They are almost surely singular to each other.
First of all, here is a basic observation from measure theory that is useful in the proof. Consider two random measures, $M_1$ and $M_2$ on the same space - say, $[0,1]$, depending on some randomness $\omega \in \Omega$. With these measures one associates measures $\mathsf{Q}_1$, $\mathsf{Q}_2$ on $\Omega \times [0,1]$, defined as
$$\mathsf{Q}_i(d\omega, dt) := \mathsf{P}(d\omega) M_i(\omega, dt)$$
In other words, $M_i$ can be obtained from $\mathsf{Q}_i$ by disintegration w.r.t. the $t \in [0,1]$ coordinate. On the other hand, one can do the disintegration w.r.t. the other coordinate, i.e. $\omega$, and define the corresponding functions $t \mapsto \mathsf{Q}_i(t)$ with values in measures on $\Omega$, as follows:
$$\mathsf{Q}_i(d\omega, dt) = \mathsf{Q}_i(t, d\omega) \mu(dt),$$
where $\mu$ is a measure on $[0,1]$, such that $\mathsf{E} M_i \ll \mu$. The basic observation that I'm talking about is this:
The following are equivalent:
- $M_1 \ll M_2$ almost surely
- $\mathsf{Q}_1 \ll \mathsf{Q}_2$ (as measures on $\Omega \times [0,1]$)
- $\mathsf{Q}_1(t) \ll \mathsf{Q}_2(t)$ (as measures on $\Omega$) for $\mu$-almost all $t$
Now we turn back to our problem. Consider the random measure $M := \delta_I$, and the measure $M_1 := \mathsf{E} \left[ M \middle| \tilde B \right]$. $M_1$ is exactly the conditional distribution of the maximum of $B + \tilde B$ given a realization of $\tilde B$. I'm proving that $M_1$ is almost surely singular, that is, singular to the deterministic measure $M_2 := \mathrm{Lebesgue}$. By the above observation, this is equivalent to saying that the distribution of $(B, \tilde B)$ according to the measure $\mathsf{Q}_1(t, d\omega)$ is singular to $\mathsf{Q}_2(t) = \mathsf{P}$, for Lebesgue-almost all $t$.
Now we describe the distribution of $(B, \tilde B)$ according to $\mathsf{Q}_1(t)$. It turns out to be easier to describe the distribution according to $\mathsf{Q}(t)$ first --- that is, the one that corresponds to $M = \delta_I$. This description (and a little bit more) is given by the Williams' path decomposition theorem (e.g. Theorem 4.9 in Revuz-Yor). It follows from there that the distribution of the process $\frac{1}{\sqrt 2}(B(t) + \tilde B(t) - B(t + \cdot) - \tilde B(t + \cdot))$ according to $\mathsf{Q}(t)$ is absolutely continuous w.r.t. that of the Bessel(3) process started from $0$. $\frac{1}{\sqrt 2} (B - \tilde B)$ is still the Brownian motion according to $\mathsf{Q}_1(t)$, since it was $\mathsf{P}$-independent of $\frac{1}{\sqrt 2} (B + \tilde B)$ and $M_1$ is measurable w.r.t. $\frac{1}{\sqrt 2}(B + \tilde B)$.
Now, another purely measure-theoretic observation: the fact that $M_1 = \mathsf{E} \left[ M \middle| \tilde B \right]$ corresponds, in terms of the $\mathsf{Q}$ measures, to the following: the distribution of $\tilde B$ according to $\mathsf{Q}_1(t)$ is the same as that according to $\mathsf{Q}(t)$. But from the paragraph above it follows that $\tilde B(t + \cdot) - \tilde B(\cdot)$, according to $\mathsf{Q}_1(t)$, can be decomposed as $\frac{1}{\sqrt 2}$ times the sum of a Brownian motion and an independent Bessel(3) process (up to absolute continuity).
It remains to prove that if $B$ is a Brownian motion and $U$ is an independent Bessel(3) process starting from $0$ then the distribution of $\frac{1}{\sqrt 2}(B + U)$ is singular to that of the Brownian motion. There are many ways to do that; the simplest one that I know is the following. The Bessel(3) process is essentially "the Brownian motion conditioned to stay positive", in particular, its distribution is the limit of Brownian motions conditioned on some convex sets. This implies that it's strictly log-concave (w.r.t. the covariance of the Brownian motion) --- i.e. "more log-concave than the Gaussian distribution of the Brownian motion". It follows from the inequality of Harge (Theorem 1.1 in Harge "A convex/log-concave correlation inequality for Gaussian measure and an application to abstract Wiener spaces") that for any linear functional $\psi$ on $C[0,1]$ we have $\mathsf{E} (\psi(U - \mathsf{E} U))^2 \le \mathsf{E} \psi(B)^2$; by approximation, this extends to measurable linear functionals, i.e. stochastic integrals of deterministic $L^2$ functions. In other words, for any deterministic function $f \in L^2[0,1]$ the stochastic integral $\intop f d (U - \mathsf{E} U)$ "converges" (i.e. is well-defined as an appropriate limit...). For the Brownian motion, however, $\intop f d B$ converges without the need to recenter $B$. Since $\mathsf{E} U(t) \sim t^{1/2}$ is not in the $W^{1,2}$ Sobolev space, this means that there are some functionos $f \in L^2$, such that $\intop f d B$ is well-defined bub $\intop f d U$ is not --- thus, $\intop f d ((B + U) / \sqrt 2)$ is also not defined. This proves that the law of $(B + U) / \sqrt 2$ is singular to that of $B$.
We can re-write the problem in terms of $W(t)$ alone, or, even better, in terms of the drifted Brownian motion $\tilde W(t) = W(t) - M t$, where $M$ is the supremum of $|b(s)|$.
Define
$$ B(t) = -1 - \int_0^t b(s) ds - M t ,$$
so that $X(t) = \tilde W(t) - B(t)$ up to time
$$ \tau = \inf \{ t > 0 : \tilde W(t) \leqslant B(t) \} . $$
Fix $t_0 > 0$ and define
$$ \sigma = \inf \{ t \in (0, t_0] : \tilde W(t) \leqslant B(t_0) \} . $$
Since $B$ is a non-increasing function, we clearly have $\sigma \geqslant \tau$, and hence the measure
$$\mu(dx) = \mathbb P(t_0 < \tau, \tilde W(t_0) - B(t_0) \in dx)$$
is dominated by the measure
$$\nu(dx) = \mathbb P(t_0 < \sigma, \tilde W(t_0) - B(t_0) \in dx) .$$
The latter is, however, just the distribution at time $t_0$ of the drifted Brownian motion $\tilde W(t) - B(t_0)$, killed upon hitting $0$. As you write in the statement of the problem, this is known to have a density function continuously vanishing at zero, and hence $\mu(dx)$ also has a density function continuously vanishing at zero.
It remains to note that $\mu$ is precisely the distribution of $X(t_0)$, up to an extra atom at $0$.
Remark: A more general approach to the problem, which seems to work also when $b(s)$ is an (adapted) stochastic process rather than a deterministic function, would involve showing first that the distribution of $Y(t)$ — and thus also that of $X(t)$ — has a bounded density function (save for an atom at $0$), and then using Chapman–Kolmogorov equation and a comparison argument similar to the one given above to conclude that the density function of the distribution of $X(t)$ goes to zero at $0$.
Best Answer
It appears that a proof is given in
Choi, ByoungSeon; Roh, JeongHo, On the trivariate joint distribution of Brownian motion and its maximum and minimum, Stat. Probab. Lett. 83, No. 4, 1046-1053 (2013). ZBL1266.60142.