2. Optimal stopping of a Brownian excursion

Let $W=\{W_{t}\}_{t\geq 0}$ be a standard Brownian motion.

Definition 2.1. A non-negative continuous process $X=\{X_{t}\}_{0\leq t \leq 1}$ is a (scaled) Brownian excursion process if it satisfies the SDE

(2.1) \begin{equation} {\mathrm{d}} X_{t}=\biggl(\dfrac{1}{X_{t}}-\dfrac{X_{t}}{1-t}\biggr)\,{\mathrm{d}} t + {\mathrm{d}} W_{t}, \quad 0<t<1, \end{equation}

subject to intial condition $X_{0}=0$ . See Pitman [Reference Pitman8] for more details.

Note that the ${{1}/{X_t}}$ term in the drift prevents the process from hitting zero before time 1, and the ${{-X_t}/{(1-t)}}$ term makes sure that it returns to zero at time 1. In this section we are interested in the optimal stopping problem for $X=\{X_{s}\}_{t\leq s\leq 1}$ , a Brownian excursion started at $(t,X_{t})=(t,x)$ , with the identity payoff. The associated value function is

(2.2) \begin{equation} \bar{V}(t,x)=\sup\nolimits_{t\leq \tau\leq 1}\mathbb{E}_{t,x}[X_{\tau}],\end{equation}

and the goal is to find the optimal stopping time $\tau^{*}$ for which the supremum in (2.2) is attained. Given the Markovian structure of the problem, we expect that the optimal stopping rule will be of threshold form $\tau^* = \tau_c$ , where

\begin{equation*} \tau_c = \inf\{s\geq t\colon X_{s}\geq c(s)\},\end{equation*}

where $c(\! \cdot\! )$ denotes the optimal stopping boundary. The region $\{ (t,x) \in (0,1) \times [0,c(t)) \}$ is called the continuation region, and $\{ (t,x) \in (0,1) \times [c(t),\infty)\}$ is called the stopping region.

Let $V=(V(t,x))_{ (t,x) \in [0,1)\times[0,\infty) }$ be a candidate for the solution of the optimal stopping problem, expressed as a function of the starting point. We want to describe the properties which V must possess. Later we will introduce a candidate solution $V^*$ which satisfies these properties. Finally, we show that for each $(t,x) \in [0,1)\times[0,\infty)$ we have $V^*(t,x)=\bar{V}(t,x)$ .

When $X_{t}$ is in the stopping region, i.e. $X_{t}\geq c(t)$ , we expect $V(t,X_{t})=X_{t}$ . Thus we only need to find $V(t, X_{t})$ when $X_{t}$ is in the continuation region. Assuming that V is sufficiently smooth, we deduce from Itô’s formula and (2.1) that

\begin{equation*} \begin{split} {\mathrm{d}} V(t, X_{t})=\biggl[V_{t}(t,X_{t}) +V_{x}(t,X_{t})\biggl(\dfrac{1}{X_{t}}-\dfrac{X_{t}}{1-t}\biggr) +\dfrac{1}{2}V_{xx}(t, X_{t})\biggr]\,{\mathrm{d}} t+V_{x}(t,X_{t})\,{\mathrm{d}} W_{t}. \end{split}\end{equation*}

Since the value function evolves as a martingale in the continuation region $x<c(t)$ , we obtain the HJB equation

(2.3) \begin{equation} V_{t}(t,x) +\biggl(\dfrac{1}{x}-\dfrac{x}{1-t}\biggr)V_{x}(t,x) + \dfrac{1}{2}V_{xx}(t, x) = 0 \end{equation}

with boundary conditions

(2.4) \begin{align} V(t,c(t))&= c(t), \end{align}

(2.5) \begin{align} V_{x}(t,c(t))&= 1, \end{align}

(2.6) \begin{align} V(t,0)&<\infty. \end{align}

Here (2.4) and (2.5) are the value matching condition and smooth fit condition respectively, and (2.6) ensures that the value function is finite when $x=0$ .

We make the ansatz $c(t)=C\sqrt{1-t}$ , together with

(2.7) \begin{equation} V(t,x) = \sqrt{1-t}f\biggl(\dfrac{x}{\sqrt{1-t}}\biggr),\end{equation}

where C is a constant and $f\colon \mathbb{R}_+ \to \mathbb{R}_+$ a function to be determined. In this way, setting $y={{x}/{\sqrt{1-t}}}$ , the problem given in (2.3)–(2.6) is transformed into

(2.8) \begin{equation} f''(\,y)+ \biggl( \dfrac{2}{y}-y \biggr)\,f'(y)-f\,(\,y)=0\end{equation}

on $y<C$ , with boundary conditions

(2.9) \begin{align} f(C)&=C , \end{align}

(2.10) \begin{align} f'(C)&=1 , \end{align}

and such that $f(0)<\infty$ . It is easily seen that ${{1}/{y}}$ is a solution to (2.8) and then that a second solution is given by

\begin{align*}\frac{1}{y}\int_{0}^{y}{\mathrm{e}}^{{{t^{2}}/{2}}}\,{\mathrm{d}} t.\end{align*}

Then the general solution to the ODE in (2.8) is

(2.11) \begin{equation} f\,(\,y)= \dfrac{A}{y}+ \dfrac{B}{y}\int_{0}^{y}{\mathrm{e}}^{{{t^{2}}/{2}}}\,{\mathrm{d}} t.\end{equation}

The finiteness condition at zero implies $A=0$ . From (2.9) and (2.10) we find that coefficient B satisfies

(2.12) \begin{equation} B = \dfrac{C^{2}}{\int_{0}^{C}{\mathrm{e}}^{{{t^{2}}/{2}}}\,{\mathrm{d}} t},\end{equation}

where C solves

(2.13) \begin{equation} C = 2 {\mathrm{e}}^{-{{C^2}/{2}}} \int_0^C {\mathrm{e}}^{t^2/2} \,{\mathrm{d}} t.\end{equation}

Indeed, C is the unique solution of (2.13) in $(0,\infty)$ . To see this, define $h\colon [0,\infty) \to \mathbb{R}$ by $h(c)= 2\int_{0}^{c}{\mathrm{e}}^{{{t^{2}}/{2}}}\,{\mathrm{d}} t-c{\mathrm{e}}^{{{c^{2}}/{2}}}$ . Note that $h'(c) = {\mathrm{e}}^{{{c^{2}}/{2}}}(1-c^2)$ so that h is increasing on (0, 1) and decreasing on $(1,\infty)$ . Uniqueness of a positive root of h in $(1,\infty)$ (and the fact that the root is in (1, 2)) follows from the fact that $h(0)=0$ , $h(1)= 2\int_{0}^{1}{\mathrm{e}}^{{{t^{2}}/{2}}}\,{\mathrm{d}} t-\sqrt{{\mathrm{e}}}> 2 - \sqrt{{\mathrm{e}}}>0$ , and, from the convexity of ${\mathrm{e}}^{t^2/2}$ ,

\begin{align*}h(2) < 2 \frac{1 +{\mathrm{e}}^2}{2} - 2{\mathrm{e}}^2 < 0.\end{align*}

Numerical results show that the root is given by $C \approx 1.50339538$ .

Figure 1.

A sample path $X(\omega)= (X_t(\omega))_{0 \leq t \leq 1}$ of a Brownian excursion and the optimal stopping boundary $c(s)=C\sqrt{1-s}$ , where $C \approx 1.50339538$ . Also shown is the optimal stopping time $\tau^*(\omega) = \inf \{ u \in (0,1)\colon X_u(\omega) \geq C\sqrt{(1-u)} \}$ .

Figure 1 Long description

A line graph depicting a sample path of a Brownian excursion and the optimal stopping boundary, with the optimal stopping time marked. The x-axis represents the variable s ranging from 0 to 1, and the y-axis represents the variable X sub s ranging from 0 to 1.6. The graph includes a solid line representing the function c(s) equals C times the square root of 1 minus s. A dashed vertical line marks the optimal stopping time at approximately 0.45 on the x-axis. The sample path fluctuates around the optimal stopping boundary, showing peaks and troughs. All values are approximated.

The next result is technical and the proof is postponed to the Appendix.

Putting together (2.7) with (2.11) for the choice $A=0$ and B given by (2.12) gives the candidate value function $V^*=(V^{*}(t,x))_{ (t,x) \in [0,1)\times[0,\infty) }$ :

(2.14) \begin{equation} V^{*}(t,x)= \begin{cases} \displaystyle C^{2}\biggl(\int_{0}^{C}{\mathrm{e}}^{{{t^{2}}/{2}}}\,{\mathrm{d}} t\biggr)^{-1}\biggl(\dfrac{1-t}{x}\biggr) \int_{0}^{{{x}/{\sqrt{1-t}}}}{\mathrm{e}}^{{{t^{2}}/{2}}} \,{\mathrm{d}} t &\text{if $ x<c(t)$,} \\[9pt] x &\text{if $ x\geq c(t)$,} \end{cases}\end{equation}

where $c(t)= C\sqrt{1-t}$ and C solves (2.13). The next step is to show that the candidate value function $V^{*}(t,x)$ is indeed the true solution to the optimal stopping problem introduced in (2.2).

Theorem 2.1. The candidate value function $V^{*}$ given in (2.14) coincides with the value function $\bar{V}$ defined in (2.2). Moreover, the associated stopping time $\tau^*$ , given by

\begin{equation*} \tau^{*}=\inf\{s\geq t\;\colon X_{s}\geq c(s)\}, \end{equation*}

where $c(s)=C\sqrt{1-s}$ and $C \in (0,\infty)$ is the unique solution to (2.13), is optimal for the problem started at $(t,x)$ .

Proof. Fix $t_0 \in [0,1)$ . We show that $V^{*}(t_0,x)=\bar{V}(t_0,x)$ by first proving $V^{*}(t_0,x)\geq \bar{V}(t_0,x)$ , and then proving the reverse inequality.

For $t \geq t_0$ , let $M = \{ M_t \}_{t \geq t_0}$ be given by $M_{t}=V^{*}(t,X_{t})$ . Itô’s formula gives us

\begin{equation*} {\mathrm{d}} M_{t}=V^{*}_{t}(t,X_{t})\,{\mathrm{d}} t+V_{x}^{*}(t,X_{t})\,{\mathrm{d}} X_{t}+\dfrac{1}{2}V_{xx}^{*}(t,X_{t})\,{\mathrm{d}} \langle X\rangle_{t}. \end{equation*}

Therefore, recalling (2.1), we find

\begin{equation*} {\mathrm{d}} M_{t}=\biggl[V_{t}^{*}(t,X_{t})+\biggl(\dfrac{1}{X_{t}}-\dfrac{X_{t}}{1-t}\biggr)V_{x}^{*}(t,X_{t}) +\dfrac{1}{2}V_{xx}^{*}(t,X_{t})\biggr]\,{\mathrm{d}} t+V_{x}^{*}(t,X_{t})\,{\mathrm{d}} W_{t}.\end{equation*}

Since $V^{*}(t,x)$ solves (2.3) when $x<c(t)$ , the drift term vanishes in the continuation region, i.e. $X_{t}<c(t)$ . In contrast, when X is in the stopping region, i.e. $X_t\geq c(t)$ , we have $V^{*}(t,x)=x$ , and the drift term simplifies. Putting the cases together,

(2.15) \begin{equation} {\mathrm{d}} M_{t}= g(t,X_t)\mathbb{I}\{X_{t}\geq c(t)\}\,{\mathrm{d}} t +V_{x}^{*}(t,X_{t})\,{\mathrm{d}} W_{t},\end{equation}

where

\begin{equation*} g(t,x)= x^{-1} \biggl( 1 -\dfrac{x^2}{1-t} \biggr).\end{equation*}

Note that for each $t \in [0,1)$ , $g(t,\cdot)$ is decreasing in x and therefore, for every $x\geq c(t)$ , $g(t,x)\leq g(t,c(t))$ . Moreover, since $C>1$ ,

\begin{equation*} g(t,c(t))= g(t,C\sqrt{1-t})=\biggl(\dfrac{1}{C}-C\biggr)\sqrt{1-t}<0.\end{equation*}

It follows that

\begin{align*}\int_{t_0}^\cdot \biggl(\frac{1}{X_{t}}-\frac{X_{t}}{1-t}\biggr)\mathbb{I}\{X_{t}\geq c(t)\}\,{\mathrm{d}} t\end{align*}

is a decreasing process.

The process $\int_{t_0}^\cdot V^{*}_{x}\,{\mathrm{d}} W_{t}$ is a local martingale: we want to argue that $\int_{t_0}^{\cdot}V^{*}_{x}(s,X_s)\,{\mathrm{d}} W_{s}$ is a martingale. This will follow if $V^*_x$ is bounded, and in the stopping region this is immediate since $V^*_x \equiv 1$ . In the continuation region we have

\begin{align*}V^{*}(t,x) = \sqrt{1-t}\,f\biggl(\frac{x}{\sqrt{1-t}}\biggr){,}\end{align*}

where

\begin{align*} f\,(\,y)= \frac{B}{y}\int_{0}^{y}{\mathrm{e}}^{{{t^{2}}/{2}}}\,{\mathrm{d}} t\end{align*}

and B and C are given in (2.12) and (2.13). Then

\begin{align*}V^{*}_{x}(t,x) \,=\, f' \biggl(\frac{x}{\sqrt{1-t}}\biggr).\end{align*}

From Lemma 2.1 we conclude that f is a convex function, and thus $f' \in [f'(0),f'(C)]=[0,1]$ on [0, C]. Hence $V_{x}^{*}=f'$ is bounded, and $\int_{t_0}^\cdot V_{x}^{*}(s,x)\,{\mathrm{d}} W_{s}$ is a martingale.

Putting together the results of the last paragraph, $M=\{M_{t}\}_{t\geq t_0}$ is a supermartingale. Moreover, $V^*(t,x) \geq x$ ; this follows by definition on $x \geq c(t)$ and from Lemma 2.1 on the continuation region. Then $\mathbb{E}_{t_0,x}[X_\tau] \leq \mathbb{E}_{t_0,x}[V^*(\tau,X_\tau)]$ . Further, by the Optional Stopping Theorem, $\mathbb{E}_{t_0,x}[V^{*}(\tau,X_{\tau})]\leq V^{*}(t_0,x)$ , and then for any stopping time $\tau$ taking values in $[t_0,1]$ , $ \mathbb{E}_{t_0,x}[X_{\tau}]\leq \mathbb{E}_{t_0,x}[V^{*}(\tau,X_{\tau})]\leq V^{*}(t_0,x).$ It follows that

(2.16) \begin{equation} \bar{V}(t_0,x)=\sup\nolimits_{t_0 \leq \tau \leq 1}\mathbb{E}_{t_0,x}[X_{\tau}]\leq V^{*}(t_0,x).\end{equation}

Now we show the reverse inequality $\bar{V}(t_0,x) \geq V^*(t_0,x)$ . If $X_{t_0} \geq c(t_0)$ , then taking $\tau = t_0$ we have $\bar{V}(t_0,x) \geq x = V^*(t_0,x)$ . It remains to consider the case $X_{t_0} < c(t_0)$ . Set $\tau^{*} = \inf\{s\geq t_0\colon X_{s}\geq c(s)\}$ . It follows that $X_{t\wedge\tau^{*}}\leq c(t)$ almost surely. Then from (2.15) we conclude that $V^{*}(t\wedge \tau^*,X_{t\wedge\tau^{*}})$ is a martingale. Using the Optional Stopping Theorem again, we have

\begin{align*}V^{*}(t_0,x)=\mathbb{E}_{t_0,x}[V^{*}(\tau^{*},X_{\tau^{*}})]=\mathbb{E}_{t,x}[X_{\tau^{*}}]{,}\end{align*}

where the second equality comes from the fact that $V^{*}(t,c(t))=c(t)$ . Hence

\begin{equation*} \bar{V}(t_0,x)=\sup\nolimits_{t_0 \leq \tau \leq 1}\mathbb{E}_{t_0,x}[X_{\tau}]\geq \mathbb{E}_{t_0,x}[X_{\tau^{*}}]= V^{*}(t_0,x).\end{equation*}

Combining this result with (2.16), we have $\bar{V}(t_0,x) = V^*(t_0,x)$ , and $\tau^*$ is optimal.

3. Optimal stopping of an $\alpha$ -dimensional Bessel bridge

A Brownian excursion can also be viewed as a three-dimensional Bessel bridge. In this section we are interested in extending the problem of Section 2 to allow for an $\alpha$ -dimensional Bessel bridge as the underlying stochastic process, and to include more general objective functions of power law form.

At least when $\alpha$ is an integer, the $\alpha$ -dimensional Bessel process corresponds to the radial part of $\alpha$ -dimensional Brownian motion. For $\alpha\geq 2$ , $\alpha$ -dimensional Brownian motion started at the origin never returns to zero. Correspondingly, an $\alpha$ -dimensional Bessel process started at the origin never hits zero after time zero (provided $\alpha \geq 2$ ). Similarly, when $\alpha\geq 2$ , an $\alpha$ -dimensional Bessel bridge is strictly positive on time-interval (0, 1). However, for $\alpha \in (0,2)$ , both an $\alpha$ -dimensional Bessel process and an $\alpha$ -dimensional Bessel bridge return to zero countably many times in (0, 1), as well as at time 1 for the Bessel bridge. This can cause complications in the study of the Bessel process; for example, when $0 < \alpha < 1$ the $\alpha$ -dimensional Bessel process does not have a representation via a stochastic differential equation. Instead we work with the squared Bessel process, and the squared Bessel bridge. Bessel processes and squared Bessel processes are discussed in detail in Revuz and Yor [Reference Revuz and Yor9]; see Chapter 11 and especially Exercise 3.11.

Definition 3.1. Suppose $\alpha>0$ . A non-negative continuous process $Q=\{Q_{t}\}_{0\leq t\leq 1}$ is an $\alpha$ -dimensional squared Bessel bridge process if it satisfies the SDE

\begin{equation*} {\mathrm{d}} Q_{t}=\biggl(\alpha - \dfrac{2Q_{t}}{1-t} \biggr)\,{\mathrm{d}} t+ 2 \sqrt{Q}_t \,{\mathrm{d}} W_{t}, \quad 0<t<1, \end{equation*}

subject to $Q_0=0$ .

Remark 3.1. Given a process Q as in Definition 3.1, we can define an $\alpha$ -dimensional Bessel bridge $X= (X_t)_{0 \leq t \leq 1}$ by $X_t = \sqrt{Q_t}$ . If $\alpha \geq 2$ so that $Q>0$ on (0, 1), we can apply Itô’s formula to deduce that X solves the SDE

\begin{align*}{\mathrm{d}} X_t = \biggl( \frac{\alpha-1}{2X_t} - \frac{X_t}{1-t} \biggr) \,{\mathrm{d}} t + {\mathrm{d}} W_t,\end{align*}

and when $\alpha=3$ we recognize the Brownian excursion of Definition 2.1. When $\alpha \geq 2$ we can continue to work with this representation, but when $\alpha \in [1,2)$ we need to deal with the times when $X=0$ , and when $\alpha \in (0,1)$ , X cannot be written as the solution of an SDE, hence our re-specification of the problem via the squared Bessel process.

Our goal is to investigate the optimal stopping problem for the $\alpha$ -dimensional Bessel bridge with power-law payoff $\phi(x)=x^n$ . For $\alpha \in (0,\infty)$ and $n \in (0,\infty)$ , define $\bar{V}(t,x)=\sup_{t\leq\tau\leq 1}\mathbb{E}_{t,x}[X_{\tau}^{n}]$ . Then $\bar{V}(t,x)=\bar{U}(t,x^2)$ , where

(3.1) \begin{equation} \bar{U}(t,q)=\sup\nolimits_{t\leq\tau\leq 1}\mathbb{E}_{t,q}[Q_{\tau}^{n/2}],\end{equation}

where Q is the squared Bessel bridge introduced in Definition 3.1. As in the last section, we aim to find the optimal stopping time $\tau^{*}$ in which the supremum in (3.1) is attained. For the problem started at time $t_0 \in (0,1)$ we expect the optimal stopping rule to be of threshold form,

\begin{equation*} \tau^{*}=\inf\{s\geq t_0\colon X_{s}\geq \sqrt{z(s)} \} = \inf\{s\geq t_0\colon Q_{s}\geq z(s)\},\end{equation*}

where $z(\!\cdot\!)$ is the optimal boundary (for the squared Bessel bridge). Following the Brownian excursion case, in the continuation region $\mathcal{C}_Q =\{ (t,q) \subseteq [0,1) \times [0,z(t)) \}$ we have that for a candidate solution $U=(U(t,q))_{ (t,q) \in [0,1) \times [0,\infty)}$ ,

\begin{equation*} {\mathrm{d}} U(t, Q_{t})=\biggl[U_{t}(t,Q_{t}) +U_{q}(t,Q_{t})\biggl(\alpha-\dfrac{2Q_{t}}{1-t}\biggr) +2 Q_t U_{qq}(t, Q_{t})\biggr]\,{\mathrm{d}} t+ 2 \sqrt{Q_t} U_{q}(t,Q_{t})\,{\mathrm{d}} W_{t},\end{equation*}

and on ${\mathcal C}_Q$ we expect that U solves the PDE

(3.2) \begin{equation} U_{t}(t,q) +\biggl(\alpha-\dfrac{2q}{1-t}\biggr)U_{q}(t,q) + 2q U_{qq}(t, q) = 0 \end{equation}

subject to

\begin{align*}U(t,z(t))= z(t)^{n/2}\quad\text{and}\quad U_{q}(t,z(t))= \frac{n}{2} z(t)^{{{n}/{2}}-1}.\end{align*}

We make the ansatz $z(t)=Z(1-t)$ for $Z \in (0,\infty)$ a positive constant (so that $Z(1-t)$ is a downward sloping straight line passing through (1, 0)) and

\begin{align*}V(t,x)=(1-t)^{{{n}/{2}}}f\biggl(\frac{x}{\sqrt{1-t}}\biggr){,}\end{align*}

or equivalently

(3.3) \begin{equation} U(t,q) = (1-t)^{{{n}/{2}}}g\biggl(\dfrac{q}{1-t}\biggr)\end{equation}

where f, g are functions to be determined such that $f(x)=g(x^2)$ . On $[0,\sqrt{Z}]$ we have that f solves the ODE $ y\,f''(\,y)+(\alpha-1-y^{2})\,f'(y)-n\,y\,f\,(\,y)=0, $ subject to $ f(\sqrt{Z}) = Z^{n/2}, f'(\sqrt{Z})=nZ^{(n-1)/2},$ which is equivalent to

(3.4) \begin{equation} 4yg''(y)+ 2(\alpha-y)g'(y)-ng(y)=0, \end{equation}

subject to

\begin{align*}g(Z) = Z^{n/2}\quad\text{and}\quad g'(Z) = \frac{n}{2}Z^{{{n}/{2}}-1}.\end{align*}

In general there does not seem to be a closed-form solution to (3.4), so we look for a solution in the form of a power series expansion. However, as well as the Brownian excursion with linear payoff $(\alpha=3,n=1)$ , there are other cases where we have explicit solutions, and before moving on to the general case we record those here.

Motivated by the Brownian excursion example, consider $G(y) = y^\theta$ . Then G solves (3.4) provided $2\theta(\theta-1) + \alpha \theta = 0$ and $-2 \theta - n=0$ . Ruling out the case $\theta=0$ , we find that G is a solution provided $n = - 2 \theta = \alpha-2$ , so that there is an explicit solution if $n=\alpha-2$ and then the solution is given by $G(y) = y^{-{{n}/{2}}}$ . This solution explodes at 0, but we can find another solution H of the form

\begin{align*} H(y) = y^{-{{n}/{2}}} \int_0^y {\mathrm{e}}^{{{s}/{2}}} s^{{{n}/{2}}-1} \dfrac{{\mathrm{d}} s}{2} = y^{1-{{\alpha}/{2}}} \int_0^y {\mathrm{e}}^{{{s}/{2}}} s^{{{\alpha}/{2}}-2} \dfrac{{\mathrm{d}} s}{2}, \quad n = \alpha - 2, \end{align*}

for which H and H ^′ are strictly positive and bounded at zero. Choosing an appropriate multiple of H, we can construct a solution to (3.4) which can be used as the basis for the candidate solution of the optimal stopping problem.

Another example from the literature where an explicit solution is known is the case of a reflecting Brownian bridge ( $\alpha=1,n=1$ ); see Ekström and Wanntorp [Reference Ekström and Wanntorp5]. In this case $G(y)= {\mathrm{e}}^{{y}/{2}}$ solves (3.4). Motivated by this example, consider $G(y) = y^\theta {\mathrm{e}}^{y/2}$ . We find that G solves (3.4) provided either $(\theta=0,\alpha=n)$ or $(n=2,\alpha=2 - 2\theta)$ . In particular, when $\alpha=n$ we find that the candidate value function is given by

\begin{align*}{U(t,q)=(1-t)^{n/2} n^{n/2}\exp\biggl({\frac{(q-\sqrt{n})}{2(1-t)}}\biggr)\quad\text{and}\quad Z=\sqrt{n}.}\end{align*}

For the Brownian bridge we recover that the candidate value function is

\begin{align*}V^*(t,x) = (1-t)^{1/2}\exp\biggl({\frac{x^2-1}{2(1-t)}}\biggr)\end{align*}

and the candidate optimal stopping rule is $\tau^* = \inf \{ t \geq t_0 \colon X_t \geq \sqrt{1-t} \}$ . Further, when $n=2$ , we find that $G(y)= y^{1 - {{\alpha}/{2}}} {\mathrm{e}}^{y/2}$ is a solution to (3.4), and the candidate value function (at least when $\alpha>2$ ) is built from the second solution to (3.4),

\begin{align*} H(y) = E y^{1- \alpha/2} {\mathrm{e}}^{y/2} \int_0^y {\mathrm{e}}^{-v/2} v^{-2 + {{\alpha}/{2}}} \dfrac{{\mathrm{d}} v}{2} \end{align*}

for some constant $E>0$ .

Returning to the general case, we look for a solution in power series form. Let $\psi(y)=\sum_{k=0}^{\infty}A_{k}y^{k}$ , where $A_0=1$ so that $\psi(0)=1$ . Then we have

\begin{align*}\psi'(y)=\sum_{k=1}^{\infty}k\,A_{k}y^{k-1}\quad\text{and}\quad \psi''(y)=\sum_{k=2}^{\infty}k(k-1)A_{k}y^{k-2}.\end{align*}

Substituting these back into (3.4), we obtain a recursive equation for the coefficients $(A_k)_{k \geq 0}$ ,

\begin{equation*}A_{k+1}=\dfrac{(2k+n)}{2(k+1)(2k+\alpha)} A_k, \quad A_0=1, \end{equation*}

which has solution

(3.5) \begin{equation} A_{k}=\dfrac{\Gamma({{\alpha}/{2}})}{\Gamma({{n}/{2}})}\dfrac{\Gamma(k+{{n}/{2}})}{\Gamma(k+{{\alpha}/{2}})2^{k}k!},\end{equation}

where $\Gamma$ is the Gamma function defined as

\begin{align*}\Gamma(z)=\int_{0}^{\infty}{\mathrm{e}}^{-x}x^{z-1}\,{\mathrm{d}} x.\end{align*}

Given $\psi$ , we can define a second solution $\psi_2$ to (3.4) of the form

\begin{align*} \psi_2(y) = \psi(y) \int_0^y \dfrac{ s^{-\alpha/2} {\mathrm{e}}^{s/2} }{\psi(s)^{2}} \dfrac{{\mathrm{d}} s}{2}, \end{align*}

and the general solution to (3.4) is

(3.6) \begin{equation}g(y) = E_1 \psi(y) + E_2 \psi_2(y).\end{equation}

Note that $\psi_2(y) \sim y^{1- \alpha/2}$ for small y. We want a solution for which g ^′ is bounded on [0, Z] and g ^″ is finite at 0 and for which the boundary conditions hold at Z. Together, the first set of conditions implies that $E_2=0$ and the boundary conditions at Z imply that Z solves $Z \psi'(Z) = ({{n}/{2}}) \psi(Z)$ . This is equivalent to $F_{\alpha,n} (Z)=0$ , where

(3.7) \begin{equation} F_{\alpha,n}(z) = \sum_{k=0}^{\infty} (2k-n)A_{k}z^{k}.\end{equation}

The next result follows from Lemma A.1 in the Appendix.

Figure 2.

An example of $F_{\alpha,n}(z)$ , where $\alpha=3$ and $n=2$ , and its corresponding root $Z_{3,2}$ , which satisfies $Z_{3,2}> {{(3+2-2)}/{2}}=1.5$ . Note that the figure confirms the results in Corollary 3.1 and Proposition 3.1 for this parameter combination; the issue is to prove these results for general parameters.

Figure 2 Long description

A line graph showing the function F subscript 3, 2 (z) with z on the x axis ranging from 0 to 3 and F subscript 3, 2 (z) on the y axis ranging from -2 to 1. The graph depicts a curve that starts at -2 on the y axis and gradually increases, crossing the x axis at approximately z equals 2.5. A dashed horizontal line is present at y equals 0, and a dashed vertical line is present at z equals 2.5, indicating the root of the function. All values are approximated.

Recall that the root $Z_{\alpha,n}$ defines the candidate optimal boundary to the optimal stopping problem for the $\alpha$ -dimensional squared Bessel process with payoff $\phi(q) = q^{n/2}$ via $\tau^* = \inf \{ t \in (0,1) \colon Q_t \geq Z(1-t) \}.$ Moreover, the coefficient $E_1$ in (3.6) can be computed using $Z^{n/2} = g(Z) = E_1 \psi(Z)$ to give

\begin{equation*} E_1 = \dfrac{Z_{\alpha,n}^{n/2}}{\sum_{k=0}^{\infty}A_{k}Z_{\alpha,n}^{k}}.\end{equation*}

Therefore we can define a candidate value function $U^*$ using (3.3) and (3.6) of the form

(3.8) \begin{equation} U^{*}(t,q)= \begin{cases} \biggl(\dfrac{Z_{\alpha,n}^{n/2}}{\sum_{k=0}^{\infty}A_{k}Z_{\alpha,n}^{k}}\biggr)(1-t)^{{{n}/{2}}}\sum_{k=0}^{\infty}A_{k}\biggl(\dfrac{q}{{1-t}}\biggr)^{k} & \text{if $ q < Z_{\alpha,n} (1-t)$,}\\[9pt] q^{n/2} &\text{if $ q \geq Z_{\alpha,n} (1-t) $,}\end{cases}\end{equation}

where $(A_{k})_{k \geq 0}$ are given by (3.5) and $Z_{\alpha,n}$ is the unique solution to (3.7).

Recall that $g(y) = E_1 \psi(y) = E_1 \sum_{k \geq 0} A_k y^k$ on [0, Z].

Proof. We have

\begin{align*}2y g'(y) - ng(y) = \frac{E_1}{2}F_{\alpha,n}(y) < 0\ \text{on [0, ${Z}$).}\end{align*}

Then, for $z \in (0,Z)$ , we have

\begin{align*}\ln \frac{g(Z)}{g(z)} = \int_z^Z \frac{g'(y)}{g(y)} \,{\mathrm{d}} y < \frac{n}{2} \int_z^Z \frac{1}{y}\,{\mathrm{d}} y = \ln \frac{Z^{n/2}}{z^{n/2}}{,}\end{align*}

and then using $g(Z) = Z^{n/2}$ we find $g(z) \geq z^{n/2}$ on $[0,Z]$ .

The next result is important for the main theorem. However, the proof is long and intricate and so it is postponed to the Appendix. Note that when $\alpha+n \leq 2$ there is nothing to prove.

Theorem 3.1. The candidate value function $U^{*}$ given in (3.8) coincides with the value function $\bar{U}$ defined in (3.1). Moreover, the associated stopping time $\tau^*$ given by

\begin{equation*} \tau^{*}=\inf\{s\geq t\colon Q_{s}\geq z(s) \}, \end{equation*}

where $z(s)=Z_{\alpha,n}(1-s)$ and $Z_{\alpha,n}$ is the root given in Corollary 3.1, is optimal for the problem started at $(t,x)$ .

Proof. The proof follows the proof of Theorem 2.1 and here we only describe the necessary changes.

Fix $t_0$ , and for $t \geq t_0$ let $M_{t}=U^{*}(t,Q_{t})$ . Recall that we chose a power series expansion such that g, g ^′, and g ^″ are all well-defined at 0. Hence $U^*_q$ is bounded on [0, Z] and $U^*_{qq}$ exists on [0, Z) so that we can apply Itô’s formula. We have

\begin{equation*} {\mathrm{d}} M_{t}= h(t,Q_t)\mathbb{I}\{Q_{t}\geq z(t) \}\,{\mathrm{d}} t +U_{q}^{*}(t,Q_{t}) 2 \sqrt{Q_t} \,{\mathrm{d}} W_{t},\end{equation*}

where

\begin{equation*} h(t,q)= n q^{{{n}/{2}}-1}\biggl( \dfrac{\alpha + n - 2}{2} - \dfrac{q}{1-t}\biggr).\end{equation*}

Since $Z_{\alpha,n} \geq {{(\alpha+n-2)}/{2}}$ , we deduce that $h \leq 0$ on the stopping region $Q_t \geq Z_{\alpha,n} (1-t)$ . It follows that $A = \{ A_t \}_{t_0 \leq t \leq 1}$ given by

\begin{align*}A_{t}\;:\!=\; \int_{t_0}^{t}h(s,Q_{s})\mathbb{I}\{Q_{s}\geq z(s)\}\,{\mathrm{d}} s\end{align*}

is a decreasing process, null at $t_0$ .

Define the local martingale $N= \{ N_t \}_{t_0 \leq t \leq 1}$ by

\begin{align*}N_{t}=\int_{t_0}^{t}U^{*}(s,Q_{s}) 2 \sqrt{Q_s} \,{\mathrm{d}} W_{s}.\end{align*}

Then $M_t = U^*(t_0,Q_{t_0}) + A_t +N_t$ . Since $M \geq 0$ , we have $N \geq -U^*(t_0,Q_{t_0})-A_t \geq -U^*(t_0,Q_{t_0})$ and therefore the local martingale N is bounded below. Hence N is a supermartingale. It follows that

\begin{align*}\mathbb{E}[M_{t}\mid \mathcal{F}_{s}]=U^*(t_0,Q_{t_0}) + \mathbb{E}[A_{t}+N_{t}\mid \mathcal{F}_{s}] \leq U^*(t_0,Q_{t_0}) + A_{s}+ N_{s}=M_{s}\end{align*}

so that M is a supermartingale. By Lemma 3.1, $g(z) \geq z^{n/2}$ on [0, Z] and hence $U^*(t,Q_t) \geq Q_t^{n/2}$ on $[0,\infty)$ , and it follows that for any stopping time $\tau$ with $\tau \geq t_0$ ,

\begin{align*} \mathbb{E}_{t_0,x}[Q_\tau^{n/2}] \leq \mathbb{E}_{t_0,x} [U^*(\tau,Q_\tau)] \leq M_{t_0} = U^*(t_0,Q_{t_0}) .\end{align*}

Thus $\bar{U}(t_0,q) \leq U^*(t_0,q)$ .

The remaining details which show that M is a martingale for $t_0 \leq t \leq \tau^*$ and hence that $\bar{U}(t_0,q) \geq U^*(t_0,q)$ follow exactly as in Theorem 2.1, using that $\sqrt{y}g'(y)$ is bounded on [0, Z] for our chosen g.