$ as $\varepsilon \rightarrow 0$, and the claim is proven.