Joint Gaussian Calculations for Brownian Increments

From Brownian Values to Increment Calculations

Brownian motion is often described through its values at fixed times, but many calculations concern increments over intervals that overlap only partly. Their dependence is determined by the part of the intervals they share. Once the covariance is known, the joint Gaussian law gives both an independence test and a formula for conditioning.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. In particular, $B_0=0$, its paths are continuous, its increments are independent over disjoint time intervals, and $B_t-B_s$ is normal with mean $0$ and variance $t-s$ for $0\le s<t$. For $0\le a\le b$, write $\Delta B_{[a,b]}=B_b-B_a$; its variance is $b-a$. We will use that any finite vector of Brownian values is jointly Gaussian, as established in the finite-dimensional laws of Brownian motion. Linear combinations of such values, including increments, are therefore jointly Gaussian as well.

Definition: The overlap length of $[a,b]$ and $[c,d]$, where $a\le b$ and $c\le d$, is $\ell([a,b],[c,d])=\max\{0,\min(b,d)-\max(a,c)\}$. It is zero for disjoint interiors; sharing only an endpoint also gives overlap length zero.

Covariance Is the Length of the Overlap

Theorem: For any $0\le a\le b$ and $0\le c\le d$, $$ \operatorname{Cov}(\Delta B_{[a,b]},\Delta B_{[c,d]}) =\ell([a,b],[c,d]). $$ The two increments are jointly Gaussian. Consequently, they are independent if and only if their overlap length is zero.

Proof. From the Brownian covariance kernel, $\operatorname{Cov}(B_s,B_t)=\min(s,t)$. Bilinearity of covariance gives $\operatorname{Cov}(B_b-B_a,B_d-B_c)=\min(b,d)-\min(b,c)-\min(a,d)+\min(a,c)$. For nonnegative $u,v$, $\min(u,v)$ is the length of the intersection of $[0,u]$ and $[0,v]$. Applying this identity to the four terms shows that the displayed difference is the length of the intersection of $(a,b]$ and $(c,d]$, which is exactly $\ell([a,b],[c,d])$. The increments are jointly Gaussian because they are linear combinations of the jointly Gaussian Brownian values at the four endpoints. For jointly Gaussian random variables $X,Y$, zero covariance implies independence: their joint characteristic function is $\exp(-\tfrac12(s^2\operatorname{Var}(X)+2st\operatorname{Cov}(X,Y)+t^2\operatorname{Var}(Y)))$; when the covariance is zero, it factors into the marginal characteristic functions. Conversely, independent variables with finite second moments have zero covariance. Thus independence holds exactly when the overlap length is zero, including intervals of length zero, whose increments are constant zero. $\square$

The covariance formula also covers nested intervals. For example, an increment on a short interval contained in a longer interval has covariance equal to the variance of the shorter increment. Intervals that merely touch at an endpoint share no positive length and give independent increments. These statements concern fixed deterministic intervals.

Worked Example: Partly Overlapping Intervals

Set $X=B_3-B_0$ and $Y=B_4-B_1$. Find their covariance matrix and correlation.

The interval lengths are $3-0=3$ and $4-1=3$, so $\operatorname{Var}(X)=3$ and $\operatorname{Var}(Y)=3$. The intervals $[0,3]$ and $[1,4]$ overlap on $[1,3]$, of length $3-1=2$. The theorem gives $\operatorname{Cov}(X,Y)=2$, hence

\operatorname{Cov}\begin{pmatrix}X\\Y\end{pmatrix} = \begin{pmatrix}3&2\\2&3\end{pmatrix}, \qquad \operatorname{Corr}(X,Y)=\frac{2}{\sqrt{3\cdot3}}=\frac23.

Check: Write $X=B_1+(B_3-B_1)$ and $Y=(B_3-B_1)+(B_4-B_3)$. The three increments on $[0,1]$, $[1,3]$, and $[3,4]$ are independent, with variances $1,2,1$. Thus each of $X,Y$ has variance $1+2=3$, and their only shared random term has variance $2$, agreeing with the covariance matrix. Since the covariance is not zero, these jointly Gaussian increments are not independent.

Worked Example: Disjoint Increments and Conditioning

Let $U=B_2-B_1$ and $V=B_5-B_3$. Find their joint law and the conditional law of $U$ given $V$.

The interval lengths are $2-1=1$ and $5-3=2$. The intervals have no overlap, so the covariance is zero. The joint Gaussian covariance matrix is therefore $\begin{pmatrix}1&0\\0&2\end{pmatrix}$, and $U,V$ are independent. In particular, for almost every conditioning value $v$, the conditional law of $U$ given $V=v$ is $N(0,1)$.

Check: The joint characteristic function is $\exp(-\tfrac12(s^2+2t^2))$, which factors as $\exp(-s^2/2)\exp(-2t^2/2)$, the two marginal characteristic functions. This confirms independence and the stated conditional law. A conditional law describes conditioning through a regular conditional distribution; it does not assert that the event $V=v$ has positive probability.

Conditioning a Gaussian Vector on Another Gaussian Vector

The scalar calculation extends to several observed increments. If $X$ is a column vector of increments and $Y$ is another column vector of increments, write their covariance matrices in blocks. The conditioning formula requires that the covariance matrix of $Y$ be invertible. The conditional covariance of $X$ may be singular: in that case the conditional Gaussian law is concentrated on a lower-dimensional set, or even at a single point.

Theorem: Let $X\in\mathbb{R}^m$ and $Y\in\mathbb{R}^n$ be jointly Gaussian, centered random vectors. Suppose $\Sigma_{YY}=\operatorname{Cov}(Y,Y)$ is positive definite, and set $\Sigma_{XY}=\operatorname{Cov}(X,Y)$, $\Sigma_{XX}=\operatorname{Cov}(X,X)$. Then, almost surely as a conditional distribution, $$ X\mid Y=y\ \sim\ N\!\left(\Sigma_{XY}\Sigma_{YY}^{-1}y,\, \Sigma_{XX}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}\right). $$ A singular covariance in the displayed conditional law is permitted and means a degenerate Gaussian distribution.

Proof. Define $A=\Sigma_{XY}\Sigma_{YY}^{-1}$ and $R=X-AY$. Direct covariance calculation gives $\operatorname{Cov}(R,Y)=\Sigma_{XY}-A\Sigma_{YY}=0$. The pair $(R,Y)$ is jointly Gaussian, since it is a linear transformation of the jointly Gaussian vector $(X,Y)$. The characteristic-function factorization used above, applied to the two vector blocks, shows that zero cross-covariance makes $R$ independent of $Y$. Also, $\operatorname{Cov}(R,R)=\Sigma_{XX}-A\Sigma_{YX}-\Sigma_{XY}A^{\mathsf T}+A\Sigma_{YY}A^{\mathsf T}$. Using $A=\Sigma_{XY}\Sigma_{YY}^{-1}$, symmetry of $\Sigma_{YY}$, and $\Sigma_{YX}=\Sigma_{XY}^{\mathsf T}$, the last three terms combine to $-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}$. Thus $R$ is centered Gaussian with the covariance in the theorem and is independent of $Y$. Since $X=AY+R$, the conditional distribution given $Y=y$ is the distribution of $Ay+R$, which has the stated mean and covariance. This reasoning remains valid if the covariance of $R$ is singular, because a linear transformation of a Gaussian vector is still Gaussian, possibly degenerate. $\square$

Worked Example: One Overlapping Increment Given Another

Let $X=B_5-B_1$ and $Y=B_4-B_0$. Find the conditional distribution of $X$ given $Y=y$.

Both increments have variance $5-1=4$ and $4-0=4$, respectively. Their intervals overlap on $[1,4]$, whose length is $4-1=3$; hence $\operatorname{Cov}(X,Y)=3$. The scalar conditioning formula gives $\mathbb{E}[X\mid Y=y]=(3/4)y$ and $\operatorname{Var}(X\mid Y)=4-(3\cdot3)/4=4-9/4=7/4$. Therefore $X\mid Y=y\sim N(3y/4,7/4)$.

For instance, at the conditioning value $y=2$, the conditional mean is $(3/4)\cdot2=3/2$, while the conditional variance remains $7/4$. This is a value of the conditional-distribution formula, not conditioning on an event of positive probability.

Check: The residual $R=X-(3/4)Y$ has covariance with $Y$ equal to $3-(3/4)\cdot4=0$, so it is independent of $Y$. Its variance is $4-2(3/4)(3)+(3/4)^2(4)=4-9/2+9/4=7/4$. Thus $X=(3/4)Y+R$ gives precisely the conditional mean and variance above.

Worked Example: Conditioning on Two Adjacent Increments

Set $X=B_4-B_0$, $Y_1=B_2-B_0$, and $Y_2=B_4-B_2$. Find the conditional law of $X$ given $(Y_1,Y_2)$.

The two observed intervals have lengths $2$ and $2$, and they have zero overlap length. Thus $\Sigma_{YY}=\begin{pmatrix}2&0\\0&2\end{pmatrix}$, which is positive definite, and its inverse is $\begin{pmatrix}1/2&0\\0&1/2\end{pmatrix}$. The variance of $X$ is $4$. Its covariance with $Y_1$ is the overlap length $2$, and its covariance with $Y_2$ is also $2$. Consequently $\Sigma_{XY}=(2,2)$, and the conditional mean is $(2,2)\begin{pmatrix}1/2&0\\0&1/2\end{pmatrix}\begin{pmatrix}y_1\\y_2\end{pmatrix}=y_1+y_2$. The conditional variance is $4-(2,2)\begin{pmatrix}1/2&0\\0&1/2\end{pmatrix}\begin{pmatrix}2\\2\end{pmatrix}=4-4=0$. Thus $X\mid(Y_1,Y_2)=(y_1,y_2)$ is the point mass at $y_1+y_2$.

Check: The identity $B_4-B_0=(B_2-B_0)+(B_4-B_2)$ gives $X=Y_1+Y_2$ pathwise. For example, if the observed values are $y_1=1$ and $y_2=-1$, the conditional value is $1+(-1)=0$, exactly as the identity requires. A zero conditional variance here is not an error; the two observations determine $X$ completely.

A Common Covariance Error

A frequent mistake is to take the covariance of increments to be the covariance of their right-endpoint Brownian values, without subtracting the contributions from the left endpoints. For example, for $X=B_3-B_1$ and $Y=B_4-B_2$, using only $\operatorname{Cov}(B_3,B_4)=3$ would give the incorrect answer $3$. The correct calculation is

\operatorname{Cov}(X,Y) =\min(3,4)-\min(3,2)-\min(1,4)+\min(1,2) =3-2-1+1=1.

This also agrees with the overlap $[2,3]$, whose length is $3-2=1$. The variances of both increments are $2$, so their covariance matrix is $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and their correlation is $1/\sqrt{2\cdot2}=1/2$. The positive covariance reflects the shared interval, but it is smaller than either variance because only one unit of time is shared.

For calculations with many increments, an efficient procedure is to list each interval length on the diagonal and each pairwise overlap length off the diagonal. The resulting covariance matrix, together with joint Gaussianity, contains the information needed for linear conditional distributions and independence tests. When conditioning on a vector, check that its covariance matrix is invertible before using the matrix formula; when its conditional covariance is zero or singular, interpret the answer as a degenerate Gaussian law rather than forcing a positive density.

Check Your Understanding

Use interval lengths and overlaps first, then apply the Gaussian formulas.

For $B_4-B_1$ and $B_6-B_3$, calculate both variances, their covariance, and their correlation. Are they independent?
Let $U=B_3-B_2$ and $V=B_6-B_4$. Find the conditional distribution of $U$ given $V$, and state why independence follows.
For $X=B_5-B_0$ and $Y=B_4-B_1$, calculate the mean and variance of the conditional law of $X$ given $Y=y$.
Explain why $B_3-B_1$ and $B_4-B_2$ are not independent, even though neither interval contains the other.
In the vector conditioning formula, what does a zero conditional variance mean, and why does it not invalidate the formula?