Bismut-Elworthy-Li Formulae for Bessel Processes

Abstract

In this article we are interested in the differentiability property of the Markovian semi-group corresponding to the Bessel processes of nonnegative dimension. More precisely, for all δ ≥ 0 and T > 0, we compute the derivative of the function $x \mapsto P^{\delta }_{T} F (x)$, where $(P^{\delta }_{t})_{t \geq 0}$ is the transition semi-group associated to the δ-dimensional Bessel process, and F is any bounded Borel function on $\mathbb {R}_{+}$. The obtained expression shows a nice interplay between the transition semi-groups of the δ—and the (δ + 2)-dimensional Bessel processes. As a consequence, we deduce that the Bessel processes satisfy the strong Feller property, with a continuity modulus which is independent of the dimension. Moreover, we provide a probabilistic interpretation of this expression as a Bismut-Elworthy-Li formula.

Appendix

In this Appendix, we prove Proposition 6.3. Recall that we still denote by (ρ _t(x))_t,x≥0 the process $(\tilde {\rho }^{\delta }_{t}(x))_{t,x \geq 0}$ constructed in Proposition 6.2.

Lemma 6.4

For all rational numbers 𝜖, γ > 0, let:

$$\displaystyle \begin{aligned} \mathbb{U}_{\gamma}^{\epsilon} := [0, T_{\epsilon}(\gamma)) \times (\gamma, +\infty) \end{aligned}$$

and set:

$$\displaystyle \begin{aligned} \mathbb{U} := \underset{\epsilon, \gamma \in \mathbb{Q}^{*}_{+}}{ \bigcup} \mathbb{U}_{\gamma}^{\epsilon}. \end{aligned}$$

Then, a.s., the function (t, x)↦ρ _t(x) is continuous on the open set $\mathbb {U}$.

Proof

By patching, it suffices to prove that, a.s., the function (t, x)↦ρ _t(x) is continuous on each $\mathbb {U}_{\gamma }^{\epsilon }$, where $\epsilon , \gamma \in \mathbb {Q}^{*}_{+}$.

Fix $\epsilon , \gamma \in \mathbb {Q}^{*}_{+}$, and let $x, y \in (\gamma , + \infty ) \cap \mathbb {Q}$. We proceed to show that, a.s., for all t ≤ s < T _𝜖(γ) the following inequality holds:

$$\displaystyle \begin{aligned} |{\rho}_{t}(x) - {\rho}_{s}(y)| \leq |x-y| \exp\left( \frac{|\delta-1|}{2\epsilon^{2}} t \right) + \frac{|\delta-1|}{2\epsilon} |s-t| + |B_{s} - B_{t}| . \end{aligned} $$

(6.29)

Since T _𝜖(γ) < T ₀(γ), a.s., for all t ≤ s ≤ T _𝜖(γ), we have:

$$\displaystyle \begin{aligned} \forall \tau \in [0,t], \qquad {\rho}_{\tau}(x) = x + \frac{\delta - 1}{2} \int_{0}^{\tau} \frac{du}{{\rho}_{u}(x)} + B_{\tau} \end{aligned}$$

as well as

$$\displaystyle \begin{aligned} \forall \tau \in [0,s], \qquad {\rho}_{\tau}(y) = y + \frac{\delta - 1}{2} \int_{0}^{\tau} \frac{du}{{\rho}_{u}(y)} + B_{\tau} \end{aligned}$$

and hence:

$$\displaystyle \begin{aligned} \forall \tau \in [0,t], \qquad |\rho_{\tau}(x) - \rho_{\tau}(y)| \leq |x-y| + \frac{|\delta-1|}{2} \int_{0}^{\tau}\frac{ |\rho_{u}(x) - \rho_{u}(y)|}{\rho_{u}(x) \rho_{u}(y)} du . \end{aligned}$$

By the monotonicity property of ρ, we have, a.s., for all t, s as above and u ∈ [0, s]:

$$\displaystyle \begin{aligned} {\rho}_{u}(x) \wedge {\rho}_{u}(y) \geq {\rho}_{u}(\gamma) \geq \epsilon \end{aligned} $$

(6.30)

so that:

$$\displaystyle \begin{aligned} \forall \tau \in [0,t], \quad |{\rho}_{\tau}(x) - {\rho}_{\tau}(y)| \leq |x-y| + \frac{|\delta-1|}{2} \int_{0}^{\tau}\frac{ |{\rho}_{u}(x) - {\rho}_{u}(y)|}{\epsilon^{2}} du, \end{aligned}$$

which, by Grönwall’s inequality, implies that:

$$\displaystyle \begin{aligned} |{\rho}_{t}(x) - {\rho}_{t}(y)| \leq |x-y| \exp \left( \frac{|\delta-1|}{2\epsilon^{2}} t \right). \end{aligned} $$

(6.31)

Moreover, we have:

$$\displaystyle \begin{aligned} {\rho}_{s}(y) - {\rho}_{t}(y) = \frac{\delta-1}{2} \int_{t}^{s} \frac{du}{{\rho}_{u}(y)} + B_{s} - B_{t} \end{aligned}$$

which, by (6.30), entails the inequality:

$$\displaystyle \begin{aligned} |{\rho}_{s}(y) - {\rho}_{t}(y)| \leq \frac{|\delta-1|}{2\epsilon} |s-t| + |B_{s} - B_{t}| . \end{aligned} $$

(6.32)

Putting inequalities (6.31) and (6.32) together yields the claimed inequality (6.29). Hence, we have, a.s., for all rationals x, y > γ and all t ≤ s < T _𝜖(γ):

$$\displaystyle \begin{aligned} |{\rho}_{t}(x) - {\rho}_{s}(y)| \leq |x-y| \exp \left( \frac{|\delta-1|}{2\epsilon^{2}} t \right) + \frac{\delta-1}{2} |s-t| + |B_{s} - B_{t}| \end{aligned}$$

and, by density of $\mathbb {Q} \cap (\gamma , + \infty )$ in (γ, +∞), this inequality remains true for all x, y > γ. Since, a.s., t↦B _t is continuous on $\mathbb {R}_{+}$, the continuity of ρ on $\mathbb {U}_{\gamma }^{\epsilon }$ is proved.

Corollary 6.5

Almost-surely, we have:

$$\displaystyle \begin{aligned} \forall x \geq 0, \quad \forall t \in [0, T_{0}(x)), \qquad {\rho}_{t}(x) = x + \frac{\delta - 1}{2} \int_{0}^{t} \frac{du}{{\rho}_{u}(x)} + B_{t} . \end{aligned} $$

(6.33)

Remark 6.15

We have already remarked in Sect. 6.3 that, for all fixed x ≥ 0, the process (ρ _t(x))_t≥0 satisfies the SDE (6.10). By contrast, the above Corollary shows the stronger fact that, considering the modification $\tilde {\rho }$ of the Bessel flow constructed in Proposition 6.2 above, a.s., for each x ≥ 0, the path $(\tilde {\rho }_{t}(x))_{t \geq 0}$ still satisfies relation (6.10).

Proof

Consider an almost-sure event $\mathbb {A} \in \mathbb {F}$ as in Remark 6.7. On the event $\mathbb {A}$, for all $r \in \mathbb {Q}_{+}$, we have:

$$\displaystyle \begin{aligned} \forall t \in [0, T_{0}(r)), \qquad {\rho}_{t}(r) = r + \frac{\delta-1}{2}\int_{0}^{t} \frac{du}{{\rho}_{u}(r)} + B_{t}. \end{aligned}$$

Denote by $\mathbb {B} \in \mathbb {F}$ any almost-sure event on which ρ satisfies the monotonicity property (6.15). We show that, on the event $\mathbb {A} \cap \mathbb {B}$, the property (6.33) is satisfied.

Suppose $\mathbb {A} \cap \mathbb {B}$ is fulfilled, and let x ≥ 0. Then for all $r \in \mathbb {Q}$ such that r ≥ x, we have:

$$\displaystyle \begin{aligned} \forall t \geq 0, \qquad \rho_{t} (x) \leq \rho_{t}(r) \end{aligned}$$

so that T ₀(r) ≥ T ₀(x). Hence, for all t ∈ [0, T ₀(x)), we have in particular t ∈ [0, T ₀(r)), so that:

$$\displaystyle \begin{aligned}{\rho}_{t}(r) = r + \frac{\delta-1}{2}\int_{0}^{t} \frac{du}{{\rho}_{u}(r)} + B_{t} . \end{aligned}$$

Since, for all u ∈ [0, t], ρ _u(r) ↓ ρ _u(x) as r ↓ x with $r \in \mathbb {Q}$, by the monotone convergence theorem, we deduce that:

$$\displaystyle \begin{aligned} \int_{0}^{t} \frac{du}{{\rho}_{u}(r)} \longrightarrow \int_{0}^{t} \frac{du}{{\rho}_{u}(x)} \end{aligned}$$

as r ↓ x with $r \in \mathbb {Q}$. Hence, letting r ↓ x with $r \in \mathbb {Q}$ in the above equation, we obtain:

$$\displaystyle \begin{aligned} {\rho}_{t}(x) = x + \frac{\delta-1}{2}\int_{0}^{t} \frac{du}{{\rho}_{u}(x)} + B_{t} . \end{aligned}$$

This yields the claim.

One of the main difficulties for proving Proposition 6.3 arises from the behavior of ρ _t(x) at t = T ₀(x). However we will circumvent this problem by working away from the event t = T ₀(x). To do so, we will make use of the following property.

Lemma 6.5

Let δ < 2 and x ≥ 0. Then the function y↦T ₀(y) is a.s. continuous at x.

Proof

The function y↦T ₀(y) is nondecreasing over $\mathbb {R}_{+}$. Hence, if x > 0, it has left- and right-sided limits at x, T ₀(x ⁻) and T ₀(x ⁺), satisfying:

$$\displaystyle \begin{aligned} T_{0}(x^{-}) \leq T_{0}(x) \leq T_{0}(x^{+}). \end{aligned} $$

(6.34)

Similarly, if x = 0, there exists a right-sided limit T ₀(0⁺) satisfying T ₀(0) ≤ T ₀(0⁺). Suppose, e.g., that x > 0. Then we have:

$$\displaystyle \begin{aligned} \mathbb{E} \left( e^{- T_{0}(x^{+})} \right) \leq \mathbb{E} \left(e^{- T_{0}(x)} \right) \leq \mathbb{E} \left(e^{- T_{0}(x^{-})} \right). \end{aligned} $$

(6.35)

Now, by the scaling property of the Bessel processes (see, e.g., Remark 3.7 in [14]), for all y ≥ 0, the following holds:

$$\displaystyle \begin{aligned} (y \rho_{t}(1))_{t \geq 0} \overset{(d)}{=} (\rho_{y^{2}t}(y))_{t \geq 0}, \end{aligned}$$

so that $T_{0}(y) \overset {(d)}{=} y^{2} T_{0}(1)$. Therefore, using the dominated convergence theorem, we have:

$$\displaystyle \begin{aligned} \mathbb{E} \left( e^{- T_{0}(x^{+})} \right) &= \underset{y \downarrow x}{\lim} \ \mathbb{E} \left( e^{- T_{0}(y)} \right) \\ &= \underset{y \downarrow x}{\lim} \ \mathbb{E} \left( e^{- y^{2} T_{0}(1)} \right) \\ &= \mathbb{E} \left( e^{- x^{2} T_{0}(1)} \right) \\ &= \mathbb{E} \left( e^{- T_{0}(x)} \right). \end{aligned} $$

Similarly, we have $\mathbb {E} \left ( e^{- T_{0}(x^{-})} \right ) = \mathbb {E} \left ( e^{- T_{0}(x)} \right )$. Hence the inequalities (6.35) are actually equalities; recalling the original inequality (6.34), we deduce that T ₀(x ⁻) = T ₀(x) = T ₀(x ⁺) a.s.. Similarly, if x = 0, we have T ₀(0) = T ₀(0⁺) a.s.

Before proving Proposition 6.3, we need a coalescence lemma, which will help us prove that the derivative of ρ _t at x is 0 if t > T ₀(x):

Lemma 6.6

Let x, y ≥ 0, and let τ be a nonnegative $(\mathbb {F}_{t})_{t \geq 0}$ -stopping time. Then, almost-surely:

$$\displaystyle \begin{aligned}\rho_{\tau}(x) = \rho_{\tau}(y) \quad \Rightarrow \quad \forall s \geq \tau, \ \rho_{s}(x) = \rho_{s}(y) . \end{aligned}$$

Proof

On the event {ρ _τ(x) = ρ _τ(y)}, the processes $(X^{\delta }_{t}(x))_{t \geq 0} := (\rho _{t}(x)^{2})_{t \geq 0}$ and $(X^{\delta }_{t}(y))_{t \geq 0} := (\rho _{t}(y)^{2})_{t \geq 0}$ both satisfy, on [τ, +∞), the SDE:

$$\displaystyle \begin{aligned} X_{t} = \rho_{\tau}(x)^{2} + 2 \int_{\tau}^{t} \sqrt{X_{s}} dB_{s} + \delta (t-\tau) . \end{aligned}$$

By pathwise uniqueness of this SDE (see [10, Theorem (3.5), Chapter IX]), we deduce that, a.s. on the event {ρ _τ(x) = ρ _τ(y)}, X _t(x) = X _t(y), hence ρ _t(x) = ρ _t(y) for all t ≥ τ.

Now we are able to prove Proposition 6.3.

Proof (Proof of Proposition 6.3)

Let t > 0 and x > 0 be fixed. First remark that:

$$\displaystyle \begin{aligned} \mathbb{P}(T_{0}(x) = t) = 0. \end{aligned}$$

Indeed, if δ > 0, then:

$$\displaystyle \begin{aligned} \mathbb{P} ( T_{0}(x) = t ) \leq \mathbb{P} ( \rho_{t}(x) =0) \end{aligned}$$

and the RHS is zero since the law of ρ _t(x) has no atom on $\mathbb {R}_{+}$ (it has density $p^{\delta }_{t}(x,\cdot )$ w.r.t. Lebesgue measure on $\mathbb {R}_{+}$, where $p^{\delta }_{t}$ was defined in Eq. (6.12) above). On the other hand, if δ = 0, then 0 is an absorbing state for the process ρ, so that, for all s ≥ 0:

$$\displaystyle \begin{aligned} \mathbb{P}(T_{0}(x) \leq s) = \mathbb{P} ( \rho_{s}(x) =0) \end{aligned}$$

and the RHS is continuous in s on $\mathbb {R}_{+}$, since it is given by $\exp (-\frac {x^{2}}{2s})$ (see [10, Chapter XI, Corollary 1.4]). Hence, also in the case δ = 0 the law of T ₀(x) has no atom on $\mathbb {R}_{+}$. Hence, a.s., either t < T ₀(x) or t > T ₀(x).

First suppose that t < T ₀(x). A.s., the function y↦T ₀(y) is continuous at x, so there exists a rational number y ∈ [0, x) such that t < T ₀(y); since, by Remark (6.7), t↦ρ _t(y) is continuous, there exists $\epsilon \in \mathbb {Q}_{+}^{*}$ such that t < T _𝜖(y). By monotonicity of z↦ρ(z), for all s ∈ [0, t] and z ≥ y, we have:

$$\displaystyle \begin{aligned}\rho_{s}(z) \geq \rho_{s}(y) \geq \epsilon . \end{aligned}$$

Hence, recalling Corollary 6.5, for all s ∈ [0, t] and $h \in \mathbb {R}$ such that |h| < |x − y|:

$$\displaystyle \begin{aligned} \rho_{s}(x+h) = x + h + \int_{0}^{s} \frac{\delta-1}{2}\frac{du}{\rho_{u}(x+h)} + B_{s}. \end{aligned}$$

Hence, setting $\eta ^{h}_{s}(x):=\frac {\rho _{s}(x+h) - \rho _{s}(x)}{h}$,we have:

$$\displaystyle \begin{aligned} \forall s \in [0,t], \qquad \eta^{h}_{s}(x) = 1 - \frac{\delta - 1}{2} \int_{0}^{t} \frac{\eta^{h}_{u}(x)}{\rho_{u}(x) \rho_{u}(x+h)} du \end{aligned}$$

so that:

$$\displaystyle \begin{aligned} \eta^{h}_{t}(x) = \exp \left( \frac{1-\delta}{2} \int_{0}^{t} \frac{ds}{\rho_{s}(x) \rho_{s}(x+h)} \right).\end{aligned} $$

Note that, for all s ∈ [0, t] and $h \in \mathbb {R}$ such that |h| < |x − y|, we have $(s, x + h) \in [0, T_{\epsilon }(y)) \times (y, + \infty ) \subset \mathbb {U}$. Hence, by Lemma 6.4, we have, for all s ∈ [0, t]

$$\displaystyle \begin{aligned} \rho_{s}(x+h) \underset{h \to 0}{\longrightarrow} \rho_{s}(x) \end{aligned}$$

with the domination property:

$$\displaystyle \begin{aligned} \frac{1}{\rho_{s}(x) \rho_{s}(x+h)} \leq \epsilon^{-2}\end{aligned} $$

valid for all |h| < |x − y|. Hence, by the dominated convergence theorem, we deduce that:

$$\displaystyle \begin{aligned} \eta^{h}_{t}(x) \underset{h \to 0}{\longrightarrow} \exp \left( \frac{1-\delta}{2} \int_{0}^{t} \frac{ds}{\rho_{s}(x)^{2}} \right) \end{aligned}$$

which yields the claimed differentiability of ρ _t at x.

We now suppose that t > T ₀(x). Since the function y↦T ₀(y) is a.s. continuous at x, a.s. there exists $y > x, \ y \in \mathbb {Q}$, such that t > T ₀(y). By Remark (6.7), the function t↦ρ _t(y) is continuous, so that $\rho _{T_{0}(y)}(y) = 0$. By monotonicity of z↦ρ(z), we deduce that, for all z ∈ [0, y], we have:

$$\displaystyle \begin{aligned} \rho_{T_{0}(y)}(z) = 0 . \end{aligned}$$

By Lemma 6.6, we deduce that, leaving aside some event of probability zero, all the trajectories (ρ _t(z))_t≥0 for $z \in [0,y]\cap \mathbb {Q}$ coincide from time T ₀(y) onwards. In particular, we have:

$$\displaystyle \begin{aligned} \forall z \in [0,y]\cap \mathbb{Q}, \qquad \rho_{t}(z) = \rho_{t}(x) . \end{aligned}$$

Since, moreover, the function z↦ρ _t(z) is nondecreasing, we deduce that it is constant on the whole interval [0, y]:

$$\displaystyle \begin{aligned} \forall z \in [0,y], \qquad \rho_{t}(z) = \rho_{t}(x) . \end{aligned}$$

In particular, the function z↦ρ _t(z) has derivative 0 at x. This concludes the proof.