Multidimensional Girsanov and Correlated Brownian Motion

Why the Multidimensional Formula Uses a Dot Product

With several Brownian coordinates, a change of measure must account for every coordinate and for their joint covariance structure. The scalar Girsanov formula extends by replacing the scalar product in its exponent with a Euclidean dot product. The resulting shift changes the drift of each coordinate, while leaving the Brownian covariance matrix unchanged. This is essential when a vector SDE is driven by correlated noise: the drift correction depends on the diffusion matrix, but the covariance of the noise does not change.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a $d$-dimensional Brownian motion $\mathbf{B}_t=(B_t^1,\ldots,B_t^d)^\mathsf{T}$. Its coordinates are independent standard Brownian motions under $\mathbb{P}$, so $[B^i,B^j]_t=\delta_{ij}t$, where $\delta_{ij}=1$ if $i=j$ and $0$ otherwise. Vector stochastic integrals below are Ito integrals: for a progressively measurable vector process $\theta$, $\int_0^t\theta_s^\mathsf{T}\,d\mathbf{B}_s$ is defined when $\mathbb{E}\int_0^T\|\theta_s\|^2\,ds<\infty$; equivalently, each coordinate is in the corresponding square-integrable Ito-integrand class.

Definition: Let $\theta_t\in\mathbb{R}^d$ be progressively measurable and satisfy Novikov’s condition on $[0,T]$: $$ \mathbb{E}\exp\left(\frac12\int_0^T\|\theta_s\|^2\,ds\right)<\infty. $$ The associated density process is $$ Z_t=\exp\left(-\int_0^t\theta_s^\mathsf{T}\,d\mathbf{B}_s-\frac12\int_0^t\|\theta_s\|^2\,ds\right),\qquad 0\le t\le T. $$

Novikov’s condition is a sufficient condition for this stochastic exponential to be a true martingale, not merely a local martingale. It also implies $\mathbb{E}\int_0^T\|\theta_s\|^2\,ds<\infty$: for $x\ge0$, $x\le 2e^{x/2}$, and the condition gives a finite expectation of the exponential. Thus the integrals defining $Z$ are well defined in the stated class. The Novikov criterion itself is a standard result established in the earlier material on stochastic exponentials.

The Multidimensional Drift-Change Theorem

Theorem: Under the conditions above, $Z_t$ is a positive $\mathbb{P}$-martingale with $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$ for $s\le t\le T$, and $\mathbb{E}[Z_t]=1$. Define the probability measure $\mathbb{Q}$ on $\mathcal{F}_T$ by $d\mathbb{Q}=Z_T\,d\mathbb{P}$, and set $$ \widetilde{\mathbf{B}}_t=\mathbf{B}_t+\int_0^t\theta_s\,ds. $$ Then $\widetilde{\mathbf{B}}$ is a $d$-dimensional Brownian motion under $\mathbb{Q}$, with $[\widetilde B^i,\widetilde B^j]_t=\delta_{ij}t$. If an $m$-dimensional process satisfies $$ d\mathbf{X}_t=\mathbf{b}_t\,dt+\Sigma_t\,d\mathbf{B}_t, $$ then, under $\mathbb{Q}$, it satisfies $$ d\mathbf{X}_t=(\mathbf{b}_t-\Sigma_t\theta_t)\,dt+\Sigma_t\,d\widetilde{\mathbf{B}}_t. $$ Here $\mathbf{b}_t$ is the drift vector and $\Sigma_t$ is an $m\times d$ diffusion matrix. The identity is understood whenever the displayed stochastic integrals exist; for example, locally square-integrable progressively measurable diffusion coefficients suffice.

Proof. First, apply Ito’s formula to the exponential defining $Z$. For $g(x)=e^x$, $g'(x)=g(x)$ and $g''(x)=g(x)$; the exponent has stochastic differential $-\theta_t^\mathsf{T}d\mathbf{B}_t-\tfrac12\|\theta_t\|^2dt$, and its quadratic variation is $\|\theta_t\|^2dt$. The explicit $\tfrac12g''\,dt$ correction cancels the negative half-norm term, giving $dZ_t=-Z_t\theta_t^\mathsf{T}d\mathbf{B}_t$. Novikov’s criterion makes this positive local martingale a true martingale. Consequently $Z_t=\mathbb{E}[Z_T\mid\mathcal{F}_t]$, $\mathbb{E}[Z_t]=1$, and $\mathbb{Q}$ is a probability measure equivalent to $\mathbb{P}$ on $\mathcal{F}_T$, since $Z_T>0$ almost surely.

For each coordinate $i$, product integration gives $$ d(Z_t\widetilde B_t^i)=\widetilde B_t^i\,dZ_t+Z_t\,d\widetilde B_t^i+d[Z,\widetilde B^i]_t. $$ The finite-variation part of $\widetilde B^i$ is $\int_0^t\theta_s^i\,ds$, so $d[Z,\widetilde B^i]_t=d[Z,B^i]_t=-Z_t\theta_t^i\,dt$. Substituting $dZ_t=-Z_t\sum_j\theta_t^j\,dB_t^j$ and $d\widetilde B_t^i=dB_t^i+\theta_t^i\,dt$, the two $dt$ terms cancel. Thus $Z\widetilde B^i$ is a $\mathbb{P}$-local martingale. The density-process criterion (Bayes’ formula, applied after localization) then implies that $\widetilde B^i$ is a $\mathbb{Q}$-local martingale. This argument asserts only a local martingale property at this stage; a local martingale need not be a true martingale without additional integrability.

Adding a finite-variation process does not change quadratic covariation. Hence $$ [\widetilde B^i,\widetilde B^j]_t=[B^i,B^j]_t=\delta_{ij}t. $$ The multidimensional Levy characterization says that a continuous local martingale starting at zero with these covariations is a standard $d$-dimensional Brownian motion relative to the filtration. This standard characterization, applied under $\mathbb{Q}$, proves the Brownian claim. Finally $d\mathbf{B}_t=d\widetilde{\mathbf{B}}_t-\theta_tdt$; substitution into the SDE gives the asserted drift $\mathbf{b}_t-\Sigma_t\theta_t$. Its $i$-th component is $b_t^i-\sum_{j=1}^d\Sigma_t^{ij}\theta_t^j$. $\square$

The signs follow directly from the density convention: the exponent contains $-\int\theta^\mathsf{T}d\mathbf{B}$, so the Brownian motion under $\mathbb{Q}$ is $\mathbf{B}+\int\theta\,dt$. Reversing the sign in the density reverses both shifts. The notation $dB_t^i\,dB_t^j=\delta_{ij}dt$, $dB_t^i\,dt=0$, and $dt\,dt=0$ summarizes the covariation calculation; it is not the ordinary chain rule.

Worked Example: A Two-Coordinate Density

Take $d=2$, $T=2$, and the constant kernel $\theta=(\tfrac12,-1)^\mathsf{T}$. Verify Novikov’s condition and write the density and shifted Brownian coordinates.

First, $\|\theta\|^2=(\tfrac12)^2+(-1)^2=\tfrac14+1=\tfrac54$. Therefore $$ \mathbb{E}\exp\left(\frac12\int_0^2\|\theta\|^2\,ds\right) =\exp\left(\frac12\cdot\frac54\cdot2\right)=e^{5/4}<\infty. $$ The density at time $2$ is $$ Z_2=\exp\left(-\frac12B_2^1+B_2^2-\frac12\cdot\frac54\cdot2\right) =\exp\left(-\frac12B_2^1+B_2^2-\frac54\right). $$ The new Brownian coordinates are $\widetilde B_t^1=B_t^1+\tfrac12t$ and $\widetilde B_t^2=B_t^2-t$. Their covariations remain $[\widetilde B^1]_t=t$, $[\widetilde B^2]_t=t$, and $[\widetilde B^1,\widetilde B^2]_t=0$, because adding deterministic linear functions does not alter quadratic covariation.

Check: the exponent’s compensator at $t=2$ is $-\tfrac12\int_0^2(1/4+1)\,ds=-\tfrac12\cdot\tfrac54\cdot2=-\tfrac54$, agreeing with the density above. By the theorem, $\mathbb{E}[Z_2]=1$; this is also the true-martingale conclusion required to define $\mathbb{Q}$.

Correlated Brownian Coordinates

The theorem starts with independent Brownian coordinates. Correlated Brownian motion can be represented as a linear transformation of such a vector. Let $L$ be a fixed $d\times d$ matrix and define $\mathbf{W}_t=L\mathbf{B}_t$. Its covariance matrix per unit time is $A=LL^\mathsf{T}$, since $[W^i,W^j]_t=A_{ij}t$. Under the measure change, $\widetilde{\mathbf{W}}_t=L\widetilde{\mathbf{B}}_t$ has the same covariance matrix, while $$ \mathbf{W}_t=\widetilde{\mathbf{W}}_t-\int_0^tL\theta_s\,ds. $$ Thus the drift shift in the correlated coordinates is $L\theta$, not $\theta$ itself.

Corollary: Suppose $L$ is invertible and write the desired drift shift in the $\mathbf{W}$ coordinates as $\boldsymbol{\eta}_t=L\theta_t$. Then the density exponent can be written $$ -\int_0^t\boldsymbol{\eta}_s^\mathsf{T}A^{-1}\,d\mathbf{W}_s -\frac12\int_0^t\boldsymbol{\eta}_s^\mathsf{T}A^{-1}\boldsymbol{\eta}_s\,ds, \qquad A=LL^\mathsf{T}. $$ Under the resulting measure, $\widetilde{\mathbf{W}}=\mathbf{W}+\int\boldsymbol{\eta}_sds$ has covariance matrix $A$ per unit time.

Proof. Since $L$ is invertible, $\theta=L^{-1}\boldsymbol{\eta}$. Also $d\mathbf{W}=L\,d\mathbf{B}$, so $d\mathbf{B}=L^{-1}d\mathbf{W}$. Therefore $$ \theta^\mathsf{T}d\mathbf{B} =\boldsymbol{\eta}^\mathsf{T}L^{-\mathsf{T}}L^{-1}d\mathbf{W} =\boldsymbol{\eta}^\mathsf{T}(LL^\mathsf{T})^{-1}d\mathbf{W} =\boldsymbol{\eta}^\mathsf{T}A^{-1}d\mathbf{W}. $$ The same matrix identity gives $\|\theta\|^2=\boldsymbol{\eta}^\mathsf{T}A^{-1}\boldsymbol{\eta}$. Substituting these equalities into the density proves its stated form. The covariance claim follows from $\widetilde{\mathbf{W}}=L\widetilde{\mathbf{B}}$ and the theorem. $\square$

Worked Example: A Correlated Pair

Let $L=\begin{pmatrix}1&0\\[2pt]3/5&4/5\end{pmatrix}$, so $\mathbf{W}=L\mathbf{B}$. Find the covariance matrix and the independent-coordinate kernel that produces a shift $\boldsymbol{\eta}=(1,2)^\mathsf{T}$ in $\mathbf{W}$.

Multiplication gives $$ A=LL^\mathsf{T} =\begin{pmatrix}1&0\\3/5&4/5\end{pmatrix} \begin{pmatrix}1&3/5\\0&4/5\end{pmatrix} =\begin{pmatrix}1&3/5\\3/5&1\end{pmatrix}. $$ Thus $[W^1,W^2]_t=\tfrac35t$, while each coordinate has quadratic variation $t$. The matrix $L$ is invertible, and solving $L\theta=\boldsymbol{\eta}$ gives $\theta_1=1$ and $\tfrac35\theta_1+\tfrac45\theta_2=2$. Hence $\tfrac45\theta_2=\tfrac75$, so $\theta_2=\tfrac74$.

The covariance is unchanged after the measure change, and $\widetilde{\mathbf{W}}=\mathbf{W}+\int(1,2)^\mathsf{T}ds$. For a check, $A^{-1}=\tfrac{25}{16}\begin{pmatrix}1&-3/5\\-3/5&1\end{pmatrix}$. Therefore $A^{-1}\boldsymbol{\eta}=(-5/16,35/16)^\mathsf{T}$, and $\boldsymbol{\eta}^\mathsf{T}A^{-1}\boldsymbol{\eta}=-5/16+70/16=65/16$. The independent-coordinate kernel has squared norm $1^2+(7/4)^2=1+49/16=65/16$, verifying the quadratic term in the density.

Applying the Formula to a Vector SDE

For an $m$-dimensional state driven by $d$ Brownian coordinates, the diffusion matrix maps a Brownian drift shift into a state-space drift shift. In particular, the $i$-th component of $\Sigma\theta$ is $\sum_{j=1}^d\Sigma^{ij}\theta^j$. The covariance matrix of the noise in the state coordinates is $\Sigma\Sigma^\mathsf{T}\,dt$, and remains the same after changing from $\mathbf{B}$ to $\widetilde{\mathbf{B}}$. This is a change of measure, not a claim of strong or weak uniqueness for the SDE. The formula changes the drift representation; uniqueness requires separate hypotheses on the coefficients.

Worked Example: Drift Adjustment and State Covariance

Consider a two-dimensional SDE with drift $\mathbf{b}=(4,-1)^\mathsf{T}$, diffusion matrix $\Sigma=\begin{pmatrix}1&2\\-1&1\end{pmatrix}$, and Girsanov kernel $\theta=(\tfrac12,-1)^\mathsf{T}$. Compute the new drift and verify the covariance.

First, $$ \Sigma\theta =\begin{pmatrix}1&2\\-1&1\end{pmatrix} \begin{pmatrix}1/2\\-1\end{pmatrix} =\begin{pmatrix}1/2-2\\-1/2-1\end{pmatrix} =\begin{pmatrix}-3/2\\-3/2\end{pmatrix}. $$ Consequently $\mathbf{b}-\Sigma\theta=(4-(-3/2),-1-(-3/2))^\mathsf{T}=(11/2,1/2)^\mathsf{T}$. Under $\mathbb{Q}$, the SDE has drift $(11/2,1/2)^\mathsf{T}$ and diffusion matrix $\Sigma$.

The covariance matrix is $$ \Sigma\Sigma^\mathsf{T} =\begin{pmatrix}1&2\\-1&1\end{pmatrix} \begin{pmatrix}1&-1\\2&1\end{pmatrix} =\begin{pmatrix}5&1\\1&2\end{pmatrix}. $$ The same product applies under $\mathbb{Q}$ because the diffusion matrix is unchanged and $\widetilde{\mathbf{B}}$ has identity covariance per unit time. Check: substituting $d\mathbf{B}=d\widetilde{\mathbf{B}}-\theta\,dt$ into the original equation gives drift $\mathbf{b}-\Sigma\theta=(11/2,1/2)^\mathsf{T}$, as computed componentwise.

Common Mistakes and When the Formula Applies

Wrong sign. If the density is written with the negative exponent $-\int\theta^\mathsf{T}d\mathbf{B}$, then $\widetilde{\mathbf{B}}=\mathbf{B}+\int\theta\,dt$, and the SDE drift becomes $\mathbf{b}-\Sigma\theta$. For the preceding example, incorrectly using $\mathbf{b}+\Sigma\theta$ would give $(5/2,-5/2)^\mathsf{T}$, not $(11/2,1/2)^\mathsf{T}$. Substitution into the SDE shows the error: the finite-variation term from $d\mathbf{B}=d\widetilde{\mathbf{B}}-\theta dt$ is $-\Sigma\theta dt$.

Changing the drift does not remove correlation. The covariance matrix is determined by quadratic covariation, and adding finite-variation terms cannot alter it. If the original Brownian coordinates have identity covariance, their shifted versions still have identity covariance; a linear transformation $L$ retains covariance $LL^\mathsf{T}$. A covariance-matrix expression involving $A^{-1}$ requires $A$ to be positive definite. If $L$ is singular, use the underlying Brownian representation and its kernel rather than the inverse-matrix formula.

Do not confuse a local martingale with a martingale. The product calculation in the proof first gives a local martingale. Novikov’s condition is what ensures that the density $Z$ is a true martingale and defines a probability measure. The shifted process is identified as Brownian under that measure by its local-martingale property and quadratic covariations. These Ito-integral statements use the Ito convention; a Stratonovich integral has a separate conversion correction and is not substituted directly into the density formula.

Check Your Understanding

In each question, assume the integrands satisfy the stated integrability conditions and any density used satisfies Novikov’s condition.

For a three-dimensional kernel $\theta$, write the density exponent and the shifted Brownian vector.
If $d\mathbf{X}=\mathbf{b}\,dt+\Sigma\,d\mathbf{B}$, what is the $i$-th component of its drift under the changed measure?
For $\mathbf{W}=L\mathbf{B}$, explain why its covariance matrix per unit time is $LL^\mathsf{T}$ both before and after the change of measure.
When is the expression $\boldsymbol{\eta}^\mathsf{T}A^{-1}\boldsymbol{\eta}$ for the density’s quadratic term available?
Explain why the local-martingale calculation for $Z\widetilde B^i$ alone does not establish that $Z$ is a true martingale.