Brownian Motion Axioms and Standard Examples

The Defining Conditions

Brownian motion is a continuous-time random process whose evolution has two complementary features: its paths are continuous, while its increments over disjoint time intervals are independent Gaussian random variables. The definition is deliberately expressed through increments. It specifies how the process changes over an interval without requiring that its value at a later time be independent of its earlier values.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. A real-valued process $(B_t)_{t\ge0}$, with state space $S=\mathbb{R}$ and continuous time index $t\ge0$, is a standard Brownian motion if it satisfies the following conditions.

Definition: (1) $B_0=0$ almost surely. (2) The sample paths are continuous almost surely: $\mathbb{P}(\{\omega:t\mapsto B_t(\omega)\text{ is continuous on }[0,\infty)\})=1$. (3) The process has independent increments: for any finite collection of pairwise disjoint time intervals, the increments over those intervals are mutually independent. (4) For every $0\le s<t$, $B_t-B_s$ has the Gaussian distribution $N(0,t-s)$.

The variance in condition (4) is the length of the time interval. In particular, an increment over an interval of length $h>0$ has mean $0$ and variance $h$. When $s=t$, the increment is identically zero; it can be regarded as the degenerate Gaussian $N(0,0)$. The independence condition concerns increments, not the values $B_s$ and $B_t$, which generally are not independent.

“Almost surely continuous” allows a null set of outcomes on which a chosen process version may fail to have continuous paths. It does not mean merely that $t\mapsto B_t$ is continuous in probability, or that each $B_t$ has a smoothly changing distribution. Those are different statements. The four conditions together are the definition used here; in particular, a process with Gaussian increments but dependent increments is not standard Brownian motion.

Two Consequences of the Axioms

Proposition: If $(B_t)_{t\ge0}$ is standard Brownian motion, then $B_t\sim N(0,t)$ for every $t>0$, and $B_0=0$ almost surely.

Proof. The initial condition gives $B_0=0$ almost surely. For $t>0$, write $B_t=B_0+(B_t-B_0)$. Since $B_0=0$ almost surely, $B_t=B_t-B_0$ almost surely. The Gaussian-increment condition applied to $s=0$ gives $B_t-B_0\sim N(0,t)$. Therefore $B_t\sim N(0,t)$. $\square$

Proposition: For any $0=t_0<t_1<\cdots<t_n$, the increments $B_{t_k}-B_{t_{k-1}}$, $1\le k\le n$, are mutually independent, and the $k$th increment has distribution $N(0,t_k-t_{k-1})$. Their joint density at $(x_1,\ldots,x_n)\in\mathbb{R}^n$ is $$ \prod_{k=1}^n \frac{1}{\sqrt{2\pi(t_k-t_{k-1})}}\exp\!\left(-\frac{x_k^2}{2(t_k-t_{k-1})}\right). $$

Proof. The intervals $[t_{k-1},t_k]$ have disjoint interiors, and Brownian independent increments assert mutual independence of the corresponding increments. The Gaussian-increment condition gives the $k$th increment the distribution $N(0,t_k-t_{k-1})$; its interval length is positive by the strict ordering of the times. Each such distribution has the density displayed for its factor. Independence means that the joint density is the product of the individual densities, which gives the formula. $\square$

The second proposition is a direct way to check a finite list of increments. It does not say that the values $B_{t_1},\ldots,B_{t_n}$ are independent: each value is a sum of increments from time $0$. The axioms describe those increments, while path continuity supplies a separate sample-path requirement.

Worked Checks of the Definition

Worked Example: Negating a Brownian Motion

Suppose $(B_t)_{t\ge0}$ is standard Brownian motion on $(\Omega,\mathcal{F},\mathbb{P})$, and define $X_t=-B_t$. Check that $(X_t)_{t\ge0}$, with state space $\mathbb{R}$ and time index $t\ge0$, is standard Brownian motion.

First, $X_0=-B_0=0$ almost surely. If a path $t\mapsto B_t(\omega)$ is continuous, then $t\mapsto -B_t(\omega)$ is continuous, so the paths of $X$ are continuous almost surely.

For any $0\le s<t$, $X_t-X_s=-(B_t-B_s)$. If $Z\sim N(0,t-s)$, then $-Z\sim N(0,t-s)$, since the centered Gaussian density is unchanged by replacing its argument $x$ with $-x$. Thus $X_t-X_s\sim N(0,t-s)$.

Finally, negating each of a collection of independent random variables preserves their independence. Indeed, for Borel sets $A_1,\ldots,A_m$, the event $\{-Y_k\in A_k\}$ equals $\{Y_k\in -A_k\}$, where $-A_k=\{-x:x\in A_k\}$. If the $Y_k$ are independent, the probability of the intersection of these events is the product of their probabilities. Applying this to Brownian increments verifies independent increments for $X$. All four conditions hold.

Worked Example: Multiplying by a Constant

For a fixed real constant $c$, define $X_t=cB_t$ on the same probability space, with state space $\mathbb{R}$ and time index $t\ge0$. Determine exactly when $X$ is standard Brownian motion.

The paths are continuous almost surely and $X_0=0$ almost surely for every $c$. For $s<t$, $$ X_t-X_s=c(B_t-B_s)\sim N\!\left(0,c^2(t-s)\right). $$ Multiplication by the fixed constant acts separately on each increment, so independent Brownian increments remain independent; this includes $c=0$, when every resulting increment is constant. The required Gaussian increment law is $N(0,t-s)$. Since $t-s>0$, this law agrees with $N(0,c^2(t-s))$ exactly when $c^2=1$. Therefore $X$ is standard Brownian motion if and only if $c=1$ or $c=-1$.

Check with an interval. Take $s=1$ and $t=3$. The increment of $X$ has variance $c^2(3-1)=2c^2$, while the required variance is $3-1=2$. Equality requires $2c^2=2$, hence $c^2=1$, in agreement with the general calculation.

Worked Example: Adding a Linear Drift

Let $c\in\mathbb{R}$ and define $X_t=B_t+ct$, with state space $\mathbb{R}$ and time index $t\ge0$, on the probability space supporting $B$. Check the axioms and decide whether $X$ is standard Brownian motion.

The starting value is $X_0=B_0+c\cdot0=0$ almost surely. Adding the continuous function $t\mapsto ct$ to a continuous path preserves continuity. For $0\le s<t$, however, $$ X_t-X_s=(B_t-B_s)+c(t-s)\sim N\!\left(c(t-s),t-s\right). $$ Adding a deterministic constant to a Gaussian variable shifts its mean by that constant and leaves its variance unchanged. Independent increments are also preserved: each $X$-increment is the corresponding $B$-increment plus a deterministic number, and translating independent random variables preserves their independence. The increment has the required centered law for every $s<t$ exactly when $c(t-s)=0$ for every such pair. This is equivalent to $c=0$. Thus a nonzero linear drift preserves the initial value, path continuity, and independent increments, but violates the required Gaussian increment mean.

Numerical check. For $c=\tfrac12$, $s=1$, and $t=3$, the interval length is $3-1=2$. The increment has mean $\tfrac12\cdot2=1$ and variance $2$, so its distribution is $N(1,2)$, not $N(0,2)$. When $c=0$, the same interval has increment law $N(0,2)$, as required.

Worked Example: Squaring the Brownian Path

Define $Y_t=B_t^2$, with state space $\mathbb{R}$ and time index $t\ge0$. Check whether $Y$ is standard Brownian motion.

Since $Y_0=B_0^2=0$ almost surely and squaring is a continuous function, the paths of $Y$ are continuous almost surely. But for the interval from $0$ to $1$, the increment is $Y_1-Y_0=B_1^2$. The one-time law derived from the axioms gives $B_1\sim N(0,1)$, so $B_1^2\ge0$ almost surely. Also, a standard normal variable has probability zero of being exactly $0$, so $\mathbb{P}(B_1^2=0)=0$.

A variable with distribution $N(0,1)$ has probability $\tfrac12$ of being negative, by symmetry of its density about zero. Therefore $B_1^2$ cannot have distribution $N(0,1)$. The Gaussian-increment condition fails, so $Y$ is not standard Brownian motion. This one failed condition is enough; there is no need to test every other axiom once a counterexample to one condition has been found.

Why Each Condition Matters

The examples separate the roles of the four conditions. Negation preserves the centered Gaussian increment laws and independence, so it gives another standard Brownian motion. Multiplication by a constant preserves the other structural properties but changes the increment variance unless the constant is $1$ or $-1$. Linear drift shifts the increment mean. Squaring preserves the initial value and path continuity but destroys the required Gaussian increment law.

A common mistake is to check only that increments are Gaussian and independent, while overlooking their exact mean and variance. For the drifted process in the preceding example, a mistaken calculation might write $$ (B_t+\tfrac12t)-(B_s+\tfrac12s)\ \overset{\text{wrong}}{\sim}\ N(0,t-s). $$ That drops the deterministic term. Subtracting the two process values gives $$ (B_t-B_s)+\tfrac12(t-s)\sim N\!\left(\tfrac12(t-s),t-s\right). $$ At $s=1,t=3$, the mean is $\tfrac12(2)=1$, so the increment is $N(1,2)$, not $N(0,2)$. Always subtract the process values first, then identify both the mean and the variance of the resulting increment.

These axioms are useful because they provide a precise checklist for proposed processes. Path continuity alone is not enough, and Gaussian one-time marginals alone are not enough. For a standard Brownian motion, the initial value, path regularity, independence across disjoint intervals, and the full increment law must all be checked.

Check Your Understanding

For each question, identify which defining condition is being used and show the relevant increment calculation.

If $X_t=3B_t$, what is the distribution of $X_t-X_s$, and which Brownian motion condition fails?
For $X_t=B_t-2t$, compute the mean and variance of $X_4-X_1$. Is this increment law the one required for standard Brownian motion?
Why does $B_t^2$ fail the Gaussian-increment condition on the interval from $0$ to $1$?
Explain why negating each increment preserves independence, and state why it does not change the centered Gaussian increment law.
What is the distribution of $B_5$, and which axioms give that conclusion?