2. Preliminaries

Let X be a diffusion on $\mathcal{C}(E,\mathcal{E})$ , the canonical space of all continuous functions on E with an associated $\sigma$ -field of Borel subsets of E. $X_0 = x$ almost surely (a.s.) under $\mathrm{P}_x$ , and we denote by $\mathrm{P}\,:\!=\mathrm{P}_0$ the measure under which $X_0 = 0$ a.s. The filtration $(\mathcal{F}_t^X)_{t\geq 0}$ generated by X is augmented to satisfy the usual conditions. By $\theta$ we denote the shift operator. We assume that the state space E of the process X is a subset of the real line, and that $0\in E$ in the sense that X attains 0 during the lifetime with a positive probability. If the diffusion is recurrent we may follow the description of Revuz and Yor [Reference Revuz and Yor20, Chapter XII] for excursions of recurrent processes. Otherwise we may follow the more general approach of [Reference Blumenthal2].

Let U be the subspace of the canonical space consisting of functions $u\,:\, E \rightarrow \mathbb{R}$ such that $u(0) = 0$ , $R(u) = \inf\{t>0\,:\, u(t) = 0\}$ , and $u(t) = 0$ for all $t\geq R(u)$ . If the process is transient, R may be infinite. Set $\mathcal{U}$ to be the $\sigma$ -algebra generated by the coordinate mappings. Let $\delta \equiv 0$ . By $U^{\delta}$ we denote the space $U\cup\{\delta\}$ , and $\mathcal{U}^{\delta} = \sigma(\mathcal{U},\{\delta\})$ . If the process is recurrent, an associated point process is a Poisson point process (PPP) $(e_s)_{s> 0}$ . The excursion measure on the path space (a subspace of the canonical space) will be denoted Ŷ after [Reference Blumenthal2, Chapter III, Theorem 3.24]. It is well known [Reference Revuz and Yor20, Chapter XII, Theorem 4.1] that the coordinate process restricted to (0, R) is, under Ŷ, a homogeneous strong Markov process with a transition semi-group of a process killed at 0 and an entrance law given by

\begin{align*} \eta_t({\rm d} y) = g_{y0}(t)m({\rm d} y) , \qquad t>0,\end{align*}

where $g_{y0}$ denotes the density of the first hitting time of 0 under $\mathrm{{P}}_y$ , and m is the speed measure of the original Markov process (see [Reference Salminen, Vallois and Yor22, Theorem 2]). For example, the entrance law of Brownian motion with non-negative drift $\mu$ (with speed measure given by $m({\rm d} y) = 2{\rm e}^{2\mu y}{\rm d} y$ ) is given on $(0,\infty)$ by

\begin{align*} \eta_t({\rm d} y) = \frac{y}{\sqrt{2\pi t^3}}{\rm e}^{-\frac{(y+\mu t)^2}{2t}}2{\rm e}^{2\mu y}{\rm d} y = y\sqrt{\frac{2}{\pi t^3}}{\rm e}^{-\frac{(y-\mu t)^2}{2t}}{\rm d} y ,\end{align*}

(see [Reference Borodin and Salminen3, pp. 127, 295]). It is also well known that the distribution of R under Ŷ is given on $\mathbb{R}_+$ by $ \mathrm{\widehat{P}}(R> t) = \int_0^{\infty}g_{y0}(t)m({\rm d} y).$ Define $ \sigma_0 = \inf{\{t\,:\, X_t = 0\}}$ and assume that 0 is regular in the sense that $\mathrm{P}(\sigma_0 = 0) = 1$ . By $L^{(a)}(X)$ we denote the local time of X at $a\in E$ , normalized in line with the setup of excursion theory, that is,

\begin{align*}\int_0^{t}h(X_s)\,{\rm d} s = \int_E h(a)L^{(a)}_t(X)m({\rm d} a),\end{align*}

for any non-negative measurable function h (see [Reference Salminen, Vallois and Yor22, Section 1(v)]). We denote $L(X) \,:\!= L^{(0)}(X)$ . In all considerations below, $p_t(\cdot,\cdot)$ denotes a transition density with respect to the speed measure m. In particular, it is well known that

(2) \begin{align} \mathrm{E}(L_t(X)) = \int_0^tp_u(0,0)\,{\rm d} u\end{align}

[Reference Borodin and Salminen3, Chapter II, Section 2(d)]. Since we will use the Tanaka formula we need to adjust a normalizing constant. Namely, if

(3) \begin{align} {\rm d}|X_t| = {\rm sgn}(X_t){\rm d} X_t + {\rm d} L^*_t(X),\end{align}

then it is well known that there exists a strictly positive constant $c^*$ such that $L^*(X) = c^*L(X)$ . The notation $c^*$ will be reserved explicitly for this constant. Below, it will be called a local time constant. If X is a standard Brownian motion, then $c^* = \frac{1}{2}$ . In particular, the formula for $c^*$ follows from (2) and (3):

\begin{align*} c^* = \bigg(\mathrm{E}\bigg(|X_t| -\int_0^t{\rm sgn}(X_s)\,{\rm d} X_s \bigg)\bigg)\bigg(\int_0^tp_u(0,0)\,{\rm d} u\bigg)^{-1},\end{align*}

where $t>0$ , and it follows that the right-hand side of the last formula does not depend on t.

Let us sum up the problem of normalization of local time. If s is given then the speed measure is affected by this choice to preserve $\frac{{\rm d}}{{\rm d}m}\frac{{\rm d}}{{\rm d}s} = \mathcal{A}$ (and $\mathcal{A}$ denotes the generator of a diffusion). Since for any Borel A we have $\mathrm{P}_x(X_t\in A) = \int_Ap_{t}(x,z)m({\rm d} z)$ , it is clear that the last choice affects $p_t$ and, by the identity $\mathrm{E}_x(L_t) = \int_0^tp_u(x,0)\,{\rm d} u$ , it determines the normalizing constant of local time L. So, if we set L to satisfy (2), then we define a local time constant $c^*$ by $L^*(X) = c^*L(X)$ .

3. Brownian motion with drift reflecting at 0

Let $(B_t)$ be a Brownian motion considered on $\mathcal{C}([0,\infty),\mathcal{B})$ , the canonical space of all continuous functions on $[0,\infty)$ . Let $X_t = B_t + \mu t$ , $\mu\geq 0$ . If $\mu=0$ the process is recurrent. If $\mu>0$ it is well known that under $\mathrm{P}_x$ the process X is a transient diffusion, and for any $x>0$ we have $\mathrm{P}_x(\sigma_0<\infty) = {\rm e}^{-2\mu x}$ . This implies that the total local time at 0 of the process X, i.e. $L_{\infty}(X)$ , is finite a.s., and the associated point process of excursions cannot be Poisson (see [Reference Blumenthal2, Chapter III, Section (g)]). However, the setup for the excursion path space $(U,\mathcal{U})$ for the non-recurrent case is, except for the fact that R can now be infinite, the same as in the recurrent one. It turns out that the effective substitute of the PPP is obtained in this case by considering the conditional probability given $\{\tau_t < \infty\}$ , where $(\tau_t)$ denotes the inverse of the local time L(X), i.e. $\tau_t = \inf\{s>0\,:\, L_s(X) > t\}$ (obviously $\tau_{t-} = \inf\{s>0\,:\, L_s(X) \geq t\}$ ). Thus, the fundamental excursion formula [Reference Blumenthal2, Chapter III, (3.27)] and the Markov property under the excursion measure remain valid (see [Reference Blumenthal2, Chapter III, Section (g)]). In case the process is transient, the natural decomposition of the excursion measure is

\begin{align*} \mathrm{\widehat{P}}_1(A) = \mathrm{\widehat{P}}(A\cap\{R = \infty\}), \quad \mathrm{\widehat{P}}_2(A) = \mathrm{\widehat{P}}(A\cap\{R < \infty\}), \quad A\in\mathcal{B}(\mathbb{R}_+).\end{align*}

If $\mu >0$ and $P_0(t,x,A)$ denotes the transition function of the Brownian motion with drift $\mu$ killed at 0, then it is well known that for the entrance law $\{\eta_t, t>0\}$ and $\Delta(x) = \mathrm{P}_x(\sigma_0 = \infty)$ , the coordinate process $(X_t,t>0)$ under the measure $\mathrm{\widehat{P}}_1$ is a homogeneous Markov process with transition function $Q^1_t(x,{\rm d} z) = P_0(t,x,\Delta(z){\rm d} z)/\Delta(x)$ and entrance law $\Delta(z)\eta_t({\rm d} z)$ (see [Reference Blumenthal2, Chapter III, Theorem 3.29]). Relative to $\mathrm{\widehat{P}}_2$ the coordinate process is a time-homogeneous Markov process with transition function $Q^2_t(x,{\rm d} z) = P_0(t,x,(1-\Delta(z)){\rm d} z)/(1-\Delta(x))$ and entrance law $(1-\Delta(z))\eta_t({\rm d} z)$ .

Consider now functionals of Brownian motion with drift reflecting at 0. Since we consider excursions from 0 we will assume that the process starts from 0. It is well known that such a process is realized by $|X|$ , where X is a strong solution of the stochastic differential equation

(4) \begin{align} {\rm d} X_t = {\rm d} B_t + \mu \, {\rm sgn}(X_t){\rm d} t, \qquad X_0 = 0,\end{align}

and B is a standard Brownian motion (see [Reference Peskir13]). We will consider functionals of $|X|$ under the excursion measure of X denoted by $\widehat{\mathrm{P}}$ . We will see in Remark 3.2 that there is a simple relation between the excursion measures for X and $|X|$ . If $\mu = 0$ then $X = B$ . Recall the distributional Lévy’s theorem

\begin{align*} (S-B,S)_{t\geq 0} \stackrel{{\rm law}}{=} (|B|, L^*(B))_{t\geq 0} ,\end{align*}

(see [Reference Revuz and Yor20, Chapter VI, Theorem 2.3]). We will use the distributional Lévy’s theorem given by Graversen and Shiryaev [Reference Graversen and Shiryaev9, Theorem 1] and extended by Peskir [Reference Peskir13]. We have

(5) \begin{align} (S^{({-}\mu)}-B^{({-}\mu)},S^{({-}\mu)})_{t\geq 0} \stackrel{{\rm law}}{=} (|X|, L^*(X))_{t\geq 0},\end{align}

where B is a standard Brownian motion, $B^{({-}\mu)}_t = B_t - \mu t$ , $S^{({-}\mu)}_t = \sup_{u\leq t}B^{({-}\mu)}_u$ , and $L^*$ is a local time compatible with the Tanaka formula. We will further assume that $\mu\geq 0$ .

We will now present a description of the integral functionals of reflecting Brownian motion with non-negative drift under the excursion measure $\widehat{\mathrm{P}}$ , and for $\mu>0$ under the restricted measures $\widehat{\mathrm{P}}_1$ and $\widehat{\mathrm{P}}_2$ . Let $\sigma_z = \inf\{t>0\,:\, X_t =z\}$ , $z\geq 0$ . For a non-negative, Borel function f on $[0,\infty)$ , $\lambda > 0$ , and $x\geq 0$ , define

(6) \begin{align} h^{(\mu)}_{\lambda, f}(x) &= \mathrm{E}_x\exp{\bigg({-}\lambda\int_0^{\sigma_0}f(|X_s|)\,{\rm d} s\bigg)}.\end{align}

Notice that the generator of $|X|$ is $\mathcal{A}\,f = \frac12 \frac{{\rm d}^2f}{{\rm d}x^2} + \mu\frac{{\rm d}f}{{\rm d}x}$ , $x >0$ , and $\mathcal{A}\,f(0) = \frac12\, f''(0) + \mu\,f'(0)$ . The domain of $\mathcal{A}$ is $\{\,f\,:\, \,f,\mathcal{A}\,f\in\mathcal{C}_{\rm b}[0,\infty), f'(0+) = 0\}$ , where $\mathcal{C}_{\rm b}(E)$ is a set of continuous and bounded functions on E (see [Reference Borodin and Salminen3, p. 129]). Thus, by considering the classical Dirichlet problem on a finite interval we find that whenever $h^{(\mu)}_{\lambda, f}\in\mathcal{C}^2(0,\infty)$ and f is locally integrable then $h^{(\mu)}_{\lambda, f}$ solves

\begin{align*} \frac12\frac{{\rm d}^2u(x)}{{\rm d}x^2} + \mu\frac{{\rm d}u(x)}{{\rm d}x} & = \lambda f(x)u(x), \qquad x\in(0,\infty),\\[5pt] u(0) &= 1.\end{align*}

Proposition 3.1. Let f be a non-negative Borel function on $[0,\infty)$ . If

\begin{align*} \int_0^{\infty}{\rm e}^{-\mu y}f(y)m({\rm d} y) < \infty\end{align*}

then $h^{(\mu)}_{\lambda, f}$ given by (6) has the first right-hand side derivative in 0 and $(h^{(\mu)}_{\lambda, f})'$ is locally bounded on $[0,\infty)$ .

Proof.Clearly $x\mapsto h^{(\mu)}_{\lambda, f}(x)$ is bounded, so the assertion is true on $[\delta,\infty)$ for any fixed $\delta >0$ . Thus it is enough to prove the proposition in $(0,\delta)$ . For x small and positive we have

(7) \begin{align} \frac{h^{(\mu)}_{\lambda, f}(x) - 1}{x}\approx -\frac{\lambda}{x}\ \mathrm{E}_x\int_0^{\sigma_0}f(X_s)\,{\rm d} s.\end{align}

Observe that, by monotone convergence,

\begin{align*} \mathrm{E}_x\int_0^{\sigma_0}f(X_s)\,{\rm d} s = \lim_{z\rightarrow\infty}\mathrm{E}_x\int_0^{\sigma_0\wedge\sigma_z}f(X_s)\,{\rm d} s.\end{align*}

Since $s(y) = (1-{\rm e}^{-2\mu y})/y$ is a scale function of $|X|$ (see [Reference Borodin and Salminen3, p. 129]), we have, from [Reference Revuz and Yor20, Chapter VII, Corollary 3.8],

\begin{align*}\lim_{z\rightarrow\infty}\mathrm{E}_x\int_0^{\sigma_0\wedge\sigma_z}f(X_s)\,{\rm d} s = 2\mu \, s(x) \int_0^{\infty}{\rm e}^{-\mu y}f(y)m({\rm d} y).\end{align*}

We finish the proof by taking $x\rightarrow 0$ in (7).

We define the integral functional

\begin{align*} A_{R}^f(u) = \int_0^{R(u)}f(|u(s)|)\,{\rm d} s,\end{align*}

where f is some given non-negative, Borel function on $[0,\infty)$ . Obviously, in the above setup $u\mapsto A_{R}^f(u)$ is measurable. The integral functional of $f(|X|)$ will be denoted in the same manner by $A^{f}$ , that is, for any $t>0$ ,

\begin{align*} A^{f}_t = \int_0^tf(|X_s|)\,{\rm d} s, \qquad A^f_{\infty} = \lim_{t\rightarrow\infty}A^{f}_t .\end{align*}

For $t>0$ we denote by $g_t$ the last 0 of X before the moment t, that is, $ g_t = \sup\{s \leq t\,:\, X_s = 0\}.$ Since $t\mapsto g_t$ is non-decreasing, there exists a limit $\widehat{g} = \lim_{t\rightarrow\infty}g_t$ , and since X may be transient, $\widehat{g}$ may be finite a.s.

Theorem 3.1. Under hypothesis A,

(8) \begin{align} \mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A_R^f}\big) = -c^*(h^{(\mu)}_{\lambda, f})'(0),\end{align}

where $c^*$ is a local time constant. Moreover, for $\mu >0$ ,

(9) \begin{align}\quad \mathrm{\widehat{P}}_1\big(1 - {\rm e}^{-\lambda A_R^f}\big) &= -c^*(h^{(\mu)}_{\lambda, f})'(0)\frac{\mathrm{E}\big[{\rm e}^{-\lambda A^{f}_{\widehat{g}}}-{\rm e}^{-\lambda A^{f}_{\infty}}\big]}{\mathrm{E}\big[1 - {\rm e}^{-\lambda A^{f}_{\infty}}\big]},\end{align}

(10) \begin{align}\mathrm{\widehat{P}}_2\big(1 - {\rm e}^{-\lambda A_R^f}\big) &= -c^*(h^{(\mu)}_{\lambda, f})'(0)\frac{\mathrm{E}\big[1-{\rm e}^{-\lambda A^{f}_{\widehat{g}}}\big]}{\mathrm{E}\big[1 - {\rm e}^{-\lambda A^{f}_{\infty}}\big]}.\quad\end{align}

Proof. Fix $t>0$ . Let $d_t = \inf\{s\geq t\,:\, X_s = 0 \}$ . Note that we have $d_t = t+ \sigma_0 \circ \theta_t$ . For fixed $t>0$ and $\lambda > 0$ , we write the integral $\mathrm{E}\int_0^{d_t} {\rm e}^{-\lambda\int_0^sf(|X_v|)dv}f(|X_s|)\,{\rm d} s$ in terms of excursions of X. Recall that L(X) is the process of the local time of X at 0, and $\tau_t = \inf\{s\,:\, L_s(X)> t\}$ is its inverse. Both processes are used to parametrize excursions of X from the point 0. We have

(11) \begin{align} \mathrm{E}\big(1 - {\rm e}^{-\lambda A^f_{d_t}}\big) &= \lambda\mathrm{E}\int_0^{d_t}{\rm e}^{-\lambda A^f_u}f(|X_u|)\,{\rm d} u\end{align}

\begin{align} &= \mathrm{E}\sum_{\tau_{s-}\leq t}\int_{\tau_{s-}}^{\tau_s}{\rm e}^{-\lambda\int_0^rf(|X_v|){\rm d} v}f(|X_r|)\,{\rm d} r\notag\\[3pt] &= \lambda\mathrm{E}\sum_{\tau_{s-}\leq t}{\rm e}^{-\lambda A^{f}_{\tau_{s-}}}\bigg(\int_{0}^{\sigma_0}{\rm e}^{-\lambda A^f_u}f(|X_u|)\,{\rm d} u\bigg)\circ\theta_{\tau_{s-}}\notag\\[3pt] &= \mathrm{E}\sum_{\tau_{s-}\leq t}{\rm e}^{-\lambda A^{f}_{\tau_{s-}}}\big(1 - {\rm e}^{-\lambda A^f_{\sigma_0}}\big)\circ\theta_{\tau_{s-}}.\notag\end{align}

The excursion formula [Reference Blumenthal2, Chapter III, (3.27)] applied to the last expression yields

(12) \begin{align} \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{d_t}}) = \bigg(\mathrm{E}\int_0^t{\rm e}^{-\lambda A^f_s}\,{\rm d} L_s(X)\bigg)\mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big).\end{align}

Recall that regardless of whether the process is recurrent or transient, there exists a $\sigma$ -finite measure Ŷ on the excursion path space U such that (12) holds (see [Reference Blumenthal2, Chapter III, Section (g)]). We now compute $\mathrm{E}\int_0^t{\rm e}^{-\lambda A^f_s}\,{\rm d} L_s(X)$ . Using equality in law (5) we write

\begin{align*}\mathrm{E}\int_0^t{\rm e}^{-\lambda A^f_s}\,{\rm d} L^*_s(X) = \mathrm{E}\int_0^t{\rm e}^{-\lambda \int_0^sf(S^{({-}\mu)}_u-B^{({-}\mu)}_u){\rm d} u}\,{\rm d} S^{({-}\mu)}_s.\end{align*}

It follows from (4) and Tanaka’s formula that $\left\langle |X|\right\rangle_t = t$ , so for $g\in\mathcal{C}^2([0,\infty))$ ,

\begin{align*} &\mathrm{E} g\big(S^{({-}\mu)}_t-B^{({-}\mu)}_t\big){\rm e}^{-\lambda\int_0^tf\big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big){\rm d} u}\\[3pt] &\quad = g(0) + \mathrm{E}\int_0^tg'(S^{({-}\mu)}_u-B^{({-}\mu)}_u){\rm e}^{-\lambda\int_0^uf(S^{({-}\mu)}_v-B^{({-}\mu)}_v){\rm d} v}\,{\rm d} \big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big)\\[3pt] &\qquad+ \mathrm{E}\int_0^t{\rm e}^{-\lambda\int_0^uf\big(S^{({-}\mu)}_v-B^{({-}\mu)}_v\big){\rm d} v}\bigg[\frac12g''\big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big)- \lambda (\kern1.5ptfg)\big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big)\bigg]\,{\rm d} u,\end{align*}

which can be rewritten in the form

(13) \begin{align} &\mathrm{E}g\big(S^{({-}\mu)}_t-B^{({-}\mu)}_t\big){\rm e}^{-\lambda\int_0^tf\big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big){\rm d} u}\\[3pt] & \quad = g(0) + \mathrm{E}\int_0^tg'(S^{({-}\mu)}_u-B^{({-}\mu)}_u){\rm e}^{-\lambda\int_0^uf(S^{({-}\mu)}_v-B^{({-}\mu)}_v){\rm d} v}\,{\rm d} S^{({-}\mu)}_u\notag\\[3pt] &\qquad-\mathrm{E}\int_0^tg'(S^{({-}\mu)}_u-B^{({-}\mu)}_u){\rm e}^{-\lambda\int_0^uf(S^{({-}\mu)}_v-B^{({-}\mu)}_v){\rm d} v}\,{\rm d} B_u\notag\\[3pt] &\qquad+ \mathrm{E}\int_0^t{\rm e}^{-\lambda\int_0^uf\big(S^{({-}\mu)}_v-B^{({-}\mu)}_v\big){\rm d} v}\bigg[\frac12g''\big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big)\notag\\[3pt] &\qquad+\mu g'\big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big) - \lambda (\kern1.5ptfg)\big(S^{({-}\mu)}_u-B^{({-}\mu)}_u\big)\bigg]\,{\rm d} u.\notag\end{align}

Since $f\geq 0$ , to eliminate (on the right-hand side of the last equality) the expectation of the stochastic integral, it is enough to assume that gʹ is locally bounded and use some standard localization arguments. Since it is assumed that $(h^{(\mu)}_{\lambda, f})'$ is locally bounded, we can set $g = h^{(\mu)}_{\lambda, f}$ to obtain, from (13) and from the fact that $h^{(\mu)}_{\lambda, f}$ solves the underlying Dirichlet problem,

\begin{multline*}\mathrm{E}h^{(\mu)}_{\lambda, f} \big(S^{({-}\mu)}_t-B^{({-}\mu)}_t\big){\rm e}^{-\lambda\int_0^tf(S^{({-}\mu)}_u-B^{({-}\mu)}_u){\rm d} u}\\ = 1 + \mathrm{E}\int_0^t(h^{(\mu)}_{\lambda, f})'(S^{({-}\mu)}_u-B^{({-}\mu)}_u){\rm e}^{-\lambda\int_0^uf(S^{({-}\mu)}_v-B^{({-}\mu)}_v){\rm d} v}\,{\rm d} S^{({-}\mu)}_u.\end{multline*}

The measure ${\rm d} S^{({-}\mu)}_u$ is non-zero on the set $\{ u\,:\, S^{({-}\mu)}_u-B^{({-}\mu)}_u = 0\}$ , so the last equality can be written in the form

\begin{multline*}\mathrm{E}h^{(\mu)}_{\lambda, f}\big(S^{({-}\mu)}_t-B^{({-}\mu)}_t\big) {\rm e}^{-\lambda\int_0^tf(S^{({-}\mu)}_u-B^{({-}\mu)}_u){\rm d} u}\\ = 1 + (h^{(\mu)}_{\lambda, f})'(0)\mathrm{E}\int_0^t{\rm e}^{-\lambda\int_0^uf(S^{({-}\mu)}_v-B^{({-}\mu)}_v){\rm d} v}\,{\rm d} S^{({-}\mu)}_u.\end{multline*}

Hence, using (5) and $L^* = c^*L$ , we have

(14) \begin{align}\mathrm{E}h^{(\mu)}_{\lambda, f}(|X_t|){\rm e}^{-\lambda\int_0^tf(|X_u|){\rm d} u} = 1 + c^*(h^{(\mu)}_{\lambda, f})'(0)\mathrm{E}\int_0^t{\rm e}^{-\lambda\int_0^uf(|X_v|){\rm d} v}\,{\rm d} L_u(X).\end{align}

We compute the left-hand side of (12). Since $|X|$ is Markov,

\begin{align*} \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{d_t}}) = \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{t} -\lambda (A^f_{\sigma_0})\circ\theta_t}) = \mathrm{E}\big(1 - {\rm e}^{-\lambda A^f_{t}}h^{(\mu)}_{\lambda, f}(|X_t|)\big).\end{align*}

Inserting the last expression and (14) in (12) yields

\begin{align*} -c^*(h^{(\mu)}_{\lambda, f})'(0)\mathrm{E}\big(1 - {\rm e}^{-\lambda A^f_{t}}h^{(\mu)}_{\lambda, f}(|X_t|)\big) = \big(1 - {\rm e}^{-\lambda A^f_{t}}h^{(\mu)}_{\lambda, f}(|X_t|)\big)\mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big),\end{align*}

which finishes the proof of (8). To prove (9), for $t>0$ and $\mu >0$ we consider the sum

(15) \begin{align} I = \textrm{E}\sum_{\tau_{s-}\leq t}\textbf{1}_{\{\sigma_0\circ\theta_{\tau_{s-}} = \infty\}}{\rm e}^{-\lambda A^{f}_{\tau_{s-}}}\big[1-{\rm e}^{-\lambda A^f_u}\big]\circ\theta_{\tau_{s-}},\end{align}

which by the master excursion formula yields

\begin{align*} I = \bigg(\mathrm{E}\int_0^t{\rm e}^{-\lambda A^{X}_{s}}\,{\rm d} L_s(X)\bigg)\mathrm{\widehat{P}}_1\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big).\end{align*}

On the other side, following [Reference Blumenthal2, Chapter III, Theorem 3.29], we observe that under the sum in (15) there can be at most one excursion starting at some $\tau_{s-}\in \{r\,:\, \tau_r > \tau_{r-}\}$ such that $\sigma_0\circ\theta_{\tau_{s-}} = \infty$ . Moreover, if this $\tau_{s-}$ appears after the moment t, then $d_t < \infty$ and the sum under expectation in I is zero. If $\tau_{s-}\leq t$ , then $d_t = \infty$ , $\tau_{s-} = g_t$ , and $\{\sigma_0\circ\theta_{\tau_{s-}} = \infty\} = \{ d_t = \infty\}$ . As a result, we can rewrite I as

\begin{align*} I = \mathrm{E}\textbf{1}_{\{d_t = \infty\}}{\rm e}^{-\lambda A^f_{g_t}}\big[1 - {\rm e}^{-\lambda A^f_{\sigma_0}\circ\theta{g_t}}\big] = \mathrm{E}\textbf{1}_{\{d_t = \infty\}}\big[{\rm e}^{-\lambda A^f_{g_t}} - {\rm e}^{-\lambda A^f_{d_t}}\big],\end{align*}

since $d_t = g_t + \sigma_0\circ\theta{g_t}$ . So, what we actually obtained from the two different forms of I is

(16) \begin{align}\mathrm{E}\textbf{1}_{\{d_t = \infty\}}\big[{\rm e}^{-\lambda A^f_{g_t}} - {\rm e}^{-\lambda A^f_{d_t}}\big] = \bigg(\mathrm{E}\int_0^t{\rm e}^{-\lambda A^{X}_{s}}\,{\rm d} L_s(X)\bigg)\mathrm{\widehat{P}}_1\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big).\end{align}

By (8), we have

\begin{align*} -c^*(h^{(\mu)}_{\lambda, f})'(0)\mathrm{E}\int_0^t{\rm e}^{-\lambda A^{X}_{s}}\,{\rm d} L_s(X) = \mathrm{E}\big[1 - {\rm e}^{-\lambda A^f_{d_t}}\big],\end{align*}

so we can rewrite (16) as

\begin{align*} -c^* (h^{(\mu)}_{\lambda, f})'(0)\mathrm{E}\textbf{1}_{\{d_t = \infty\}}\big[{\rm e}^{-\lambda A^f_{g_t}} - {\rm e}^{-\lambda A^f_{d_t}}\big] = \mathrm{E}\big[1 - {\rm e}^{-\lambda A^f_{d_t}}\big]\mathrm{\widehat{P}}_1\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big).\end{align*}

Since both $t\rightarrow g_t$ and $t\rightarrow d_t$ are monotone and $d_t\geq t$ , we obtain from the last equality, by letting t tend to $\infty$ ,

\begin{align*}- c^*(h^{(\mu)}_{\lambda, f})'(0)\mathrm{E}\big[{\rm e}^{-\lambda A^f_{\widehat{g}}} - {\rm e}^{-\lambda A^f_{\infty}}\big] = \mathrm{E}\big[1 - {\rm e}^{-\lambda A^f_{\infty}}\big]\mathrm{\widehat{P}}_1\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big),\end{align*}

that is, (9). The equality (10) follows easily from (8) and (9), since

\begin{align*}\mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big) = \mathrm{\widehat{P}}_1\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big) + \mathrm{\widehat{P}}_2\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf)}\big).\end{align*}

This completes the proof.

Remarks 3.1. The local time constant $c^*$ in Theorem 3.1 may be computed as follows: for any $t>0$ ,

(17) \begin{align} c^* &= \frac{\phi(t) - \mu t}{\int_0^tp_u(0,0)\,{\rm d} u}, \quad \phi(t) = \int_0^{\infty}y\kappa(t,y)\,{\rm d} y - \mu\int_0^{\infty}\int_y^{\infty}\kappa(t,v)\,{\rm d} v\,{\rm d} y,\\[5pt] \kappa(t,y) &= \frac{1}{\sqrt{2\pi t}}{\rm e}^{-\frac{(y+\mu t)^2}{2t}}, \quad p_t(0,0) = \frac{1}{\sqrt{2\pi t}}{\rm e}^{-\frac{\mu^2}{2}t}-\frac{\mu}{\sqrt{\pi}}\int_{\mu\sqrt{t/2}}^{\infty}{\rm e}^{-z^2}\,{\rm d} z.\notag\end{align}

Indeed, from the definition of the local time constant

\begin{equation*} c^* = \mathrm{E}L^*_t(X)\bigg(\int_0^tp_u(0,0)\,{\rm d} u\bigg)^{-1},\end{equation*}

where from the Tanaka formula $\mathrm{E}L^*_t(X) = \mathrm{E}|X_t| -\mu t$ . Directly from the form of the transition density of $|X_t|$ (see [Reference Borodin and Salminen3, p. 130]) we have the formula for $p_t(0,0)$ and $\mathrm{E}|X_t| = \phi(t)$ , where $\phi$ is given by (17). Notice that $c^*$ does not depend on t.

Remarks 3.2. Since $L(|X|) = 2L(X)$ it follows from the proof of Theorem 3.1 that the excursion measure of $|X|$ is equal to $\frac12\widehat{\mathrm{P}}$ . The formula in (8) corresponds to the fact that for a standard Brownian motion B, the solution of the corresponding Sturm–Liouville equation determines the Lévy measure of a subordinator associated with the functional $\int_0^tf(B_s)\,{\rm d} s$ (unlike in our considerations where the functional is $\int_0^tf(|X_s|)\,{\rm d} s$ ), and hence determines the distribution of $A_R^f$ under the excursion measure (see [Reference Bertoin1, Proposition 9.3])

\begin{align*} \mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A^f_{R}}\big) = \frac12\big((g_{\lambda,f})'(0-)-(g_{\lambda,f})'(0+)\big),\end{align*}

where $g_{\lambda,f}$ is the unique bounded solution of the associated Sturm–Liouville equation.

Corollary 3.1. The density of R under the excursion measure Ŷ is given on $(0,\infty)$ by

(18) \begin{align} \mathrm{\widehat{P}}(R\in {\rm d} z) = \frac{c^*}{\sqrt{2\pi z^3}}{\rm e}^{-\frac{\mu^2}{2}z}\,{\rm d} z.\end{align}

Moreover,

\begin{align*} \int_0^{\infty}(1-{\rm e}^{-\lambda z})\frac{1}{\sqrt{2\pi z^3}}{\rm e}^{-\frac{\mu^2z}{2}}\,{\rm d} z = \mu + \sqrt{\mu^2 + 2\lambda}, \qquad \lambda >0.\end{align*}

Proof.It follows from (1), and we will also prove it independently in Corollary 4.2, that

\begin{align*}\mathrm{\widehat{P}}(R\in {\rm d} z) = \lim_{x\downarrow 0}\frac{c^*}{x}g_{x0}(z)\,{\rm d} z.\end{align*}

We fix $\lambda > 0$ and set $f(x) = \textbf{1}_{(0,\infty)}(x)$ , $x \geq 0$ . To simplify the notation set $h_{\lambda} \,:\!= h^{(\mu)}_{\lambda, f}$ . It follows that, for $x\geq 0$ ,

\begin{align*} h_{\lambda}(x) = \mathrm{E}_x{\rm e}^{-\lambda\sigma_0} = {\rm e}^{-\mu x - x \sqrt{\mu^2 + 2\lambda}}.\end{align*}

The formula in (18) follows since we may conclude by inverting the Laplace transform that

\begin{align*} g_{x0}(z) = \frac{x}{\sqrt{2\pi z^3}}{\rm e}^{-\mu x -\frac{\mu^2 z}{2}-\frac{x^2}{2z}}.\end{align*}

The second assertion follows from Theorem 3.1, since we have $-h'_{\!\lambda}(0) = \mu + \sqrt{\mu^2 + 2\lambda}$ .

Theorem 3.1 enables us to describe the distribution of the occupation time of excursions under $\widehat{\mathrm{P}}$ . Assume that $\mu = 0$ , so $|X|$ is a recurrent process. For a given $\alpha > 0$ we define the occupation time for excursions on the path space U as $T_{\alpha}:U \rightarrow [0,\infty)$ given by

\begin{align*} T_{\alpha}(u) = \int_0^{R(u)}\textbf{1}_{[\alpha,\infty)}(u(s))\,{\rm d} s,\end{align*}

and $\widetilde{T}_{\alpha}:U \rightarrow [0,\infty)$ by

\begin{align*} \widetilde{T}_{\alpha}(u) = \int_0^{R(u)}\textbf{1}_{(0,\alpha]}(u(s))\,{\rm d} s.\end{align*}

Clearly, both $T_{\alpha}$ and $\widetilde{T}_{\alpha}$ are measurable. Moreover, $T_{\alpha}(u) = A_R(\kern1.5ptf_{\alpha})(u)$ for $u\in U$ and $f_{\alpha}(x)= \textbf{1}_{[\alpha,\infty)}(x)$ , $x \geq 0$ , and $\widetilde{T}_{\alpha}(u) = A_R(g_{\alpha})(u)$ for $u\in U$ and $g_{\alpha}(x)= \textbf{1}_{(0,\alpha]}(x)$ , $x \geq 0$ . Recall that

\begin{align*} \Gamma(a,x) = \int_x^{\infty}t^{a-1}{\rm e}^{-t}\,{\rm d} t, \qquad a>0, \quad x\geq 0.\end{align*}

Proposition 3.2. For $\alpha > 0$ , $x\geq 0$ we have, on $(0,\infty)$ ,

(19) \begin{align} \widehat{\mathrm{P}}(T_{\alpha}\in {\rm d} y) = \frac{1}{4\sqrt{\pi}\alpha^2}{\rm e}^{y/(2\alpha)^2}\bigg[\Gamma\bigg(\frac12,\frac{y}{2\alpha^2}\bigg) -2\frac{\alpha}{\sqrt{2y}}\Gamma\bigg(1,\frac{y}{2\alpha^2}\bigg) \bigg]\,{\rm d} y.\end{align}

Moreover, if $\alpha \geq x$ then

(20) \begin{align} \widehat{\mathrm{P}}(\widetilde{T}_{\alpha}>t) = \frac{1}{\sqrt{ 2\pi t}}\sum_{n=1}^{\infty}({-}1)^{n}\big[{\rm e}^{-2\alpha^2n^2/t}- {\rm e}^{-2\alpha^2(n-1)^2/t}\big], \qquad t>0.\end{align}

Proof.Recall that $c^* = \frac{1}{2}$ for $\mu = 0$ . We first prove (19). We have, from [Reference Borodin and Salminen3, Section 2.4.1, p. 200],

\begin{align*} h_{\lambda}(x) = \mathrm{E}_x{\rm e}^{-\lambda \int_0^{\sigma_0}\textbf{1}_{(\alpha,\infty)}(B_s){\rm d} s} = \textbf{1}_{[\alpha,\infty)}(x)\frac{{\rm e}^{-x\sqrt{2\lambda}}}{1+\alpha\sqrt{2\lambda}\,} + \textbf{1}_{(0,\alpha]}(x)\frac{1+(\alpha-x)\sqrt{2\lambda}}{1+\alpha\sqrt{2\lambda}}.\end{align*}

Thus, $ h'_{\lambda}(0) = -\frac{\sqrt{2\lambda}}{1+\alpha\sqrt{2\lambda}\,},$ and it follows from Theorem 3.1 that

(21) \begin{align} \mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A_R(\kern1.5ptf_{\alpha})}\big) = \frac{1}{\sqrt{2}\,} \, \frac{\sqrt{\lambda}}{1+\alpha\sqrt{2\lambda}\,}.\end{align}

On the other side,

\begin{align*} \frac{\sqrt{\lambda}}{1+\alpha\sqrt{2\lambda}\,} &= \frac{1}{\sqrt{2}\alpha}\int_0^{\infty}\big(1 - {\rm e}^{-\alpha\sqrt{2\lambda}t}\big){\rm e}^{-t}\,{\rm d} t\\[3pt] &= \frac{1}{\sqrt{2}\alpha}\int_0^{\infty}\int_0^{\infty}\big(1 - {\rm e}^{-\alpha^2\lambda t^2/(2v)}\big)\frac{1}{\sqrt{\pi v}\,}{\rm e}^{-v-t}\,{\rm d} v\,{\rm d} t\\[3pt] &= \int_0^{\infty}\big(1 - {\rm e}^{-\lambda y}\big)\frac{\alpha}{2\sqrt{\pi y^3}\,}\bigg[\int_0^{\infty}t{\rm e}^{-t -\frac{\alpha^2t^2}{2y}}\,{\rm d} t \bigg]\,{\rm d} y =\!:\, I,\end{align*}

where in the last equality we used Fubini’s theorem. We compute the inner integral in the last expression:

(22) \begin{align} I & = \int_0^{\infty}\big(1 - {\rm e}^{-\lambda y}\big)\frac{\alpha}{2\sqrt{\pi y^3}\,}\frac{{\rm e}^{y/(2\alpha^2)}}{4(\alpha^2/(2y))^{\frac32}} \nonumber \\[3pt] & \quad \times \bigg[\Gamma\bigg(\frac12,y/(2\alpha^2)\bigg)-2\sqrt{\alpha^2/(2y)}\Gamma\big(1,y/(2\alpha^2)\big)\bigg]\,{\rm d} y.\end{align}

Equation (19) follows now from inserting (22) into (21). To prove (20), we use the formula [Reference Borodin and Salminen3, Section 2.5.1, p. 201]

\begin{align*} h_{\lambda}(x) = \mathrm{E}_x{\rm e}^{-\lambda\int_0^{\sigma_0}\textbf{1}_{[0,\alpha]}(B_s){\rm d} s} = \frac{\cosh\big(\sqrt{2\lambda}(\alpha - x)\big)}{\cosh\big(\sqrt{2\lambda}\alpha\big)}, \qquad x \geq 0, \quad \alpha\geq x.\end{align*}

Since $ h'_{\!\lambda}(0) = -\tanh(\alpha\sqrt{2\lambda})\sqrt{2\lambda},$ we have, from Theorem 3.1,

(23) \begin{align}\mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda\widetilde{T}_{\alpha}}\big) = \frac{1}{2} \, \tanh(\alpha\sqrt{2\lambda})\sqrt{2\lambda}.\end{align}

Observe that $\widetilde{T}_{\alpha} \leq R(u)$ . It is well known that $\mathrm{\widehat{P}}(R>t) = \frac{1}{\sqrt{\pi t}\,}$ for any $t>0$ (see [Reference Blumenthal2, p. 112]). As a result, $\mathrm{\widehat{P}}(\widetilde{T}_{\alpha}>t) < \infty$ , and from Fubini’s theorem,

(24) \begin{align} \mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda\widetilde{T}_{\alpha}}\big) = \lambda \int_0^{\infty}{\rm e}^{-\lambda t}\mathrm{\widehat{P}}(\widetilde{T}_{\alpha}>t)\,{\rm d} t.\end{align}

On the other side, it is well known that for any $p>0$

(25) \begin{align} \frac{\tanh(\alpha p)}{p} = \int_0^{\infty}{\rm e}^{-p t}\psi_{\alpha}(t){\rm d} t,\end{align}

where $ \psi_{\alpha}(t) = \sum_{n=1}^{\infty}({-}1)^{n-1}\textbf{1}_{[2\alpha(n-1),2\alpha n]}(t) ,$ (see [Reference Zhurov, Levitin and Polyanin24, table of inverse Laplace transforms – hyperbolic functions, p. 11]. Using $\mathrm{E}\exp{\big({-}\frac{p^2}{4\gamma_{\frac12}}\big)} = {\rm e}^{-p}$ ( $\gamma_{\frac12}$ being the gamma random variable with parameter $\frac12$ ), we obtain that

(26) \begin{align} \int_0^{\infty}{\rm e}^{-pt}\psi_{\alpha}(t)\,{\rm d} t = \int_0^{\infty}{\rm e}^{-\frac{p^2}{2}t}\frac{1}{\sqrt{2\pi}\sqrt{t^3}\,}\bigg[\int_0^{\infty}{\rm e}^{-\frac{w^2}{2t}}\psi_{\alpha}(w)w\,{\rm d} w\bigg]\,{\rm d} t.\end{align}

Joining (23), (24), (25), and (26) yields

\begin{align*} \frac{1}{2}\int_0^{\infty}{\rm e}^{-\lambda t}\, \mathrm{\widehat{P}}(\widetilde{T}_{\alpha}>t)\,{\rm d} t = \frac{1}{2}\int_0^{\infty}{\rm e}^{-\lambda t}\frac{1}{\sqrt{2\pi}\sqrt{t^3}\,}\bigg[\int_0^{\infty}{\rm e}^{-\frac{w^2}{2t}}\psi_{\alpha}(w)w\,{\rm d} w\bigg]\,{\rm d} t.\end{align*}

Since $|\psi_{\alpha}(t)| \leq \sum_{n=1}^{\infty}\textbf{1}_{[2\alpha(n-1),2\alpha n]}(t) = 1,$ we have, from Fubini’s theorem,

\begin{align*}\mathrm{\widehat{P}}(\widetilde{T}_{\alpha}>t) &= \sqrt{1/(2\pi t^3)}\sum_{n=1}^{\infty}\int_{2\alpha(n-1)}^{2\alpha n}({-}1)^{n-1}w{\rm e}^{-\frac{w^2}{2t}}\,{\rm d} w\\[3pt] &= \sqrt{1/(2\pi t^3)}\sum_{n=1}^{\infty}({-}1)^nt\big[{\rm e}^{-2\alpha^2n^2/t} - {\rm e}^{-2\alpha^2(n-1)^2/t}\big],\end{align*}

which completes the proof.

4. Itó diffusions

Under some additional assumptions we are able to extend the result of Theorem 3.1 on the two subclasses of Itó diffusions. Precisely, consider one-dimensional Itô diffusion X with values in $E\subseteq \mathbb{R}$ satisfying the stochastic differential equation

(27) \begin{align} {\rm d} X_t = \sigma(X_t){\rm d} B_t + \mu(X_t){\rm d} t,\end{align}

where B is a Brownian motion, $X_0 = x\in E$ under $\mathrm{P}_x$ a.s., and $\sigma$ and $\mu$ are locally bounded measurable functions on E such that (27) has a solution unique in probability. For this we may assume the Engelbert–Schmidt conditions, i.e. that $\sigma\neq 0$ and $\frac{1}{\sigma^2}, \frac{\mu}{\sigma^2}$ are locally integrable (see [Reference Karatzas and Shreve12, Proposition 5.15]). We assume that $0\in E$ , and if m denotes the speed measure of X we demand that $m(\{0\}) = 0$ . We assume that X does not explode during lifetime. It is well known [Reference Borodin and Salminen3, Chapter II, pp. 4] that in the above setup X has a jointly continuous transition density $p_t(x,y)$ with respect to the speed measure m, i.e.

\begin{equation*}\mathrm{P}_x(X_t\in A) = \int_A p_t(x,y)m({\rm d} y) \end{equation*}

for every Borel subset A of E. Recall that $\sigma_z = \inf\{t>0\,:\, X_t = z\}$ for $z\in E$ , and notice that it is a terminal stopping time, so on $\{\sigma_z>t\}$ it is equal to $\sigma_z = t + \sigma_z\circ\theta_t$ . Since the excursion path space is a subspace of the canonical space, the same stopping time $\sigma_z$ may be defined on the space of excursions (see the discussion in [Reference Revuz and Yor20, p. 491]). Notice that R and $\sigma_0$ are terminal stopping times. Ŷ denotes the excursion measure on the path space, associated to excursions of X, and by $\widehat{X}$ we denote the coordinate process killed under Ŷ at R, so $\widehat{X}_t(u) = u(t)$ for $t < R$ (see [Reference Revuz and Yor20, p. 482]). For a positive Borel function f and $z\in E$ define

(28) \begin{align} g^{(z)}_{\lambda, f}(x) &= \mathrm{E}_x\exp{\bigg({-}\lambda\int_0^{\sigma_0\wedge\sigma_z}f(|X_s|)\,{\rm d} s\bigg)},\\[3pt] g^{(0)}_{\lambda, f}(x) &= \mathrm{E}_x\exp{\bigg({-}\lambda\int_0^{\sigma_0}f(|X_s|)\,{\rm d} s\bigg)}.\nonumber\end{align}

Let $\mathcal{A}$ denote the generator of X. Consider the Dirichlet problem

(29) \begin{align} \mathcal{A}g(x) &= \lambda f(x)g(x), \qquad x\in (0,z) ,\\[3pt] g(0) &= g(z) = 1 , \notag\end{align}

where $z>0$ and f is a given Borel and non-negative function on E. It is well known that under some mild conditions $g^{(z)}_{\lambda, f}$ solves the Dirichlet problem (29). It is enough to assume that f is non-negative and continuous (see [Reference Karatzas and Shreve12, Proposition 7.2, Chapter 5] and [Reference Revuz and Yor20, Proposition 3.1, Chapter VII]), but the assumptions on f can be relaxed. For instance, f may be chosen locally $L^{d}$ for some natural d (see the discussion on sufficient assumptions in [Reference Chen and Zhao4, Reference Freidlin8, Reference Zhang23]). For a non-negative Borel function f on E, recall the definition of the integral functional $A_{R}^f(u) = \int_0^{R(u)}f(|u(s)|)\,{\rm d} s$ . The assumptions which enable us to formulate the version of Theorem 3.1 for $E = [0,\infty)$ are gathered in the following hypothesis:

Definition. [Hypothesis B] We say that hypothesis B is satisfied for a non-negative Borel function f on E if

1. 1. 0 is regular and instantaneously reflecting,

2. 2. for every $z\in E$ the function $g^{(z)}_{\lambda, f}$ is $\mathcal{C}^2(0,\infty)$ , solves (29), and has the first right derivative in 0,

3. 3. the function $(g^{(z)}_{\lambda, f})'$ is locally bounded.

An example of a process satisfying the above conditions is a squared Bessel process with index $\mu> -1$ (see [Reference Revuz and Yor20, Chapter XI]). Another example is a radial Ornstein–Uhlenbeck process [Reference Borodin and Salminen3, Appendix I, 26] or a Pearson diffusion ([Reference Forman and Sorensen7]). Alternatively, we may define a diffusion reflecting at 0 as a solution of

\begin{align*} {\rm d} X_t = \sigma(X_t){\rm d} B_t + \mu(X_t){\rm d} t + {\rm d} L_t(X),\end{align*}

where $\sigma$ and $\mu$ are some well measurable coefficients and L(X) denotes the local time of X at 0. One may check that the last approach is analogous to what we will present in the following. If $E = \mathbb{R}$ we have the following set of assumptions:

Theorem 4.1. Let f be Borel and non-negative, $\lambda >0$ , and $z\in E$ . Under either hypothesis B or Bʹ we have

(30) \begin{align} \mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A_{R\wedge \sigma_z}^f}\big) = -c^*(g^{(z)}_{\lambda, f})'(0),\end{align}

where $c^*$ is a local time constant. In particular,

(31) \begin{align} \mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A_{R}^f}\big) = -c^*(g^{(0)}_{\lambda, f})'(0).\end{align}

Proof. Let $z\in E$ ( $E= [0,\infty)$ under B and $E= \mathbb{R}$ under Bʹ). We follow the same idea as the proof of Theorem 3.1. Beside stopping at $\sigma_z$ , the difference is the use of the Tanaka formula instead of extended Levy’s theorem. The terminal property of $\sigma_z$ enables the use of summation trick (11):

\begin{align*} \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{d_t\wedge \sigma_z}}) &= \lambda\mathrm{E}\int_0^{d_t}\textbf{1}_{\{u<\sigma_z\}}{\rm e}^{-\lambda A^f_u}f(|X_u|)\,{\rm d} u\\[3pt] &= \lambda\mathrm{E}\sum_{\tau_{s-}\leq t\wedge \sigma_z}{\rm e}^{-\lambda A^{f}_{\tau_{s-}}}\bigg(\int_0^{\sigma_0}\textbf{1}_{\{u<\sigma_z\}}{\rm e}^{-\lambda A^f_u}f(|X_u|)\,{\rm d} u\bigg)\circ\theta_{\tau_{s-}}.\end{align*}

The master formula of excursions yields

(32) \begin{align} \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{d_t\wedge \sigma_z}}) = \mathrm{E}\bigg(\int_0^{t\wedge \sigma_z}{\rm e}^{-\lambda A^f_s}\,{\rm d} L_s(X)\bigg)\mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A^f_{R\wedge \sigma_z}}\big).\end{align}

Let g be a $\mathcal{C}^2(E)$ function with both derivatives locally bounded. Let $\lambda > 0$ . By the Itó and Tanaka formulas,

\begin{align*} g(|X_t|){\rm e}^{-\lambda\int_0^tf(|X_s|){\rm d} s} &= g(0) \\[2pt] & \quad + \int_0^t{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\bigg(g'(|X_s|)\,{\rm d} |X_s| + \frac12 g''(|X_s|)\sigma^2(X_s)\bigg)\,{\rm d} s\\[2pt] & \quad -\lambda\int_0^tf(|X_s|)g(|X_s|){\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} s\\[2pt] & = g(0) + \int_0^t{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}g'(|X_s|)\big( {\rm sgn}(X_s)(\sigma(X_s)\,{\rm d} B_s\\[2pt] &\quad + \mu(X_s)\,{\rm d} s) + {\rm d} L^*_s(X)\big) \\[2pt] & \quad + \frac12 \int_0^t{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}g''(|X_s|)\sigma^2(X_s)\,{\rm d} s\\[2pt] & \quad -\lambda\int_0^tf(|X_s|)g(|X_s|){\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} s,\end{align*}

where clearly $L^*(X)$ denotes the process of local time of X at 0 compatible with the Tanaka formula. A standard localization argument shows that under hypothesis B we have

(33) \begin{align} &\mathrm{E}g(|X_{t\wedge \sigma_z}|){\rm e}^{-\lambda\int_0^{t\wedge \sigma_z}f(|X_s|){\rm d} s}\\[3pt] &\quad = g(0) + \mathrm{E}\int_0^{t\wedge \sigma_z}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}g'(|X_s|)( {\rm sgn}(X_s)\mu(X_s)\,{\rm d} s) + {\rm d} L^*_s(X))\notag\\[3pt] & \qquad + \frac12 \int_0^{t\wedge \sigma_z}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}g''(|X_s|)\sigma^2(X_s)\,{\rm d} s\notag\\[3pt] & \qquad -\lambda\int_0^{t\wedge \sigma_z}f(|X_s|)g(|X_s|){\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} s,\notag\end{align}

so that

\begin{align*} &\mathrm{E}g(|X_{t\wedge \sigma_z}|){\rm e}^{-\lambda\int_0^{t\wedge \sigma_z}f(|X_s|){\rm d} s} = g(0) + g'(0)\mathrm{E}\int_0^{t\wedge \sigma_z}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} L^*_s(X) \\[3pt] &\quad + \mathrm{E}\int_0^{t\wedge \sigma_z}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\bigg(\frac12g''(|X_s|)\sigma^2(|X_s|) + g'(|X_s|)\mu(|X_s|) - \lambda f(|X_s|)g(|X_s|)\bigg)\,{\rm d} s\\[3pt] &\quad + \mathrm{E}\int_0^{t\wedge \sigma_z}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}g'(|X_s|)\mu(|X_s|)({\rm sgn}(X_s) -1)\,{\rm d} s,\end{align*}

and the last term of the above equality is easily seen to be 0. We choose $g = g^{(z)}_{\lambda, f}$ defined by (28), and since it is a solution of the associated Dirichlet problem, we obtain

(34) \begin{equation} \mathrm{E}g^{(z)}_{\lambda, f}(|X_{t\wedge \sigma_z}|) {\rm e}^{-\lambda\int_0^{t\wedge \sigma_z}f(|X_s|){\rm d} s}= 1 + (g^{(z)}_{\lambda,f})'(0)\mathrm{E}\int_0^{t\wedge \sigma_z}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} L^*_s(X).\end{equation}

Under hypothesis Bʹ we rewrite (33) with $\sigma_z\wedge\sigma_{-k}$ , $k>0$ , in place of $\sigma_z$ and observe that the identities ${\rm sgn} (x) \mu(x) = \mu(x)$ , $\sigma^2(x) = \sigma^2(|x|)$ and the choice of $g = g^{(z)}_{\lambda,f}$ yield

\begin{align*} & \mathrm{E}g^{(z)}_{\lambda, f}(|X_{t\wedge \sigma_z\wedge\sigma_{-k}}|) {\rm e}^{-\lambda\int_0^{t\wedge \sigma_z\wedge\sigma_{-k}}f(|X_s|)\,{\rm d} s} \\[3pt] &\quad = 1 + (g^{(z)}_{\lambda,f})'(0)\mathrm{E}\int_0^{t\wedge \sigma_z\wedge\sigma_{-k}}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} L^*_s(X).\end{align*}

Letting $k\rightarrow\infty$ we obtain that the equality (34) holds under Bʹ. Since $L^* = c^* L$ , we finally have

(35) \begin{equation} \mathrm{E}g^{(z)}_{\lambda, f}(|X_{\sigma_z\wedge t}|){\rm e}^{-\lambda\int_0^{\sigma_z\wedge t}f(|X_s|){\rm d} s} = 1 + c^*(g^{(z)}_{\lambda,f})'(0)\mathrm{E}\int_0^{\sigma_z\wedge t}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} L_s(X).\end{equation}

Using the Markov property ( $|X|$ is a Markov process under B and Bʹ) and the terminal property of $\sigma_z$ , we write

\begin{align*} \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{d_t\wedge \sigma_z}}) &= \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{d_t\wedge \sigma_z}})(\textbf{1}_{\{\sigma_z\geq t\}} + \textbf{1}_{\{\sigma_z < t\}})\\ & = \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{t +\sigma_0\circ\theta_t}})\textbf{1}_{\{\sigma_z\geq t\}}\textbf{1}_{\{\sigma_z\circ\theta_t\geq \sigma_0\circ\theta_t\}}\\ & \quad + \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{t + \sigma_z\circ\theta_t}})\textbf{1}_{\{\sigma_z\geq t\}} \textbf{1}_{\{\sigma_z\circ\theta_t< \sigma_0\circ\theta_t\}} \end{align*}

\begin{align*} & \quad + \mathrm{E}(1 - {\rm e}^{-\lambda A^f_{\sigma_z}}g^{(z)}_{\lambda, f}(z))\textbf{1}_{\{\sigma_z < t\}}\\ & = \mathrm{E}\big(1 - {\rm e}^{-\lambda A^f_{t}}g^{(z)}_{\lambda, f}(|X_t|)\big)\textbf{1}_{\{\sigma_z\geq t\}} + \mathrm{E}\big(1 - {\rm e}^{-\lambda A^f_{\sigma_z}}g^{(z)}_{\lambda, f}(z)\big)\textbf{1}_{\{\sigma_z < t\}}\\ & = \mathrm{E}\big(1 - {\rm e}^{-\lambda A^f_{t\wedge \sigma_z}}g^{(z)}_{\lambda, f}(|X_{t\wedge \sigma_z}|)\big),\end{align*}

so that

\begin{align*} \mathrm{E}{\rm e}^{-\lambda A^f_{d_t\wedge \sigma_z}} = \mathrm{E}{\rm e}^{-\lambda A^f_{t\wedge \sigma_z}}g^{(z)}_{\lambda, f}(|X_{t\wedge \sigma_z}|).\end{align*}

Inserting the last equality into (35) and comparing it with (32) finally yield

\begin{align*}\mathrm{\widehat{P}}\big(1 - {\rm e}^{-\lambda A^f_{R\wedge \sigma_z}}\big) = -c^*(g^{(z)}_{\lambda,f})'(0),\end{align*}

where we used $\mathrm{E}\int_0^{t}{\rm e}^{-\lambda\int_0^sf(|X_u|){\rm d} u}\,{\rm d} L_s(X) > 0$ , which is true by (32). Formula (31) follows from (30) by the monotone convergence and Lebesgue theorems.

Theorem 4.2. Let $z \in E$ . Assume hypothesis B or Bʹ holds for f. Then

\begin{align*} \mathrm{\widehat{P}}\big(A^f_{R\wedge \sigma_z} \in {\rm d} v\big) = \lim_{x\downarrow 0}\frac{c^*}{x}\mathrm{P}_x\bigg(\int_0^{\sigma_0\wedge \sigma_z}f(|X_u|)\,{\rm d} u \in {\rm d} v\bigg),\end{align*}

where the limit is in the sense of weak convergence.

Proof.Under either B or Bʹ, for any $\lambda >0$ we have, by (30),

\begin{align*} \int_0^{\infty}(1-{\rm e}^{-\lambda v})\mathrm{\widehat{P}}\big(A^f_{R\wedge \sigma_z} \in {\rm d} v\big) = \lim_{x\downarrow 0}\int_0^{\infty}(1-{\rm e}^{-\lambda v})\frac{c^*}{x}\mathrm{P}_x\bigg(\int_0^{\sigma_0\wedge \sigma_z}f(|X_u|)\,{\rm d} u \in {\rm d} v\bigg),\end{align*}

so the proof will be done if we show that we may apply the monotone convergence theorem. Recall that $c^*>0$ . It is enough to conclude that the function $x\mapsto \frac{1}{x}\mathrm{P}_x\big(\int_0^{\sigma_0\wedge \sigma_z}f(|X_u|)\,{\rm d} u \in {\rm d} v\big)$ is monotone on some interval $(0,\epsilon)$ . We will prove that $x\mapsto \frac{1}{x}\mathrm{E}_x\big(1-{\rm e}^{-\lambda\int_0^{\sigma_0\wedge \sigma_z}f(|X_u|){\rm d} u}\big)$ is monotone. Due to the assumptions on $g^{(z)}_{\lambda,f}$ we may write, for $x>0$ ,

\begin{align*} g^{(z)}_{\lambda,f}(x) = 1 + x(g^{(z)}_{\lambda,f})'(0) + \frac12\int_0^x(x-v) (g^{(z)}_{\lambda,f})''(v)\,{\rm d} v,\end{align*}

so that

\begin{align*} \frac{1}{x}\mathrm{E}_x\big(1-{\rm e}^{-\lambda\int_0^{\sigma_0\wedge \sigma_z}f(|X_u|){\rm d} u}\big) = -(g^{(z)}_{\lambda,f})'(0) -\frac12\int_0^x\bigg(1-\frac{v}{x}\bigg) (g^{(z)}_{\lambda,f})''(v)\,{\rm d} v.\end{align*}

If we show that $g^{(z)}_{\lambda,f}$ is convex on some $(0,\epsilon)$ then we will have there

\begin{align*}\frac{{\rm d}}{{\rm d}x}\bigg(\int_0^x\bigg(1-\frac{v}{x}\bigg) (g^{(z)}_{\lambda,f})''(v)\,{\rm d} v\bigg) = \int_0^x\frac{v}{x^2}(g^{(z)}_{\lambda,f})''(v)\,{\rm d} v > 0.\end{align*}

For this, let $0<a<b$ and $I = (a,b)$ , and define $\sigma_I = \sigma_a\wedge\sigma_b$ . Let $x\in I$ and define $g^{(z,I)}_{\lambda,f}(x) = \mathrm{E}_x{\rm e}^{-\lambda\int_0^{\sigma_I}\textbf{1}_{\{u < \sigma_z\}}f(|X_u|){\rm d} u}$ . Let $\alpha\in(0,1)$ . We choose $J = (c,d)$ such that $a<c< x< d < b$ , and such that $\alpha = \mathrm{P}_x(\sigma_c<\sigma_d)$ (and define $\sigma_J$ and $g^{(\sigma_z,J)}_{\lambda,f}$ analogously). We have, by the terminal property of $\sigma_z$ and the strong Markov property,

\begin{align*} g^{(z,I)}_{\lambda, f}(x) &= \mathrm{E}_x\exp{\bigg({-}\lambda\bigg(\int_0^{\sigma_J}\textbf{1}_{\{s< \sigma_z\}}\,f(|X_s|)\,{\rm d} s + \bigg(\int_0^{\sigma_I}\textbf{1}_{\{s< \sigma_z\}}\,f(|X_s|)\,{\rm d} s\bigg)\circ\theta_{\sigma_J}\bigg)\bigg)}\\[3pt] &= \mathrm{E}_x\bigg[\exp{\bigg({-}\lambda\int_0^{\sigma_J}\textbf{1}_{\{s< T\}}f(|X_s|)\,{\rm d} s\bigg)}g^{(T,I)}_{\lambda, f}(X_{\sigma_J})\bigg]\\[3pt] &\leq g^{(z,I)}_{\lambda, f}(c)\mathrm{P}_x(\sigma_c < \sigma_d) + g^{(z,I)}_{\lambda, f}(d)\mathrm{P}_x(\sigma_d < \sigma_c) = \alpha \, g^I_{\lambda,f}(c) + (1-\alpha) \, g^I_{\lambda,f}(d).\end{align*}

So, we conclude that $g^{(z,I)}_{\lambda,f}$ is convex on every such I. So is $g^{(z)}_{\lambda,f}$ , since it is an extension of $g^I_{\lambda,f}$ for $a\downarrow 0$ and $b\uparrow\infty$ .

Corollary 4.1. Let $z\in E$ and assume that hypothesis B or Bʹ holds for both functions f and $h(x) = \boldsymbol{1}_{(0,\infty)}(x)\big(\lambda f(x) + \frac{\gamma}{\lambda}\big)$ , where f is non-negative and Borel, and $\lambda>0$ , $\gamma>0$ . Then

\begin{align*} \mathrm{\widehat{P}}\big(A^f_{R\wedge \sigma_z} \in {\rm d} v, R\wedge \sigma_z\in {\rm d} v\big) = \lim_{x\downarrow 0}\frac{c^*}{x}\mathrm{P}_x\bigg(\int_0^{\sigma_0\wedge \sigma_z}f(|X_u|)\,{\rm d} u \in {\rm d} v,\, \sigma_0\wedge \sigma_z \in {\rm d} v\bigg).\end{align*}

Proof.It follows that for any non-negative Borel $f_0$ we have, from Theorem 4.2,

\begin{align*}\mathrm{\widehat{P}}f_0\big(A^f_{R\wedge \sigma_z}\big) = \lim_{x\downarrow 0}\frac{c^*}{x}\mathrm{E}_x\ f_0\bigg(\int_0^{\sigma_0\wedge \sigma_z}f(|X_u|)\,{\rm d} u\bigg).\end{align*}

We set $f_0(x) = \exp\{-\lambda x\}$ and consider $A^h_R$ . By a standard Laplace transform argument we conclude that $(A^f_{R\wedge \sigma_z},R\wedge \sigma_z)$ has, under Ŷ, the same distribution as $(\int_0^{\sigma_0\wedge \sigma_z}f(|X_s|)\,{\rm d} s,\, \sigma_0\wedge \sigma_z)$ under $\lim_{x\downarrow 0}\frac{c^*}{x}\mathrm{P}_x$ .

Remark 1. Recall that the limit describing Ŷin [Reference Pitman and Yor17] was taken with respect to the scale function. Observe that for Itó diffusion defined by (27) we can choose $s(x) = \int_0^x\exp{\{-2\int_0^y\mu(z)\sigma^{-2}(z)\,{\rm d} z\}}\,{\rm d} y$ , so $s(0) =0$ and $s'(0) = 1$ . As a result, identity (1) may be rewritten as

\begin{align*} \mathrm{\widehat{P}}(A) &= \lim_{x\downarrow 0} \frac{ \mathrm{P}_x(A)}{s(x)} = \lim_{x\downarrow 0} \frac{ \mathrm{P}_x(A)}{x}\frac{x}{s(x)} = \frac{1}{s'(0)}\lim_{x\downarrow 0} \frac{ \mathrm{P}_x(A)}{x}\\[3pt] &= \lim_{x\downarrow 0} \frac{ \mathrm{P}_x(A)}{x}.\end{align*}

Notice that Ŷ depends on the choice of s.

Corollary 4.1 gives the explicit form of the distribution of R.

Corollary 4.2. Under the assumptions of Corollary 4.1, we have, from Theorem 4.2,

\begin{align*} \mathrm{\widehat{P}}(R\in {\rm d} z) = \lim_{x\downarrow 0}\frac{c^*}{x}g_{x0}(z)\,{\rm d} z,\end{align*}

so the distribution of R under the excursion measure is retrieved from the density of the first hitting time of 0.

Last, but not least, we present the absolute-continuity relationship between functionals on excursion spaces associated with local martingales. If X is a local martingale and satisfies (27), then $\mu(x) = 0$ for all $x\in E$ .

Theorem 4.3. Let $E = [0,\infty)$ . Assume that we are given two different excursion measures Ŷ and ŷ associated with two reflected at 0 Markov processes denoted by $X^{(1)}$ and $X^{(2)}$ respectively. Assume that hypothesis B holds for both processes. If $X^{(1)}$ and $X^{(2)}$ are local martingales, and $m{'}_{\!\kern-1pt1}$ , $m{'}_{\!\kern-1pt2}$ denote the densities with respect to the Lebesgue measure of their speed measures respectively, then for any $b>0$ and $\widehat{X}$ denoting the coordinate process killed at R we have

\begin{align*} c_2^*\, \mathrm{\widehat{P}}\bigg(\int_0^{\sigma_b\wedge R}f(\widehat{X}_s)m{'}_{\!\kern-1pt2}(\widehat{X}_s)\,{\rm d} s\bigg) = c_1^*\, \mathrm{\widehat{Q}}\bigg(\int_0^{\sigma_b\wedge R}f(\widehat{X}_s)m{'}_{\!\kern-1pt1}(\widehat{X}_s)\,{\rm d} s\bigg),\end{align*}

where $c_1^*, c_2^*$ are associated local time constants and f is a non-negative Borel measurable function.

Proof.It is clear that if $X^{(1)}$ and $X^{(2)}$ are local martingales, then they are in natural scale, i.e. their scale functions are up to multiplicative constants the identity functions. Consider the first process and the excursion measure Ŷ. By Theorem 4.2,

\begin{align*} \mathrm{\widehat{P}}\int_0^{\sigma_b\wedge R}f(\widehat{X}_s)\,{\rm d} s = \lim_{x\downarrow 0}\frac{c^*_1}{x}\mathrm{E}_x\int_0^{\sigma_b\wedge \sigma_0}f(X^{(1)}_s)\,{\rm d} s.\end{align*}

We use [Reference Revuz and Yor20, Corollary 3.8, Chapter VII] to write, for $0<a<x<b$ and $T = \sigma_a\wedge\sigma_b$ ,

\begin{align*} \mathrm{E}_x\int_0^{T}f(X^{(1)}_s)\,{\rm d} s = \int_{(a,b)}G(x,y)\,f(y)m{'}_{\!\kern-1pt1}(y)\,{\rm d} y,\end{align*}

where $G(x,y) = \frac{(x\wedge y - a)(b - x\vee y)}{b-a}$ . Simple algebra shows that

\begin{align*} \mathrm{E}_x\int_0^{T}f(X^{(1)}_s)\,{\rm d} s = \frac{b-x}{b-a}\int_a^x(y-a)\,f(y)m{'}_{\!\kern-1pt1}(y)\,{\rm d} y + \frac{x-a}{b-a}\int_x^b(b-y)\,f(y)m{'}_{\!\kern-1pt1}(y)\,{\rm d} y.\end{align*}

Letting $a\downarrow 0$ we obtain

\begin{align*}\lim_{x\downarrow 0}\frac{1}{x}\mathrm{E}_x\int_0^{\sigma_b\wedge \sigma_0}f(X^{(1)}_s)\,{\rm d} s &= \lim_{x\downarrow 0}\bigg[\frac{b-x}{xb}\int_0^xyf(y)m{'}_{\!\kern-1pt1}(y)\,{\rm d} y + \frac{1}{b}\int_x^b(b-y)\,f(y)m{'}_{\!\kern-1pt1}(y)\,{\rm d} y\bigg]\\[3pt] &= \frac{1}{b}\int_0^b(b-y)\,f(y)m{'}_{\!\kern-1pt1}(y)\,{\rm d} y.\end{align*}

Repeating the same procedure for ŷ yields $ \mathrm{\widehat{Q}}\int_0^{\sigma_b\wedge R}f(\widehat{X}_s)\,{\rm d} s = \frac{c^*_2}{b}\int_0^b(b-y)\,f(y)m{'}_{\!\kern-1pt2}(y)\,{\rm d} y,$ so taking $h_1 = f\ m{'}_{\!\kern-1pt2}$ and $h_2 = f \ m{'}_{\!\kern-1pt1}$ we obtain

\begin{align*}\mathrm{\widehat{P}}\int_0^{\sigma_b\wedge R}h_1(\widehat{X}_s)\,{\rm d} s = \frac{c_1^*}{b}\int_0^b(b-y)\,f(y)m{'}_{\!\kern-1pt1}(y)m{'}_{\!\kern-1pt2}(y)\,{\rm d} y = \frac{c_1^*}{c_2^*}\mathrm{\widehat{Q}}\int_0^{\sigma_b\wedge R}h_2(\widehat{X}_s)\,{\rm d} s,\end{align*}

completing the proof.