Brownian motion reflection principle (hitting probabilities conditioned on path) This is a revamped version of a question I asked previously (which I have since removed as it was poorly posed). I am considering the SDE $\mathrm{d}X_t=3X_t^{\frac23}\mathrm{d}B_t+3X_t^{\frac13}\mathrm{d}t$, which I know admits the non-unique solutions $X_t=B_t^3$ and
$$\tilde{X}_t=\begin{cases}0&t<\tau_a\\B_t^3&\tau_a\leq t\end{cases}$$
where $\tau_a=\inf\{t>a:B_t=0\}$.  I am asked to find $\mathbb{P}(X_{a+t}=0|B_a=1)$ as well as $\mathbb{P}(\tilde{X}_{a+t}=0|B_a=1)$. Intuitively, this feels like translated Brownian motion and I know I need to use reflection principle about $B_a=1$, but I am unsure of how I should begin my solution to approach the condition of the underlying Brownian path or where I should include the reflection principle. Usually when I think of the reflection principle I think of CDFs, but these events are singletons which leaves me quite confused. Any guidance is appreciated!
 A: Answer to the previous version of question for case when $\tilde{X}_t = 0$ for $\tau_a < t$.
Statement: $\mathbb{P}(X_{a+t}=0|B_a=1) = 0$ and $\mathbb{P}(\tilde{X}_{a+t}=0|B_a=1) = I_{t > a} 2(1 - \Phi(\frac{1}{\sqrt{t-a}}))$.
Proof.
Put $Y_t = B_{a+t} - B_a$ and let $\mathcal{F}_t, t \ge 0$ be the natural filtration of $B_t$. Hence $Y_t$ and $\mathcal{F}_{\le a}$ are independent and $Y_t$ is a Brownian motion - Markov property of Wiener process.
Thus
\begin{gather}
P(X_{a+t} = 0 | B_a = 1) = P(B_{a+t}^3 = 0 | B_a = 1) = \\
=  P((B_{a+t}-B_a+B_a)^3= 0 | B_a = 1) = \\
= P((Y_t+1)^3 = 0 | B_a = 1) = P((Y_t+1)^3 = 0)
\end{gather}
by independence of $Y_t$ and $\mathcal{F}_{\le a}$. It follows that
$$P(X_{a+t} = 0 | B_a = 1) =  P((Y_t+1)^3 = 0) = P(Y_t = -1) = P(B_t = -1) = 0.$$
Further, $P(\tilde{X}_{a+t}=0|B_a=1) = P(B_{a+t}^3 I_{\tau_a \ge t} = 0| B_a = 1)$.
We know that
\begin{gather}
\tau_a = \inf(t>a: B_t = 0) = [t = u+a] = a + \inf(u>0: B_{u+a}=0) =\\ = a + \inf(u >  0: B_{u+a} -B_a = -B_a) = a + \inf(u>0: Y_u = -B_a).
\end{gather}
If $B_a = 1$ then $\tau_a =  a + \inf(u>0: Y_u = -1) = a + \tilde{\tau}_{-1}$.
Thus
\begin{gather}
P(\tilde{X}_{a+t}=0|B_a=1) = P(B_{a+t}^3 I_{\tau_a \ge t} = 0| B_a = 1) = \\
= P( (B_{a+t} - B_a + B_a)^3  I_{\tau_a \ge t} = 0 |B_a =1) = \\ \nonumber
P( (Y_t + 1)^3  I_{\tilde{\tau}_{-1} \ge t-a} = 0 |B_a =1) = P( (Y_t + 1)^3  I_{\tilde{\tau}_{-1} \ge t-a} = 0)
\end{gather}
by independence,  because $\tilde{\tau}_{-1}$ is  measurable with respect to the natural filtration of $Y_t$.
Put $A = \{(Y_t+1)^3 = 0 \}$, $B = \{I_{\tilde{\tau}_{-1} \ge t-a} = 0 \}$. Hence $P(A) = 0$ and
\begin{gather}
P(\tilde{X}_{a+t}=0|B_a=1) =  P( (Y_t + 1)^3  I_{\tilde{\tau}_{-1} \ge t-a} = 0) \\
= P(A \cup B) = P(B) = P(\tilde{\tau}_{-1} < t-a) = \\
= P( \inf(u>0: Y_u = -1) < t-a)  = P( \inf(u>0: -Y_u = 1) < t-a)  
\end{gather}
But $-Y_{t}$ is a Brownian motion, thus
$$P(\tilde{X}_{a+t}=0|B_a=1) =  P( \inf(u>0: B_u = 1) < t-a)  = P(\tau_1^* < t-a),$$
where $\tau_s^* =  \inf(u>0: B_u = s) $.
For all $s > 0$, $t > 0$ we have (I use standard properties of maximum of Brownian motion)
\begin{gather}
P(\tau_s^* < t) = P( \max_{u \in [0,t]} W_u > s) = P(|W_t| \ge s) = P(|N(0,t)| \ge s) = \\ =
P(\sqrt{t}|N(0,1)| \ge s) = P(|N(0,1)| \ge \frac{s}{\sqrt{t}}) = 2P(N(0,1) \ge \frac{s}{\sqrt{t}}) = \\ 
= 2(1 - \Phi(\frac{s}{\sqrt{t}})).
\end{gather}
If $s > 0 $ and $t \le 0$ then $P(\tau_s^* < t)= 0$.
Hence
$$\mathbb{P}(\tilde{X}_{a+t}=0|B_a=1) = I_{t > a} 2(1 - \Phi(\frac{1}{\sqrt{t-a}})).$$
Do you have any questions?
