Abstract Wiener spaces for pinned processes (e.g., Brownian Bridge)

Question

In introductions to abstract Wiener spaces, the sample paths usually form a Banach space; so, in particular, the sum of two sample paths is a valid sample path and also an element of the Banach space. Then the Gaussian measure is constructed as a measure on this Banach space, the Cameron–Martin space is a subspace thereof, etc.

However, when the stochastic process is pinned to some non-zero value, then the addition of two sample paths no longer yields a valid sample path, hence the sample paths do not form a Banach space and the whole construction breaks down—so how are abstract Wiener spaces constructed for such pinned processes?

A simple example of this is a Brownian bridge, which is typically constructed to be pinned to zero at both time 0 and some time $T > 0$, but what if it is pinned to some value $b \neq 0$ at time $T$? Is the only way out to take the zero-pinned Wiener space and add $b\tau/T$ to every sample path? So then in a sense there is no true Wiener space for non-zero-pinned processes (since this additional term takes it out of the Banach space)? Also, how do I know that a linear drift is correct instead of some other drift function?

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Note: I substantially modified this question after realizing that I had originally kind of missed the point.

Answer 1

A Gaussian process can be thought of as a "standard Gaussian" on its Cameron-Martin space $\mathcal{M}$. That is, given an orthonormal basis $\psi_1,\psi_2, \dots$ of the Cameron-Martin space, the process is given by the sum $f=\sum_i \xi_i \psi_i$, where $\xi_i$ are i.i.d. standard Gaussians. Depending on the point of view, we may now think of it either as a collection of random variables $\langle f,\psi\rangle$ indexed by $\mathcal{M}$, or say that it converges almost surely in a suitable Banach space $\mathcal{B}$. In either case, the distribution is independent of the choice of the basis.

On the Cameron-Martin space of the Browninan motion (say, on $[0,T]$), the evaluation $f\mapsto f(t),$ $t\in (0,T)$, is a bounded linear functional, thus $f(t)=\langle f,\psi^{(t)}\rangle$ for some function $\psi^{(t)}$. It is not hard to see that in fact, $\psi^{(t)}(x)=\min(x,t)$. We can then take $\psi_1=\frac{1}{\sqrt{t}}\psi^{(t)}$ and complete the basis. What this gives us is a disintegration of our Gaussian measure with respect to the value of $f(t)$: namely, conditionally on $f(t)$, the distribution of $f$ is given by $f(t)\psi^{(t)}/t$ plus an independent Gaussian on the orthogonal complement or $\{\psi^{(t)}\}$. This orthogonal complement is exactly the space of all functions in $\mathcal{M}$ vanishing at $t$. You can define the corresponding Banach space $\mathcal{B}_0\subset \mathcal{B}$, and the "pinned" (i.e., conditioned) process will live in the closed affine subspace $f(t)\psi^{(t)}/t+\mathcal{B}_0$ of $\mathcal{B}$.

This construction works for any Gaussian process and any bounded linear functional, or even, more generally, any decomposition $\mathcal{M}=\mathcal{M_1}\oplus\mathcal{M_2}$. Of course, the fact that $\psi^{(t)}$ happens to be a linear function on $(0,t)$ is special to the Brownian motion.