Joint Gaussian Laws of Wiener Integrals

From Individual Gaussianity to a Joint Law

A deterministic Wiener integral is Gaussian, but knowing the distributions of several integrals one at a time does not, by itself, determine their dependence. The joint law also depends on how the integrands overlap in $L^2$. The key fact is that every linear combination of deterministic Wiener integrals is itself a deterministic Wiener integral. Its variance is the squared $L^2$ norm of the corresponding linear combination of integrands.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to $(\mathcal{F}_t)$. Fix $T<\infty$. An admissible Ito integrand on $[0,T]$ is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Each deterministic measurable function $f$ with $\int_0^T f(s)^2\,ds<\infty$ is such an integrand. Write $I(f)=\int_0^T f(s)\,dB_s$, where this is the Ito integral.

We use the earlier result on Gaussianity of deterministic Wiener integrals: for every deterministic $f\in L^2([0,T])$, $I(f)$ has the centered normal distribution with variance $\int_0^T f(s)^2\,ds$. When this variance is zero, the statement means $I(f)=0$ almost surely. The Ito isometry gives $\mathbb{E}[I(f)^2]=\int_0^T f(s)^2\,ds$. Applying it to $f+g$ and $f-g$, and subtracting, gives the covariance identity $\mathbb{E}[I(f)I(g)]=\int_0^T f(s)g(s)\,ds$.

Definition: A random vector $X=(X_1,\ldots,X_n)$ is jointly Gaussian if every real linear combination $\sum_{j=1}^n a_jX_j$, for $a\in\mathbb{R}^n$, has a (possibly degenerate) normal distribution. Its covariance matrix has entries $\operatorname{Cov}(X_i,X_j)$. A degenerate normal distribution is allowed; its covariance matrix need not be invertible.

The Joint Gaussian Theorem

Theorem: Let $f_1,\ldots,f_n\in L^2([0,T])$ be deterministic, and set $X_j=I(f_j)$. Then $X=(X_1,\ldots,X_n)$ is a centered jointly Gaussian vector with covariance matrix $C$ given by $$ C_{ij}=\mathbb{E}[X_iX_j]=\int_0^T f_i(s)f_j(s)\,ds,\qquad 1\le i,j\le n. $$ Its characteristic function is $\mathbb{E}[\exp(iu^\mathsf{T}X)]=\exp(-\tfrac12u^\mathsf{T}Cu)$ for $u\in\mathbb{R}^n$. The matrix $C$ is positive semidefinite and may be singular.

Proof. Each $f_j$ is an admissible integrand because it is deterministic, hence progressively measurable, and has finite squared integral. The covariance identity above gives $C_{ij}=\int_0^T f_i(s)f_j(s)\,ds$. For any $a=(a_1,\ldots,a_n)\in\mathbb{R}^n$, linearity of the Ito integral gives $\sum_{j=1}^n a_jX_j=I(\sum_{j=1}^n a_jf_j)$. The integrand $\sum_j a_jf_j$ is deterministic and belongs to $L^2([0,T])$, since $\int_0^T(\sum_j a_jf_j)^2\,ds\le(\sum_j a_j^2)(\sum_j\int_0^T f_j^2\,ds)<\infty$. By the stated Gaussianity result, this linear combination is centered normal with variance $\int_0^T(\sum_j a_jf_j(s))^2\,ds=a^\mathsf{T}Ca$. Thus $X$ is jointly Gaussian by definition. Taking $a=u$, the characteristic function of this centered normal linear combination at argument $1$ is $\exp(-\tfrac12u^\mathsf{T}Cu)$, which proves the formula. Finally, $a^\mathsf{T}Ca=\int_0^T(\sum_j a_jf_j(s))^2\,ds\ge0$ for every $a$, so $C$ is positive semidefinite. The proof does not require strict positivity: if a nonzero $a$ gives zero integral, the corresponding linear combination is zero almost surely and $C$ is singular. $\square$

The theorem identifies the entire joint distribution, not just its individual marginal distributions. In particular, two coordinates with zero covariance are independent here. This implication uses joint Gaussianity; zero covariance alone does not imply independence for arbitrary random variables.

Proof of the independence assertion. For two jointly Gaussian centered variables with zero covariance, the joint characteristic function at $(u,v)$ is $\exp(-\tfrac12(u^2C_{11}+v^2C_{22}))$. This is the product of the individual characteristic functions $\exp(-u^2C_{11}/2)$ and $\exp(-v^2C_{22}/2)$. A joint characteristic function that factors into the marginal characteristic functions characterizes independence. The argument includes zero variances: a centered variable with variance zero equals zero almost surely and is independent of every random variable. $\square$

Worked Covariance Calculations

Worked Example: Two Overlapping Integrands

On $[0,1]$, let $X=I(f)$ and $Y=I(g)$, where $f(s)=1$ and $g(s)=s$. Find their joint law’s mean vector and covariance matrix, then check the variance of $X-2Y$.

Both means are zero by the deterministic-integral Gaussianity result. The entries of the covariance matrix are inner products: $\operatorname{Var}(X)=\int_0^1 1\,ds=1$, $\operatorname{Cov}(X,Y)=\int_0^1 s\,ds=\tfrac12$, and $\operatorname{Var}(Y)=\int_0^1 s^2\,ds=\tfrac13$. Therefore

C=\begin{pmatrix}1&\tfrac12\\[2pt]\tfrac12&\tfrac13\end{pmatrix}, \qquad \mathbb{E}\begin{pmatrix}X\\Y\end{pmatrix} =\begin{pmatrix}0\\0\end{pmatrix}.

The determinant is $1\cdot\tfrac13-\tfrac12\cdot\tfrac12=\tfrac13-\tfrac14=\tfrac1{12}>0$, so this covariance matrix is nonsingular. The linear combination $X-2Y=I(1-2s)$ is centered normal. Directly from its integrand, its variance is $\int_0^1(1-2s)^2\,ds=1-2+\tfrac43=\tfrac13$. Using the covariance matrix instead gives $\operatorname{Var}(X-2Y)=1+4\cdot\tfrac13-4\cdot\tfrac12=1+\tfrac43-2=\tfrac13$. The two calculations agree.

Worked Example: Orthogonal Integrands Give Independent Integrals

On $[0,1]$, let $X=I(f)$, $Y=I(g)$, with $f(s)=1$ and $g(s)=\sqrt{3}(2s-1)$. Compute their covariance matrix and determine whether $X$ and $Y$ are independent.

First, $\operatorname{Var}(X)=\int_0^1 1\,ds=1$. The cross term is $\operatorname{Cov}(X,Y)=\sqrt{3}\int_0^1(2s-1)\,ds=\sqrt{3}[s^2-s]_0^1=0$. For the other variance, $\operatorname{Var}(Y)=3\int_0^1(2s-1)^2\,ds =3\int_0^1(4s^2-4s+1)\,ds =3(\tfrac43-2+1)=1$. Thus $C$ is the $2$ by $2$ identity matrix, and the joint characteristic function is $\exp(-\tfrac12(u^2+v^2))=\exp(-u^2/2)\exp(-v^2/2)$. Since the vector is jointly Gaussian, the factorization proves that $X$ and $Y$ are independent standard normal variables. As a check, their covariance computed from the matrix is $0$, and each diagonal entry gives variance $1$.

Worked Example: A Singular Joint Gaussian Vector

On $[0,1]$, take $X_1=I(1)$, $X_2=I(s)$, and $X_3=I(1+s)$. Compute the covariance matrix and exhibit its degeneracy.

The six distinct inner products are $\int_0^1 1\,ds=1$, $\int_0^1s\,ds=\tfrac12$, $\int_0^1(1+s)\,ds=\tfrac32$, $\int_0^1s^2\,ds=\tfrac13$, $\int_0^1s(1+s)\,ds=\tfrac12+\tfrac13=\tfrac56$, and $\int_0^1(1+s)^2\,ds=1+1+\tfrac13=\tfrac73$. Consequently

C=\begin{pmatrix} 1&\tfrac12&\tfrac32\\ \tfrac12&\tfrac13&\tfrac56\\ \tfrac32&\tfrac56&\tfrac73 \end{pmatrix}.

At the level of integrands, $1+s=1+s$, so linearity gives $X_3=X_1+X_2$ almost surely. Accordingly $C$ is singular: the nonzero vector $(1,1,-1)^\mathsf{T}$ satisfies $C(1,1,-1)^\mathsf{T}=0$. For example, its first row gives $1+\tfrac12-\tfrac32=0$, its second gives $\tfrac12+\tfrac13-\tfrac56=0$, and its third gives $\tfrac32+\tfrac56-\tfrac73=0$. This relation is consistent with the theorem: a jointly Gaussian vector need not have a density on all of $\mathbb{R}^3$. The check is the exact almost-sure identity $X_1+X_2-X_3=I(1+s-1-s)=I(0)=0$.

What the Covariance Matrix Records

The Gram-matrix form $C_{ij}=\langle f_i,f_j\rangle_{L^2}$ makes several features of the joint law visible.

The diagonal entries are squared norms, so they are nonnegative and give the marginal variances.
The off-diagonal entries record signed overlap. Orthogonality of two integrands gives zero covariance, and in this jointly Gaussian setting it also gives independence.
Linear dependence among the integrands produces an almost-sure linear relation among their integrals, hence a singular covariance matrix.
Conversely, if $a^\mathsf{T}Ca=0$, then $\int_0^T(\sum_j a_jf_j)^2\,ds=0$. Thus $\sum_j a_jf_j=0$ almost everywhere, and the corresponding linear combination of integrals is zero almost surely.

The last point follows because a nonnegative function with integral zero vanishes almost everywhere. It explains precisely when degeneracy occurs, including the case that one of the integrands is zero almost everywhere.

Common Pitfall: Gaussian Marginals Do Not Suffice

A frequent error is to infer joint Gaussianity merely because every coordinate is normally distributed. For a concrete counterexample, let $Z$ be standard normal and let $\varepsilon$ be independent of $Z$, with probabilities $\mathbb{P}(\varepsilon=1)=\mathbb{P}(\varepsilon=-1)=\tfrac12$. Set $U=Z$ and $V=\varepsilon Z$. Both $U$ and $V$ are standard normal: conditionally on either value of $\varepsilon$, $V$ has the same symmetric standard normal law. Also $\mathbb{E}[UV]=\mathbb{E}[\varepsilon Z^2]=0$. Nevertheless, $|U|=|V|$ almost surely, so the pair is not a pair of independent standard normals; indeed its joint law is concentrated on the two lines $v=u$ and $v=-u$. The zero covariance does not rescue joint Gaussianity. For deterministic Wiener integrals, it is the linear-combination argument in the theorem—not marginal Gaussianity alone—that establishes the joint Gaussian law.

When applying the theorem, calculate the covariance entry by entry from the integrands and then check the result using at least one linear combination. In particular, do not assume that a covariance matrix is invertible: an integrand relation may force exact dependence, while the joint Gaussian conclusion remains valid.

Check Your Understanding

Use the deterministic-integral theorem and the $L^2$ inner product to answer the following.

For $X=I(1)$ and $Y=I(s^2)$ on $[0,1]$, compute $\operatorname{Var}(X)$, $\operatorname{Cov}(X,Y)$, and $\operatorname{Var}(Y)$.
Explain why the covariance matrix of any finite vector of deterministic Wiener integrals is positive semidefinite, including when some integrands vanish almost everywhere.
Suppose $f_3=2f_1-f_2$ almost everywhere. What almost-sure relation holds among $I(f_1),I(f_2),I(f_3)$, and what does it imply about their covariance matrix?
For two deterministic integrands with zero $L^2$ inner product, which additional fact about their integrals is needed before concluding independence?
Why does knowing that each coordinate has a normal marginal distribution not by itself establish that a random vector is jointly Gaussian?