Proving Brownian Increments Are Independent From the Gaussian Covariance

If your course defines Brownian motion using independent increments, the property is already an assumption. If it starts with a centred Gaussian process with covariance

\min(s,t)

, prove independence by computing zero covariance between disjoint increments and using their joint Gaussian distribution.

Prove joint Gaussianity, then calculate the covariance

Assume $B$ is centred Gaussian with $E[B_sB_t]=\min(s,t)$ and continuous paths. Take two increments on disjoint intervals, with $a<b\le c<d$ . By bilinearity,

\begin{aligned} \operatorname{Cov}(B_b-B_a,B_d-B_c) &=\min(b,d)-\min(b,c)-\min(a,d)+\min(a,c)\\ &=b-b-a+a=0. \end{aligned}

For arbitrary

0=t_0<t_1<\cdots<t_n

, the vector of increments is a linear transformation of the jointly Gaussian vector of Brownian values. Its covariance matrix is diagonal by the calculation above. Its diagonal entry for interval

i

\operatorname{Var}(B_{t_i}-B_{t_{i-1}}) =t_i+t_{i-1}-2t_{i-1}=t_i-t_{i-1}.

The joint characteristic function therefore factors into the product of the one-dimensional characteristic functions. This proves the increments are mutually independent and each has a normal distribution with the stated variance. Also

B_0=0

almost surely because its variance is zero. Together with the assumed continuity, these establish the standard Brownian properties from the Gaussian-process definition.

For a list of increments, collect all their endpoints into one vector of Brownian values. That vector is jointly Gaussian by the process definition. The increment vector is a linear transformation of it, with one

+1

and one

-1

per row. So zero covariance between each pair gives a diagonal covariance matrix for a jointly Gaussian vector and hence mutual independence. Pairwise independence alone would not prove the full assertion for a general process. This is where the Gaussian assumption does essential work.

Touching intervals versus overlapping intervals

Consider $D_1=B_2-B_1$ and $D_2=B_4-B_2$ . Their covariance is $2-2-1+1=0$ . They are jointly Gaussian, so $D_1\sim N(0,1)$ and $D_2\sim N(0,2)$ are independent. In contrast, $B_2-B_1$ and $B_3-B_{1.5}$ overlap for half a time unit. Their covariance is $2-1.5-1+1=0.5$ , and they are dependent.

To connect this to filtration questions, take

u\le s<t

. The covariance of

B_t-B_s

with

B_u

u-u=0

. The increment is therefore independent of every finite vector of Brownian values observed by

s

. The events determined by such finite vectors generate the natural filtration. A standard extension argument, using the pi–lambda theorem, gives independence from that entire filtration.

The argument also works for any ordered collection of disjoint intervals with gaps. Each off-diagonal covariance cancels by the same four-term calculation. Joint Gaussianity is essential here. For a general process, two increments can be uncorrelated and still share dependence through their magnitudes.

Proving Brownian Increments Are Independent From the Gaussian Covariance

Prove joint Gaussianity, then calculate the covariance

Touching intervals versus overlapping intervals

Explore a Brownian path at different resolutions

References and next steps