Brownian Difference Quotients at a Fixed Time

Difference Quotients at a Fixed Time

For a differentiable function, increments over a short interval are approximately proportional to the interval length. Brownian increments instead have standard deviation proportional to the square root of the interval length. Dividing by the interval length therefore magnifies the fluctuations as the interval shrinks. We will make this precise for a fixed deterministic time, and distinguish what follows from a one-time distribution calculation from the stronger pathwise conclusion given by the local law of the iterated logarithm.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ carrying a standard Brownian motion $(B_t)_{t\ge0}$, with its natural filtration $\mathcal{F}_t=\sigma(B_s:0\le s\le t)$. Thus $B_0=0$, paths are continuous almost surely, and increments over disjoint time intervals are independent centered normal random variables whose variances equal the interval lengths. Fix a deterministic time $t\ge0$. For $h>0$, define the right-hand difference quotient

D_h(t)=\frac{B_{t+h}-B_t}{h}.

A finite right derivative at $t$ means that $D_h(t)$ has a finite real limit as $h\downarrow0$, for that sample path. If $t>0$, existence of a two-sided derivative would in particular imply existence of this right-hand limit. The fixed-time question is whether that limit can be finite with positive probability.

Proposition (Gaussian scaling of the quotient): For every fixed $t\ge0$ and $h>0$, $D_h(t)$ has distribution $N(0,1/h)$. In particular, for each fixed $K\ge0$, $$ \mathbb{P}(|D_h(t)|\le K)=2\Phi(K\sqrt h)-1\longrightarrow 0 \quad\text{as }h\downarrow0, $$ where $\Phi$ is the standard normal distribution function.

Proof. By the Brownian increment property, $B_{t+h}-B_t$ is normal with mean $0$ and variance $h$. Dividing a centered normal random variable by $h$ divides its variance by $h^2$, so $D_h(t)$ is centered normal with variance $h/h^2=1/h$. Equivalently, $\sqrt{h}\,D_h(t)$ has the standard normal distribution. Therefore $\mathbb{P}(|D_h(t)|\le K)=\mathbb{P}(|Z|\le K\sqrt h)=\Phi(K\sqrt h)-\Phi(-K\sqrt h)$, where $Z$ is standard normal. Symmetry gives the displayed expression. As $h\downarrow0$, $K\sqrt h\to0$; continuity of $\Phi$ and $\Phi(0)=1/2$ give the limit $0$, including when $K=0$. $\square$

This proposition says that the quotients are not tight near any bounded interval: their probability of lying in any fixed bounded interval tends to zero. It does not say, from the one-time distributions alone, that the quotient tends to $+\infty$ or $-\infty$ along every sample path. The quotients at different values of $h$ are dependent. We will use the small-ball probability to rule out a finite limit, and then use the local law of the iterated logarithm for a stronger pathwise statement.

Worked Example: A Single Interval Length

Fix $t$ and take $h=\tfrac14$. The increment $B_{t+1/4}-B_t$ has distribution $N(0,\tfrac14)$, so its standard deviation is $\sqrt{1/4}=\tfrac12$. Dividing by $1/4$ gives $$ D_{1/4}(t)=4(B_{t+1/4}-B_t)\sim N(0,4). $$ The variance check is $4^2\cdot\tfrac14=4$, as required by the formula $1/h=4$.

For example, the probability that the quotient lies between $-1$ and $1$ is $\mathbb{P}(|D_{1/4}(t)|\le1)=2\Phi(1/2)-1$: standardizing a $N(0,4)$ variable divides the endpoints by its standard deviation $2$. The same result follows directly from the general formula, since $K\sqrt h=1\cdot\sqrt{1/4}=\tfrac12$. This calculation is a fixed-$h$ probability, not a claim about a pathwise limit.

Why a Finite Derivative Is Impossible at a Fixed Time

Theorem (No finite right derivative at a fixed time): For every fixed deterministic $t\ge0$, $$ \mathbb{P}\left(\lim_{h\downarrow0}D_h(t)\text{ exists as a finite real number}\right)=0. $$ Consequently, at every fixed $t>0$, the probability that the two-sided derivative exists as a finite number is also zero.

Proof. Suppose, for contradiction, that the event of a finite right derivative has positive probability. On that event, $D_h(t)$ converges as $h\downarrow0$ to a finite random variable $D$; define $D=0$ off the event. For any sequence $h_m\downarrow0$, the quotients $D_{h_m}(t)$ then converge almost surely on the derivative event. To use a contradiction without conditioning on that event, it is enough to note that on the event, the quotients are eventually bounded by some finite, possibly path-dependent, bound. Equivalently, for a direct argument, let $A$ be the positive-probability event of finite convergence and choose an integer $K$ such that the subset $A_K\subseteq A$, on which $|D_h(t)|\le K$ for all sufficiently small $h$, has positive probability. Such a $K$ exists because $A$ is the countable union of these subsets.

For each $m$, choose a deterministic $h_m>0$ small enough that $h_m\downarrow0$ and the bound defining $A_K$ applies for all sufficiently large $m$ on $A_K$. More formally, the random threshold for “sufficiently small” can be covered by the countable events on which it is at least $1/r$; one of these intersections with $A_K$ has positive probability. Call that intersection $A_{K,r}$. For all sufficiently large $m$ with $h_m<1/r$, it is contained in $\{|D_{h_m}(t)|\le K\}$. But the proposition gives $\mathbb{P}(|D_{h_m}(t)|\le K)\to0$, contradicting $\mathbb{P}(A_{K,r})>0$. Thus the event of a finite right derivative has probability zero. A finite two-sided derivative at $t>0$ entails a finite right derivative, so its event is a subset of a probability-zero event. $\square$

The argument uses the fact that the derivative, if it exists, is finite and hence the quotients are eventually bounded on that event. The proposition rules out such boundedness with positive probability by testing one common bound $K$ as $h$ shrinks. It is important to keep this quantifier straight: for each fixed $K$, the probability of $|D_h(t)|\le K$ tends to zero. The proof does not replace this by an unjustified claim about a supremum over all $K$.

Worked Example: Shrinking Along a Geometric Sequence

Let $h_n=4^{-n}$, for integers $n\ge1$. At each $n$, the quotient has distribution $N(0,1/h_n)=N(0,4^n)$. For any fixed $K\ge0$, the proposition gives $$ \mathbb{P}(|D_{h_n}(t)|\le K)=2\Phi(K\sqrt{4^{-n}})-1 =2\Phi(K\,2^{-n})-1. $$ The equality $\sqrt{4^{-n}}=2^{-n}$ follows because $n$ is an integer.

For $K=2$, the first two probabilities are $2\Phi(1)-1$ and $2\Phi(1/2)-1$: at $n=1$, $2\cdot2^{-1}=1$, and at $n=2$, $2\cdot2^{-2}=\tfrac12$. In general, $2\,2^{-n}\to0$, so continuity of $\Phi$ shows that the probability tends to zero. As a check, the variance at $n=2$ is $4^2=16$, and the standard deviation is $4$; the interval $[-2,2]$, when standardized, becomes $[-1/2,1/2]$, giving exactly $2\Phi(1/2)-1$.

This sequence illustrates the shrinking small-ball probabilities. It does not by itself justify treating the quotients for different $n$ as independent or asserting an almost-sure limit from their marginal distributions.

Conditional Scaling and Dependence Across Scales

The scaling calculation also holds conditionally on the Brownian history up to time $t$. Brownian increments after $t$ are independent of $\mathcal{F}_t$, so for each $h>0$ the conditional law of $D_h(t)$ given $\mathcal{F}_t$ is the same centered normal law $N(0,1/h)$. This describes the distribution of a future quotient when the past is known; it does not make quotients at different future interval lengths independent.

Worked Example: A Conditional Small-Ball Probability

Take $t=2$ and $h=\tfrac19$. The increment $B_{2+1/9}-B_2$ is independent of $\mathcal{F}_2$ and has distribution $N(0,\tfrac19)$. Dividing by $1/9$ gives $D_{1/9}(2)\sim N(0,9)$, since $(1/9)/(1/9)^2=9$. Thus, almost surely, its conditional probability of lying in $[-1,1]$ is $$ \mathbb{P}(|D_{1/9}(2)|\le1\mid\mathcal{F}_2) =2\Phi(1\cdot\sqrt{1/9})-1 =2\Phi(1/3)-1. $$ To check by standardization, the quotient has standard deviation $3$, so the endpoints $-1$ and $1$ become $-1/3$ and $1/3$, producing the same probability. The conditional expression is an almost-sure equality of conditional probabilities, not a statement about conditioning on a particular past event.

For a concrete dependence check, take $0<h\le k$. The increments over $[t,t+h]$ and $[t,t+k]$ overlap, and their covariance is $h$, because the shared part has length $h$. Therefore $$ \operatorname{Cov}(D_h(t),D_k(t)) =\frac{h}{hk}=\frac1k. $$ For example, with $h=\tfrac14$ and $k=1$, the covariance is $1$, while the variances are $4$ and $1$; the correlation is $1/(2\cdot1)=\tfrac12$. Thus even these two quotients are not independent. Their nested increments are one reason that marginal Gaussian scaling should not be misread as a pathwise assertion across every scale.

Pathwise Oscillation from the Local Law of the Iterated Logarithm

The Gaussian calculation proves that a finite derivative has probability zero. The earlier material on the local law of the iterated logarithm gives a stronger description. We use that standard result here, without reproving it: for Brownian motion started at zero, $$ \limsup_{h\downarrow0}\frac{B_h}{\sqrt{2h\log\log(1/h)}}=1, \qquad \liminf_{h\downarrow0}\frac{B_h}{\sqrt{2h\log\log(1/h)}}=-1 \quad\text{almost surely}. $$ The expressions are considered for sufficiently small $h$, so that the logarithms are defined and the denominator is positive. For a fixed deterministic $t$, the process $B_{t+h}-B_t$, indexed by $h\ge0$, is itself a standard Brownian motion. Hence the same local law applies at that fixed time.

Since $$ D_h(t)= \frac{B_{t+h}-B_t}{\sqrt{2h\log\log(1/h)}} \sqrt{\frac{2\log\log(1/h)}{h}}, $$ and the second factor tends to infinity, the local law implies $\limsup_{h\downarrow0}D_h(t)=+\infty$ and $\liminf_{h\downarrow0}D_h(t)=-\infty$, almost surely. For the positive conclusion, choose scales along which the first factor tends to $1$; it is eventually positive and bounded away from zero, while the second factor diverges. The negative conclusion follows by choosing scales along which the first factor tends to $-1$. This pathwise oscillation rules out a finite right derivative even more directly.

This remains a fixed-time statement: for each deterministic $t$, the claimed event has probability one. It does not, by itself, prove a single probability-one statement simultaneously for every time $t$. A result about differentiability at all points requires additional reasoning and is treated separately. The distinction between “for each fixed time” and “for all times on one event” is essential when quantifiers involve uncountably many times.

Common pitfall. From $\operatorname{Var}(D_h(t))=1/h$, one may correctly conclude that the variance grows without bound, but it is incorrect to conclude from variance alone that $D_h(t)\to+\infty$ almost surely. The centered Gaussian quotient is symmetric: for every $h>0$, its probabilities of being positive and negative are both $1/2$. Also, quotients at nested scales are dependent, as the covariance calculation above shows. The correction is to use the small-ball formula to disprove convergence to any finite value, and to use the local law of the iterated logarithm when asserting the stronger almost-sure oscillation to both signs of infinity.

Check Your Understanding

For each question, identify whether it concerns a fixed-$h$ distribution, a limit in probability, or a pathwise statement.

For fixed $h>0$, what are the mean and variance of $D_h(t)$? Derive them from the increment distribution.
For a fixed $K>0$, calculate the limit of $\mathbb{P}(|D_h(t)|\le K)$ as $h\downarrow0$.
Why does a finite right derivative on an event of positive probability contradict the small-ball probability calculation?
For $0<h\le k$, compute $\operatorname{Cov}(D_h(t),D_k(t))$ using the overlap of the two Brownian increments.
What additional pathwise conclusion about the limsup and liminf follows from the local law of the iterated logarithm?

Brownian Difference Quotients at a Fixed Time

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