I have a stochastic integral of the form \begin{align*} X(t) = \int_0^t h(v) W(v) dv \end{align*} where $W(v)$ is the standard Brownian motion and $h(v)$ is a positive, integrable function. While $X(t)$ is a mean-zero Gaussian process, it can be shown, for $s < t$, \begin{align*} \text{Cov}(X(s), X(t) - X(s)) = \int_{s}^{t}\int_{0}^{s} h(u) h(v) (u \wedge v) du dv > 0 \end{align*} and therefore $X(t)$ does not have independent increments.
In my problem, I am interested in the distribution of $\sup_{t \in [0,\tau]}|X(t)|$ for fixed $\tau$. If $X(t)$ had independent increments, then I could express $X(t) = W(\alpha(t))$, where $\alpha(t) = \text{Var}(X(t))$, and derive my distribution through $\sup_{t \in [0,\tau]}|W(t)|$, which is well-known. Alas, this is not the case here.
Some ideas:
Derive the distribution of $\sup_{t \in [0,\tau]}|X(t)|$ directly. While there is literature on this subject, they all depend on tail approximations, while my problem is very not in the tails. Since I can compute the exact form of the covariance function, perhaps my specific form allots for the exact distribution to be calculated, although I'm not technically savvy to do so myself. Suggestions welcome.
I can transform $g(X_t)$ to get it into the form $W(\alpha(t))$. This paper discusses one class of transformations, and it appears that I could apply the techniques here, although it does not appear that I can back-transform on my original scale. Is my reasoning correct? And are there any more modern techniques of transformation?