The local scale of Brownian increments
Brownian paths are continuous, but continuity alone does not describe how quickly their increments shrink near a given time. The local law of the iterated logarithm gives a sharper answer: for a fixed time, the increment is almost surely of order \(\sqrt{2h\log\log(1/h)}\) along its largest positive and negative fluctuations as the time increment \(h\) decreases to zero. The extra iterated-logarithm factor is essential; it distinguishes the almost-sure extremes from the usual square-root scale.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying standard Brownian motion \((B_t)_{t\geq 0}\). Thus \(B_0=0\), the paths are continuous almost surely, and increments over disjoint time intervals are independent centered normal random variables whose variances equal the lengths of their intervals. Fix a deterministic time \(t\geq 0\). For \(0<h<e^{-1}\), define the normalized right increment \(R_t(h)=(B_{t+h}-B_t)/\sqrt{2h\log\log(1/h)}\). The restriction on \(h\) ensures that \(\log\log(1/h)>0\), so the denominator is real and positive. The time \(t\) is fixed, not selected from the observed path.
The two limits describe distinct kinds of oscillation. The limsup says that positive increments come arbitrarily close to the upper scale, while the liminf says that negative increments come arbitrarily close to its negative. Neither assertion says that every sufficiently small increment is positive or negative. Instead, positive and negative fluctuations both continue to occur at arbitrarily small time scales.
What the limits imply
Here are useful precise translations of the theorem. If a real-valued function \(r(h)\) has limsup \(1\) as \(h\downarrow0\), then for every \(\varepsilon>0\), (i) \(r(h)<1+\varepsilon\) for all sufficiently small \(h\), and (ii) \(r(h)>1-\varepsilon\) for arbitrarily small \(h\). The first assertion follows from the definition of limsup as the limiting supremum over shrinking neighborhoods; the second follows because otherwise the limsup would be at most \(1-\varepsilon\). Applying these facts to \(R_t\), and applying the corresponding facts for liminf, gives the following bounds.
For completeness, the eventual lower bound comes from the liminf: because \(\liminf_{h\downarrow0}R_t(h)=-1\), values below \(-1-\varepsilon\) cannot occur arbitrarily close to zero. The infinitely-often lower statement follows because values all at least \(-1+\varepsilon\) sufficiently near zero would force the liminf to be at least \(-1+\varepsilon\). The upper statements follow in the same way from the limsup. These arguments concern strict inequalities; replacing the threshold by a non-strict one also gives valid, weaker bounds.
The absolute oscillation has the same sharp constant:
Proof. At every admissible \(h\), the ratio inside the limsup is \(|R_t(h)|\), so it is nonnegative. Since \(\limsup R_t(h)=1\), for every \(\varepsilon>0\) there are arbitrarily small \(h\) with \(R_t(h)>1-\varepsilon\); for these \(h\), \(|R_t(h)|>1-\varepsilon\). Thus the limsup of \(|R_t(h)|\) is at least \(1\). Also, the eventual bounds \(-1-\varepsilon<R_t(h)<1+\varepsilon\) imply \(|R_t(h)|<1+\varepsilon\) for all sufficiently small \(h\). The limsup is therefore at most \(1+\varepsilon\) for every \(\varepsilon>0\), hence at most \(1\). Combining both inequalities proves the result. \(\square\)
For a fixed \(t>0\), the left-increment statement gives an analogous absolute-oscillation limit on the left. The two one-sided statements hold simultaneously with probability one: each holds on an event of probability one, and the intersection of two probability-one events still has probability one. These are statements at a fixed deterministic time. They do not, by themselves, assert that one probability-one event works for every time \(t\).
Worked applications of the normalization
Worked Example: Thresholds Along an Exponential Sequence
Fix \(t\geq0\) and take \(h_n=e^{-n}\), where \(n\) is an integer at least \(2\). Compute the normalization and state what the LIL guarantees along this sequence and across all small \(h\).
First, \(1/h_n=e^n\), so \(\log(1/h_n)=n\) and \(\log\log(1/h_n)=\log n\). Also \(2h_n=2e^{-n}\). Therefore the normalizing factor at \(h_n\) is \(\sqrt{2e^{-n}\log n}\), and the corresponding normalized increment is \((B_{t+e^{-n}}-B_t)/\sqrt{2e^{-n}\log n}\).
For \(\varepsilon=\tfrac14\), the LIL says that for all sufficiently small \(h\), the normalized increment is less than \(1+\tfrac14=\tfrac54\) and greater than \(-1-\tfrac14=-\tfrac54\). It also says that there are arbitrarily small \(h\) at which the normalized increment exceeds \(1-\tfrac14=\tfrac34\), and arbitrarily small \(h\) at which it is below \(-1+\tfrac14=-\tfrac34\).
Check: The sequence \(e^{-n}\) decreases to zero. The all-small-\(h\) bounds therefore hold at \(h_n\) for every sufficiently large \(n\). But the LIL as stated does not guarantee that the infinitely-often threshold crossings occur specifically at \(h_n\); it guarantees them at arbitrarily small time increments, not at every chosen sequence.
Worked Example: Bounding the Absolute Increment
Again fix \(t\geq0\). Use the absolute-oscillation result to give an eventual bound with tolerance \(\tfrac12\), and find the corresponding scale explicitly.
Substituting \(\varepsilon=\tfrac12\) into the absolute consequence gives, almost surely, for all sufficiently small \(h>0\), $$ |B_{t+h}-B_t|< \left(1+\tfrac12\right)\sqrt{2h\log\log(1/h)} =\tfrac32\sqrt{2h\log\log(1/h)}. $$ The coefficient is \(1+\varepsilon=\tfrac32\). For the matching lower-exceedance statement, the absolute ratio is greater than \(1-\tfrac12=\tfrac12\) for arbitrarily small \(h\). Thus there are arbitrarily small \(h\) such that \(|B_{t+h}-B_t|>\tfrac12\sqrt{2h\log\log(1/h)}\).
Check: The absolute ratio is nonnegative, and its limsup is \(1\). The upper coefficient \(\tfrac32\) is greater than \(1\), while the infinitely-often coefficient \(\tfrac12\) is less than \(1\), as required by the limsup statement.
Worked Example: Oscillations on Both Sides of a Fixed Time
Let \(t=2\). State the local conclusions for increments approaching \(2\) from the right and from the left, and use a common tolerance of \(\tfrac12\).
On the right, the normalized increment is \((B_{2+h}-B_2)/\sqrt{2h\log\log(1/h)}\). On the left, for \(0<h<2\), it is \((B_2-B_{2-h})/\sqrt{2h\log\log(1/h)}\). Both denominators are positive when \(h<e^{-1}\), and \(h<2\) ensures the left time \(2-h\) is nonnegative. The theorem gives limsup \(1\) and liminf \(-1\) for each normalized increment, almost surely. Since there are only two probability-one statements to intersect, all four limits hold simultaneously with probability one.
With \(\varepsilon=\tfrac12\), each normalized increment is eventually between \(-\tfrac32\) and \(\tfrac32\). For each of the two sides, normalized values exceed \(\tfrac12\) at arbitrarily small \(h\), and fall below \(-\tfrac12\) at arbitrarily small \(h\). The bounds concern increments from the fixed value \(B_2\); the left increment is written \(B_2-B_{2-h}\), not \(B_{2-h}-B_2\), so its sign convention agrees with the right increment.
Check: The left and right limits use the same denominator because both intervals have length \(h\). Reversing the order in the left numerator would multiply that normalized increment by \(-1\), swapping its limsup and liminf.
Why the iterated logarithm matters
The usual Brownian increment scale \(\sqrt{h}\) describes the size of a single increment in distribution, but it does not capture the almost-sure largest fluctuations over arbitrarily small time scales around a fixed time. The LIL says the sharp local scale includes the factor \(\sqrt{2\log\log(1/h)}\). The factor diverges as \(h\downarrow0\), although very slowly. In particular, the absolute increment divided by \(\sqrt h\) has infinite limsup almost surely: writing it as the product of the LIL-normalized absolute increment and \(\sqrt{2\log\log(1/h)}\), the first factor exceeds \(\tfrac12\) at arbitrarily small \(h\), while the second tends to infinity. This is an oscillation statement, not a claim about a derivative.
Common mistake: omitting the iterated logarithm. A wrong normalization would be \((B_{t+h}-B_t)/\sqrt{h}\), followed by the claim that its limsup is a finite constant. The LIL gives instead $$ \frac{|B_{t+h}-B_t|}{\sqrt h} = \frac{|B_{t+h}-B_t|}{\sqrt{2h\log\log(1/h)}}\, \sqrt{2\log\log(1/h)}. $$ The first factor has limsup \(1\), and is therefore greater than \(\tfrac12\) at arbitrarily small \(h\); the second factor tends to infinity. Consequently the limsup of the left side is infinite. The correction is to include \(\sqrt{2\log\log(1/h)}\) when expressing the sharp almost-sure local oscillation scale.
A second caution is the quantifier in the theorem. A limsup of \(1\) gives an eventual upper bound with any constant strictly greater than \(1\), and infinitely-often values above any constant strictly less than \(1\). It does not say that every sufficiently small normalized increment is close to \(1\), or that a prescribed sequence of increments attains the limsup. Keeping these distinctions explicit is essential when turning the theorem into a pathwise bound.
Check Your Understanding
Use the theorem’s limits and their definitions to answer the following questions.
- For a fixed \(t\geq0\), write the two limits for the normalized right increment and explain what each says about the signs of small increments.
- For \(\varepsilon=\tfrac13\), give the eventual upper and lower bounds for the normalized right increment, and state the two infinitely-often threshold conclusions.
- Why does the absolute-oscillation limsup equal \(1\), rather than merely being at most \(1\)?
- For a fixed \(t>0\), what numerator gives the left-increment convention used here, and what changes if its order is reversed?
- Explain why a limsup equal to \(1\) does not guarantee threshold crossings along every preselected sequence \(h_n\downarrow0\).