Brownian Path Oscillations and Continuity Bounds

Continuity alone says that sufficiently short increments are small. For Brownian motion, we can ask how small they are, uniformly over an interval, and how often unusually large increments occur. Gaussian tail estimates and the two Borel–Cantelli lemmas answer these questions. The final part introduces finite variation and localisation, which will be needed to compare Brownian paths with integrators from ordinary calculus.

For any continuous function \(f:[0,1] \to \mathbb R\), the modulus of continuity of \(f\) is defined as the quantity

\[\begin{aligned} \omega_f (\delta) = \sup_{s,t \in [0,1]:|t-s|<\delta} |f(t)-f(s)| ~. \end{aligned}\]

For Brownian motion, we seek matching upper and lower bounds for this modulus. The upper bound must control many increments simultaneously; the lower bound can use independence on disjoint intervals. Let \(\{B_t\}_{t \geq 0}\) denote a Brownian motion.

We first obtain an upper bound with an unspecified constant. For a dyadic interval \(I=[k2^{-n},(k+1)2^{-n}]\), the exponential maximal inequality gives

\[\mathbb P\left(\sup_{t\in I}|B_t-B_{k2^{-n}}|>K\sqrt{n2^{-n}}\right)\leq2e^{-K^2n/2}.\]

There are \(2^n\) such intervals in \([0,1]\). Choose \(K\) so that \(K^2/2>\log2\); the union bounds are then summable. Almost surely, all sufficiently fine dyadic intervals satisfy the indicated bound. If \(2^{-(n+1)}<t-s\leq2^{-n}\), the two points lie in the same or adjacent dyadic cells. Comparing through their endpoints gives \(|B_t-B_s|\leq3K\sqrt{n2^{-n}}\). Consequently, for a deterministic constant \(C\), almost surely

\[|B_t-B_s|\leq C\sqrt{|t-s|\log(1/|t-s|)}\]

for all sufficiently small positive \(|t-s|\), where the threshold may depend on the path.

To get a lower bound with the sharp constant, fix \(0<c<\sqrt2\), put \(h_n=e^{-n}\), and consider the \(N_n=\lfloor e^n\rfloor\) disjoint increments of length \(h_n\) in \([0,1]\). Each exceeds \(c\sqrt{nh_n}\) with probability \(p_n=\mathbb P(Z>c\sqrt n)\), where \(Z\) is standard normal. Integration of the Gaussian density over \([x,x+1/x]\) gives \(p_n\geq c_1n^{-1/2}e^{-c^2n/2}\) for large \(n\). Independence within each row therefore gives

\[\mathbb P\left(\max_{0\leq k<N_n}(B_{(k+1)h_n}-B_{kh_n})\leq c\sqrt{nh_n}\right) =(1-p_n)^{N_n}\leq\exp\left[-c_2n^{-1/2}e^{(1-c^2/2)n}\right].\]

These probabilities are summable. Thus a qualifying increment exists for every sufficiently large \(n\), almost surely. This proves the required lower bound, including the simultaneous quantifier over arbitrarily small intervals.

For the sharp upper bound, a finer grid avoids losing a fixed constant when endpoints are approximated. Fix \(q>1\), \(0<\delta<1/4\), and \(c>\sqrt{1+2\delta}\). Set \(h_n=q^{-n}\), and use a grid of spacing \(\delta h_n\), including the endpoint (1). There are at most \(C_\delta/h_n\) pairs of grid points whose separation is at most \((1+2\delta)h_n\). A Gaussian tail bound and a union bound give

\[\mathbb P\left(\max_{\substack{u,v\text{ on the grid}\\|u-v|\leq(1+2\delta)h_n}} |B_u-B_v|>c\sqrt{2h_n\log(1/h_n)}\right) \leq C_\delta h_n^{c^2/(1+2\delta)-1}.\]

The exponent is positive, so this is summable in \(n\). Approximate the endpoints of any interval of length at most \(h_n\) by grid points at distance at most \(\delta h_n\). The coarse bound already proved controls the two endpoint errors. Almost surely, for all sufficiently large \(n\),

\[\sup_{|t-s|\leq h_n}|B_t-B_s| \leq c\sqrt{2h_n\log(1/h_n)}+2C\sqrt{\delta h_n\log(1/(\delta h_n))}.\]

For \(h_n/q<h\leq h_n\), divide by \(\sqrt{2h\log(1/h)}\). The resulting limiting upper bound is at most \(\sqrt q(c+\sqrt2 C\sqrt\delta)\). Take a countable sequence of choices with \(\delta\downarrow0\), \(c\downarrow1\), and \(q\downarrow1\). Together with the lower bound this proves Lévy’s continuity theorem:

\[\limsup_{h\downarrow0}\sup_{0\leq t\leq1-h}\frac{|B_{t+h}-B_t|}{\sqrt{2h\log(1/h)}}=1\quad\text{almost surely}.\]

1. Lack of Monotonicity

A Brownian motion is not monotone on any interval \([a,b]\) such that \(0 < a < b < \infty\), almost surely. To see this: by way of contradiction, suppose for a fixed interval \([a,b]\), \(B_s\) is monotone. If this is true, then for any \(n\) subintervals \(a=a_1 < a_2< \dots < a_{n+1} = b\), then the increments \(B_{a_{i+1}} - B_{a_i}\) do not change signs for \(1 \leq i \leq n\). Independence between these increments implies that the probability all increments have the same sign equals \(2 \times 2^{-n}\), Therefore, the probability of monotonicity on \([a,b]\) converges to zero as \(n \to \infty\), and this contradicts the initial statement. The conclusion holds for all intervals \([a,b]\) if we take a union over these intervals with rational endpoints. \(\blacksquare\)

2. Deviation

Brownian motion has arbitrarily large positive and negative fluctuations relative to \(\sqrt t\):

\[\limsup_{t\to\infty}\frac{B_t}{\sqrt t}=+\infty, \qquad\liminf_{t\to\infty}\frac{B_t}{\sqrt t}=-\infty\quad\text{almost surely}.\]

Both statements follow immediately from the law of the iterated logarithm below, applied to \(B\) and \(-B\). In particular, the negative fluctuation concerns a limit inferior at positive times; Brownian motion here is indexed by \(t\geq0\).

3. Law of Iterated Logarithm

The law of the iterated logarithm identifies the largest fluctuations at infinity. It requires two different geometric grids: a ratio near one for the upper bound, and a large ratio for independent increments in the lower bound. This also explains the proof given earlier. Write \(H(t)=\sqrt{2t\log\log t}\) for sufficiently large \(t\).

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 positive limsup. Apply the same proof to \(-B\) to obtain the negative liminf. The upper bound also controls the absolute value. Thus

\[\limsup_{t\to\infty}\frac{B_t}{\sqrt{2t\log\log t}}=1, \qquad\liminf_{t\to\infty}\frac{B_t}{\sqrt{2t\log\log t}}=-1\quad\text{almost surely}.\]

4. Ancillary Facts

On a filtered probability space \((\Omega, \mathcal F, \{\mathcal F_t\}_{t \geq 0} \mathop{\mathrm{\mathbf{P}\!\,}})\), a stopping time is a random variable \(T: \Omega \to [0,\infty]\) such that \(\{T \leq t\} \in \mathcal F_t\) for all \(t \geq 0\). The stopped \(\sigma\)-field is defined as

\[\begin{aligned} \mathcal F_T = \{A \in \mathcal F_\infty : A \cap \{T \leq t\} \in \mathcal F_t \,\,\,,\,\, \forall t \geq 0 \} ~. \end{aligned}\]

For a continuous function \(a:[0,T]\to\mathbb R\) with \(a(0)=0\), finite variation means that the sums \(\sum_i|a(t_i)-a(t_{i-1})|\) are bounded over all finite partitions of \([0,T]\). Equivalently, there is a finite signed Borel measure \(\mu\), with \(\mu(\{0\})=0\), such that \(a(t)=\mu([0,t])\). The associated Stieltjes integral is \(\int_0^T f(s)\,da(s)=\int_{[0,T]}f\,d\mu\) whenever \(\int|f|\,d|\mu|<\infty\). Its total variation is

\[V_T(a)=\int_0^T|da(s)|=|\mu|([0,T]) =\sup_{\pi}\sum_i|a(t_i)-a(t_{i-1})|.\]

The sums converge to this total variation along every sequence of partitions whose mesh tends to zero; nested partitions are not required:

\[\lim_{n\to\infty}\sum_i|a(t_i^n)-a(t_{i-1}^n)|=V_T(a).\tag{1}\]

Every sum is at most \(V_T(a)\). For the reverse inequality, choose a fixed partition \(0=s_0<\cdots<s_m=T\) with variation sum larger than \(V_T(a)-\varepsilon\). When the mesh of the \(n\)-th partition is sufficiently small, choose distinct ordered points \(u_j^n\) from it with \(u_j^n\to s_j\), including the endpoints. The triangle inequality, applied to each group of consecutive intervals between these points, gives

\[\sum_i|a(t_i^n)-a(t_{i-1}^n)|\geq\sum_{j=1}^m|a(u_j^n)-a(u_{j-1}^n)| \longrightarrow\sum_{j=1}^m|a(s_j)-a(s_{j-1})|>V_T(a)-\varepsilon.\]

Continuity supplies the displayed limit. Letting \(\varepsilon\downarrow0\) proves the claim and avoids treating unrelated partitions as a filtration.

For processes, finite variation means finite variation on every compact time interval. Work with a filtration satisfying the usual conditions. A continuous adapted process \(M\), with \(M_0\in L^1\), is a continuous local martingale if there are stopping times \(T_n\uparrow\infty\) almost surely such that every stopped process \(M^{T_n}\) is a uniformly integrable martingale. This convention allows a nonzero initial value. One can always include a deterministic cutoff in a localising sequence.¹

Remark 1. Every continuous martingale is a local martingale: \(M^n_t=\mathbb{E}[M_n\mid\mathcal F_{t\wedge n}]\) is uniformly integrable. Stopping a local martingale preserves the property, as does replacing a reducing sequence \(T_n\) by \(T_n\wedge S_n\), where \(S_n\uparrow\infty\) are stopping times. For two local martingales, the minimum of their reducing sequences reduces any linear combination. These facts follow from optional sampling for uniformly integrable martingales.

A nonnegative continuous local martingale is a supermartingale. Indeed, Fatou’s lemma first gives \(\mathbb{E}M_t\leq\mathbb{E}M_0<\infty\), and conditional Fatou gives, for \(s\leq t\),

\[\mathbb{E}[M_t\mid\mathcal F_s] \leq\liminf_n\mathbb{E}[M_{t\wedge T_n}\mid\mathcal F_s] =\liminf_nM_{s\wedge T_n}=M_s.\]

If instead \(|M_t|\leq Z\) for all \(t\), outside one null set, with \(Z\in L^1\), conditional dominated convergence preserves equality in this argument. The family \(\{M_t:t\geq0\}\) is dominated by \(Z\), hence uniformly integrable. Thus \(M\) is a uniformly integrable martingale.

Finally, \(\tau_n=\inf\{t\geq0:|M_t|\geq n\}\) is a stopping time by continuity and adaptation, and \(\tau_n\uparrow\infty\) since each path is bounded on compact intervals. The stopped process is a local martingale and \(|M_{t\wedge\tau_n}|\leq n+|M_0|\). The preceding domination argument shows that these hitting times reduce \(M\). This proof uses only localisation and conditional convergence, not a decomposition of an arbitrary semimartingale.

Example 2 (Supremum of “Centred” Brownian Motion). Suppose \(\{B_t\}_{t \geq 0}\) is a Brownian motion starting from zero, and \(\mu > 0\). We can derive the distribution of the supremum of a “centred” Brownian motion

\[\begin{aligned} \sup_{t \geq 0} B_t - \mu t ~. \end{aligned}\]

It is clear that

\[\begin{aligned} \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} B_t - \mu t \geq y \right) = 1 \quad \text{if } y \leq 0 ~. \end{aligned}\]

For \(y > 0\), observe that

\[\begin{gathered} \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} B_t - \mu t \geq y \right) = \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} B_{(\frac{1}{2\mu})^2 t} - \mu \left(\frac{1}{2\mu} \right)^2 t \geq y \right) \\ = \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} 2\mu B_{(\frac{1}{2\mu})^2 t} - \frac{t}{2} \geq 2 \mu y \right) = \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} B_t - \frac{t}{2} \geq 2 \mu y \right) = \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} e^{B_t - \frac{t}{2}} \geq e^{2 \mu y} \right) ~. \end{gathered}\]

This means that, setting \(M_t = e^{B_t - \frac{t}{2}}\) for all \(t \geq 0\), we have that \(\{M_t\}_{t \geq 0}\) is a nonnegative martingale with continuous sample path. Since \(\frac{B_t}{t} \xrightarrow{\rm a.s.} 0\) as \(t \to \infty\), we have

\[\begin{aligned} \lim\limits_{t \to \infty} B_t - \frac{t}{2} = \lim\limits_{t \to \infty} t \left( \frac{B_t}{t} - \frac12 \right) = -\infty \quad \rm a.s. ~. \end{aligned}\]

For \(a=e^{2\mu y}>1\), let \(\tau_a=\inf\{t:M_t=a\}\). The stopped martingale \(M^{\tau_a}\) is bounded by \(a\), so optional sampling and dominated convergence give \(1=\mathbb{E} M_{t\wedge\tau_a}\to a\mathbb P(\tau_a<\infty)\). On \(\{\tau_a=\infty\}\), the limit is zero, while on \(\{\tau_a<\infty\}\) it is \(a\). Since \(M_t\to0\), a level \(a>1\) is attained whenever the supremum is at least \(a\). Therefore

\[\begin{aligned} \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} B_t - \mu t \geq y \right) = \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} M_t \geq e^{2\mu y} \right) = e^{-2\mu y} ~. \end{aligned}\]

This means

\[\begin{aligned} \mathop{\mathrm{\mathbf{P}\!\,}}\left( \sup_{t \geq 0} B_t - \mu t \leq y \right) = \begin{cases} 1 - e^{-2\mu y} & \text{ for } y \geq 0 ~, \\ 0 & \text{ otherwise} \, . \end{cases} \end{aligned}\]

So, \(\sup_{t \geq 0} B_t - \mu t\) is distributed according to the exponential distribution with parameter \(2\mu\).

Almost sure divergence of the localising sequence means \(T_n(\omega)\to\infty\) outside a single null set.↩︎

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