Proving the Cameron-Martin Theorem by Gaussian Approximation

From Gaussian Coordinates to a Path-Space Density

Brownian motion has infinitely many random coordinates, but any finite collection of its orthogonal Gaussian coordinates has an elementary translation density. The Cameron-Martin proof by Gaussian approximation makes this finite-dimensional calculation do the main work: approximate the shift derivative by its first $N$ coordinates, pass to the limit in the likelihoods, and then verify that the coordinates generate the Borel sets of continuous paths. The last step matters. An identity on a sigma-field completed under Wiener measure alone does not automatically give an identity for a translated path law.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, and fix $T>0$. Write $C_0([0,T])$ for the continuous real-valued paths on $[0,T]$ that start at zero, and let $\mu$ be the law of $(B_t)_{0\le t\le T}$ on this space. For a deterministic progressively measurable integrand $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, the Ito integral $\int_0^T H_s\,dB_s$ is defined in $L^2(\mathbb{P})$. In particular, the integrands used below are deterministic functions in $L^2([0,T])$.

Definition: A Cameron-Martin shift on $[0,T]$ is a deterministic function $h$ of the form $h(t)=\int_0^t g(s)\,ds$, where $g=h'\in L^2([0,T])$. Its squared Cameron-Martin norm is $\|h\|_H^2=\int_0^T h'(s)^2\,ds$. The translated path is $B+h$, meaning $t\mapsto B_t+h(t)$.

The target identity, for every bounded Borel function $F:C_0([0,T])\to\mathbb{R}$, is

\mathbb{E}[F(B+h)] = \mathbb{E}\!\left[ F(B)\exp\!\left( \int_0^T h'(s)\,dB_s-\frac12\int_0^T h'(s)^2\,ds \right)\right].

Equivalently, the law of $B+h$ has density $\exp(\int_0^T h'\,dB-\frac12\|h\|_H^2)$ with respect to Wiener measure. The sign is fixed by the displayed expectation identity: translating the argument of $F$ by $+h$ puts the positive stochastic integral in the density on the unshifted path.

Finite Gaussian Coordinates

Choose a Haar orthonormal basis $(e_j)_{j\ge1}$ of $L^2([0,T])$, including the normalized constant function. Each $e_j$ is a step function on a dyadic partition. Define $Z_j=\int_0^T e_j(s)\,dB_s$ and $a_j=\int_0^T e_j(s)h'(s)\,ds$. The Ito isometry and joint Gaussianity of deterministic Brownian integrals give independent standard normal variables $Z_j$. Parseval's identity gives $\sum_{j\ge1}a_j^2=\|h'\|_{L^2}^2<\infty$.

Lemma: For $N\ge1$, set $L_N=\exp(\sum_{j=1}^N a_jZ_j-\frac12\sum_{j=1}^N a_j^2)$. Then $\mathbb{E}[L_N]=1$, $\mathbb{E}[L_N^2]=\exp(\sum_{j=1}^N a_j^2)$, and, for any bounded Borel function $\varphi:\mathbb{R}^N\to\mathbb{R}$, $$ \mathbb{E}[\varphi(Z_1+a_1,\ldots,Z_N+a_N)] =\mathbb{E}[\varphi(Z_1,\ldots,Z_N)L_N]. $$

Proof. The vector $(Z_1,\ldots,Z_N)$ has density $(2\pi)^{-N/2}\exp(-\frac12\sum_{j=1}^N z_j^2)$. In the integral for the left side, substitute $y_j=z_j+a_j$. The resulting density at $y$, divided by the standard normal density at $y$, is \[ \exp\!\left(\sum_{j=1}^N a_jy_j-\frac12\sum_{j=1}^N a_j^2\right). \] This proves the identity. Taking $\varphi=1$ gives $\mathbb{E}[L_N]=1$. The normal moment-generating formula $\mathbb{E}[e^{uZ_j}]=e^{u^2/2}$, applied to the independent coordinates, gives $\mathbb{E}[L_N^2]=\exp(\sum_{j=1}^N a_j^2)$. $\square$

The finite likelihoods converge strongly enough to pass expectations to the limit. Indeed, for $M\ge N$, independence of the coordinates after $N$ and the same normal moment calculation give $\mathbb{E}[L_ML_N]=\mathbb{E}[L_N^2]$. Therefore \[ \mathbb{E}[(L_M-L_N)^2] =\exp\!\left(\sum_{j=1}^M a_j^2\right)-\exp\!\left(\sum_{j=1}^N a_j^2\right). \] The right side tends to zero as $M,N\to\infty$, because the partial sums of $\sum_j a_j^2$ converge. Thus $L_N$ converges in $L^2(\mathbb{P})$, and hence in $L^1(\mathbb{P})$, to a random variable $L$. In particular, $\mathbb{E}[L]=1$.

For each $N$, $\mathbb{E}[L_{N+1}\mid\sigma(Z_1,\ldots,Z_N)]=L_N$, because the new factor has expectation one and is independent of the first $N$ coordinates. This verifies the martingale property of $(L_N)$ with respect to its coordinate filtration: each variable is integrable and its conditional expectation is the preceding variable. The limit has the stated exponential form. The partial sums $\sum_{j=1}^N a_jZ_j$ converge in $L^2(\mathbb{P})$ to $\int_0^T h'(s)\,dB_s$, by the Ito isometry and completeness of the Haar basis. Also, $\sum_{j=1}^N a_j^2\to\|h'\|_{L^2}^2$. Consequently, \[ L=\exp\!\left(\int_0^T h'(s)\,dB_s-\frac12\|h'\|_{L^2}^2\right) \quad\text{almost surely}. \] For example, convergence in probability of the exponents gives convergence in probability of their exponentials, while $L_N\to L$ in $L^2$, so the two limits agree almost surely.

Why These Coordinates Determine the Whole Path

For a Haar step function $e_j$, its integral against a continuous path is a finite linear combination of path increments. Thus $Z_j$ can be read as a Borel function of the Brownian path, not merely as an abstract random variable defined up to a Wiener-null set. More importantly, the coordinates recover every dyadic increment. The indicator of any dyadic interval belongs to a finite-dimensional dyadic step-function space, and that space is spanned by the constant function and finitely many Haar functions. Its Brownian integral is the increment over that interval; its path integral against $h$ is the corresponding increment of $h$.

Proposition: On $C_0([0,T])$, the sigma-field generated by the Haar coordinate functionals is the Borel sigma-field.

Proof. Each Haar coordinate functional on paths is a finite linear combination of evaluations at partition points, so it is Borel. This proves that the coordinate sigma-field is contained in the Borel sigma-field. Conversely, for every dyadic time $t$, the indicator of $[0,t]$ is in a finite dyadic step-function space. Its integral against a path $x\in C_0([0,T])$ is $x(t)$, and its expansion in that finite Haar space expresses $x(t)$ as a finite linear combination of Haar coordinates of $x$. Hence evaluation at every dyadic time is measurable with respect to the coordinate sigma-field. Dyadic times are dense in $[0,T]$. For any fixed $x\in C_0([0,T])$ and any $r>0$, continuity gives \[ \{y:\|y-x\|_\infty<r\} = \bigcup_{\substack{q\in\mathbb{Q}\\0<q<r}} \ \bigcap_{t\in D} \{y:|y(t)-x(t)|\le q\}, \] where $D$ is the countable set of dyadic times in $[0,T]$. To check this equality, the supremum of $|y(t)-x(t)|$ over $D$ equals its supremum over $[0,T]$, by continuity; that supremum is less than $r$ exactly when it is at most some rational $q<r$. Thus open balls are in the coordinate sigma-field. Since $C_0([0,T])$ is separable, open sets are countable unions of open balls from a countable base. Every open set, and therefore every Borel set, is coordinate-measurable. $\square$

This argument gives generation on the path space itself. It avoids relying on a reconstruction that holds only almost surely under $\mu$, which would not by itself control exceptional sets under the translated law.

The Cameron-Martin Density Formula

Theorem: If $h(0)=0$, $h$ is absolutely continuous, and $h'\in L^2([0,T])$, then for every bounded Borel $F:C_0([0,T])\to\mathbb{R}$, $$ \mathbb{E}[F(B+h)] = \mathbb{E}\!\left[F(B) \exp\!\left(\int_0^T h'(s)\,dB_s-\frac12\int_0^T h'(s)^2\,ds\right)\right]. $$

Proof. First take $F$ to depend on the first $N$ Haar coordinates: $F(x)=\varphi(Z_1(x),\ldots,Z_N(x))$ for bounded Borel $\varphi$. Since each Haar function is a step function, its coordinate on the translated path is $Z_j(B+h)=Z_j(B)+a_j$. The finite Gaussian lemma gives $\mathbb{E}[F(B+h)]=\mathbb{E}[F(B)L_N]$. Also, the $L^2$ convergence above implies $\mathbb{E}[F(B)L_N]\to\mathbb{E}[F(B)L]$, since \[ |\mathbb{E}[F(B)(L_N-L)]| \le \|F\|_\infty\,\mathbb{E}[|L_N-L|]\longrightarrow0. \] Thus the desired identity holds for every bounded function measurable with respect to finitely many coordinates. These cylinder events form a collection closed under finite intersections and generate the Borel sigma-field by the proposition. Both sides define finite measures on that sigma-field: the left side is the law of $B+h$, and the right side is the measure $A\mapsto\mathbb{E}[\mathbf{1}_{\{B\in A\}}L]$, which has total mass $\mathbb{E}[L]=1$. The uniqueness theorem for finite measures agreeing on a generating intersection-closed class extends their equality to all Borel sets. Integration of bounded Borel $F$ against these equal measures proves the formula. $\square$

Worked Densities and Checks

Worked Example: A Constant-Slope Shift

Take $T=2$ and $h(t)=3t$, so $h'(t)=3$. The derivative is square-integrable, and its energy is \[ \int_0^2 h'(s)^2\,ds=\int_0^2 9\,ds=18. \] The Ito integral of the constant integrand is $\int_0^2 3\,dB_s=3B_2$. Therefore the density is $L=\exp(3B_2-9)$. The shift moves the endpoint from $B_2$ to $B_2+6$.

Check: $B_2$ is normal with mean $0$ and variance $2$. Hence \[ \mathbb{E}[L]=e^{-9}\mathbb{E}[e^{3B_2}] =e^{-9}e^{3^2\cdot2/2}=e^{-9}e^9=1. \] Under the density, the endpoint has mean $\mathbb{E}[B_2L]=6$: differentiating the normal moment-generating function gives $\mathbb{E}[B_2e^{3B_2}]=6e^9$, and multiplication by $e^{-9}$ gives $6$, matching the endpoint shift.

Worked Example: A Piecewise-Linear Shift

On $[0,1]$, set $h'(s)=2$ for $0\le s<\tfrac12$ and $h'(s)=-1$ for $\tfrac12\le s\le1$. Thus $h(\tfrac12)=2\cdot\tfrac12=1$, and $h(1)=1-1\cdot\tfrac12=\tfrac12$. Its energy is \[ \int_0^1 h'(s)^2\,ds=4\cdot\tfrac12+1\cdot\tfrac12=\tfrac52. \] Its Ito integral is $2(B_{1/2}-B_0)-(B_1-B_{1/2})=3B_{1/2}-B_1$. The density is therefore $L=\exp(3B_{1/2}-B_1-\tfrac54)$.

Check: the two increments $U=B_{1/2}$ and $V=B_1-B_{1/2}$ are independent normals with variances $\tfrac12$ each. The exponent's random part is $2U-V$, with variance $2^2\cdot\tfrac12+(-1)^2\cdot\tfrac12=\tfrac52$. Thus $\mathbb{E}[L]=e^{-5/4}e^{(5/2)/2}=1$. Under the translated path, the two increment means are $1$ and $-\tfrac12$, exactly the increments of $h$.

Worked Example: A Quadratic Shift

Take $T=1$ and $h(t)=t^2$. Then $h'(s)=2s$, and the energy is \[ \int_0^1 (2s)^2\,ds=4\left[\frac{s^3}{3}\right]_0^1=\frac43. \] The density is $L=\exp(\int_0^1 2s\,dB_s-\tfrac23)$. The integral can also be written $\int_0^1 2s\,dB_s=2B_1-2\int_0^1B_s\,ds$. This follows by summing the ordinary product increments for $sB_s$: the deterministic-time and Brownian cross-increment sums vanish in the limit, leaving the displayed integration-by-parts identity.

Check: the stochastic integral is centered normal with variance $\int_0^1(2s)^2ds=\tfrac43$, by the Ito isometry and Gaussianity of deterministic Brownian integrals. Its exponential moment is $e^{(4/3)/2}=e^{2/3}$, so $\mathbb{E}[L]=e^{-2/3}e^{2/3}=1$. The shift's endpoint is $h(1)=1$, consistent with the formula's translation of the full path, not just of its endpoint.

Common Pitfall: A Sign Error and a Sigma-Field Gap

A tempting but incorrect density for the expectation $\mathbb{E}[F(B+h)]$ is $\exp(-\int h'\,dB-\frac12\|h'\|_H^2)$. For the constant-slope example, that candidate would weight $B_2$ by $e^{-3B_2-9}$, giving weighted mean $-6$, whereas $B_2+6$ has mean $6$. The correct sign is positive; the finite-dimensional substitution $y=z+a$ fixes it.

A second error is to prove the identity only for coordinate events and then assert that these generate the path sigma-field up to Wiener-null sets. That assertion is not enough: a translated law might assign positive probability to a set that is null under Wiener measure. Here the Haar-coordinate argument proves generation of the Borel sigma-field on $C_0([0,T])$ itself, so the measure-extension step applies to the translated law without an unverified null-set assumption.

Check Your Understanding

Use the finite-coordinate argument and the path-space theorem to answer the following.

For $h(t)=ct$ on $[0,T]$, write the density and verify directly from the normal moment-generating function that it has expectation one.
Why does $\sum_j\langle h',e_j\rangle^2<\infty$ imply that the finite likelihoods converge in $L^2$?
Explain why Haar coordinates determine increments over dyadic intervals, and how continuity then determines the path.
For $h(t)=t^2$ on $[0,1]$, calculate its Cameron-Martin energy and the variance of $\int_0^1 2s\,dB_s$.
In the density identity, which sign precedes $\int_0^T h'(s)\,dB_s$, and what endpoint-mean check detects the opposite sign?