The Cameron-Martin Theorem and Translation Density

Translating Brownian Paths

Let $T>0$, and let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carry a standard Brownian motion $B_t$. On the space $C_0([0,T])$ of continuous real-valued paths starting at zero, Wiener measure $\mu$ is the probability law of $B$. A deterministic path $h$ translates a path $x$ to $x+h$. The central question is whether this translation changes Wiener measure to an equivalent measure, one that has the same null sets.

For the stochastic integral below, $h'$ is deterministic, hence progressively measurable. Whenever $\int_0^T |h'(t)|^2\,dt<\infty$, it belongs to $\mathcal{L}^2$: $\mathbb{E}\int_0^T |h'(t)|^2\,dt<\infty$. Thus $\int_0^T h'(t)\,dB_t$ is the Ito integral. The derivative $h'$ is the almost-everywhere derivative with respect to time, not a derivative of a Brownian path.

Definition: The Cameron-Martin space on $[0,T]$ is $$ H_0^1([0,T])=\left\{h:h(0)=0,\ h\text{ is absolutely continuous, and }\int_0^T|h'(t)|^2\,dt<\infty\right\}. $$ Its squared norm is $\|h\|_{H_0^1}^2=\int_0^T|h'(t)|^2\,dt$. For a continuous $h$ with $h(0)=0$, define the translated measure $\mu_h$ by $\mu_h(A)=\mu(\{x:x+h\in A\})$ for Borel sets $A\subseteq C_0([0,T])$.

The Cameron-Martin Theorem

Theorem (Cameron-Martin): Let $h\in C_0([0,T])$. If $h\in H_0^1([0,T])$, then $\mu_h$ and $\mu$ are equivalent and $$ \frac{d\mu_h}{d\mu}(x)= \exp\left(\int_0^T h'(t)\,dx_t-\frac12\int_0^T|h'(t)|^2\,dt\right) \quad\text{for }\mu\text{-almost every }x. $$ Here $\int h'\,dx$ denotes the Ito integral under Wiener measure. If $h\notin H_0^1([0,T])$, then $\mu_h$ and $\mu$ are mutually singular: there is a Borel set having full measure for one and zero measure for the other.

The density’s sign is tied to the definition of translation: $\mu_h$ is the law of $B+h$, not of $B-h$. For intuition, in one dimension a normal random variable shifted by $m$ has density ratio $\exp(mx-m^2/2)$ relative to a standard normal. The theorem gives the path-space counterpart, with the stochastic integral encoding the whole shift.

The Cameron-Martin theorem is a standard Gaussian-measure result. Its finite-dimensional Gaussian approximation proof is given in a subsequent part of the course; the singularity alternative and the identification of the path-space density require that approximation argument. The results proved here verify the density’s normalization and consequences once the theorem supplies the formula.

Why the Exponential Is a Probability Density

Proposition: If $h\in H_0^1([0,T])$ and $$ Z_t=\exp\left(\int_0^t h'(s)\,dB_s-\frac12\int_0^t|h'(s)|^2\,ds\right), $$ then $Z_t$ is a true martingale on $[0,T]$, and $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$ for $0\le s\le t\le T$.

Proof. The deterministic-integrand Ito integral $I_t=\int_0^t h'(u)\,dB_u$ is centered Gaussian, and its increments over disjoint time intervals are independent. For $s\le t$, write $I_t-I_s=\int_s^t h'(u)\,dB_u$ and $v_{s,t}=\int_s^t|h'(u)|^2\,du$. This increment is independent of $\mathcal{F}_s$ and has distribution $N(0,v_{s,t})$. The normal moment-generating formula gives $$ \mathbb{E}\left[\exp\left(I_t-I_s-\frac12v_{s,t}\right)\middle|\mathcal{F}_s\right] =\exp\left(\frac12v_{s,t}-\frac12v_{s,t}\right)=1. $$ Since $Z_t=Z_s\exp(I_t-I_s-\tfrac12v_{s,t})$, the conditional expectation is $Z_s$. Also $\mathbb{E}[Z_t]=1$, so $Z_t$ is integrable at each time and is a true martingale, not merely a local martingale. $\square$

In particular, the terminal exponential has expectation one and is strictly positive almost surely. Consequently, when it is the Radon-Nikodym density, the translated and original measures are equivalent: for any Borel set $A$, $\int_A Z_T\,d\mu=0$ if and only if $\mu(A)=0$. This verifies the equivalence assertion in the Cameron-Martin case, once the theorem identifies $Z_T$ as the density.

Corollary: For $p\in\mathbb{R}$, $$ \mathbb{E}[Z_T^p]=\exp\left(\frac{p^2-p}{2}\|h\|_{H_0^1}^2\right). $$

Proof. The random variable $I_T=\int_0^T h'(t)\,dB_t$ is normal with mean zero and variance $v=\|h\|_{H_0^1}^2$. Therefore $Z_T^p=\exp(pI_T-\tfrac p2v)$, and the normal moment-generating formula gives $\mathbb{E}[Z_T^p]=\exp(\tfrac12p^2v-\tfrac12pv)$, as claimed. $\square$

Worked Densities and the Absolute-Continuity Test

Worked Example: A Constant-Rate Shift

On $[0,2]$, let $h(t)=\tfrac32t$. Find the translation density and check its normalization. The path is absolutely continuous, $h(0)=0$, and $h'(t)=\tfrac32$. Its squared Cameron-Martin norm is $$ \int_0^2|h'(t)|^2\,dt=\int_0^2\frac94\,dt=\frac92. $$ Thus the theorem gives $$ Z_2=\exp\left(\frac32B_2-\frac94\right). $$ Indeed, $\tfrac12\|h\|_{H_0^1}^2=\tfrac94$, and $B_2\sim N(0,2)$. Hence $$ \mathbb{E}[Z_2]=e^{-9/4}\mathbb{E}[e^{(3/2)B_2}] =e^{-9/4}e^{(1/2)(9/4)(2)}=1. $$ Under the probability measure with density $Z_2$ relative to $\mathbb{P}$, the path $B$ has the law of a Brownian path translated by $h$. For a direct check at time $t$, the joint Gaussian moment-generating formula gives $\mathbb{E}^{\mathbb{Q}}[e^{uB_t}]=\exp(\tfrac32ut+\tfrac12u^2t)$ for $0\le t\le2$; this is the moment-generating function of $N(\tfrac32t,t)$, as expected. The measure $\mathbb{Q}$ here is defined by this density; $\mathbb{P}$ remains the original measure.

Worked Example: A Piecewise-Linear Shift

On $[0,2]$, take $h(t)=2t$ for $0\le t\le1$, and $h(t)=3-t$ for $1\le t\le2$. The values agree at $t=1$, so $h$ is continuous and absolutely continuous, with derivative $2$ on $(0,1)$ and $-1$ on $(1,2)$. Therefore $$ \|h\|_{H_0^1}^2=\int_0^1 4\,dt+\int_1^2 1\,dt=4+1=5. $$ The Ito integral is $$ \int_0^2h'(t)\,dB_t=2(B_1-B_0)-(B_2-B_1)=3B_1-B_2, $$ so the density is $Z_2=\exp(3B_1-B_2-\tfrac52)$. For the normalization check, $3B_1-B_2=2B_1-(B_2-B_1)$, a sum of independent centered normal variables with variance $4\cdot1+1\cdot1=5$. Its exponential moment is $\mathbb{E}[e^{3B_1-B_2}]=e^{5/2}$, and hence $\mathbb{E}[Z_2]=e^{-5/2}e^{5/2}=1$. The shifted means also match the path: the covariance of $B_t$ with $3B_1-B_2$ is $2t$ for $t\le1$, and $3-t$ for $1\le t\le2$, which gives the mean $h(t)$ under the density tilt.

Worked Example: A Continuous Shift That Is Singular

On $[0,1]$, let $h(t)=\sqrt{t}$. This is continuous and $h(0)=0$, but for $t>0$ its derivative is $h'(t)=1/(2\sqrt{t})$. The required energy diverges: $$ \int_0^1|h'(t)|^2\,dt=\int_0^1\frac{1}{4t}\,dt=\infty. $$ For example, the integral from $\varepsilon$ to $1$ is $-\tfrac14\log\varepsilon$, which tends to infinity as $\varepsilon$ decreases to zero. Thus $h\notin H_0^1([0,1])$, and the Cameron-Martin theorem says that $\mu_h$ and $\mu$ are mutually singular. There is no Radon-Nikodym density of $\mu_h$ with respect to $\mu$. Continuity of a shift alone is not enough for absolute continuity.

Interpreting the Density

For a measurable path functional $F$ for which either side is integrable, the Radon-Nikodym identity gives $$ \mathbb{E}[F(B+h)]=\mathbb{E}\left[F(B)\exp\left(\int_0^T h'(t)\,dB_t-\frac12\|h\|_{H_0^1}^2\right)\right]. $$ The left side averages the functional over shifted Brownian paths. The right side averages it over ordinary Brownian paths, weighted by the likelihood of the translation. For example, choosing $F(x)=1$ recovers $\mathbb{E}[Z_T]=1$. Choosing $F(x)=e^{u x(t)}$ gives a further check: the right side is the moment-generating function of a normal variable with variance $t$ and mean $h(t)$, namely $\exp(uh(t)+u^2t/2)$, which agrees with $B_t+h(t)$.

The criterion is global: it uses the integrated squared derivative over the whole time interval. A shift can be smooth on every interval $[\varepsilon,T]$ and still fail the criterion because its energy diverges near zero, as in the square-root example. Conversely, when the energy is finite, the density is positive, has expectation one, and has finite moments of every real order. The exponential is therefore a genuine likelihood ratio, not just a formal expression.

Common Pitfalls

Reversing the shift changes the sign. The theorem above defines $\mu_h(A)=\mu(\{x:x+h\in A\})$, the law of $B+h$. With this convention, the exponent contains $+\int h'\,dB$. Confusing this with the law of $B-h$ reverses the stochastic-integral sign.

Finite energy is the exact condition, not merely a convenient sufficient assumption. If $h\in H_0^1$, translation gives equivalent measures and the displayed density. If $h\notin H_0^1$, the theorem gives mutual singularity, not just failure of the particular exponential formula. Neither conclusion follows from continuity alone.

Do not omit part of the time interval when factoring the density. If a partition $0=t_0<\cdots<t_n$ is used to factor a terminal exponential, it must have $t_n=T$ for the listed intervals to cover $[0,T]$. If $t_n<T$, the factor on $[t_n,T]$ must also be included. For deterministic $h'$, that tail factor is independent of the earlier interval factors and has expectation one, but it is still part of $Z_T$.

Shift $h$	Energy test	Conclusion
Absolutely continuous, $h(0)=0$	$\int_0^T\|h'(t)\|^2\,dt<\infty$	Equivalent measures; exponential density applies
Continuous, $h(0)=0$, but not in $H_0^1$	Energy is infinite or absolute continuity fails	Translated and Wiener measures are mutually singular

Check Your Understanding

For each question, use the translation convention and finite-time interval stated above.

On $[0,3]$, take $h(t)=t$. Compute its Cameron-Martin norm squared and write the terminal density.
For a deterministic $h'\in L^2([0,T])$, explain why the exponential density process is a true martingale rather than only a local martingale.
On $[0,1]$, determine whether $h(t)=t^{3/4}$ has finite Cameron-Martin energy. State the measure relation implied by the theorem.
For the shift $h(t)=at$ on $[0,T]$, calculate the variance of $\int_0^T h'(t)\,dB_t$ and check the normalization of the exponential density.
Explain why a partition ending at $t_n<T$ does not by itself give all the factors in the terminal density $Z_T$.

Shift \(h\)	Energy test	Conclusion
Absolutely continuous, \(h(0)=0\)	\(\int_0^T\|h'(t)\|^2\,dt<\infty\)	Equivalent measures; exponential density applies
Continuous, \(h(0)=0\), but not in \(H_0^1\)	Energy is infinite or absolute continuity fails	Translated and Wiener measures are mutually singular