Brownian Motion: Markov Properties and Path Continuity

Brownian motion combines a simple distributional rule with a strong path requirement: increments over disjoint intervals are independent Gaussian variables, and the paths are continuous. These properties lead to two kinds of questions. How does the process restart after an observation time, and how rapidly can a continuous path fluctuate? The Markov property and the continuity estimates below answer these questions.

We begin with the probability space and filtration against which adaptation and independence are measured. Fix a probability space \((\Omega, \mathcal F, \mathop{\mathrm{\mathbf{P}\!\,}})\), where \(\mathcal F\) is a sigma-field on \(\Omega\). Let \(\{\mathcal F_t\}_{t \geq 0}\) be a filtration.

A process \(X\) is jointly measurable if \((\omega,t)\mapsto X_t(\omega)\) is measurable for \(\mathcal F\otimes\mathscr B(\mathbb R_+)\). It is progressively measurable if, for each \(T\geq0\), its restriction to \(\Omega\times[0,T]\) is measurable for \(\mathcal F_T\otimes\mathscr B([0,T])\). These are different requirements. An adapted process with continuous paths is progressive: its left-endpoint step approximations on each finite interval are measurable for this smaller product sigma-field and converge pointwise.

A process \(\{B_t\}_{t \geq0}\) is called an one-dimensional Brownian motion if

\(B_t\) is \(\mathcal F_t\)-measurable for all \(t \geq 0\),
\(B_t-B_s \perp \!\!\! \perp\mathcal F_s\) for all \(s \leq t\),
\(B_t - B_s \sim \mathcal{N}(0,t-s)\) for all \(s \leq t\),
\(B_0 = 0\) almost surely,
\(B_t\) has continuous paths for all \(t\geq0\).

We can extend this definition to an arbitrary \(d\)-dimension: \((B^{(1)}, \dots, B^{(d)})\) is a \(d\)-dimensional Brownian motion if each coordinate \(B^{(i)}\) is a one-dimensional Brownian motion, and the coordinate processes are mutually independent. The natural terminal sigma-field is \(\mathcal F^{B}_\infty=\sigma(B_t:t\geq0)\). For a general filtration, \(\mathcal F_\infty=\sigma(\bigcup_{t\geq0}\mathcal F_t)\) can be larger; independent information may also be available. These sigma-fields coincide when we use the natural filtration, with the corresponding completion when null sets are added.

By definition, here are some facts about a Brownian motion. If \(B\) is a Brownian motion starting at zero, then, for \(a > 0\), \(Y_t = a B_{t/a^2}\) is also a Brownian motion starting at zero. To see this, we fix \(\{\mathcal F_t\}_{t \geq 0}\) to be a filtration for \(\{B_t\}_{t \geq 0}\). Let \(\mathcal G_t = \mathcal F_{t/a^2}\) be a filtration for \(\{Y_t\}\). Clearly, \(\{Y_t\}\) by definition has continuous sample paths, is \(\mathcal G_t\)-measurable for all \(t \geq 0\), and \(Y_0 = 0\). If \(s<t\), \(Y_t - Y_s = a(B_{t/a^2} - B_{s/a^2}) \sim \mathcal{N}(0,t-s)\), because \(\mathop{\mathrm{Var}}(Y_t - Y_s) = a^2 \left( \frac{t-s}{a^2} \right) = t-s\). The independence \(Y_t-Y_s \perp \!\!\! \perp\mathcal F_{s/a^2}\) follows from definition.

Brownian motion is a Gaussian process. For \(0=t_0<t_1<\cdots<t_n\), the variables

\[Z_i=\frac{B_{t_i}-B_{t_{i-1}}}{\sqrt{t_i-t_{i-1}}},\qquad 1\leq i\leq n,\]

are independent standard normals. Since \(B_{t_j}=\sum_{i=1}^j\sqrt{t_i-t_{i-1}}Z_i\), every finite-dimensional vector is jointly Gaussian.

Conversely, suppose \(B\) has continuous paths, all finite-dimensional vectors are jointly Gaussian, \(\mathbb{E}B_t=0\), and \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). Use its natural filtration \(\mathcal F_t=\sigma(B_r:r\leq t)\). The covariance formula gives \(B_0=0\) almost surely and \(B_t-B_s\sim\mathcal{N}(0,t-s)\). Its covariance with every finite vector \((B_{r_1},\ldots,B_{r_m})\), \(r_j\leq s\), is zero. Joint Gaussianity therefore gives independence from that entire vector. A generating-class argument extends this to independence from \(\mathcal F_s\), proving the Brownian property.

Remark 1. The completed natural Brownian filtration is already right-continuous. Let \(\mathcal F_t^{00}=\sigma(B_s:s\leq t)\), let \(\mathcal{N}\) contain all subsets of null sets in \(\mathcal F_\infty^{00}\), and put \(\mathcal F_t^0=\sigma(\mathcal F_t^{00}\cup\mathcal{N})\). Then \(\bigcap_{\varepsilon>0}\mathcal F_{t+\varepsilon}^0=\mathcal F_t^0\). One way to verify this is to start with a bounded continuous function \(Y\) of finitely many Brownian coordinates. Independent Gaussian increments give its conditional expectation at time \(s\) by keeping the coordinates already observed and integrating the future increments against their Gaussian law. As \(s\downarrow t\), path continuity and continuity of these Gaussian expectations give \(\mathbb{E}[Y\mid\mathcal F_s^0]\to\mathbb{E}[Y\mid\mathcal F_t^0]\) in \(L^1\). Bounded cylinder functions are dense in \(L^1(\mathcal F_\infty^0)\), and conditional expectation is an \(L^1\) contraction, so the same limit holds for every integrable \(Y\). The reverse martingale convergence theorem identifies the limit as the conditional expectation on \(\bigcap_{s>t}\mathcal F_s^0\). Applying this to indicators in that intersection proves equality modulo null sets, hence equality after completion. Completion does not change independence, so \(B\) remains Brownian for this filtration.

Remark 2. Equality of finite-dimensional distributions does not imply that two arbitrary processes are indistinguishable, or transfer path regularity without additional hypotheses. For example, let \(\Omega=[0,1]\) with Lebesgue measure and define

\[\begin{aligned} X_t (\omega) &= 0 ~~,~~ \forall t \,,\, \omega ~, \\ Y_t (\omega) &= \begin{cases} 1 & \text{if } t=\omega ~, \\ 0 & \text{otherwise} ~.\end{cases} \end{aligned}\]

For each fixed \(t\), \(X_t=Y_t\) almost surely, and the same holds jointly at every finite collection of fixed times. Nevertheless, every path of \(X\) is continuous and every path of \(Y\) has a spike. The two processes have the same law on the cylinder sigma-field; the set of continuous functions is not determined by that sigma-field on the unrestricted space of all functions. This is not an example of two Brownian motions with different laws.

If instead \(X\) and \(Y\) are measurable random elements of \(\mathscr C[0,t_0]\) with the uniform norm, equal finite-dimensional distributions do determine the same law. Its Borel sigma-field is generated by evaluations at rational times: by continuity a uniform distance is the supremum of the distances at those times, and separability reduces open sets to countable unions of balls. Equality on finite intersections of evaluation events therefore extends to every Borel set.

1. Markov Properties

The Markov property describes the process after a fixed time; the strong Markov property allows that time to be chosen from the observed path. Let \(B\) be Brownian motion for a filtration satisfying the usual conditions. For a deterministic \(u\geq0\), the process \(Y_t=B_{u+t}-B_u\) is Brownian motion for \(\mathcal G_t=\mathcal F_{u+t}\) and is independent of \(\mathcal F_u\). This follows by applying independent increments to each finite vector of future increments.

Now let \(T\) be an almost surely finite stopping time. The process \(Y_t=B_{T+t}-B_T\) is Brownian motion for \(\mathcal G_t=\mathcal F_{T+t}\), independent of \(\mathcal F_T\). To prove the assertion, fix \(0<t_1<\cdots<t_m\), a bounded continuous function \(f:\mathbb R^m\to\mathbb R\), and \(A\in\mathcal F_T\). Approximate \(T\) from above by \(T_n=2^{-n}(\lfloor2^nT\rfloor+1)\). Then \(A\cap\{T_n=k2^{-n}\}\in\mathcal F_{k2^{-n}}\), and independent increments give

\[\begin{aligned} &\mathbb{E}\left[f(B_{T_n+t_1}-B_{T_n},\ldots,B_{T_n+t_m}-B_{T_n})\mathbb1_A\right]\\ &=\sum_{k\geq1}\mathbb{E}\left[f(B_{k2^{-n}+t_1}-B_{k2^{-n}},\ldots,B_{k2^{-n}+t_m}-B_{k2^{-n}}) \mathbb1_{A\cap\{T_n=k2^{-n}\}}\right]\\ &=\mathbb{E} f(B_{t_1},\ldots,B_{t_m})\,\mathbb P(A). \end{aligned}\]

Path continuity and bounded convergence allow \(n\to\infty\). A monotone-class argument extends the identity to bounded Borel functions. It identifies the finite-dimensional laws and independence from \(\mathcal F_T\). To obtain the increment-independence condition relative to \(\mathcal G_s\), apply the same argument to the stopping time \(T+s\). Adaptation and continuity are inherited from \(B\), so all parts of the Brownian definition hold.

Reflection at a level

The strong Markov property also gives the reflection identity used in the later local-time calculations. Fix \(a>0\), set \(T_a=\inf\{s\geq0:B_s=a\}\), and fix \(t>0\). On \(\{T_a<t\}\), the process after \(T_a\) is an independent Brownian motion. Conditional on \(\mathcal F_{T_a}\), its value after the positive time \(t-T_a\) has a symmetric normal distribution. Formally one applies the finite-stopping-time result to \(T_a\wedge t\) and conditions on this random duration. Hence

\[\mathbb P(B_t\geq a,T_a<t)=\frac12\mathbb P(T_a<t).\]

Continuity implies \(\{B_t\geq a\}\subseteq\{T_a\leq t\}\), and \(\mathbb P(T_a=t)=0\) because \(\{T_a=t\}\subseteq\{B_t=a\}\). Therefore

\[\mathbb P\left(\sup_{s\leq t}B_s\geq a\right) =\mathbb P(T_a\leq t)=2\mathbb P(B_t\geq a) =2\bigl[1-\Phi(a/\sqrt t)\bigr].\]

In particular \(T_a<\infty\) almost surely by letting \(t\uparrow\infty\). For \(S_t=\sup_{s\leq t}B_s\), this also gives \(\mathbb P(S_t\leq a)=2\Phi(a/\sqrt t)-1\) for \(a\geq0\), using continuity in \(a\) at zero. Reflection of \(-B\) gives the corresponding negative-level result.

2. Continuity of Paths

A function \(f:[0,1]\to\mathbb R\) is \(\alpha\)-Hölder continuous if \(|f(t)-f(s)|\leq C|t-s|^\alpha\) for some finite \(C\) and all \(s,t\). Brownian paths have this property almost surely for every \(0<\alpha<1/2\). We prove it by controlling all dyadic intervals at once.

For \(\lambda>0\), independent normal increments give

\[\mathbb E[e^{\lambda(B_t-B_s)}\mid\mathcal F_s]=e^{\lambda^2(t-s)/2},\]

so \(E_t=\exp(\lambda B_t-\lambda^2t/2)\) is a nonnegative martingale with \(E_0=1\). Let \(\tau_x=\inf\{t:B_t\geq x\}\) for \(x>0\). Bounded-time optional sampling gives \(\mathbb E E_{\tau_x\wedge h}=1\), while on \(\{\tau_x\leq h\}\), continuity gives \(E_{\tau_x}\geq e^{\lambda x-\lambda^2h/2}\). Thus

\[\mathbb P\left(\sup_{t\leq h}B_t\geq x\right)\leq e^{-\lambda x+\lambda^2h/2}.\]

Choose \(\lambda=x/h\) to obtain \(e^{-x^2/(2h)}\). Applying the same argument to \(-B\) gives a factor two for the absolute maximum. Define

\[A_n=\bigcup_{k=0}^{2^n-1} \left\{\sup_{k2^{-n}\leq t\leq(k+1)2^{-n}}|B_t-B_{k2^{-n}}|>2^{-n\alpha}\right\}.\]

Stationary increments and the union bound give

\[\mathbb P(A_n)\leq2^{n+1}\exp\left(-\frac12 2^{n(1-2\alpha)}\right).\]

This sequence is summable. Borel–Cantelli therefore says that, almost surely, \(A_n\) does not occur for any sufficiently large \(n\). If \(2^{-(n+1)}<t-s\leq2^{-n}\), the two times lie in the same or neighbouring dyadic intervals at level \(n\). Comparing each value with the relevant endpoints gives

\[|B_t-B_s|\leq3\,2^{-n\alpha}\leq3\,2^\alpha|t-s|^\alpha.\]

This handles short intervals. Boundedness of each continuous path on \([0,1]\) supplies a finite constant for the remaining intervals. A countable intersection over exponents approaching \(1/2\) gives all smaller positive exponents.

The law of the iterated logarithm gives a sharper scale at a fixed endpoint and at infinity:

\[\limsup_{t\to\infty}\frac{|B_t|}{\sqrt{2t\log\log t}}=1, \qquad \limsup_{t\downarrow0}\frac{|B_t|}{\sqrt{2t\log\log(1/t)}}=1 \quad\text{almost surely}.\tag{1}\]

For the upper bound at infinity, fix \(\varepsilon>0\) and choose \(q>1\) close enough to one that \((1+\varepsilon)^2/q>1\). For all sufficiently large \(n\), the maximal estimate above gives

\[\mathbb P\left(\sup_{s\leq q^n}|B_s|>(1+\varepsilon) \sqrt{2q^{n-1}\log\log q^{n-1}}\right) \leq2\bigl((n-1)\log q\bigr)^{-(1+\varepsilon)^2/q}.\]

The probabilities are summable. On \(q^{n-1}\leq t\leq q^n\), the denominator \(\sqrt{2t\log\log t}\) is at least its value at the left endpoint, so Borel–Cantelli gives an eventual upper bound of \(1+\varepsilon\) for the ratio.

For the lower bound, take \(0<\varepsilon<1\) and now choose a new, sufficiently large \(q\) so that

\[a=\frac{(1-\varepsilon)^2}{1-q^{-1}}<1, \qquad 2q^{-1/2}<\frac{\varepsilon}{2}.\]

The events

\[E_n=\left\{B_{q^{n+1}}-B_{q^n}>(1-\varepsilon) \sqrt{2q^{n+1}\log\log q^{n+1}}\right\}\]

are independent. For a standard normal variable \(Z\) and \(x\geq1\), integrating its density over \([x,x+1/x]\) gives \(\mathbb P(Z>x)\geq c x^{-1}e^{-x^2/2}\), with an absolute \(c>0\). Standardizing the increment therefore yields, for large \(n\),

\[\mathbb P(E_n)\geq c_q\frac{(n+1)^{-a}}{\sqrt{\log(n+1)}}.\]

This series diverges because \(a<1\), so \(E_n\) occurs infinitely often. The already proved upper bound gives eventually

\[|B_{q^n}|\leq2\sqrt{2q^n\log\log q^n} \leq\frac{\varepsilon}{2}\sqrt{2q^{n+1}\log\log q^{n+1}}.\]

On infinitely many of the events \(E_n\), subtraction now gives

\[B_{q^{n+1}}\geq\left(1-\frac{3\varepsilon}{2}\right) \sqrt{2q^{n+1}\log\log q^{n+1}}.\]

Taking a countable sequence \(\varepsilon\downarrow0\) proves the assertion at infinity. For the assertion at zero, define \(\widehat B_t=tB_{1/t}\) for \(t>0\) and \(\widehat B_0=0\). The upper bound just proved implies continuity at zero. Its covariance is \(\min(s,t)\), and it is centered Gaussian, so the Gaussian characterization above makes it Brownian motion. Apply the result at infinity to \(\widehat B\) and substitute \(t=1/u\). This proves the second assertion. ◻

Le Gall, Jean-François. 2016. Brownian Motion, Martingales, and Stochastic Calculus. 1st ed. 2016. Graduate Texts in Mathematics 274. Cham: Springer International Publishing : Imprint: Springer. https://doi.org/10.1007/978-3-319-31089-3.

Reading: Le Gall (2016), Brownian Motion, Martingales, and Stochastic Calculus.