Proving the Strong Markov Property by Approximation

Why Approximate a Stopping Time?

Brownian motion has independent increments after every fixed time: once time $s$ is fixed, the future increments are independent of the information available by $s$. The strong Markov property says that the same fresh-start description holds when $s$ is replaced by a stopping time, a random time whose occurrence can be determined from information available up to that time. The proof here turns the random time into a sequence of discrete-valued stopping times, applies independent increments at each of their possible values, and then uses continuous paths to pass to the limit.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with a filtration $(\mathcal{F}_t)_{t\ge0}$. Assume $(B_t)_{t\ge0}$ is Brownian motion with respect to this filtration: it is adapted, has continuous paths, and for each deterministic $s\ge0$, the increments after $s$ are independent of $\mathcal{F}_s$ and have the usual Brownian distributions. For a stopping time $\tau$, its information set is $\mathcal{F}_\tau=\{A\in\mathcal{F}:A\cap\{\tau\le t\}\in\mathcal{F}_t\text{ for every }t\ge0\}$. We first prove the result for bounded stopping times.

Definition: The shifted process after $\tau$ is $B^\tau_t=B_{\tau+t}-B_\tau$, for $t\ge0$. The strong Markov property asserts that, conditional on $\mathcal{F}_\tau$, this whole shifted process has the same law as Brownian motion started at $0$, independently of $\mathcal{F}_\tau$.

Upward Rounding Gives Stopping Times

For a bounded stopping time $\tau$, define its upward dyadic approximation by

\tau_n=2^{-n}\left\lceil 2^n\tau\right\rceil,\qquad n\ge0.

Thus $\tau_n$ is the first point on the grid $\{k2^{-n}:k=0,1,2,\ldots\}$ that is at least $\tau$. It satisfies $\tau\le\tau_n$, $0\le\tau_n-\tau<2^{-n}$, and $\tau_n\downarrow\tau$. The strict upper bound holds even when $\tau$ lies on the grid: in that case the error is zero.

Lemma: If $\tau$ is a bounded stopping time, then each $\tau_n$ is a finite-valued stopping time and $\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}$.

Proof. Write $\delta=2^{-n}$. If $t$ lies between grid points $k\delta$ and $(k+1)\delta$, then $\{\tau_n\le t\}=\{\tau_n\le k\delta\}=\{\tau\le k\delta\}$, which belongs to $\mathcal{F}_{k\delta}\subseteq\mathcal{F}_t$. Hence $\tau_n$ is a stopping time. Boundedness of $\tau$ means that $\tau_n$ takes only finitely many grid values. Finally, if $\tau\le\sigma$ are stopping times and $A\in\mathcal{F}_\tau$, then $A\cap\{\sigma\le t\}=A\cap\{\tau\le t\}\cap\{\sigma\le t\}\in\mathcal{F}_t$: on $\{\sigma\le t\}$, $\tau\le t$, and each event on the right is measurable at time $t$. Thus $A\in\mathcal{F}_\sigma$; take $\sigma=\tau_n$. $\square$

Worked Example: Rounding a Truncated Hitting Time

Let $T_a=\inf\{t\ge0:B_t=a\}$ for a fixed level $a$, and set $\tau=T_a\wedge1$. This is a bounded stopping time. Consider a path on which $\tau=5/12$. At grid width $1/4$, the upward approximation is $\lceil(1/4)^{-1}(5/12)\rceil/4=\lceil5/3\rceil/4=1/2$, with error $1/2-5/12=1/12<1/4$. At grid width $1/16$, it is $\lceil16(5/12)\rceil/16=\lceil20/3\rceil/16=7/16$, with error $7/16-5/12=1/48<1/16$.

These values decrease toward $5/12$, while each approximation remains at or after $\tau$. For any continuous Brownian path, this implies $B_{\tau_n+t}-B_{\tau_n}\to B_{\tau+t}-B_\tau$ for every fixed $t\ge0$. The numerical path value illustrates the approximation; the proof applies pathwise on every continuous path.

The Discrete-Valued Case

The key calculation is available when the stopping time has only finitely many possible values. For a vector of times $0\le t_1<\cdots<t_m$ and a bounded continuous function $g:\mathbb{R}^m\to\mathbb{R}$, the desired identity is

\mathbb{E}\!\left[g\!\left(B_{\tau+t_1}-B_\tau,\ldots,B_{\tau+t_m}-B_\tau\right)\middle|\mathcal{F}_\tau\right] = \mathbb{E}\!\left[g(B_{t_1},\ldots,B_{t_m})\right].

Proof. First suppose $\tau$ takes finitely many deterministic values $s_1,\ldots,s_r$. For any $A\in\mathcal{F}_\tau$, the event $A_j=A\cap\{\tau=s_j\}$ belongs to $\mathcal{F}_{s_j}$, by the definition of $\mathcal{F}_\tau$ and taking the difference of the events at $s_j$ and just before $s_j$. On $A_j$, the vector of increments after $\tau$ is the vector of increments after $s_j$. Brownian independent increments therefore give \[ \mathbb{E}\!\left[\mathbf{1}_{A_j}g(B_{s_j+t_1}-B_{s_j},\ldots,B_{s_j+t_m}-B_{s_j})\right] =\mathbb{P}(A_j)\mathbb{E}[g(B_{t_1},\ldots,B_{t_m})]. \] Summing over $j=1,\ldots,r$ gives the same identity with $A$ in place of $A_j$. Since it holds for every $A\in\mathcal{F}_\tau$, it is exactly the stated conditional-expectation identity. $\square$

The same reasoning applies to each $\tau_n$. To pass to $\tau$, take any $A\in\mathcal{F}_\tau$. The lemma puts $A$ in $\mathcal{F}_{\tau_n}$, so the discrete-valued identity applies with this same event $A$ for every $n$. The event must be held fixed while the random variables in the identity converge; this is why the inclusion of information sets matters.

Strong Markov Property by Approximation

Theorem: Let $(B_t)_{t\ge0}$ be Brownian motion with respect to $(\mathcal{F}_t)_{t\ge0}$, and let $\tau$ be a bounded stopping time. Then, for every $m\ge1$, $0\le t_1<\cdots<t_m$, and bounded continuous $g:\mathbb{R}^m\to\mathbb{R}$, $$ \mathbb{E}\!\left[g\!\left(B_{\tau+t_1}-B_\tau,\ldots,B_{\tau+t_m}-B_\tau\right)\middle|\mathcal{F}_\tau\right] =\mathbb{E}\!\left[g(B_{t_1},\ldots,B_{t_m})\right]\quad\text{almost surely}. $$ Consequently, the entire shifted path $B^\tau$ is independent of $\mathcal{F}_\tau$ and has Brownian law.

Proof. Fix $A\in\mathcal{F}_\tau$ and set $Y_n=(B_{\tau_n+t_1}-B_{\tau_n},\ldots,B_{\tau_n+t_m}-B_{\tau_n})$ and $Y=(B_{\tau+t_1}-B_\tau,\ldots,B_{\tau+t_m}-B_\tau)$. By the discrete-valued result and $\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}$, \[ \mathbb{E}[\mathbf{1}_A g(Y_n)]=\mathbb{P}(A)\mathbb{E}[g(B_{t_1},\ldots,B_{t_m})]. \] Almost surely, every Brownian path is continuous and $\tau_n\to\tau$, so $Y_n\to Y$. Continuity of $g$ gives $g(Y_n)\to g(Y)$. Since $g$ is bounded, bounded convergence gives \[ \mathbb{E}[\mathbf{1}_A g(Y)]=\mathbb{P}(A)\mathbb{E}[g(B_{t_1},\ldots,B_{t_m})]. \] This holds for every $A\in\mathcal{F}_\tau$, which proves the conditional identity. Bounded continuous test functions determine the distribution of a finite-dimensional real vector, so the shifted process has Brownian finite-dimensional distributions independent of $\mathcal{F}_\tau$. Finally, $B^\tau$ has continuous paths. On the space of continuous paths, the Borel sets are generated by evaluations at nonnegative rational times; the standard $\pi$-$\lambda$ uniqueness theorem therefore extends the finite-dimensional independence identity to the whole path. $\square$

If $\tau$ is finite almost surely but not bounded, apply the bounded result to $\tau\wedge r$, for positive integers $r$. For $A\in\mathcal{F}_\tau$, the event $A\cap\{\tau\le r\}$ belongs to $\mathcal{F}_{\tau\wedge r}$. On $\{\tau\le r\}$, the shifted paths at $\tau\wedge r$ and $\tau$ agree. The bounded identity applied on this event gives the desired identity restricted to $\{\tau\le r\}$; letting $r\to\infty$ and using bounded convergence proves the result for finite-almost-surely $\tau$. This localization does not assert a value for the shifted process on $\{\tau=\infty\}$.

Worked Example: A Two-Valued Stopping Time

Define $\tau=1/3$ if $B_{1/3}\ge0$, and $\tau=2/3$ otherwise. This is a stopping time: before $1/3$ the event $\{\tau\le t\}$ is empty; for $1/3\le t<2/3$ it is $\{B_{1/3}\ge0\}\in\mathcal{F}_t$; and for $t\ge2/3$ it is the whole space. Since $B_{1/3}$ is a centered nondegenerate normal variable, $\mathbb{P}(\tau=1/3)=\mathbb{P}(\tau=2/3)=1/2$. The equality case has probability zero.

Take $h=1/4$ and $u=2$. For any $A\in\mathcal{F}_\tau$, the events $A_1=A\cap\{\tau=1/3\}$ and $A_2=A\cap\{\tau=2/3\}$ belong respectively to $\mathcal{F}_{1/3}$ and $\mathcal{F}_{2/3}$. Independent increments and the characteristic function of a centered normal variable give \[ \mathbb{E}\!\left[\mathbf{1}_{A_j}e^{\,2i(B_{\tau+h}-B_\tau)}\right] =\mathbb{P}(A_j)e^{-2^2h/2} =\mathbb{P}(A_j)e^{-1/2},\qquad j=1,2. \] Adding these equations yields $\mathbb{E}[e^{\,2i(B_{\tau+h}-B_\tau)}\mid\mathcal{F}_\tau]=e^{-1/2}$ almost surely. The exponent is checked using $2^2(1/4)/2=1/2$, the characteristic function of a centered normal increment with variance $h=1/4$.

Worked Example: Two Future Increments After a Hitting Time

Let $\tau=T_1\wedge1$, where $T_1=\inf\{t\ge0:B_t=1\}$. This is a bounded stopping time. Define $Y_1=B_{\tau+1}-B_\tau$ and $Y_2=B_{\tau+2}-B_{\tau+1}$. The strong Markov property says that $(Y_1,Y_2)$, conditional on $\mathcal{F}_\tau$, has the law of two successive unit-time Brownian increments. Each increment is $N(0,1)$, and the two are independent. Therefore \[ \mathbb{P}(Y_1>0,\ Y_2>0\mid\mathcal{F}_\tau) =\tfrac12\cdot\tfrac12=\tfrac14\quad\text{almost surely}. \] The calculation uses symmetry of each centered normal distribution and independence of disjoint Brownian increments. As a check, integrating the conditional probability gives the same unconditional probability, $\mathbb{P}(Y_1>0,Y_2>0)=1/4$.

Why Rounding Down Is a Pitfall

Continuity alone does not prove a strong Markov identity. The approximating times must also preserve the relationship between the information at $\tau$ and the information at the approximating time. Upward rounding gives $\tau\le\tau_n$, hence $\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}$; it allows the same event $A\in\mathcal{F}_\tau$ in every discrete-time calculation.

Common mistake: Round down and assume the resulting time is a stopping time because it is close to $\tau$. For the two-valued stopping time in the previous example, rounding down to the grid of width $1/2$ gives $\tau_{\mathrm{down}}=0$ when $B_{1/3}\ge0$ and $\tau_{\mathrm{down}}=1/2$ otherwise. Thus $\{\tau_{\mathrm{down}}\le0\}=\{B_{1/3}\ge0\}$, which is not in $\mathcal{F}_0$ for the usual Brownian filtration. The rounded-down time is not a stopping time. The correction is to round upward: then the approximation is a stopping time, lies after $\tau$, and converges to it, so both the information inclusion and the pathwise limit needed in the proof hold.

Check Your Understanding

Use the approximation argument to answer each question.

For $\tau_n=2^{-n}\lceil2^n\tau\rceil$, what is the strict upper bound on $\tau_n-\tau$, and when is the difference zero?
Why does $\tau\le\tau_n$ imply $\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}$?
In the two-valued example, compute the conditional characteristic function of $B_{\tau+1/2}-B_\tau$ at $u=1$.
Which convergence theorem permits passing from the discrete approximations to the limit for bounded continuous $g$?
Why does rounding down fail to give a stopping time in the example with values $1/3$ and $2/3$?