The process $X(t)=\int\_0^t B(s) ds+B(t)$ is a centered continuous Gaussian process. Therefore it defines a Gaussian measure on $C[0,T]$. Therefore there is a Cameron-Martin space with Cameron-Martin norm. I can compute the covariance and get some complicated expression.


Is there a clean expression for the Cameron-Martin norm associated to $X$?



What is the Cameron-Martin norm associated to $X(t)=\int_0^t B(s) ds+B(t)$?
I will show that for $x\in\mathcal C^1$, $x$ is in the Cameron-Martin space $\mathcal H$ and
$$ |x|\_\mathcal{H}^2 = \int\_0^T\left(x'(t)-\int\_0^t\mathbf e^{-(t-s)}x'(s)\mathrm ds\right)^2\mathrm dt. $$
Expanding the product and using Fubini's theorem, it can be rewritten as
$$ |x|\_\mathcal{H}^2 = \int\_0^T|x'(t)|^2\mathrm dt + \int\_0^T\int\_0^Tk(s,t)x'(s)x'(t)\mathrm ds\mathrm dt $$
for some explicit kernel $k$. I claim that with slightly more effort, one can show that $\mathcal H=\mathrm H^1$ *as vector spaces*, and that the above expression stays true for $x'\in\mathrm L^2$ the weak derivative. Hopefully this is clean enough for your purposes.


**The Cameron-Martin space.**
Define $\overline F:\mathcal C\to\mathcal C$ as the continuous map sending $B$ to $X$; formally, $F(b):t\mapsto\int\_0^tb\_sds + b\_t$. Set $i\_B:\mathrm H^1\to\mathcal C$ the usual Cameron-Martin inclusion for Brownian motion.¹ The important observation is that the continuous map $\overline F\circ i\_B:\mathrm H^1\to\mathcal C$ is injective. Let us postpone the proof to the end of the answer.


Define $\mathcal H$ as the image of $\mathrm H^1$ under $\overline F\circ i\_B$, together with the inner product coming from $\mathrm H^1$ (we use the injectivity here). In other words, $\mathcal H$ is the set of all $\overline F\circ i\_B(b)$ for $b\in\mathrm H^1$, and $|\overline F\circ i\_B(b)|\_\mathcal{H}^2=|b|\_{\mathrm H^1}^2$. It is obviously canonically isomorphic to $\mathrm H^1$, through some $F:\mathrm H^1\to\mathcal H$.


I claim that the inclusion $i\_X:\mathcal H\hookrightarrow\mathcal C$ is the Cameron-Martin space of $X$. Indeed, for any $\phi,\psi\in\mathcal C^\*$, and noting that $\overline F\circ i\_B=i\_X\circ F$,
$$ \begin{align\*}
 \mathbb E[\phi(X)\psi(X)]
&= \mathbb E[\phi\circ\overline F(B)\psi\circ\overline F(B)] \\
&= \big\langle\phi\circ\overline F\circ i\_B,\psi\circ\overline F\circ i\_B\big\rangle\_{(\mathrm H^1)^\*} \\
&= \big\langle\phi\circ i\_X\circ F,\psi\circ i\_X\circ F\big\rangle\_{(\mathrm H^1)^\*} \\
&= \big\langle\phi\circ i\_X,\psi\circ i\_X\big\rangle\_\mathcal{H^\*}.
 \end{align\*} $$


**Expression for the norm.**
Suppose $x\in\mathcal C$ is actually of class $\mathcal C^1$, and define
$$ b:t\mapsto
 \int\_0^t\mathbf e^{-(t-s)}x'(s)\mathrm ds
 = \mathbf e^{-t}\int\_0^t\mathbf e^sx'(s)\mathrm ds. $$
Then $b$ is clearly in $\mathrm H^1$, and in fact we will show $x=F(b)$, so $x\in\mathcal H$ and $|x|\_\mathcal{H}=|b|\_{\mathrm H^1}$ and we have the expected expression for the Cameron-Martin norm.


The fact that $x=F(b)$ follows from computing derivatives:
$$ \frac{\mathrm d}{\mathrm dt}\big(F(b)(t)-x(t)\big)
 = b(t) + b'(t) - x'(t)
 = 0. $$


**Injectivity of the transform.**
Since $i\_B$ is injective, we need only show that $\overline F$ is injective. Choose $b\in\mathcal C$ such that $\overline F(b)=0$. This means precisely that
$$ 0=\overline F(b)(t)=\int\_0^tb(s)\mathrm ds + b(t) $$
for all $t$. Then the function
$$ t\mapsto\mathbf e^t\int\_0^tb(s)\mathrm dt $$
is $\mathcal C^1$ with vanishing derivative hence always zero, and using the relation again,
$$ b(t) = -\int\_0^tb(s)\mathrm ds = -\mathbf e^{-t}\cdot0 = 0. $$




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¹ $\mathrm H^1$ is the Sobolev space of functions $b$ that can be written as the integral of a function $\dot b\in\mathrm L^2$.


